Solved Problems in Geophysics

Solved Problems in Geophysics

Solving problems is an indispensable exercise for mastering the theory underlying the various branches of geophysics. Without this practice, students often find it hard to understand and relate theoretical concepts to their application in real-world situations. This book is a collection of nearly 200 problems in geophysics, which are solved in detail showing each step of their solution, the equations used and the assumptions made. Simple figures are also included to help students understand how to reduce a problem to its key elements. The book begins with an introduction to the equations most commonly used in solving geophysical problems. The subsequent four chapters then present a series of exercises for each of the main, classical areas of geophysics – gravity, geomagnetism, seismology and heat flow and geochronology. For each topic there are problems with different degrees of difficulty, from simple exercises that can be used in the most elementary courses, to more complex problems suitable for graduate-level students. This handy book is the ideal adjunct to core course textbooks on geophysical theory. It is a convenient source of additional homework and exam questions for instructors, and provides students with step-by-step examples that can be used as a practice or revision aid. Elisa Buforn is a Professor of geophysics at the Universidad Complutense de Madrid (UCM) where she teaches courses on geophysics, seismology, physics, and numerical methods. Professor Buforn’s research focuses on source fracture processes, seismicity, and seismotectonics, and she is Editor in Chief of Física de la Tierra and on the Editorial Board of the Journal of Seismology. Carmen Pro is an Associate Professor at the University of Extremadura, Spain, where she has taught geophysics and astronomy for over 20 years. She has participated in several geophysical research projects and is involved in college management. Agustín Udías is an Emeritus Professor at UCM and is the author of a large number of papers about seismicity, seismotectonics, and the physics of seismic sources, as well as the textbook Principles of Seismology (Cambridge University Press, 1999). He has held positions as Editor in Chief of Física de la Tierra and the Journal of Seismology and as Vice President of the European Seismological Commission.

Solved Problems in Geophysics ELISA BUFORN Universidad Complutense, Madrid

CARMEN PRO Universidad de Extremadura, Spain

AGUSTÍN UDÍAS Universidad Complutense, Madrid

cambridge university press Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo, Delhi, Mexico City Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9781107602717 # Elisa Buforn, Carmen Pro and Agustín Udías 2012 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2012 Printed in the United Kingdom at the University Press, Cambridge A catalogue record for this publication is available from the British Library Library of Congress Cataloging-in-Publication Data Buforn, E. Solved problems in geophysics / Elisa Buforn, Carmen Pro, Agustín Udías. p. cm. Includes bibliographical references. ISBN 978-1-107-60271-7 (Paperback) 1. Geophysics–Problems, exercises, etc. I. Pro, Carmen. II. Udías Vallina, Agustín. III. Title. QC807.52.B84 2012 550.78–dc23 2011046101 ISBN 978-1-107-60271-7 Paperback Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

Contents

Preface

1 Introduction Gravity Geomagnetism Seismology Heat flow Geochronology

2 Gravity Terrestrial geoid and ellipsoid Earth’s gravity field and potential Gravity anomalies. Isostasy Tides Gravity observations

3 Geomagnetism Main field Magnetic anomalies External magnetic field Main (internal), external, and anomalous magnetic fields Paleomagnetism

4 Seismology Elasticity Wave propagation. Potentials and displacements Reflection and refraction Ray theory. Constant and variable velocity Ray theory. Spherical media Surface waves Focal parameters

5 Heat flow and geochronology Heat flow Geochronology Bibliography v

page vii 1 1 4 6 10 11 13 13 25 53 95 116 121 121 142 156 174 201 208 208 211 224 243 277 307 324 335 335 345 352

Preface

This book presents a collection of 197 solved problems in geophysics. Our teaching experience has shown us that there was a need for a work of this kind. Solving problems is an indispensable exercise for understanding the theory contained in the various branches of geophysics. Without this exercise, the student often finds it hard to understand and relate the theoretical concepts with their application to practical cases. Although most teachers present exercises and problems for their students during the course, the hours allotted to the subject significantly limit how many exercises can be worked through in class. Although the students may try to solve other problems outside of class time, if there are no solutions available this significantly reduces the effectiveness of this type of study. It helps, therefore, both for the student and for the teacher who is explaining the subject if they have problems whose solutions are given and whose steps can be followed in detail. Some geophysics textbooks, for example, F.D. Stacey, Physics of the Earth; G.D. Garland, Introduction to Geophysics; C.M. Fowler, The Solid Earth: An Introduction to Global Geophysics; and W. Lowrie, Fundamentals of Geophysics, contain example problems, and, in the case of Stacey’s, Fowler’s, and Lowrie’s textbooks, their solutions are provided on the website of Cambridge University Press. The main difference in the present text is the type of problems and the detail with which the solutions are given, and in the much greater number. All the problems proposed in the book are solved in detail, showing each step of their solution, the equations used, and the assumptions made, so that their solution can be followed without consulting any other book. When necessary, and indeed quite often, we also include figures that allow the problems to be more clearly understood. For a given topic, there are problems with different degrees of difficulty, from simple exercises that can be used in the most elementary courses, to more complex problems with greater difficulty and more suitable for teaching at a more advanced level. The problems cover all parts of geophysics. The book begins with an Introduction (Chapter 1) that includes the equations most used in solving the problems. The idea of this chapter is not to develop the theory, but rather to simply give a list of the equations most commonly used in solving the problems, at the same time as introducing the reader to the nomenclature. The next four chapters correspond to the division of the problems into the four thematic blocks that are classic in geophysics: gravity, geomagnetism, seismology, and heat flow and geochronology. We have not included problems in geodynamics, since this would depart too much from the approach we have taken, which is to facilitate comprehension of the theory through its application to specific cases, sometimes cases which are far from the real situation on Earth. Indeed, some of the problems may seem a bit artificial, but their function is to help the student practise with what has been seen in the vii

viii

Preface

theory. Neither did we want to include specific problems of geophysical prospecting as this would have considerably increased the length of the text, and moreover some of the topics that would be covered in prospecting, such as gravimetric and geomagnetic anomalies, are already included in other sections of this work. Chapter 2 contains 68 problems in gravity divided into five sections. The first section is dedicated to the terrestrial geoid and ellipsoid, proposing calculations of the parameters that define them in order to help better understand these reference surfaces. The second corresponds to calculating the gravitational field and potential for various models of the Earth, including the existence of internal structures. Gravity anomalies are dealt with in the third section, with a variety of problems to allow students to familiarize themselves with the corrections to the observed gravity, with the concept of isostasy, and with the Airy and Pratt hypotheses. The fourth section studies the phenomenon of the Earth’s tides and their influence on the gravitational field. The last section is devoted to the observations of gravity from measurements made with different types of gravimeters and the corrections necessary in each case. We also include the application of these observations to the accurate determination of different types of height. Chapter 3 contains 42 problems in geomagnetism divided into five sections. The first is devoted to the main (internal) field generated by a tilted dipole at the centre of the Earth. It includes straightforward problems that correspond to the calculation of the geomagnetic coordinates of a point and the theoretical components of the magnetic field. This section also introduces the student to the use of the principal units used in geomagnetism. The second considers the magnetic anomalies generated by different magnetized bodies and their influence on the internal field. The third section is devoted to the external field and its variation with time. In the fourth section, we propose problems of greater complexity involving the internal field, the external field, and anomalous magnetized bodies at the same time. The last section is devoted to problems in paleomagnetism. Chapter 4 contains 69 problems in seismology divided into seven sections. The first presents some simple exercises on the theory of elasticity. The second addresses the problem of the propagation of seismic energy in the form of elastic waves, resolving the problems on the basis of potentials, and calculating the components of their displacements. We study the reflection and refraction of seismic waves in the third section. The fourth is devoted to the problem of wave propagation using the theory of ray paths in a plane medium of constant and variable velocity of propagation. The fifth studies the problem of the propagation of rays in a spherical medium of either constant or variable propagation velocity, with the calculation of the travel-time curves for both plane and spherical media. The sixth section contains problems in the propagation of surface waves in layered media. The seventh section is devoted to problems of calculating the focal parameters and the mechanism of earthquakes. Chapter 5 includes 11 problems in heat flow with the propagation of heat in plane and spherical media, and seven problems in geochronology involving the use of radioactive elements for dating rocks. Finally, we provide a bibliography of general textbooks on geophysics and of specific textbooks for the topics of gravity, geomagnetism, and seismology. We have tried to include only those most recent and commonly used textbooks which are likely to be found in university libraries.

ix

Preface

In sum, the book is a university text for students of physics, geology, geophysics, planetary sciences, and engineering at the undergraduate or Master’s degree levels. It is intended to be an aid to teaching the subjects of general geophysics, as well as the specific topics of gravity, geomagnetism, seismology, and heat flow and geochronology contained in university curricula. The teaching experience of the authors in the universities of Barcelona, Extremadura, and the Complutense of Madrid highlighted the need for a work of this kind. This text is the result of the teaching work of its authors for over 20 years. Thanks are due to the generations of students over those years who, with their comments, questions, and suggestions, have really allowed this work to see the light. We are also especially grateful to Prof. Greg McIntosh who provided us with some problems on paleomagnetism, to Prof. Ana Negredo for her comments on heat flow and geochronology problems, and to Dr R.A. Chatwin who worked on translating our text into English. The text is an extension of the Spanish edition published by Pearson (Madrid, 2010).

E. BUFORN, C. PRO AND A. UDÍAS

1

Introduction

Gravity As a first approximation the Earth’s gravity is given by that of a rotating sphere. The gravitational potential of a sphere of mass M is: V ¼

GM r

where r is the position vector (Fig. A) and G the universal gravitational constant. If the sphere is rotating with angular velocity o the centrifugal potential at a point on the surface is given by 1 F ¼ o2 r2 sin2 y 2 where y is the angle that r forms with the axis of rotation. The gravity potential is their sum U ¼ V þ F. The value of the acceleration due to gravity (the gravity ‘force’) is given by the gradient of the potential: g ¼ rU The radial component of the gravity force is given by gr ¼

GM þ ro2 sin2 y r2

The potential of the Earth to a first-order approximation corresponds to that of a rotating ellipsoid, and is given by    m
r 2 2 GM a J2
a3  2 U¼ 3sin ’ 1 þ cos ’ a r 2 r 2 a

where ’ ¼ 90º y is the geocentric latitude and a the equatorial radius. The coefficient m is the ratio between the centrifugal and gravitational forces on the sphere of radius a at the equator: m¼

1

a3 o2 GM

2

Introduction

w

North Pole

P q

r a Equator

l

Fig. A

The dynamic form factor J2 is defined as J2 ¼

C A a2 M

where C and A are the moments of inertia about the axis of rotation and an equatorial axis. The flattening of the ellipsoid (the shape of the Earth to a first-order approximation) of equatorial and polar radius a and c is: a¼

a

c a

In terms of J2 and m, 3 m a ¼ J2 þ 2 2 The dynamic ellipticity is H¼

C

A C

The gravity flattening is b¼

gp

ge ge

where gp and ge are the normal values of gravity at the pole and the equator, respectively.

3

Gravity

The gravity at a point of geocentric latitude ’ = 90º   g ¼ ge 1 þ bsin2 ’

y is

The geocentric latitude of a point is the angle between the equator and the radius vector of the point. The geodetic latitude is defined as the angle between the equatorial plane and the normal to the ellipsoid surface at a point. Astronomical latitude is the angle between the equatorial plane and the observed vertical at a point. The normal or theoretical gravity at a point of geocentric latitude ’ referred to the GRS1980 reference ellipsoid is g ¼ 9:780327 ð1 þ 0:0053024 sin2 ’

0:0000059 sin2 2’Þ m s

2

The effect of the Sun and Moon on the Earth is to produce the phenomenon of the tides. If one considers more generally the tidal effect due to an astronomical body of mass M at a distance R from the centre of the Earth, one must add the corresponding potential, which, in the first-order approximation, is given by c¼

GMr2  3cos2 # 2R3

1



where r is the geocentric radius vector of the point, and # is the angle the position vector r forms with the distance vector R. Gravity anomalies, defined as Dg ¼ g –g, are the effects of the existence of anomalous masses inside the Earth. The gravity anomaly along the Z (vertical) axis at a point distance x along the horizontal axis produced by a sphere of radius R, density contrast Dr, and buried at a depth d, is given by
gðx; zÞ ¼

@Va G
M ðz þ dÞ ¼h i3=2 @z x2 þ ðz þ dÞ2

where Va is the potential produced by the anomalous spherical mass DM ¼ 4/3pR3 Dr. For problems in two dimensions, one uses the anomaly produced by an infinite horizontal cylinder at depth d, perpendicular to the plane under consideration. The anomalous potential is given by 0 1 and the anomaly by

1 B C Va ¼ 2pG
ra2 ln@qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA x2 þ ðz þ dÞ2


gðx; zÞ ¼

@Va 2pG
ra2 ðz þ dÞ ¼ @z x2 þ ðz þ dÞ2

To correct for the height above sea level at which measurements are made, one uses the concepts of the free-air and Bouguer anomalies. The free-air anomaly is
gFA ¼ g

g þ 3:086h

4

Introduction

where g is the observed gravity, h the height in metres, and the anomaly is obtained in gu (gravity units) mm s 2. The Bouguer anomaly is
g B ¼ g

g þ ð3:086

0:419rÞh

with r being the density of the plate of thickness h. To account for isostatic compensation at height in mountainous areas, one adds an isostatic correction which can be calculated assuming either the Airy or Pratt hypotheses. With the Airy hypothesis, the root t of a mountain is given by t¼

rc rM

rc

h

where rc and rM are the densities of the crust and mantle, and h is the height of the mountain. For an ocean zone, with water density ra, the anti-root is t0 ¼

rc rM

ra 0 h ra

With the Pratt hypothesis, the density contrast in a mountainous area is
r ¼ r

r0 ¼

h r Dþh 0

where D is the level of compensation, h the height of the mountain, and r0 the density at sea level. For an oceanic zone of depth h0 : r0 D ra h0 D h0 0
r ¼ r r0 r0 ¼

The isostatic correction can be calculated using a cylinder of radius a and height b, whose base is located at a distance c beneath the point, and with density contrast Dr: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

I a2 þ c 2 C ¼ 2pG
r b þ a2 þ ðc bÞ2 For mountainous zones, with the Airy hypothesis: b ¼ t, c ¼ h þ H þ t (H ¼ crustal thickness, h ¼ height of the point); and with the Pratt hypothesis: b ¼ D, c ¼ D þ h.

Geomagnetism To a first approximation, the internal magnetic field of the Earth can be approximated by a centred dipole inclined at 11.5 to the axis of rotation. The potential created by a magnetic dipole at a point distant r from its centre and forming an angle y with the axis of the dipole is F¼

Cm cos y r2

5

Geomagnetism

GNP 90º – fB l – lB 180º – l∗

90º – f

GMNP q = 90º – f∗

D∗

P

Fig. B

where C ¼ m0/4p with m0 ¼ 4p  10 7 H m 1, and m is the dipole moment in units of A m². The product Cm is given in T m3. The components of the magnetic dipole field B are: Br ¼ By ¼

@F 2Cm cos y ¼ @r r3 1 @F 2Cm sin y ¼ r @y r3

In the centred dipole approximation for the Earth’s magnetic field, the geomagnetic coordinates (f*, l*) of a point (y ¼ 90º f*) in terms of its geographic coordinates (f, l) and those of the Geomagnetic North Pole (GMNP) (fB, lB) can be calculated using the expressions of spherical trigonometry (Fig. B): sin f ¼ sin fB sin f þ cos fB cos f cosðl sin l ¼

sinðl

lB Þ

lB Þ cos f cos f

The vertical and horizontal components of the field, the geomagnetic constant B0, and the total field are given by: Z  ¼ 2B0 sin f H  ¼ B0 cos f Cm B0 ¼ 3 a qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  F ¼ H 2 þ Z 2 ¼ B0 1 þ 3sin2 f

The units used for the components of the magnetic field are the tesla T and the nanotesla nT ¼ 10 9 T. The NS (X*) and EW (Y*) components are X  ¼ H cos D Y  ¼ H sin D and the declination and inclination are given by

6

Introduction

cos fB sinðl cos f tan I  ¼ 2 tan f

sin D ¼

lB Þ

The radius vector at each point of the line of force is: r ¼ r0 cos2 f ¼ r0 sin2 y where r0 is the radius vector of the point of the line of force located at the geomagnetic equator. Magnetic anomalies are produced by magnetic materials within the Earth. The anomalous potential due to a vertical dipole buried at depth d is FA ¼

Cmcosy Cmðz þ dÞ ¼h i3=2 r2 x2 þ ðz þ dÞ2

The vertical (z) and the horizontal (x) components of the magnetic anomaly at the surface (z ¼ 0) produced by a vertical magnetic dipole at depth d are:
Z ¼
X ¼

Cmð2d 2

x2 Þ

ðx2 þ d 2 Þ5=2 3Cmxd ðx2 þ d 2 Þ5=2

The Earth is affected by an external magnetic field produced mainly by the activity of the Sun. This field is variable in time, with distinct periods of variation. The most noticeable is the diurnal variation (Sq) with a maximum at 12 noon local time. The most important nonperiodic variations are the so-called magnetic storms.

Seismology Earthquakes produce elastic waves which propagate through the interior and along the surface of the Earth. Using the plane-wave approximation, the displacements of the internal P- and S-waves (uiP and uiS) can be obtained from a scalar potential and a vector potential:
 ui ¼ ui P þ ui S ¼ ðr’Þi þ r  cj i

  ’ ¼ A exp ika gj xj at   cj ¼ Bj exp ikb gj xj bt

where A and Bj are the amplitudes, xj the coordinates of the observation point, ka and kb the wavenumbers, gj are the direction cosines defined from the azimuth az and angle of incidence i of the ray as:

7

Seismology

g1 ¼ sin i cos az g2 ¼ sin i sin az g3 ¼ cos i and a and b are the P- and S-wave velocities of propagation, respectively, defined from the Lamé coefficients (l and shear modulus m) and the density r: sffiffiffiffiffiffiffiffiffiffiffiffiffiffi l þ 2m vP ¼ a ¼ r rffiffiffi m vS ¼ b ¼ r Units used are: displacement amplitudes (u) in µm; potential amplitudes (A, Bi) in 10 wavenumber (k) in km 1; and wave velocity (a, b) in km s 1. Poisson’s ratio is defined in terms of the Lamé coefficients as s¼

3

m 2;

l 2ð l þ m Þ

The angle of polarization of S-wave e is defined as e ¼ tan

1

uSH uSV

where uSH is the amplitude of the SH component, and uSV that of the SV component. SH and SV are the horizontal and vertical components of the S-wave on the wavefront plane. The coefficients of reflection V and transmission W are given by the respective ratios between the amplitudes of the reflected or transmitted potentials and the incident potential: A A0 A0 W ¼ A0 V¼

where A0 is the amplitude of the incident wave potential, A that of the reflected potential, and A0 of the transmitted potential. Snell’s law for plane media is expressed as p¼

sin i v

and for spherical media p¼

r sin i v

where p is the ray parameter, i the angle of incidence, v the propagation velocity of the medium, and r the position vector along the ray.

8

Introduction

In the case of plane media with propagation velocity varying with depth v(z), the epicentral distance and the travel time of a ray for a surface focus are given by ðh pdz x ¼ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2 p2 0 ðh 2 dz t ¼ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2 p2 0

where  ¼ v 1 and h is the depth of maximum penetration of the ray. The variation of the epicentral distance x with the ray parameter p is given by dx ¼ dp where

2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2 B0 20 p2 B¼

dB dz pdz ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2 p2 0 B2

ðB

1 dv v dz

In spherical media with velocity varying with depth v(r), the epicentral distance, trajectory along the ray, and travel time are given by ð r0 p dr pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼2 2 r  r2 rp ð r0 dr s ¼ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 r2 rp ð r0 dr pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi t¼2 r2 rp v  2

where  ¼ rv 1, r0 is the radius at the surface of the Earth, and rp is the radius at the point of maximum penetration of the ray. The variation of the distance from the epicentre D with the ray parameter p in a spherical medium is dB ðB dr d
2 drpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2 ¼ dp ð1 B0 Þ 20 p2 B 2 Þ  2 p2 0 ð1 where r dv B¼ v dr The radial and vertical components (u1 and u3) of surface waves can be obtained from the potentials ’ and c. The transverse component (u2) is kept apart @’ @c ¼ ’;1 c;3 @x1 @x3 u2 ¼ C exp½ iksx3 þ ikðx1 ctފ @’ @c þ ¼ ’;3 þ c;1 u3 ¼ @x3 @x1 u1 ¼

9

Seismology

where c is the wave propagation velocity and ’ ¼ A exp½ ikrx3 þ ikðx1

ctފ

c ¼ B exp½ iksx3 þ ikðx1 rffiffiffiffiffiffiffiffiffiffiffiffiffi c2 r¼ 1 a2 sffiffiffiffiffiffiffiffiffiffiffiffiffi c2 1 s¼ b2

ctފ

For surface waves, c < b < a, and hence r and s are imaginary. For dispersive waves, the relationship between the phase velocity c and the group velocity U is dc U ¼cþk dk where k is the wavenumber. The position of the seismic focus is given by the coordinates of the epicentre (’0, l0) and the depth h. The time is that of the origin of the earthquake t0. The size is given by the magnitude which is proportional to the logarithm of the amplitude of the recorded waves. For surface waves this is: Ms ¼ log

A þ 1:66 log
þ 3:3 T

where A is the amplitude of ground motion in microns, T is the period in seconds, and ∆ the epicentral distance in degrees. The magnitude of the moment is given by 2 Mw ¼ log M0 3

6:1

where M0 is the seismic moment in N m (newton metres). The seismic moment is related to the displacement of the fault ∆u and its area S: M0 ¼ m
uS The mechanism of earthquakes is given by the orientation of the fracture plane (fault) defined by the angles ’ (azimuth), d (dip), and l (slip angle or rake), or by the vectors n (the normal to the fault plane) and l (the direction of slip). The elastic displacement of the waves produced by a point shear fault is   @Gki uk ðxs ; t Þ ¼ m
uðt ÞS li nj þ lj ni @xj

where Gki is the medium’s Green’s function which, for an isotropic, homogeneous, infinite medium, and P-waves in the far-field regime, is given by
1 r GPki ¼ gi gk d t 2 4pra r a

10

Introduction

The P-wave displacements are given by: uPk ðxs ; t Þ ¼


u_ ðt ÞS  m l i nj þ l j ni g i g j g k 3 4pra r

This equation can be expressed also in terms of the moment tensor Mij uPk ðxs ; t Þ ¼

M_ ij ðt Þ ggg 4pra3 r i j k

Mij is a more general representation of a point source.

Heat flow The Fourier law of heat transfer by diffusion states that the heat flux q_ is proportional to the gradient of the temperature T: q_ ¼

KrT

where K is the thermal conductivity coefficient. The units of heat flow are W m 2. The heat diffusion equation, assuming that K is constant, is given by kr2 T þ

e @T ¼ rCv @t

where Cv is the specific heat, r the density, e the heat generated per unit volume and unit time (heat sources), and k the thermal diffusivity: k¼

K rCv

If there are no heat sources, the diffusion equation is kr2 T ¼

@T @t

In the case of one-dimensional flow with periodic variation of temperature over time, one has:  rffiffiffiffiffiffi rffiffiffiffiffiffiffiffi

 o o T ð z; t Þ ¼ T0 exp zþi z þ ot 2k 2k where z is the vertical direction (positive towards the nadir) and o the angular frequency. In the case of stationary one-dimensional solutions (T constant in time) one obtains from the diffusion equation: T¼

e 2 q_ 0 z þ z þ T0 2K K

11

Geochronology

where T0 and q_ 0 are the temperature and flow at the surface (z ¼ 0). For a spherical Earth, assuming that the thermal conductivity is constant, and that the amount of heat per unit volume depends only on time, the diffusion equation takes the form: 2

@ T 2 @T @T K þ þ eðtÞ ¼ rCv @r2 r @r @t where r is the radial direction. For the stationary case, the above equation reduces to

1 d e 2 dT r ¼ r2 dr dr K Integrating twice, one has T ¼ T0 þ

e  2 R 6K

r2

where T0 is the temperature at the surface (r ¼ R).



Geochronology Geochronology is based on determining the age of a rock by measuring the decay of its radioactive elements. In a sample of radioactive material, the number of atoms that have yet to disintegrate after time t is given by nt ¼ n0 e

lt

where n0 is the initial number of atoms, and l the decay constant. The rate of decay dn/dt is the activity R, so that R ¼ R0 e

lt

where R0 is the initial activity (at t ¼ 0). The half-life (or period) of the sample is the time it takes for the activity R to fall to half its initial value. It is given by: T1= ¼ 2

0:693 l

The mean life-time t of one of the atoms that existed at the start is given by: t ¼

1 l

If a sample consists of NR radioactive nuclei and NE stable nuclei, the time to arrive at the propotion NE/NR is given by

12

Introduction

t¼



1 NE 1þ l NR

If the rubidium–strontium (Rb-Sr) method is used to date a sample, a correction must be made for the contamination of the stable 86Sr isotope relative to the radioisotope 87Sr:    87  87  87  Sr Sr Rb lt ¼ þ e 1 86 Srtotal 86 Sr initial 86 Sr    This expression corresponds to a straight line (isochrone) of slope e lt 1 and intercept  87 Sr  corresponding to the initial content 86 initial . Sr

2

Gravity

Terrestrial geoid and ellipsoid 1. Calculate the geodetic and geocentric latitudes of a point P on the ellipsoid whose radius vector is 6370.031 km, given that m ¼ 3.4425  103, GM ¼ 39.86005  1013 m3 s2, and 6356.742 km is the polar radius. Determine J2 and b. The major semi-axis (equatorial radius) is a and the minor (polar radius) is c, the geocentric latitude is ’, and the geodetic latitude is ’d (Fig. 1). The coefficient m is given by the equation m¼

o 2 a3 GM

where G ¼ 6.671011 m3 kg1 s2 is the gravitational constant, M the Earth’s mass, and the angular velocity is o = 2p/T, where T is the rotation period (T ¼ 24 h). We obtain for the semi-axis a the value a¼



1=3 mGMT 2 ¼ 6378:127 km 4p2

The Earth’s flattening a can be obtained directly since we already know a and c so a¼

a  c 6378:127  6356:742 ¼ ¼ 3:3529  103 c 6378:127

The radius vector to the point P is given by the equation r ¼ a(1  a sin2 ’ ) From this equation we can calculate the geocentric latitude ’:   6370:031 ¼ 6378:127 1  3:3539  103 sin2 ’ ’ ¼ 37 580 2200

The relation between the geocentric ’ and geodetic ’d latitudes is given by tan ’d ¼

1 ð 1  aÞ 2

tan ’

Substituting the already obtained values for a and ’ ’d ¼ 38 090 3500 13

14

Gravity

P

r

C

jd

j a

Fig. 1

The dynamic form factor J2 can be obtained from the equation 3 m a ¼ J2 þ 2 2 Then J2 ¼

2
m ¼ 1:0878  103 a 3 2

From this value we can determine the gravity flattening b using the equation 5 b ¼ m  a ¼ 5:2533  103 2 2. Taking the first-order approximation, let two points of the ellipsoid at 45 N and 30 S be situated at distances of 6367.444 km and 6372.790 km from the centre, respectively. If the normal gravity values are 9.806193 m s2 for the first and 9.793242 m s2 for the second, calculate: the flattening, gravity flattening, coefficient m, equatorial radius, polar radius, dynamic form factor, and the Earth’s mass.

Data r1 ¼ 6367.444 km r2 ¼ 6372.790 km

w1 = 45 N g1 = 9.806193 m s2 w2 = 30 S g 2 = 9.793242 m s2

The normal or theoretical gravity at a point can be expressed in terms of the normal gravity at the equator ge, the gravity flattening b, and the latitude of the point ’:     g1 ¼ ge 1 þ b sin2 ’1 g2 ¼ ge 1 þ b sin2 ’2

If we divide both expressions we obtain:

g1 1 þ b sin2 ’1 ¼ g2 1 þ b sin2 ’2

15

Terrestrial geoid and ellipsoid

From this expression we can obtain the gravity flattening, since we already know g1, g2, ’1, ’2: b¼

g1  g2 ¼ 5:297  103 g2 sin2 ’1  g1 sin2 ’2

The distance r from the centre of the ellipsoid to points on its surface can be given as a function of the flattening a, the equatorial radius a, and the latitude ’: r1 ¼ að1  a sin2 ’1 Þ r2 ¼ að1  a sin2 ’2 Þ If we divide both expressions r1 1  asin2 ’1 ¼ r2 1  asin2 ’2 Thus we obtain the value of the flattening, a ¼ 3:353  103 From this value we find the equatorial radius, a¼

r1 ¼ 6378:137 km 1  asin2 ’1

The polar radius c can be found from this value and the flattening: a¼

ac a

and

c ¼ að1  aÞ ¼ 6356:751 km

The coefficient m is obtained from a and b: 5 aþb¼ m 2

2 m ¼ ða þ bÞ ¼ 3:460  103 5

From this value we can obtain the value of the Earth’s mass M from m¼

o 2 a3 GM

2p p where T ¼ 24 hours. T Therefore

with o ¼

M¼

4p2 a3 ¼ 5:946  1024 kg T 2 Gm

3. Obtain the value of the terrestrial flattening in the first-order approximation, given that the normal gravity values for two points of the ellipsoid are: Point 1: w1 ¼ 42º 200 g1 ¼ 980.389 063 Gal Point 2: w2 ¼ 47º 300 g2 ¼ 980.854 830 Gal Take the equatorial radius to be 6378.388 km.

16

Gravity

For this problem we use the equations of Problems 1 and 2: g1 ¼ ge ð1 þ bsin2 ’1 Þ g2 ¼ ge ð1 þ bsin2 ’2 Þ If we divide these expressions: g1 1 þ bsin2 ’1 ¼ g2 1 þ bsin2 ’2 and we can solve for the gravity flattening, b: b ¼ 5:288 2675  103 Using the gravity flattening b we can determine the value of gravity at the equator, ge: ge ¼

g1 ¼ 978:043 614 Gal 1 þ bsin2 ’1

Using the following equations 5 aþb¼ m 2 3 m a ¼ J2 þ 2 2 o2 a3 m¼ GM we derive the expression ge ¼

GM ð1  bÞ þ o2 a a2

and substituting the values we obtain GM ¼ 3.986 5415  1014 m3 s2 From this value, taking T ¼ 24 hours, we obtain m¼

o 2 a3 4p2 a3 ¼ 2 ¼ 3:442 5698  103 GM T GM

And finally the Earth’s flattening is 5 a ¼ m  b ¼ 3:318 1575  103 2 4. P is a point of the terrestrial ellipsoid at latitude 60 ºS and distance to the centre of 6362.121 km. The Earth’s mass is 5.9761  1024 kg and the ratio between the polar and equatorial semi-axes is 0.9966. Taking the first-order approximation, calculate: (a) The flattening and the coefficient J2. (b) The value of normal gravity in mGal at P. As in the previous problems we use the equations given in Problems 1 and 2.

17

Terrestrial geoid and ellipsoid

(a) From the equation for the flattening (Problem 2) a¼1

c ¼ 3:4  103 a

The value of the equatorial radius a is obtained from the equation a¼

r ¼ 6 378 386 m 1  asin2 ’

The coefficient m is given by m¼

o 2 a3 4p2 a3 ¼ 2 ¼ 3:4429  103 GM T GM

The gravity flattening b is given by 5 b ¼ m  a ¼ 5:2072  103 2 The dynamic form factor J2 is found from the relation J2 ¼

2a  m ¼ 1:1190  103 3

(b) The normal gravity at that point is given by g ¼ ge ð1 þ bsin2 ’Þ ¼ 981 856:3 mGal 5. At a point P on the ellipsoid at latitude 50 ºS, the value of normal gravity is 9.810 752 m s2 and the distance to the centre of the Earth is 6365.587 km. Given that the mass of the Earth is 5.976  1024 kg and the ratio between the minor and major semi-axes is c/a ¼ 0.996 6509, calculate: (a) The flattening, equatorial radius, gravity flattening, dynamic form factor, and coefficient m. (b) The normal gravity at the equator. (c) The centrifugal force at P.

Data ’ ¼ 50 S r ¼ 6365.587 km

g ¼ 9.810 752 m s2.

(a) According to Problem 1, the flattening is given by a¼1

c ¼ 3:349  103 a

and the equatorial radius a is a¼

r ¼ 6378:122 km 1  asin2 ’

18

Gravity

Taking T ¼ 24 hours, the coefficient m is then given by 4p2 a3 ¼ 3:4425  103 T 2 GM

m¼

b and J2 can be obtained from the equations 5 b ¼ m  a ¼ 5:2571  103 2

J2 ¼

2
m a ¼ 1:0852  103 3 2

(b) The normal gravity ge at the equator is given by ge ¼

g ¼ 9:780 579 m s2 1 þ bsin2 ’

(c) The centrifugal force at point P is given by its radial and transverse components f ¼ fr er þ fy ey where, since 90  y = ’, fr ¼ o2 r sin2 y ¼ o2 r cos2 ’ ¼ 0:013 909 m s2 fy ¼ o2 r sin y cos y ¼ o2 r cos ’ sin ’ ¼ 0:016 576 m s2 6. Taking the first-order approximation, calculate the Earth’s flattening a, gravity flattening b, dynamic form factor J2, and polar radius c, given that: ge ¼ 978.032 Gal (normal gravity at the equator) a ¼ 6378.136 km (equatorial radius) GM ¼ 39:8603  1013 m3 s2 and that for a point on the ellipsoid at latitude 60 ºN the normal gravity value is 981 921 mGal. Calculate also the radius vector of this point and the gravitational potential. Assuming a first-order approximation, the expression for the normal gravity is g ¼ ge ð1 þ b sin2 ’Þ The value of b is given by b¼

g  ge ¼ 5:302  103 ge sin2 ’

The coefficient m, taking T ¼ 24 hours, is m¼

4p2 a3 ¼ 3:442  103 T 2 GM

a and J2 are determined from the equations 5 a ¼ m  b ¼ 3:303  103 2

J2 ¼

2
m a ¼ 1:055  103 3 2

19

Terrestrial geoid and ellipsoid

The polar radius c is determined from the flattening a¼

ac ) c ¼ að1  aÞ ¼ 6357:069 km a

The radius vector at the point of latitude 60 ºN is given by r ¼ að1  asin2 ’Þ ¼ 6362:335 km The gravity potential at the same point, in the first-order approximation, is found using Mac Cullagh’s formula,   
r 3 m 2 GM J2
a2  2 1 cos ’ ¼ 6:263 57  107 m2 s2 U¼ 3 sin ’  1 þ r a 2 2 r

7. Assuming that the Moon is an ellipsoid of equatorial radius 1738 km and polar radius 1737 km, with J2 = 3.8195  104 and a mass of 7.3483  1022 kg, calculate its period of rotation. First, we calculate the lunar flattening a, a¼

ac ¼ 5:7537  104 a

The coefficient m is found from the values of a and J2: 3 m a ¼ J2 þ ) m ¼ 2a  3J2 ¼ 7:5900  106 2 2 From the value of m we find the period of rotation T: 2 3 1 4p2 a3 4p a 2 )T ¼ m¼ 2 ¼ 27:32 days T GM mGM 8. Calculate, in the first-order approximation, the latitude and radius vector of a point P of the terrestrial ellipsoid for which the value of normal gravity is 979.992 Gal, given that the Earth’s mass is 5.976  1024 kg and that normal gravity for another point Q at 50º S latitude is 981.067 Gal, for the equator is 978.032 Gal, and that J2 is 1.083  103. The gravity flattening b is found from the normal gravity at Q with latitude ’1 = 50 : g1 1 ge ¼ 5:288  103 b¼ 2 sin ’Q Using the same expression and the normal gravity at P we calculate its latitude gA g 1 sin ’P ¼ e ; b 2

’P ¼ 37 590 46:5700

The Earth’s flattening is given by a¼

15 b J2 þ ¼ 3:353  103 8 4

20

Gravity

and the value of the coefficient m by m ¼ 2a  3J2 ¼ 3:457  103 From the value of m we obtain the equatorial radius a:

1 mGM 3 ¼ 6 387 062:758 m a¼ o2 where we have substituted o = 2p/T, taking T ¼ 24 hours. Finally, we find the radius vector of point P r ¼ að1  asin2 ’Þ ¼ 6 378 946:678 m 9. At a point P on the terrestrial ellipsoid of latitude 70 ºS and radius vector 6359.253 km, the value of normal gravity is 982.609 Gal. If the mass of the Earth is 5.9769  1024 kg and the equatorial radius is 6378.136 km, calculate the value of normal gravity at the Pole, the dynamic form factor, and the centrifugal force at the Pole and the equator. The flattening a is given by a¼

1
r ¼ 3:3528  103 1  a sin2 ’

Putting T ¼ 24 hours, the coefficient m is obtained from the equation m¼

o 2 a3 ¼ 3:4425  103 GM

From a and m we find the gravity flattening b: 5 b ¼ m  a ¼ 5:2535  103 2 Normal gravity at the equator is found from the value of the gravity at point P: ge ¼

g ¼ 9:780 72 m s2 1 þ b sin2 ’

From this value we find the normal gravity at the Pole: gp ¼ ge ð1 þ bÞ ¼ 9:832 10 m s2 The dynamic form factor is found from the values of a and m: 2
m a ¼ 1:0877  103 J2 ¼ 3 2 The centrifugal force is given by the expression:

f ¼ o2 r sin2 yer þ o2 r sin y cos yey At the pole, y = 90  ’ = 0 ! f = 0. At the equator, y ¼ 90º ! f = o2 a er = 0.033 73 er m s2

21

Terrestrial geoid and ellipsoid

10. Let two points of the ellipsoid be of latitudes w1 and w2, with radius vectors 6372.819 km and 6362.121 km, respectively. The ratio of the normal gravities is 0.997 37, the flattening 3.3529  103, and the gravity flattening 5.2884  103. Calculate: (a) The Earth’s mass. (b) The latitude of each point and the dynamic form factor. (a) The equatorial radius a can be obtained from the ratio of the two normal gravities
a  r  1 2 1 þ b g1 1 þ bsin ’1 aa
¼l ¼ ¼ g2 1 þ bsin2 ’2 1 þ b a  r2 aa

where l ¼ 0.997 37. We solve for a and obtain a¼

bðlr2  r1 Þ ¼ 6382:94 km ð a þ bÞ ð l  1 Þ

We calculate the mass of the Earth from the flattening and gravity flattening as in Problem 2: 2 m ¼ ða þ bÞ ¼ 3:4565  103 5

M¼

4p2 a3 ¼ 5:9653  1024 kg T 2 Gm

(b) The latitudes at each point are calculated from the radius vectors a  r1 sin2 ’1 ¼ ! ’1 ¼ 43 270 5800 aa a  r2 ! ’2 ¼ 80 330 4600 sin2 ’2 ¼ aa The dynamic form factor is obtained from the values of a, b, and m: 9 b ¼ 2a þ m  J2 2 2 J2 ¼ ð2a þ m  bÞ ¼ 1:0831  103 9 11. Let a point A have a value of gravity of 9793 626.8 gu and a geopotential number of 32.614 gpu. Calculate the gravity at a point B, knowing that the increments in dynamic and Helmert height over point A are 271.116 m and 271.456 m, respectively. Take g45 ¼ 9.806 2940 m s2. Give the units for each parameter. The dynamic heights at points A and B are given by: CA g45 C B HDB ¼ g45 HDA ¼

where C is the value of the geopotential at each point gravity for a point on the ellipsoid at 45º latitude.

N P

j¼1

gj dhj

!

and g45 the normal

22

Gravity

Subtracting both equations, HDB  HDA ¼

ðCB  CA Þ g45

Solving for CB,   CB ¼ CA þ g45 HDB  HDA ¼ 298:478 gpu

If heights are given in km and normal gravity in Gal, geopotentials are in gpu (geopotential units) 1 gpu ¼ 1 kGal m ¼ 1 Gal km The Helmert orthometric height H is given by H¼

C g þ 0:0424H

ð11:1Þ

where C is in gpu, g in Gal, and H in km. Solving for H: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g  g2 þ 4  0:0424C H¼ 2  0:0424 Since the point A is above the geoid (CA > 0), we take the positive solution, HA ¼ 33:301 m Then, the Helmert height at point B is HB ¼ HA þ
HAB ¼ 304:757 m The gravity at point B is calculated using Equation (11.1) gB ¼

CB  0:0424HB ¼ 979:382 75 Gal HB

12. Calculate the value of gravity in gravimetric units and mGal of a point on the Earth’s surface whose orthometric (Helmert) and dynamic heights are 678.612 m and 679.919 m, respectively, taking g 45 ¼ 9.806 294 m s2. The geopotential is calculated from the dynamic height, HD ¼

C ) C ¼ HD g45 ¼ 666:748 gpu g45

where HD is given in km and g45 in Gal. Knowing the geopotential, we calculate the gravity from the orthometric (Helmert) height H, using its definition, H¼

C C ) g ¼  0:0424H ¼ 982:489 34 Gal ¼ 9 824 893:4 gu g þ 0:0424H H

13. If at a point on the surface of the Earth of Helmert height 1000 m one observes a value of gravity of 9.796 235 m s2, calculate the average value of gravity

23

Terrestrial geoid and ellipsoid

between that point and the geoid along the direction of the plumb-line, and the point’s geopotential number. The mean value of gravity between a height H and the surface of the geoid is given by ð 1 H g ¼ gðzÞdz H 0 where g(z) is the value of gravity at a distance z from the geoid along the vertical path to a point of height H. This value can be obtained using the Poincaré and Prey reduction from the value of g observed at the Earth’s surface at a point of height H, gðzÞ ¼ g þ 0:0848ðH  zÞ Then ð ð 1 H 1 H gðzÞdz ¼ ½g þ 0:0848ðH  zÞdz g ¼ H 0 H 0

H 1 gz þ 0:0848 Hz  0:0424z2 0 ¼ H g ¼ g þ 0:0424 H ¼ 979:6659 Gal where g is given in Gal and H in km. The geopotential C can be obtained from the formula for the Helmert height, H¼

C ) C ¼ ðg þ 0:0424 HÞ H ¼ 979:666 gpu g þ 0:0424H

14. For two points A and B belonging to a gravity measurement levelling line, one obtained: gA = 9.801 137 6 m s2 CA = 933.316 gpu Gross increment elevation: DhBA ¼ 20:340 B Increment in dynamic height: H A D  H D ¼ 20:340 m. Given that the normal gravity at 45 º latitude is 9806 294 gu, calculate the Helmert heigth of point B. As in Problem 11, the dynamic heights at A and B are given by CA g45 CB HDB ¼ g45

HDA ¼

Subtracting both equations: HDB  HDA ¼

ðCB  CA Þ g45

Solving for CB:   CB ¼ CA þ g45 HDB  HDA ¼ 913:371 gpu

24

Gravity

The geopotential at B can be obtained from the gross increment in elevation between A and B,
g þ g  A B CB ¼ CA þ
hBA 2

and, solving for gB,

gB ¼

2ðCB  CA Þ  gA ¼ 980:103 08 Gal
hBA

Finally we calculate the orthometric Helmert height at point B, H¼

C g þ 0:0424H

Substituting the values for point B, and solving for H we obtain pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi gB  gB 2 þ 4  0:0424CB H¼ ¼ 931:875 m 2  0:0424 15. A, B, and C are points connected by a geometric levelling line. Given that the normal gravity at a latitude of 45º is 980.6294 Gal, complete the following table:

Station

Gravity (Gal)

Height Increment (m)

Geopotential Number (gpu)

Dynamic Height (m)

Helmert Height (m)

A B C

979.88696 ? 979.88665

– 0.541 ?

664.982 ? ?

? 677.577 ?

? ? 657.134

Station A Dynamic height: HDA ¼

CA ¼ 678:118 m g45

Helmert height: C g þ 0:0424H pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g  g2 þ 4  0:0424  C H¼ ¼ 678:611 m 2  0:0424 H¼

Station B The geopotential number is found from the dynamic height: CB ¼ g45 HDB ¼ 664:452 gpu

ð15:1Þ

Earth’s gravity field and potential

25

From this value and the difference in height with respect to station A we find the gravity at B:
g þ g  A B CB ¼ CA þ
hA B 2 from which we get gB = 979.877 84 Gal The Helmert height of B is found as in station A:

HB ¼ 678:077 m

Station C From the known values of gravity and Helmert height we find the geopotential number (Equation 15.1) CC ¼ gHC þ 0:0424HC2 ¼ 661:574 gpu To calculate the difference in height of C with respect to B we begin with the expression
g þ g  B C
hCB CC ¼ CB þ 2 from which


hCB ¼

2ðCC  CB Þ ¼ 2:937 m gB þ gC

The dynamic height is found directly from the geopotential number: HDC ¼

CC ¼ 674:642 m g45

The complete table is:

Station

Gravity (Gal)

Height increment (m)

Geopotential number (gpu)

Dynamic height (m)

Helmert height (m)

A B C

979.88696 979.87784 979.88665

– 0.541 2.937

664.982 664.452 661.574

678.118 677.577 674.642

678.611 678.077 657.134

Earth’s gravity field and potential 16. Suppose an Earth is formed by a sphere of radius a and density r, and within it there are two spheres of radius a/2 with centres located on the axis of rotation. The density of that of the northern hemisphere is 5r and of that of the southern hemisphere is r /5. The value of the rotation is such that m ¼ 0.1. Determine: (a) The potential U in the r3 approximation. (b) The values of gr and gu for a point on the equator in the r2 approximation.

26

Gravity

(c) The error made in (b) with respect to the exact solution. (d) The deviation of the vertical from the radial at the equator. (a) The gravitational potential is the sum of the potentials of the three spheres V ¼ V1 þ V2 þ V3 ¼

GM GM1 GM2 þ þ 0 r q q

ð16:1Þ

where r is the distance from a point P to the centre of the sphere of radius a and mass M, where M is given by 4 M ¼ pra3 3 q and q0 are the distances to the centres of the two spheres in its interior in the northern and southern hemispheres which have differential masses M1 and M2, respectively (Fig. 16). The differential masses are those corresponding to the difference in density in each case with respect to the large sphere: 4 a3 M M1 ¼ pð5r  rÞ ¼ 3 8 2 differential mass of the sphere in the northern hemisphere  a3 4
r M ¼ M2 ¼ p  r 3 5 8 10

differential mass of the sphere in the southern hemisphere The distance q can be calculated using the cosine law ffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2
a 2
a a 2 q¼ r þ cos y  2 r cos y ¼ r 1 þ 2 2 2 2r 2r Considering this expression, 1/q corresponds to one of the generating functions of the Legendre polynomials. Then 1/q, in the first-order approximation, is given by    1 1 a 1
a 2  2 ¼ 1 þ cos y þ 3cos y  1 q r 2r 2 2r

Since cos y0 =  cos y, 1/q0 is given by    1 1 a 1
a 2  2 ¼ 1  cos y þ 3cos y  1 q0 r 2r 2 2r P 5ρ

q r

r2

a/2 θ

α a r3

ρ

ρ/5

Fig. 16

q' θ'

Earth’s gravity field and potential

27

If we substitute in Equation (16.1), the potentials for each sphere are given by

 GM GM 1 a a2  V2 ¼ þ 2 cos y þ 3 3cos2 y  1 V1 ¼ r 2 r 2r 8r

 GM 1 a a2  2 V3 ¼   cos y þ 3 3cos y  1 10 r 2r2 8r

Then, the total gravity potential is the sum of the three gravitational potentials plus the potential of the centrifugal force due to the rotation: 



1 1 1 1 1 a U ¼ GM 1 þ  þ þ cos y 2 10 r 4 20 r2

   1 1 a2  1 2 2 2 2  þ 3 cos y  1 þ r o sin y 16 80 r3 2 In terms of the coefficient m, given here by m ¼

o2 a3 , GM

   1
r 2 2 GM 7 a 3 a2 1 a3  2 þ cos y þ 3cos y  1 þ m sin y U¼ a 5 r 10 r2 20 r3 2 a (b) Using this first-order approximation of the potential, the radial and tangential components of gravity at the equator, r = a and y = 90 , putting m ¼ 0.1, are @U GM ¼ 1:3 2 @r a 1 @U GM ¼ 0:3 2 gy ¼ r @y a

gr ¼

(c) To calculate exactly the value of gr at the equator we have to calculate the exact contribution of each of the three spheres plus the centrifugal force (m ¼ 0.1): gr ¼ gr1 þ gr2 þ gr3  m

GM a2

GM a2 GM gr2 ¼  2 cos a 2r2 GM cos a gr3 ¼  10r32

gr1 ¼ 

where rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 r 2 ¼ a2 þ ¼ r 3 4 and a is the angle which forms r2 and r3 with the equator (Fig. 16)

28

Gravity

sin a ¼ Then GM gr ¼  2 a

a=2 1 ¼ pffiffiffi r2 5



4 4 GM 1 þ pffiffiffi  pffiffiffi  m ¼ 1:19 2 a 5 5 25 5 0 1

GM 1 C B GM 1 pffiffiffi þ pffiffiffiA gy ¼ gy1 þ gy2 ¼ @ 5 2 5 5 2 a 10 a2 5 4

4 GM 2 2 GM ¼ 2  pffiffiffi þ pffiffiffi ¼ 0:14 2 a a 5 5 25 5

The error made in the first-order approximation with respect to the exact solution is GM GM ¼ 0:11 2 a2 a GM GM
gy ¼ ð0:3 þ 0:14Þ 2 ¼ 0:16 2 a a
gr ¼ ð1:19 þ 1:3Þ

(d) The deviation of the vertical with respect to the radial direction is given by the angle i which is determined from the gravity components gr and gy. At the equator this angle is: • Using the first order approximation tan i ¼

gy 0:3 ) i ¼ 13:0 ¼ gr 1:3

• Using the exact values tan i ¼

0:16 ) i ¼ 7:6 1:19

17. A spherical planet is formed by a sphere of radius a and density r, and inside it a sphere of radius a/2 and density 5r centred at the midpoint of the radius of the northern hemisphere. There is no rotation. (a) Determine J0, J1, and J2. (b) What is the deviation of the vertical from the radial at the equator? (a) The total gravitational potential is the sum of the potentials of the two spheres (Fig. 17) where g, is the attraction due to the potential V1 and g2 that due to the potential V2: V ¼ V1 þ V2 ¼

GM GM 0 þ r q

where r and q are the distances from a point P to the centres of the large and small spheres, respectively.

Earth’s gravity field and potential

29

P

q

5r r a /2 q

g2 i

g1

a

r

Fig. 17

As we did in Problem 16, for the small sphere of radius a/2 we take the differential mass M 0 4 a3 16 M M 0 ¼ p ð5r  rÞ ¼ pra3 ¼ 3 8 24 2 where M is the mass of the sphere of radius a and density r. For 1/q we take the first-order approximation of the Legendre polynomial, as we did in Problem 16:

 1 1 a 1
a 2  2 ¼ 1 þ cos y þ 3cos y  1 q r 2r 2 2r Then, the expression for the gravitational potential V is:

 GM GM 1 a 1 a2  2 þ þ 2 cos y þ 3cos y  1 V ¼ r 2 r 2r 4 4r3

 3 a a2  2 ¼ GM þ cos y þ 3cos y  1 2r 4r2 32r3

(b) We know that the potential can be expressed by an expansion in zonal spherical harmonics (Legendre polynomials) given in the first-order approximation by




a3 1   GM a a 2 J0 þ J1 cos y þ J2 3cos2 y  1 V ¼ a r r r 2 Comparing the two expressions we obtain,

30

Gravity

3 2 1 J1 ¼ 4 1 J2 ¼ 16 J0 ¼

The components of gravity at the surface of the large sphere (r ¼ a) are:

 @V GM 3 1 3  ¼ 2   cos y  3cos2 y  1 @r a 2 2 32

1 @V GM 1 3 gy ¼ ¼ 2  sin y  cos y sin y r @y a 4 16

gr ¼

and at the equator, y ¼ 90 : 45GM 32a2 GM gy ¼  2 4a

gr ¼ 

At the equator the deviation of the vertical with respect to the radial direction is gy 8 ¼ gr 45 i ¼ 10:08

tan i ¼

18. Suppose an Earth is formed by a sphere of radius a and density r, and within it there are two spheres of radius a/2 and density 2r with centres located on the axis of rotation in each hemisphere. If M is the mass of the sphere of radius a, calculate: (a) The potential U(r,u) and the form of the equipotential surface passing through the Poles. (b) The component gr of gravity in the first-order approximation for points on the surface. (c) Calculate gr directly at the Pole and the equator, and compare with the first-order approximation. (a) This problem is similar to Problem 16, but now the density of the two spheres is the same. The total gravity potential is the sum of the gravitational potentials of the three spheres (V, V1 and V2) plus the potential due to the rotation F: where

U ¼ V þ V1 þ V2 þ F GM r GM 0 V1 ¼ q1 GM 0 V2 ¼ q2 1 2 2 2 F ¼ o r sin y 2 V¼

ð18:1Þ

Earth’s gravity field and potential

31

q1

2r

P

r r

q2 a/2

q

q

a a

q’

2r

Fig. 18

and where M is the mass of the large sphere of radius a and M 0 the differential mass of each of the small spheres of radius a/2, r is the distance from a point P to the centre of the large sphere, and q1 and q2 the distances from P to the centres of the small spheres (Fig. 18). As in Problem 16 the differential mass is given by the difference in density between the large and the small spheres:
a 3 M 4 M 0 ¼ pð2r  rÞ ¼ 3 2 8 The inverse of the distance 1/q can be approximated by  
a 2 1   1 1 a 2 ¼ 1 þ cos y þ 3cos y  1 q1 r 2r 2r 2

and since cos y0 =  cos y

 
a 2 1   1 1 a 2 3cos y  1 ¼ 1  cos y þ q2 r 2r 2r 2

The potential of the rotation can be written in terms of the coefficient m = a3o2/GM,

GM 1 1 a3 o2 3 GM m
r 3 2 r sin2 y ¼ sin y F¼ 3 r 2 a GM r 2 a Substituting in Equation (18.1)   
r 3 m 2 GM 10 1
a 2  2 þ sin y U¼ 3cos y  1 þ r 8 8 2r a 2

ð18:2Þ

32

Gravity At the Poles, r ¼a and y = 0 , and the potential is   GM 10 2 21 GM U poles ¼ þ ¼ a 8 32 16 a The form of the equipotential surface which passes through the Poles (r ¼ a) is obtained from Equation (18.2) r¼

  
r 3 m 2 GM 10 1
a 2  þ 3cos2 y  1 þ sin y Upoles 8 8 2r a 2

Making the approximation r ¼ a in the right-hand side: r¼

   32 10 m 2 1  a þ sin y þ 3cos2 y  1 42 8 2 32

and substituting cos2 y = 1  sin2 y, we obtain 

 32 3 m 2  r ¼a 1 sin y 42 32 2

This is the equation of an ellipse with flattening a ¼ (32/42)(3/32m/2). Since there is symmetry with respect to the axis of rotation, the equipotential surface is an ellipsoid of revolution.

 At the poles: y ¼ 0 ) rp ¼ a.  32 m 3 At the equator: y ¼ 90 ) re ¼ a 1 þ  42 2 32 Depending on the value of m, we have the following cases, m 3 ¼ ) re ¼ a ) sphere 2 32 m 3 < ) re < a ) prolate ellipsoid 2 32 m 3 > ) re > a ) oblate ellipsoid 2 32

3 < a ) prolate ellipsoid m¼0)r ¼a 1 42 (b) For the gravity at the Pole, in the first-order approximation, we take the derivative of the potential (18.2) and substitute y ¼ 0 ) rp ¼ a:   @U GM 10 6 GM gr ¼ ¼ 2   ¼ 1:4375 2 @r a 8 32 a (c) The exact solution for the gravity at the pole is the sum of the attractions of the three spheres: gr ¼ 

GM GM GM GM  2 ¼ 1:5555 2 a2 2a 18a2 a

Earth’s gravity field and potential

33

At the equator we take the derivative of the potential and substitute r = a and y = 90 :   @U GM 10 3 gr ¼ ¼ 2  þ þm @r a 8 32   GM 37 GM  m ¼  2 ½1:1562  m ¼ 2 a 32 a For the exact solution we write gr ¼ 

GM 2GM  cos a þ o2 a a2 8q2

From Fig. 18 the distance q is given by a2 5 þ a2 ¼ a2 4 ffiffiffi 4 r 4 cos a ¼ 5 q2 ¼

Therefore " # rffiffiffi GM 8 4 GM gr ¼  2 1 þ  m ¼  2 ½1:1789  m a 40 5 a The approximated values are smaller than the exact solutions. 19. For the case of Problem 18, if GM = 4  103 m3 s2, a = 6  103 km, and v = 7  105 s1, calculate the values of J2, a, m, H, and b. From the definition of m we obtain m¼

a3 o2 216  1018  49  1010 ¼ ¼ 2:6  103 GM 4  1014

The value of J2 is obtained by comparing the two expressions for the potential U (Problem 18):  
a2 1   m
r 3 2 GM U¼ 1þ J2 3cos2 y  1 þ sin y r r 2 2 a  
 2   5 GM 1 a m
r 3 2 3cos2 y  1 þ sin y 1þ U¼ 4 r 40 r 2 a Then, J2 = 0.05. The flattening is obtained from the relation 3 m a ¼ J2 þ 2 2 a¼

3 2:3  103  50  103 þ ¼ 0:0765 2 2

34

Gravity

X3

2r

r a/2

X1

a 2r

Fig. 19a

The gravity flattening is given by b¼

gp  ge 1:555  1:175 ¼ ¼ 0:323 ge 1:175

where we have used the values of gravity at the Pole and equator obtained in Problem 18, and in the latter we have substituted the value obtained for the coefficient m. The dynamic ellipticity H is defined as the ratio of the moments of inertia with respect to the polar and equatorial radius (Fig. 19a): H¼

CA C

where A and C are the moments of inertia of a sphere respect to the polar and equatorial radi: (axes x1 and x3). The moment of inertia of a sphere of radius R is 2 Isph ¼ MR2 5 We have to add to the moment of inertia of the sphere of radius a the moments of inertia of the two internal spheres of radius a/2. For the C-axis (x3) we have 2 2 M a2 ¼ 0:425Ma2 IC ¼ Isph a þ 2Isph a2 ¼ Ma2 þ 2 5 58 4 For the A-axis (x1) the moment of inertia of each of the small spheres is given by (Fig. 19b)

Earth’s gravity field and potential

35

a

b

R h

Fig. 19b

I ¼ ICM þ Mh2 since in this case the A axis does not coincide with the centre of mass, where R ¼ a/2 and h ¼ a/2: Isph a=2 ¼

2 M a2 M a 2 þ ¼ 0:044Ma2 58 4 8 4

2 IA ¼ Isph a þ 2Isph a=2 ¼ Ma2 þ 2  0:044Ma2 ¼ 0:488Ma2 5 Finally H¼

C  A IC  IA 0:425  0:488 ¼ ¼ 0:147 ¼ C IC 0:425

20. Suppose an Earth is formed by a sphere of radius a and density r, and within it there is a sphere of radius a/2 and density 5r centred at the midpoint of the northernhemisphere polar radius. If m ¼ 1/8 and M is the mass of the sphere of radius a, determine: (a) The form of the equipotential surface passing through the North Pole. (b) For latitude 45º, the astronomical latitude and the deviation of the vertical from the radial. (a) The gravitational potential is the sum of the potentials for the sphere of radius a and that of the sphere of radius a/2 (Fig. 20): V ¼ V1 þ V2 ¼

GM GM1 þ r q

As in the previous problems the potential of the small sphere is given in terms of differential mass M1:

36

Gravity

i

gq

a/2

P

q g

5r

gr

fa

q

j

a

r

Fig. 20


a3 M 4 ¼ M1 ¼ pð5r  rÞ 3 2 2

and for the inverse of the distance 1/q we use the approximation  
a 2 1   1 1 a 2 ¼ 1 þ cos y þ 3cos y  1 q r 2r 2r 2

Then, the total gravitational potential is    GM 3 a 1
a2  V ¼ þ cos y þ 3cos2 y  1 r 2 4r 16 r

The total potential U is the sum of the gravitational potential V plus the potential of rotation F, where 1 F ¼ r2 o2 sin2 y 2 and using the coefficient m = o2a3/GM = 1/8, we have   
r 3 m 2 GM 3 a 1
a2  U¼ þ cos y þ 3cos2 y  1 þ sin y r 2 4r 16 r a 2 At the North Pole, y = 0 and r ¼ a, and the value of the potential is   GM 3 1 1 15GM Up ¼ þ þ ¼ a 2 4 8 8a

Earth’s gravity field and potential

37

The form of the equipotential surface is found by putting U ¼ Up:   
r 3 m 2 15GM GM 3 a 1
a2  ¼ þ cos y þ 3cos2 y  1 þ sin y 8a r 2 4r 16 r a 2

Putting r ¼ a inside the square brackets and solving for r we find   4 1 1 r ¼ a 1 þ cos y þ cos2 y 5 6 12

(b) The deviation of the vertical with respect to the radial direction is given by the angle i: tan i ¼

gy gr

To find this value we have to calculate the two components of gravity  

@U 3 1 2a 1 3a2 1 2r 2 2 ¼ GM  2  gr ¼ cos y  3 cos y  1 þ sin y @r 2r 4 r3 16a3 16 r4   1 @U GM a a2 r2 ¼  2 sin y  6 cos y sin y þ sin y cos y gy ¼ r @y r 4r 16r3 8a3 For a point on the surface we put r ¼a:   GM 19 1 11 2 gr ¼  2 þ cos y þ cos y a 16 2 16   GM 1 1 sin y þ sin y cos y gy ¼  2 a 4 4 and for latitude 45º GM 1:88 a2 GM gy ¼  2 0:30 a 0:30 ) i ¼ 9:0. Then the angle i is given by tan i ¼ 1:88 The astronomical latitude is gr ¼ 

fa ¼ 90  y  i ¼ 36:0 21. If the internal sphere of Problem 20 is located on the equatorial radius at longitude zero, find expressions for the components of gravity: gr , gu , gl. As in the previous problem the differential mass of the small sphere M1 is (Fig. 21a): M1 ¼

M 2

The total potential U is the sum of the gravitational potentials V and V1, and the potential due to rotation F. According to Fig. 21b, using the relations of spherical triangles, if ’ and l are the coordinates of the point where the potential is evaluated, then

38

Gravity

B

P q

r

y j

l = 0° A

a/2

l

Fig. 21a B 90°

l

90°– j P

y

A

Fig. 21b

cosc ¼ cos 90º cos(90º’) þ sin 90º sin(90º’) cosl cosc ¼ cos’ cosl Using the expression for 1/q as in Problem 16,  
a 2 1   1 1 a ¼ 1 þ cos c þ 3cos2 c  1 q r 2r 2r 2

The gravitational potential of the small sphere is given by    GM a a2  2 2 1 þ cos ’ cos l þ 2 3cos ’cos l  1 V1 ¼ 2r 2r 8r

The total potential U is given by:    1 2 2 2 GM 3 a a2  2 2 U¼ þ cos ’ cos l þ r o cos ’ 3cos ’cos l  1 þ r 2 4r 2 16r2

Earth’s gravity field and potential

39

Using the coefficient m and sin y = cos ’, we obtain    r2 m 2 3 a2 a2  2 2 U ¼ GM þ sin y cos l þ 3sin ycos l  1 þ 3 sin y 2r 4r2 16r3 a 2

The three components of gravity are found by differentiating U with respect to r, y, and l:    r @U 3 a2 3a2  2 2 2 gr ¼ ¼ GM  2  3 sin y cos l  3sin ycos l  1 þ msin y @r 2r 2r 16r4 a3  2  1 @U a a2 r cos y cos l þ 6 cos l sin y cos y þ m sin y cos y ¼ GM gy ¼ 4r3 16r4 r @y a3   1 @U a2 3a2 gl ¼ ¼ GM  3 sin l  4 sin y cos l sin l r sin y @l 4r 8r 22. A planet is formed by a sphere of radius a and density r, with a spherical core of density 5r and radius a/2 centred on the axis of rotation in the northern hemisphere and tangential to the equator. The planet rotates with m ¼ 1/4. For the point at coordinates (45º N, 45º E), calculate: (a) The astronomical latitude. (b) The deviation of the vertical from the radial. (c) The angular velocity of rotation that would be required for this deviation to be zero. (a) The gravitational potential is the sum of the potentials of the two spheres (Fig. 22): V ¼ V1 þ V2 ¼

GM GM 0 þ r q

ð22:1Þ

P

q

5r

g gr r j

a

r

Fig. 22

i

gq

a/2

40

Gravity

As in Problem 16, the inverse of the distance from a point P to the centre of the small sphere, 1/q, can be approximated by    1 1 a 1
a 2  ¼ 1 þ cos y þ 3cos2 y  1 q r 2r 2 2r

As in previous problems we use the differential mass of the small sphere, 4 a3 2 M 0 ¼ p ð5r  rÞ ¼ pra3 3 8 3 4 and since M ¼ pra3 , then M 0 = M/2. 3 Substituting in Equation (22.1) we obtain    GM GM 1 a 1 a2  2 V ¼ þ þ 2 cos y þ 3cos y  1 r 2 r 2r 2 4r3    3 a a2  2 þ 2 cos y þ 3cos y  1 ¼ GM 2r 4r 16r3

The total potential U is the sum of V plus the potential due to rotation F:    GM 3 a a2  1 2 þ cos y þ 3cos y  1 þ r2 o2 sin2 y U ¼V þF¼ r 2 4r 2 16r2

The components of gravity gr and gy are    @U 3 a 3a2  2 ¼ GM  2  3 cos y  3cos y  1 þ ro2 sin2 y gr ¼ @r 2r 2r 16r4   1 @U a a2 gy ¼ 6 cos y sin y þ ro2 sin y cos y ¼ GM  3 sin y  r @y 4r 16r4

ð22:2Þ ð22:3Þ

For a point on the surface of the large sphere and coordinates 45º N, 45º E, we have that y = 45 , r ¼ a, and a3 o 2 1 ¼ GM 4 Putting these values in (22.2) and (22.3), we obtain m¼

gr ¼ 1:82

GM GM and gy ¼ 0:24 2 a2 a

To find the astronomical latitude we first have to find the deviation of the vertical with respect to the radial: tan i ¼

gy ¼ 0:13 ; i ¼ 7:5 gr

The astronomical latitude is, then, f = ’  i = 45  7.5 = 37.5 . (b) The deviation of the vertical with respect to the radial direction, as already found, is i = 7.5 . (c) If we want the deviation of the vertical to be null, i ¼ 0, this implies gy = 0.

Earth’s gravity field and potential

41

writing Equation (22.3) in terms of the coefficient m, where m¼ we have GM gy ¼ 0 ¼ 2 a

a3 o2 GM

ð22:4Þ

pffiffiffi

2 2þ3 m  16 2

and solving for m gives m ¼ 0.73. Substituting in Equation (22.4) we obtain rffiffiffiffiffiffiffiffi GM 0:85 o¼ a3

23. A planet consists of a very thin spherical shell of mass M and radius a, within which is a solid sphere of radius a/2 and mass M 0 centred at the midpoint of the equatorial radius of the zero meridian. The planet rotates with angular velocity v about an axis normal to the equatorial plane. Calculate: (a) The potential at points on the surface as a function of latitude and longitude. (b) The components of the gravity vector. (c) If M 0 ¼ 10 M, what is the ratio between the tangential and radial components of gravity at the North Pole? (a) The potential U is the sum of the gravitational potentials due to the spherical shell V1, and to the interior sphere V2, plus the potential due to the rotation of the planet F (Fig. 23): U ¼ V1 þ V2 þ F 1 F ¼ o2 r2 cos2 ’ 2 GM V1 ¼ r GM 0 V2 ¼ q

ð23:1Þ

where r is the distance from a point P on the surface of the planet to its centre, q is the distance from point P to the centre of the interior sphere, and ’ the latitude of point P. Using the cosine law, rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 q ¼ r2 þ  ar cos c 4 where c is the angle between r and the equatorial radius, and its inverse can be approximated by (Problem 16)    1 1 a a2  2 ¼ 1 þ cos c þ 2 3cos c  1 ð23:2Þ q r 2r 8r Using the relation for spherical triangles cos a ¼ cos b cos c þ sin b sin c cos A

42

Gravity

l

–j 90° P r y a

y q

j a/2

x

l

Fig. 23

putting b ¼ 90º ’, c ¼90º, A ¼ l, and a ¼ c, where l is the longitude of P, then cos c = cos ’ cos l Substituting in (23.1), the potential due to the small sphere is    GM 0 a a2  2 2 1 þ cos ’ cos l þ 2 3cos ’cos l  1 V2 ¼  r 2r 8r

The total potential U is

   GM GM 0 a a2  2 2 þ 1 þ cos ’ cos l þ 2 3cos ’cos l  1 U¼ r r 2r 8r 1 þ o2 r2 cos2 ’ 2 (b) The components of the gravity vector are obtained from Equation (23.3): @U @r    GM 1 a 3a2  0 2 2 ¼  2 þ GM  2  3 cos ’ cos l  4 3cos ’cos l  1 r r r 8r

gr ¼

þ o2 rcos2 ’ 1 @U 1 @U gy ¼ ¼ r @y r @’   0 GM a a2 2 sin ’ cos l þ 6 cos ’cos l sin ’ ¼ r 2r2 8r3 þ o2 r cos ’ sin ’

ð23:3Þ

Earth’s gravity field and potential

43

  1 @U GM 0 a a2 2 gl ¼  2 cos ’ sin l  3 6cos ’ cos l sin l ¼ 8r r cos ’ @l r cos ’ 2r (c) At the North Pole, ’ = 90 and r ¼ a. Putting M 0 ¼10M and substituting in the previous equations we obtain gr ¼ 

GM GM 0 3GM 0 a2 GM  2 þ ¼ 7:25 2 a2 a 8a4 a gy ¼

GM 0 5GM ¼ 2 2a2 a

The ratio between the radial and the tangential components of gravity at the North Pole is gr ¼ 1:45 gy 24. An Earth consists of a sphere of radius a and density r, within which there are two spheres of radius a/2 centred on the axis of rotation and tangent to each other. The density of that of the northern hemisphere is 4r and that of the southern hemisphere is r/4. (a) Express the gravitational potential in terms of M (the mass of the large sphere) up to terms of 1/r3. (b) What astronomical latitude corresponds to points on the equator (without rotation)? (c) What error is made by using the 1/r3 approximation in calculating the value of gr at the equator? (a) The total gravitational potential V is the sum of the potentials of the sphere of radius a (V0) and of the two spheres of radius a/2 situated in the northern (V1) and southern (V2) hemispheres (Fig. 24): V ¼ V0 þ V1 þ V2 As in previous problems the large sphere is considered to have uniform density r and the effect of the two interior spheres is calculated using their differential masses

The potentials are

4 M ¼ pra3 3 4 a3 3M M1 ¼ pð4r  rÞ ¼ 3 8 8  a3 4
r 3M ¼ M2 ¼ p  1 3 4 32 8

44

Gravity

4r

x

r

r1 a/2 a a r2

r/4

x

Fig. 24

GM r    GM1 3GM 1 a a2  ¼ þ 2 cos y þ 3 3cos2 y  1 V1 ¼ r1 8 r 2r 8r    GM2 3GM 1 a a2  V2 ¼  2 cos y þ 3 3cos2 y  1 ¼ r2 32 r 2r 8r V0 ¼

where r1 and r2 have been calculated as in Problem 16. Then, the total gravitational potential in the 1/r3 approximation is

V ¼ GM

 41 15 a 9 a2  þ cos y þ 3cos2 y  1 2 3 32r 64 r 256 r





(b) The components of the gravity vector, taking into account that there is no rotation, are    @V 41 15a 27a2  2 gr ¼ ¼ GM   cos y  3cos y  1 @r 32r2 32r3 256r4 ð24:1Þ   1 @V 15a 27a2 gy ¼ ¼ GM  sin y  cos y sin y r @y 64r3 128r3 At the equator, r ¼ a and y = 90 and we obtain

Earth’s gravity field and potential

45

GM a2 GM gy ¼ 0:243 2 a gr ¼ 1:175

The astronomical latitude (’a) is the angle between the vertical and the equatorial plane. In our case at the equator this is given by the deviation of the vertical from the radial direction: tan ’a ¼

gy ¼ 0:207 gr

Then ’a ¼ 11:68 N (c) If we want to calculate the exact value of gr at the equator, we calculate the exact attractions of each sphere and add them: gr0 ¼ 

GM a2

gr1 ¼ 

3GM cos a 8r12

gr2 ¼ 

3GM cos a 32r22

ð24:2Þ

where r1 and r2 are the distances from the centre of each of the two interior spheres (Fig. 24): rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi a2 a 5 r 1 ¼ r 2 ¼ a2 þ ¼ 4 2 and a is the angle which r1 and r2 form with the equatorial plane: sin a ¼

a=2 1 ¼ pffiffiffi r1 5

The radial component of gravity is given by

grT ¼ gr0 þ gr1 þ gr2 ¼ 1:335

GM a2

The error we make using the approximation is gapprox  gexact ¼ 0:160

GM ; a2

that is; 16%:

25. An Earth consists of a sphere of radius a and density r within which there are two spheres of radius a/2 centred on the axis of rotation and tangent to each other. The density of that of the northern hemisphere is 2r and that of the southern hemisphere is r/2. (a) Express the potential V in terms of M (the mass of the large sphere), G, and r up to terms in 1/r3.

46

Gravity

P

q1 a/2

q2

g

r

2r x

gq

gr

r q ja

j

r

a

b

r/2

x

Fig. 25

(b) According to the value of this potential V, which astronomical latitudes correspond to the geocentric latitudes 45º N and 45º S? (c) What must the rotation period be for the astronomical and geocentric latitudes to coincide? (d) What error is made by the 1/r3 approximation in calculating the value of gr at the equator? And at the North Pole? (a) As in previous problems the effect of the interior spheres is given in terms of their differential masses (Fig. 25):
a3 M 4 ¼ M1 ¼ pð2r  rÞ 3 2 8

 4 r a 3 M M2 ¼ p  r ¼ 3 2 2 16 The distances q1 and q2 from the centre of each sphere to an arbitrary point P are found using the cosine law: a2 ar  2 cos y 4 2 2 a ar q2 ¼ r2 þ þ 2 cos y 4 2 q1 ¼ r 2 þ

Earth’s gravity field and potential

47

Using the approximation for 1/q (Problem 16), the total gravitational potential V is the sum of the potentials of the three spheres: GM GM1 GM2 þ þ r q1 q2    GM 17 3 a 1
a2  2 þ cos y þ 3cos y  1 ¼ r 16 32 r 128 r

V ¼

(b) The components of the gravity vector are given by    @V 1 17 6 a 3 a2  2 gr ¼ cos y þ 3cos y  1 ¼ GM 2 þ @r r 16 32 r3 128 r4   1 @V GM 3 a 1 a2 ¼ sin y þ 6 cos y sin y gy ¼ r @y r 32 r2 128 r3

ð25:1Þ

If the point P is at the surface, r ¼ a, then    GM 17 3 3  þ cos y þ 3cos2 y  1 gr ¼  2 a 16 16 128   GM 3 6 gy ¼  2 sin y þ cos y sin y a 32 128 At geocentric latitude 45º N, y ¼ 45º, GM a2 GM gy ¼ 0:09 2 a

gr ¼ 1:21

The deviation of the vertical with respect to the radial direction i is given by tan i ¼

gy ¼ 0:074 ) i45 ¼ 4:2 gr

According to Fig. 25, the astronomical latitude ’a can be determined from the deviation of the vertical i, ’a þ i þ ð180  ’Þ ¼ 180 ) ’a ¼ ’  i ¼ 45  4:2 ¼ 40:8 N In the same way, for geocentric latitude 45º S (y = 135 ) GM a2 GM gy ¼ 0:04 2 a

gr ¼ 0:94

gy Then tan i135 ¼ ¼ 0:043 ) i135 ¼ 2:5 gr Then the astronomical latitude is ’a ¼ 45  2:5 ¼ 47:5 ¼ 47:5 S

48

Gravity

(c) If we want the astronomical and geocentric latitudes to coincide, then the deviation of the vertical must be null, i ¼ 0º. This implies that gytotal must be zero. To do this by means of the rotation, we have to make the tangential component of the centrifugal force gyR be equal and of opposite sign to that of the gravitational potential gyV : gy total ¼ gy V þ gy R ¼ 0 ) gy V ¼ gy R The tangential component due to rotation is gyR ¼

1 @F r @y

1 where F ¼ o2 r2 sin2 y. Then 2 gyR ¼ o2 r cos y sin y For a point on the surface at latitude 45º N, r = a and y = 45 , so gy V ¼ gy R ) 0:09 GM =a2 ¼ o2 a=2 From here we can calculate the period of rotation 2p 2p ¼ pffiffiffiffiffiffiffiffiffi T¼ o 0:18

rffiffiffiffiffiffiffiffi a3 GM

For a point at latitude 45º S, r = a and y = 135 , so gyR

¼

gyV

1 GM 2p )  o2 a ¼ 0:04 2 ) T ¼ pffiffiffiffiffiffiffiffiffi 2 a 0:08

rffiffiffiffiffiffiffiffi a3 GM

(d) The value of the radial component of gravity at the equator, r ¼ a, y = 90 , by substitution in (25.1), is gr ¼ 1:04

GM a2

If we calculate the exact value by adding the contributions of the three spheres (Fig. 25) grexact ¼ grM þ gr1 þ gr2 grM ¼ 

GM a2

gr1 ¼ g1 cos b gr2 ¼ g2 cos b where

Earth’s gravity field and potential

49

a 2 cos b ¼ rffiffiffi ¼ pffiffiffi 5 5 a 4 GM 4 g1 ¼  8 5a2 GM 4 g2 ¼ 16 5a2 grexact ¼ 

GM 0:96 a2

The error in the approximation is: grerror



GM GM GM ¼ 0:96 2  1:04 2 ¼ 0:08 2 a a a

In a similar way, for a point at the North Pole, r ¼ a, y ¼ 0º: gr ¼ 1:30 GM =a2 grexact ¼ grM þ gr1 þ gr2 GM a2 GM GM gr1 ¼ 
2 ¼  2 a 2a 8 2 GM GM gr2 ¼
¼ a2 36a2 16 a þ 2

grM ¼ 

grexact ¼ 

GM 1:47 a2

The error in the approximation is grerror



GM GM GM ¼ 1:47 2  1:30 2 ¼ 0:17 2 a a a

26. A spherical Earth of radius a has a core of radius a/2 whose centre is displaced a/2 along the axis of rotation towards the North Pole. The core density is twice that of the mantle. (a) What should the period of rotation of the Earth be for the direction of the plumbline to coincide with the radius at a latitude of 45º S? (b) What are the values of J0, J1, J2, and m? (a) As in previous problems we calculate the gravitational potential by the sum of the potentials of the two spheres, using for the core the differential mass (Fig. 26): GM GM 0 V ¼ V1 þ V2 ¼ þ r q 2 4 a M M 0 ¼ pð2r  rÞ ¼ 3 8 8

50

Gravity

a/2

P

2r

q r

x

q

j

a 45°

g

r

i gr

gq

Fig. 26

As we saw in Problem 16, we use for 1/q the first-order approximation    GM GM 1 a 1 a2  2 V ¼ þ þ 2 cos y þ 3cos y  1 r 8 r 2r 2 4r3

ð26:1Þ

The total potential U is the sum of the gravitational potential V and the potential due to rotation 1 F ¼ r2 o2 sin2 y 2 U ¼ GM

   91 a a2  1 2 þ cos y þ 3cos y  1 þ r2 o2 sin2 y 8 r 16r2 64r3 2

In order that the direction of the plumb-line coincides with the radial direction, the tangential component of gravity, gy, must be null:

1 @U GM a 1 a2 1 gy ¼ sin y  6 cos y sin y ¼  r @y r 16 r2 64 r3 þ ro2 sin y cos y

For a point on the surface at latitude 45º S, the tangential component of gravity is, with r = a, y = 135 , gy ¼ 0:003

GM ao2  a2 2

Earth’s gravity field and potential

51

Putting this component equal to zero, we find the value of the period of rotation T: rffiffiffiffiffiffiffiffi GM ao2 GM 2p a3 2 gy ¼ 0:003 2  ¼ 0 ) o ¼ 0:006 3 ) T ¼ a 2 a 0:077 GM (b) The gravitational potential V of Equation (26.1) can be written as    GM 9 a a2  2 V ¼ 1þ cos y þ 3cos y  1 r 8 18r 72r2

We obtain the values of J1 and J2 by comparison with the equation 
a2  GM a V ¼ J0 þ J1 P1 þ J2 P2 r r r Since the total mass is (9/8)M, we obtain J0 ¼ 1 1 J1 ¼ 18 1 J2 ¼ 72

27. Within a spherical planet of radius a and density r there are two spherical cores of radius a/2 and density r 0 with centres located on the axis of rotation at a/2 from the planet’s centre, one in the northern hemisphere and the other in the southern hemisphere. (a) Neglecting rotation of the planet, calculate what the ratio r0 /r should be for the gravity flattening to be 1/8. (b) If the planet rotates so that m ¼ 1/16, and the ratio of the densities is that found in part (a), calculate the astronomical latitude which corresponds to the geocentric latitude 45º N. (a) Since there is no rotation the total potential U is the sum of the gravitational potentials of the three spheres (Fig. 27). As in previous problems we use the mass M of the planet with uniform density r and for the two cores the differential masses M 0 . For 1/q we use the approximation as in Problem 16:

4 a3 0 4 a3 0 r M r0 0 1 ð27:1Þ M ¼ p ðr  rÞ ¼ p ðr  rÞ ¼ 3 8 3 8 r 8 r The potential U is


a 2   GM GM 0 2 U¼ þ 2þ 3cos y  1 r r 2r

The radial components of gravity at the equator and the Pole are found by taking the derivative of the potential U:

 @U GM 2 3a2  ¼  2 þ GM 0  2  4 3cos2 y  1 gr ¼ @r r r 4r

52

Gravity On the surface r ¼ a, and at the equator y = 90 and at the Pole y = 0 , so GM GM 0 5  2 a2 a 4 GM GM 0 14 p gr ¼  2  2 a a 4 gre ¼ 

The gravity flattening is given by b¼

gp  ge 1 ¼ ge 8

By substituting the values of gravity we find the relation between M and M 0 : 7 0 5 0 1 M  2 M þ M þ 4 M 67 ¼ ) M ¼ M0 5 0 8 4 M  M 4 Putting M 0 in terms of M from Equation (27.1) we find the ratio of the densities:

4 M r0 r0  1 ) ¼ 1:48 M0 ¼ M ¼ 67 8 r r (b) For a rotating planet we add to the potential U the rotational potential, F:


a 2   GM GM 0 GM
r 3 m 2 2 þ sin y 3cos y  1 þ 2þ U¼ r r 2r r a 2

The radial and tangential components of gravity are now

 @U GM 2 3a2  0 2 gr ¼ ¼  2 þ GM  2  4 3cos y  1 @r r r 4r rm 2 þ GM 3 sin y a GM ¼  1:11 2 r 
 1 @U GM 0 a 2 ¼ 2  6 cos y sin y gy ¼ r @y r 2r GM
r 3 þ 2 m sin y cos y r a GM ¼  0:013 2 r

From Fig. 27 we see that the relation between the geocentric and astronomical latitudes is ’a ¼ ’  i where i is the deviation of the vertical with respect to the radial direction, which is given by tan i ¼

gy ¼ 0:012 ) i ¼ 0:7 gr

53

Gravity anomalies. Isostasy

gq

a/2 r’

g

x

gr

i r

j = 45°

ja a

r⬘

x

Fig. 27

Then the astronomical latitude for geocentric latitude 45º is ’a ¼ 45  0:7 ¼ 44:3

Gravity anomalies. Isostasy 28. For two-dimensional problems, the gravitational potential of an infinite horizontal cylinder of radius a is

1 2 V ¼ 2pGra ln r where r is the distance measured perpendicular to the axis. Assume that a horizontal cylinder is buried at depth d as measured from the surface to the cylinder’s axis. (a) Calculate the anomaly along a line of zero elevation on the surface perpendicular to the axis of the cylinder. (b) At what point on this line is the anomaly greatest? (c) What is the relationship between the distance at which the anomaly is half the maximum and the depth at which the cylinder is buried? (d) For a sphere of equivalent mass to produce the same anomaly, would it be at a greater or lesser depth?

54

Gravity

P

x

∆g

gr gz ∆gmax

d r

(1/2)∆gmax

a x

x1/2

x

r

Fig. 28

(a) The gravity anomaly produced by an infinite horizontal cylinder buried at depth d, with centre at x ¼ 0 (Fig. 28), is given by the derivative in the vertical direction (z-axis) of the gravitational potential V: 0 1 1 B C V ¼ 2pG
ra2 ln@qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA x2 þ ðz þ dÞ2
g ¼ gz ¼ 

@V 2p
rGa2 d ¼ @z x2 þ ðz þ dÞ2

ð28:1Þ

For points on the surface (z ¼ 0):
g ¼

2p
rGa2 d x2 þ d 2

(b) To find the point at which the anomaly has its maximum value, we take the derivative with respect to x and put it equal to zero: @
g ¼ 0 ) 2p
rGa2 d2x ¼ 0 ) x ¼ 0 @x Substituting x ¼ 0 in (28.1):
gmax ¼

2pG
ra2 d

(c) The distance at which the anomaly has a value equal to half its maximum value gives us the depth d at which the cylinder is buried:

55

Gravity anomalies. Isostasy


gmax 2pG
ra2 2pG
ra2 d ¼
g ) ¼ ) x1=2 ¼ d 2 2d x2 þ d 2 (d) The gravitational potential produced by a sphere of differential mass DM buried at depth d under x ¼ 0 is given by V ¼

The gravity anomaly is

G
M G
M ¼
1=2 r x2 þ ðz þ dÞ2


gz ¼

@V G
M ðz þ dÞ ¼
3=2 @z x2 þ ðz þ dÞ2

and for a point on the surface z ¼ 0,


gz ¼

G
Md ð x2

þ d 2 Þ3=2

The maximum value for x ¼ 0 is
gmax ¼

G
M d2

The distance at which the anomaly has half its maximum value is




G
Md G
M 3=2 ¼ 2d 2 x21=2 þ d 2

x1=2 ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 22=3  1 d ¼ 0:766d

Therefore, the sphere is at a greater depth than the cylinder. 29. At a point at latitude 42º 29 0 19 00 and height 378.7 m the value of gravity is observed to be 980 252.25 mGal. Calculate in gravimetric units (gu): (a) The free-air anomaly. (b) The Bouguer anomaly if the density of the crust is 2.65 g cm3. (a) We first calculate the normal or theoretical value of gravity given by the expression   g ¼ 9:7803268 1 þ 0:00530244sin2 ’  0:0000058sin2 2’ m s2

We substitute for ’ its value 42º 290 1900 and obtain

g ¼ 9:803 9299 m s2 The free-air anomaly, using the free-air correction, is
gFA ¼ g þ 3:086h  g ¼ 238:7 gu

56

Gravity

(b) The Bouguer anomaly is calculated from the free-air anomaly using the Bouguer correction with a crust density of 2.65 g cm3:
g B ¼ g þ 3:086h  2pGrh  g ¼
gFA  2pGrh ¼ 659:3 gu 30. An anomalous mass is formed by two equal tangent spheres of radius R, with centres at the same depth d ( d  R ) and density contrast Dr. (a) Calculate the Bouguer anomaly at the surface (z ¼ 0) produced by the mass anomaly along a profile passing through the centres of the two spheres. (b) Represent it graphically for x ¼ 0 (above the tangent point), 500, 1000, and 2000 m taking R ¼ 1 km, d ¼ 3 km, and Dr ¼ 1 g cm3. (a) For one sphere the anomaly for points on the surface (z ¼ 0) is (Problem 28)
g ¼

G
Md ð x2

þ d 2 Þ3=2

For two spheres the anomaly is the sum of the attractions of the two spheres (Fig. 30a):
g ¼

G
Md G
Md þ r13 r23

where qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx  RÞ2 þ d 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 ¼ ðx þ RÞ2 þ d 2

r1 ¼

Then

G
Md G
Md
g ¼ h i3=2 þ h i3=2 ð x  RÞ2 þd 2 ð x þ RÞ2 þd 2 P(x,0)

x

d

r2

R x

Fig. 30a

x

r1

57

Gravity anomalies. Isostasy

(b) To represent graphically the curve of the anomaly (Fig. 30b), we first find the point at which it is a maximum: 0 1 @
g @ B ¼ @h @x @x

G
Md

G
Md C i3=2 þ h i3=2 A ¼ 0 ð x  RÞ2 þd 2 ð x þ RÞ2 þd 2

h i5 h i5 ðx þ RÞ2 ðx  RÞ2 þ d 2 ¼ ðx  RÞ2 ðx þ RÞ2 þ d 2 ) x ¼ 0 maximum

Using the data given in the problem, we find the values of the anomaly for the five points, with Dr = 1 g cm3, R = 1 km, d = 3 km 4 4
M ¼ pR3
r ¼ p  109  103 ¼ 4:19  1012 kg 3 3

x (m)

Dg (gu)

0 500 1000 1500 2000

53.0 52.0 48.9 43.9 37.5

55

Anomaly (gµ)

50

45

40

–2000

Fig. 30b

–1000

0 Distance (m)

1000

2000

58

Gravity

31. At a point at geocentric latitude 45º N and height 2000 m the observed value of gravity is g ¼ 6690 000 gu. Taking the approximation that the Earth is an ellipsoid of equatorial radius a ¼ 6000 km, density ¼ 4 g cm3, J2 ¼ 103, and m ¼ 103, calculate for that point: (a) The free-air and the Bouguer anomalies. (b) The distance from the free surface to that of the sphere of radius a (precision 1 gu). (a) The volume of an ellipsoid is: 4 V ¼ pa3 ð1 þ 2aÞ 3 The flattening is a¼

3J2 m þ ¼ 2  103 2 2

and the mass is 4 Me ¼ V r ¼ pa3 ð1 þ 2aÞr ¼ 3:624  1024 kg 3 Using G = 6.67  1011 m3 kg1 s2 GM 6:67  1011  3:624  1024 ¼ ¼ 6:732 994 m s2 a2 36  1012 The value of gravity at the equator in the first-order approximation is given by

GM 3 ge ¼ 2 1 þ J2  m ¼ 6:736 361 m s2 a 2 For a point at latitude 45º N the radial component of gravity is   gr ¼ ge 1 þ b sin2 ’ ¼ 6:738 045 m s2

where we have used the value of the gravity flattening b given by 5 b ¼ m  a ¼ 0:5  103 2 The free-air correction is C FA ¼ 

2GM h ¼ 2:24  106 h m s2 ¼ 2:244h gu a3

Then, the free-air anomaly at that point is
gFA ¼ g  g þ C FA ¼ 6 690 000  6 738 045 þ 2:244  2000 ¼ 43 557 gu In order to calculate the Bouguer anomaly, we first calculate the Bouguer correction C B ¼ 2pGrh ¼ 1:676  106  h m s2 ¼ 1:676h gu

59

Gravity anomalies. Isostasy

N h a r

a

Fig. 31

Then, the Bouguer anomaly is:
gB ¼ g  g þ C AL  2pGrh ¼
gFA  C B ¼ 45 557  1:676  2000 ¼ 46 909 gu (b) If we call N the distance at the given point between the free surface and the surface of the sphere of radius a (Fig. 31), this is given by: N ¼arh where r is the radius of the ellipsoid at latitude 45º N which to a first approximation is

1 3 r ¼ a 1  asin ’ ¼ 6000 1  2  10  ¼ 5994 km 2 

2



Then, N ¼ 6000  5994  2 ¼ 4 km 32. Beneath a point A at height 400 m there exists an anomalous spherical mass of radius 200 m, density 3.5 g cm3, whose centre is 200 m below the reference level. A point B is located at a height of 200 m and a horizontal distance of 400 m from A, and a third point C is at a height of 0 m and at a horizontal distance of 800 m from A. The density of the medium above the reference level is 2.6 g cm3, and below the reference level it is 2.5 g cm3. The theoretical value of gravity is 980 000 mGal. Calculate: (a) The values of gravity at A, B, and C. (b) The Bouguer anomalies at these points.

60

Gravity

Precision 1 gu. (a) The gravity at each point is given by g ¼ g  C FA þ C B þ C am where Normal gravity: g ¼ 9800 000 gu Free-air correction: CFA ¼ 3.086 h Bouguer correction: CB ¼ 0.419 r1 h r1 ¼ 2.6 g cm3 is the density of the material above the reference level C am is the anomaly produced by the buried sphere at a point at height h and a horizontal distance x from its centre:

C

am

4 G pR3 ðrsph  r2 Þðh þ dÞ G
M ðh þ dÞ 3 ¼

3=2 3=2 ¼ 2 2 x þ ðh þ dÞ x2 þ ðh þ dÞ2

ð32:1Þ

where d is the depth to the centre from the reference level; and rsph and r2 are the densities of the sphere and of the medium where it is located, respectively. In our case: d ¼ 200 m, rsph ¼ 3.5 g cm3, and r2 ¼ 2.5 g cm3. For point A, x ¼ 0, we obtain C FA ¼ 1234 gu

C

am

C B ¼ 436 gu

4 G pR3 ðrsph  r2 Þ ¼ 6 gu ¼ 3 ¼ ðh þ d Þ2 ðh þ d Þ2 G
M

The value of gravity is gA ¼ 9 799 208 gu. At point B: CFA ¼ 617 gu CB ¼ 218 gu The anomaly produced by the sphere is calculated by Equation (32.1), substituting x ¼ xB ¼ 400 m and h ¼ hB ¼ 200 m C am ¼ 5 gu We obtain gB ¼ 9 799 606 gu. At point C: The free-air and Bouguer corrections are null, because the point is at the reference level. The anomaly due to the sphere, by substitution in Equation (32.1), x ¼ xC ¼ 800 m, and h ¼ hC ¼ 0, is C am ¼ 1 gu The value of gravity is: gC ¼ 9800 001 gu. (b) The Bouguer anomaly is given by DgB ¼ g þ CFA – CB – g

61

Gravity anomalies. Isostasy

By substitution of the values for each point we obtain that the anomalies correspond to those produced by the sphere:
gAB ¼ 6 gu
gBB ¼ 5 gu
gCB ¼ 1 gu 33. For a series of points in a line and at zero height which are affected by the gravitational attraction exerted by a buried sphere of density contrast 1.5 g cm3, the anomaly versus horizontal distance curve has a maximum of 4.526 mGal and a point of inflexion at 250 m from the maximum. Calculate: (a) The depth, anomalous mass, and radius of the sphere. (b) The horizontal distance to the centre of the sphere of the point at which the anomaly is half the maximum. (a) We know that the inflection point of the curve of the anomaly produced by a sphere buried at depth d corresponds to the horizontal distance d/2. Then d xinf ¼ ) d ¼ 2xinf ¼ 2  250 ¼ 500 m 2 The maximum value of the anomaly at x ¼ 0 is G
M ¼ 4:526 mGal ¼ 45:26 gu
gmax ¼ d2 and solving for DM DM ¼

45:26  106 m s2  5002 m2 ¼ 1:6964  1011 kg 6:67  1011 m3 s2 kg1

From this value we calculate the radius of the sphere:



1=3 4 3
M 1=3 3  1:6964  1011 kg
M ¼ pR3
r ) R ¼ ¼ ¼ 300 m 3 4p
r 4  3:14  1:5  103 kg m3 (b) In order that
g ¼ 12
gmax with z ¼ 0 we write G
Md 1 G
M ) x1=2 ¼ 383 m
3=2 ¼ 2 d2 x21=2 þ d 2

34. At a point at height 2000 m, the measured value of gravity is 9.794 815 m s2. The reference value at sea level is 9.8 m s2. The crust is 10 km thick and of density 2 g cm3, and the mantle density is 3 g cm3. Calculate: (a) The free-air, Bouguer, and isostatic anomalies. Use the Pratt hypothesis with a cylinder of radius 10 km and a 40 km depth of compensation. (b) If beneath this point there is a spherical anomalous mass of GDM ¼ 160 m3 s2 at a 2000 m depth, what should the compensatory cylinder’s density be for the compensation to be total?

62

Gravity

P

C

a

b

Fig. 34a

(a) The free-air anomaly is
gFA ¼ g  g þ C FA ¼ 9 794 815  9 800 000 þ 3:086  2000 ¼ 987 gu The Bouguer anomaly is
gB ¼ g  g þ C FA  C B ¼
gFA  0:4191rh ¼ 987  0:4191  2  2000 ¼ 689 gu To calculate the isostatic anomaly (Fig. 34a) we begin with the calculation of the isostatic correction assuming Pratt’s hypothesis and using only a vertical cylinder of radius 10 km under the point and the compensation level at 40 km. In this way, the correction consists of the gravitational attraction of a cylinder of radius a and height b at a point at distance c from the base of the cylinder, which is given by  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi I C ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2 ð34:1Þ where Dr is the contrast of densities, which according to Pratt’s hypothesis is given by
r ¼

hr0 Dþh

ð34:2Þ

where r0 is the density for a block at sea level, which in our case is formed by a crust of density 2 g cm3 and thickness 10 km over a mantle of density 3 g cm3 and thickness 30 km. For the whole 40 km we use a mean value of density r0 ¼

2  10 þ 3  30 ¼ 2:75 g cm3 40

63

Gravity anomalies. Isostasy

P h

d

rc = 2 g cm–3

10 km

D = 40 km

rM = 3 g cm–3

Fig. 34b

By substitution in (34.2) we obtain
r ¼

2  2:75 ¼ 0:13 g cm3 42

The isostatic correction (34.1) is C I ¼ 2  3:1416  6:67  1011  0:13  103 b40 þ ¼ 382 gu

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 100 þ 4  100 þ 1764c  103

Finally the isostatic anomaly is given by
gI ¼ g  g þ C FA þ C B þ C I ¼
g B þ C I ¼ 689 þ 382 ¼ 307 gu (b) If under the point considered there is an anomalous spherical mass (Fig. 34b) at depth d ¼ 2 km, the anomaly it produces is
g am ¼

G
M 160 ¼ ¼ 40 gu d2 ð2000Þ2

The total anomaly now is the Bouguer anomaly plus the anomaly due to the sphere:
g ¼ 689  40 ¼ 729 gu If the isostatic compensation is total (isostatic anomaly equal to zero), this anomaly must be compensated by the cylinder. Thus, the necessary contrast of densities Dr to do this can be calculated using expression (34.1): 729  106 ¼ 2  3:1416  6:67  1011 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  40 þ 100 þ 4  100 þ 1764  103 
r

64

Gravity

so
r ¼ 0:25 g cm3 As the mean value (crust–mantle) of the density is 2.75 g cm3, the density of the cylinder must now be r0  r ¼
r ¼ 0:25 ¼ 2:75  r ) r ¼ 2:50 g cm3 35. At a point on the Earth at height 1000 m, the observed value of gravity is 979 700 mGal. The value at sea level is 980 000 mGal. (a) Calculate the free-air and Bouguer anomalies. (b) According to the Airy hypothesis, which is the state of compensation of that height? (c) What should the depth of the root be for the compensation to be total? To calculate the compensation, use cylinders of radius 40 km, crustal thickness H ¼ 30 km, crust density 2.7 g cm3, and mantle density 3.3 g cm3. (a) The free-air anomaly is
g FA ¼ g  g þ C FA ¼ g  g þ 3:086h ¼ 86 gu The Bouguer anomaly is
gB ¼ g  g þ C FA  C B ¼
gFA  2p G r h ¼ 1046 gu (b) To calculate the isostatic anomaly according to the Airy hypothesis we first need to obtain the value of the root given by the equation t¼

rc h ¼ 4500 m rM  rc

The isostatic correction is given by  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C I ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2 Substituting the values Dr ¼ rM – rC ¼ 600 kg m3 b ¼ t ¼ 4500 m c ¼ h þ H þ t ¼ 35 500 m a ¼ 40 km we obtain: CI ¼ 409 gu the isostatic anomaly is:
gI ¼ g  g þ C FA þ C B þ C I ¼
g B þ C I ¼ 637 gu The negative value of the anomaly indicates that the zone is overcompensated.

65

Gravity anomalies. Isostasy

(c) If we want the compensation to be total, the value of the isostatic correction must be DgI ¼ DgB þ CI ¼ 0 ¼> CI ¼ – DgB ¼ 1046 gu Since the isostatic correction under the Airy hypothesis is  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C ¼ 2pG
r t þ a2 þ ðh þ H Þ2  a2 þ ðh þ H þ tÞ2 I

substituting and solving for t, we obtain t ¼ 13 068 m For a total isostatic compensation the value of the root (13 068 m) must be much larger than that corresponding to the 1000 m height, which is only 4500 m. 36. Gravity measurements are made at two points A and B of altitude 1000 m and  1000 m above the reference level, respectively, 2 km apart along a W-E profile at latitude 38.80º N. Below a point C located in the direction AB and 1 km from A is buried a sphere of radius 1 km and centre 3 km below the reference level, of density r ¼ 1.76 g cm3. Calculate: (a) The value of gravity at A and B. (b) Using the Airy assumption and neglecting the sphere, calculate the root at A and B. Crustal density rC ¼ 2.76 g cm3, mantle density rM ¼ 3.72 g cm3, a ¼ 10 km, and H ¼ 30 km. (a) The gravity observed at points A and B is given by gA ¼ g  C FA þ C B þ C am gB ¼ g þ C FA  C B þ C am where g is the theoretical gravity, CFA the free-air correction, CB the Bouguer correction, and C am the attraction due to the anomalous mass. The theoretical gravity at the observation point at latitude 38.80º N is   g ¼ 9:780 32 1 þ 0:005 3025sin2 ’  0:000 0058sin2 2’ ¼ 9:800714 m s2

The free-air and Bouguer corrections are:

C FA ¼ 3:806h ¼ 3:806  1000 ¼ 3806 gu C B ¼ 0:419rC h ¼ 0:419  2:76  1000 ¼ 1156 gu The attraction due to the spherical anomalous mass (Fig. 36) is given by C am ¼


G
Mðz þ dÞ 3=2 x2 þ ðz þ dÞ2

66

Gravity

A

XA

h

XB

C

–h B

d

r c = 2.76 g / cm3

x ρ

R

Fig. 36

For points A and B, by substitution of the values 4
M ¼ pR3
r 3 R ¼ 1000 m;
r ¼ 1000 kg m3 zA ¼ h ¼ 1000 m zB ¼ h ¼ 1000 m xA ¼ xB ¼ 1000 m d ¼ 3000 m we find CAam ¼ 16 gu CBam ¼ 50 gu Then the values of gravity at both points are gA ¼ 9800 627:9  3806 þ 1156:4  16 ¼ 9:798 048 m s2 gB ¼ 9800 627:9 þ 3806  1156:4  50 ¼ 9:803 314 m s2 (b) To calculate the value of the root under A and B according to the Airy hypothesis we use the equation t¼

rC h rM  rC

67

Gravity anomalies. Isostasy

where rC and rM are the crust and mantle densities. Then we find 2:76  1000 ¼ 2875 m 3:72  2:76 2:76  ð1000Þ ¼ 2875 m tB ¼ 3:72  2:76 tA ¼

37. At a point at latitude 43º N, the observed value of gravity is 9800 317 gu, and the free-air anomaly is 1000 gu. (a) Calculate the Bouguer anomaly. Take rC ¼ 2.67 g cm3. (b) If the isostatic compensation is due to a cylinder of radius 10 km which is beneath the point of measurement, what percentage of the Bouguer anomaly is compensated by the classical models of Airy and Pratt? (c) According to the Pratt hypothesis, what density should the cylinder have for the compensation to be total? (a) First we calculate the normal gravity at latitude 43º N:   g ¼ 978:0320 1 þ 0:005 3025sin2 ’  0:000 0058sin2 2’ Gal g ¼ 9804 385 gu

The height of the point is determined from the free-air anomaly,
g FA ¼ g  g þ 3:086h ¼ 9 800 317  9 804 385 þ 3:086h ¼ 1000 gu and solving for h, h ¼ 1642 m From this value we calculate the Bouguer anomaly
gB ¼ g  g þ 1:967h ¼ 838 gu (b) To apply the isostatic compensation using the Airy hypothesis we first calculate the root corresponding to the height h ¼ 1642 m: t ¼ 4:45h ¼ 7307 m The isostatic correction is determined using Equation (34.1) of Problem 34: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2

ð37:1Þ

where a ¼ 10 km; b ¼ t ¼ 7307 m; c ¼ t þ 30 000 þ h ¼ 38 949 m
r ¼ rM  rC ¼ 3:27  2:67 ¼ 0:6 g cm3 which results in C I ¼ 70 gu. This represents 8% of the observed Bouguer anomaly.

68

Gravity

If we use the Pratt hypothesis, the contrast of densities corresponding to h ¼ 1642 m is given by hr0 ¼ 0:043 g cm3
r ¼ Dþh where we have used r0 ¼ rC ¼ 2.67 g cm3 as the density of the crust. We now substitute in Equation (37.1), b = D = 100 km, c = D þ h = 101.642 km and the obtained value of Dr ¼ 0.043 g cm3, and obtain C I ¼ 148 gu We have to determine again the Bouguer anomaly using the density according to the Pratt hypothesis Dr0 ¼ 2:63 g cm3 Dþh
gB ¼
g AL  2pGrh ¼ 810 gu r¼

The isostatic correction corresponds now to 18% of the Bouguer anomaly. (c) If the compensation is total the isostatic correction must be equal to the Bouguer anomaly with changed sign: C I ¼ 
g B Using the Pratt hypothesis in order to calculate the density r of the cylinder under the point, we have to take into account that this density must also be the density used in the determination of the Bouguer anomaly. Then we write C I ¼ 
gB ¼ 
gFA þ 2p G r h qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2pGðr0  rÞ b þ a2 þ ðc  bÞ2  a2 þ c2 ¼ 
gFA þ 2pGrh and putting N¼ and solving for r, we obtain

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

b þ a2 þ ð c  bÞ 2  a2 þ c 2

r¼


gFA þ 2pGr0 N ¼ 2:46 g cm3 2pGðh þ N Þ

where we have used the values r0 ¼ 2.67 g cm3 and DgFA ¼ 1000 gu. 38. At a point on the Earth’s surface, a measurement of gravity gave a value of 9795 462 gu. The point is 2000 m above sea level. At sea level the crust is 20 km thick and of density rC ¼ 2 g cm3. The density of the mantle is rM ¼ 4 g cm3. (a) Calculate the free-air and Bouguer anomalies.

69

Gravity anomalies. Isostasy

(b) Calculate the isostatic anomaly according to the Airy and Pratt assumptions. Use cylinders of 10 km radius and compensation depth of 60 km. (c) Beneath the point, there is an anomalous spherical mass of GDM ¼ 1200 m3 s2. How deep is it? Take g ¼ 9.8 m s2. (a) The free-air anomaly is given by
g FA ¼ g  g þ 3:086h ¼ 9 795 462  9 800 000 þ 3:086h ¼ 1634 gu For the Bouguer anomaly we first calculate the Bouguer correction C B ¼ 0:419rh ¼ 0:419  2  2000 ¼ 1676 gu Then we obtain
g B ¼ g  g þ C FA  C B ¼
gFA  C B ¼ 42 gu (b) To calculate the isostatic anomaly according to the Airy hypothesis we determine first the value of the root corresponding to the height 2000 m: t¼

rc 2 2000 ¼ 2000 m h¼ rM  rc 42

The isostatic correction, using a single cylinder under the point, is given by qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2 where (Fig. 38a)

P

C

a

b

Fig. 38a

ð38:1Þ

70

Gravity

a ¼ 10 km;

b ¼ t ¼ 2 km;

c ¼ H þ t þ h ¼ 20 þ 2 þ 2 ¼ 24 km

Calling A the term inside the brackets in Equation (38.1) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A ¼ 2 þ 100 þ 484  100 þ 576 ¼ 0:166 km The isostatic correction is, then, given by

C I ¼ 2  3:1416  6:67  108 cm3 =gs2  2 g=cm3  166  102 cm ¼ 139 gu Finally, the isostatic anomaly using the Airy hypothesis is
g I ¼ g  g þ C FA  C B þ C I ¼ 97 gu According to the Pratt hypothesis, the regional density is given by r¼

D 60 r0 ¼ 3:33 ¼ 3:22 g cm3 Dþh 60 þ 2

where D is the compensation depth (in this problem 60 km) and for r0 (Fig. 38b) we have used the mean value of the density of the crust (2 g cm3) and of the mantle (4 g cm3) along the compensation depth 1 2 r0 ¼ 2 þ 4 ¼ 3:33 g cm3 3 3 The contrast of densities is
r ¼ 3:33  3:22 ¼ 0:11 g cm3 For the isostatic correction, using the Pratt hypothesis, the term A is now qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A ¼ 60 þ 102 þ ð62  60Þ2  102 þ 622 ¼ 7:4 km

and the correction

P x h

20 km

rC = 2 g cm–3

40 km rM = 4 g cm–3

Fig. 38b

71

Gravity anomalies. Isostasy

C I ¼ 2  3:1416  6:67  1011 m3 =kg1 s2  0:11  103 kg=m3  7:4  103 m ¼ 341 gu Since according to the Pratt hypothesis, the density of the compensating cylinder extends to the surface of the height 2000 m, we have to calculate again the Bouguer anomaly using this density (3.33 g cm3). We find for the Bouguer and isostatic anomalies the values CB¼ 2 p G r h ¼ 2699 gu DgB ¼ DgFA – CB ¼ 1065 gu DgI ¼ DgB þ CI ¼ 724 gu (c) If we assume that the isostatic anomaly is produced by a spherical anomalous mass buried under the point at a depth d under sea level its gravitational effect is given by
gmax ¼

G
M ðh þ dÞ2

¼
gI ¼ 724 gu

Solving for d we obtain h þ d ¼ 3517 m ) d ¼ 1517 m 39. At a point P at height 2000 m above sea level, a measurement is made of gravity. The crust at sea level, where gravity is 9.8 m s2, is 20 km thick and of density 3 g cm3, and the density of the mantle is 4 g cm3. Below the point P, at 2000 m depth under sea level, is an anomalous spherical mass of GDM ¼ 1200 m3 s2. (a) Neglecting the isostatic compensation, what would be the value of gravity at the point P? (b) With isostatic compensation, what now is the value of gravity at that point? Use the Airy and Pratt assumptions for the isostatic compensation (Pratt depth of compensation, 100 km) with single cylinders of 20 km radius under the point. (a) Without isostatic compensation, the gravity observed at point P is equal to the sum of the normal gravity plus the free-air and Bouguer corrections and the effect of the anomalous mass. Remember that the free-air correction has negative sign: gP ¼ g  C FA þ C B þ C am

ð39:1Þ

where, g ¼ 9800 000 gu C FA ¼ 3:086  h ¼ 3:086  2000 ¼ 6172 gu C B ¼ 0:419rh ¼ 0:419  3  2000 ¼ 2514 gu The anomaly due to the anomalous mass is given by C am ¼
gmax ¼

G
M ðh þ d Þ2

ð39:2Þ

72

Gravity

By substitution of the values, C am ¼

G
M ðh þ dÞ2

¼ 75 gu

The gravity observed a P is, then, given by gP ¼ 9 800 000  6172 þ 2514 þ 75 ¼ 9 796 417 gu (b) If there is isostatic compensation, according to the Airy hypothesis, we determine first the depth of the root, using the density of the crust rC and of the mantle rM: t¼

rC 3 h¼  2000 ¼ 6000 m rM  rC 43

The isostatic correction is

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2 I

ð39:3Þ

where a ¼ 20 km, b ¼ 6 km, c ¼ 2 þ 20 þ 6 ¼ 28 km, Dr ¼ 1 g cm3, so qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 2 3:14 6:67  1011 1103  6 þ 202 þ ð28  6Þ2  202 þ 282  103 ¼ 553 gu

Then the observed gravity at point P is gP ¼ 9796 417  553 ¼ 9795 864 gu According to the Pratt hypothesis we first determine the contrast of densities
r ¼

hr0 Dþh

where D is the level of compensation (100 km) and r0 is the mean density for the crust and mantle down to depth 100 km: r0 ¼

20  3 þ 80  4 ¼ 3:8 g cm3 100

Substituting we find
r ¼

2  3:8 ¼ 0:074 g cm3 100 þ 2

The isostatic correction is C I ¼ 2  3:14  6:67  1011  0:074  103 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 100 þ 202 þ ð102  100Þ2  202 þ 1022  103 ¼ 501 gu

73

Gravity anomalies. Isostasy

where we have used in Equation (39.3) the values a ¼ 20 km;

b ¼ 100 km;

c ¼ 100 þ 2 ¼ 102 km

The value of gravity at point P is now gP ¼ 9796 417  501 ¼ 9795 916 gu which is larger by 52 gu than using the Airy hypothesis 40. A point P is at altitude 1000 m above sea level. Beneath this point is a sphere of 1 km radius and GDM ¼ 650 m3 s2, with its centre 4 km vertically below the point P. Given that the density of the sphere is twice that of the crust and 3/2 that of the mantle, calculate: (a) The density of the sphere, crust, and mantle. (b) The value of gravity that would be observed at P for the isostatic compensation to be total including the sphere. (c) The radius of the sphere for the root to be null. Comment on the result. Use the Airy hypothesis for the isostatic compensation with H ¼ 30 km, 20 km radius of the cylinder, and theoretical gravity g ¼ 980 Gal. (a) If we know GDM we can calculate the contrast of densities between the anomalous mass and the crust 4 3G
M 3  650 G
M ¼ p
rR3 G )
r ¼ ¼ 3 4pR3 G 4  3:1416  109  6:67  1011 ¼ 2:326 g cm3 Since the density of the sphere rsph is double that of the crust rc, the densities of the crust and mantle are rsph ¼ 2rC )
r ¼ rsph  rC ¼ 2rC  rC ¼ rC ¼ 2:326 g cm3 rsph ¼ 4:652 g cm3 2 rM ¼ rsph ¼ 3:101 g cm3 3 (b) If isostatic compensation is total we have
g I ¼ 0 ¼ g þ gP þ C FA þ C B þ C I þ C am

ð40:1Þ

where g is the normal gravity, gP the observed gravity at point P, CFA the free-air correction, CB the Bouguer correction, CI the isostatic correction and C am the gravitational effect of the anomalous mass. The free-air and Bouguer corrections are given by C FA ¼ 3:806h ¼ 3:806  1000 ¼ 3086 gu C B ¼ 0:419r h ¼ 0:419  2:326  1000 ¼ 975 gu

74

Gravity

P x h

d a

rC

rsph

H

rM

t

R

Fig. 40

The effect of the spherical mass is C am ¼

G
M ðh þ d Þ

2

¼

650 2

ð4Þ  106

¼ 41 gu

We calculate the isostatic correction using the Airy hypothesis and taking into account the presence of the spherical anomalous mass. Thus, according to Fig. 40, the equilibrium between the gravity at P and at sea level far from P is given by
 4 pa2 rC H þ pa2 rM t ¼ pa2 rC ðh þ H þ t Þ þ pR3 rsph  rC 3

and solving for t:

 4
a2 hrC þ R3 rsph  rC 3 t¼ ¼ 3011 m a2 ðrM  rC Þ As in previous problems we calculate the isostatic correction using a cylinder under point P  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C I ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2

75

Gravity anomalies. Isostasy

where a ¼ 20 km, b ¼ t ¼ 3011 m, c ¼h þH þ t ¼ 34 011 m and Dr ¼ rM – rC ¼ 0.775 g cm3, resulting in C I ¼ 145 gu If compensation is total, the isostatic anomaly must be null. This implies that the Bouguer anomaly is equal, with opposite sign, to the isostatic correction
gI ¼ 0 ¼
gB  C I )
gB ¼ 145gu But the Bouguer correction can be obtained from Equation (40.1):
g I ¼ 0 ¼ g þ gP þ C FA þ C B þ C I þ C am Solving for CB we obtain C B ¼ 975 gu From the definition of the Bouguer anomaly we can find the value gP of gravity at P:
gB ¼ g P  g þ C FA  C B þ C am 145 ¼ g P  9800 000 þ 3086  975 þ 41 ) g P ¼ 9797 703 gu (c) Since the density of the sphere is greater than the density of the crust, there is an excess of gravity at P with respect to other points at sea level far from P, which must be compensated by a root of crustal material inside the mantle with negative gravitational influence. In this situation the root can never be null. 41. At 10 km beneath sea level vertically under a point P of height 2000 m there exists an anomalous spherical mass GDM ¼ 104 m3 s2. At sea level, gravity is 9800 000 gu and the crustal thickness 20 km. The density of the crust is 2 g cm3, and of the mantle 4 g cm3. Using the Airy assumption for the isostatic compensation with a cylinder of 10 km radius, calculate for that point: (a) The observed gravity. (b) The free-air, Bouguer, and isostatic anomalies. (a) For point P the Bouguer correction is C B ¼ 0:419rh ¼ 0:419  2  h ¼ 0:838h ¼ 1676 gu The gravity at point P, if there is no isostatic compensation and other effects, can be obtained from the normal gravity and the free-air and Bouguer corrections gP ¼ g  C FA þ C B ¼ 9800 000  3:086  2000 þ 0:838  2000 ¼ 9795504 gu

ð41:1Þ

Since there is an anomalous mass under point P we have to add its gravitational contribution to the gravity at P. For a spherical mass at depth h þ d under P the gravitational attraction is

76

Gravity

C am ¼

G
M 2

ðh þ d Þ

¼

104 m3 s2 ¼ 104 m s2 ¼ 100 gu 108 m2

We calculate the root corresponding to the isostatic compensation, assuming the Airy hypothesis, and taking into account the presence of the anomalous mass in the same way as in Problem 40:
 4 pa2 rC H þ pa2 rM t ¼ pa2 rC ðh þ H þ t Þ þ pR3 rsph  rC 3 ¼ pa2 rC ðh þ H þ t Þ þ Ma t¼

pa2 rC h þ Ma ¼ 2239 m pa2 ðrM  rC Þ

The isostatic correction using a cylinder is given by qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2

ð41:2Þ

where a ¼ 10 km, b ¼ t ¼ 2.239 km, c ¼ 20 þ 2 þ 2 ¼ 24.239 km, Dr ¼ 4  2 ¼ 2 g cm3, resulting in C I ¼ 154 gu The gravity at point P is the value obtained in (41.1) plus the contribution of the anomalous mass and minus the isostatic correction: g P ¼ 9795 504 þ 100  154 ¼ 9795 450 gu (b) The free-air anomaly is equal to this observed value minus the normal gravity plus the free-air correction:
gFA ¼ gP  g þ C AL Substituting the values we obtain
gFA ¼ 9795 450  9800 000 þ 3:086  2000 ¼ 1622 gu

The Bouguer anomaly is given by
g B ¼ gP  g þ C AL  C B ¼ 54 gu Finally the isostatic anomaly is the Bouguer anomaly plus the isostatic correction:
gI ¼ 54 þ 154 ¼ 100 gu This value corresponds to the gravitational contribution of the anomalous mass. 42. At a point P of height 2000 m above sea level the measured value of gravity is 979.5717 Gal. Beneath P is a sphere centred at a depth of 12 km below sea level, 1 g cm3 density, and radius 5 km. Assuming the Airy hypothesis (H ¼ 30 km, rC ¼ 2.5 g cm3), rM ¼ 3.0 g cm3), calculate the isostatic anomaly at the point in gu and mGal.

77

Gravity anomalies. Isostasy

For the compensation, assume cylinders of the same radius as the sphere. Normal gravity g ¼ 9.8 m s2. We first calculate the root t, assuming the Airy hypothesis, corresponding to the height 2000 m of point P. If the situation is of total isostatic equilibrium, we have to introduce the effect produced by the sphere in the determination of the root (Fig. 40):  4
p rsph  rC a3 þ pa2 rC h þ pa2 rC H þ pa2 rC t ¼ pa2 HrC þ pa2 rM t 3 so

 4
rC h þ a rsph  rC 3 t¼ ¼ 10 000 m r M  rC

ð42:1Þ

The negative value of t (anti-root) is due to the deficit of mass produced by the presence of the sphere (rsph < rC) under point P. The isostatic correction, as in previous problems, is calculated taking a cylinder under the point: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 2pGðrM  rC Þ b þ a2 þ ðc  bÞ2  a2 þ c2

where b ¼ t ¼ 10 000 m, a ¼ 5000, and c ¼ H þ h ¼ 32 000 m. We obtain CI ¼ 36 gu. The isostatic anomaly at P is equal to the observed gravity minus the normal gravity and the free-air, Bouguer, and isostatic corrections, and the attraction of the spherical mass:
gI ¼ g P  g þ C FA  C B  C I  C am

ð42:2Þ

The effect of the anomalous mass is given by
 4 G pa3 rsph  rC ¼ 267 gu C am ¼ 3 ðh þ d Þ2 By substitution in (42.2)
g I ¼ 9 795 717  9 800 000 þ 3:086  2000  0:419  2:5  2000 þ 267  36 ¼ 25 gu 43. A point A on the Earth’s surface is at an altitude of 2100 m above sea level. Calculate: (a) The value of gravity at A if the isostatic anomaly is 2.5 mGal. Assume the Airy hypothesis (rC ¼ 2.6 g cm3, rM ¼ 3.3 g cm3, H ¼ 30 km). (b) If the previous value had been measured with a Worden gravimeter of constant 0.301 82 mGal/division giving a reading of 630.6, calculate the value of gravity at another point B at which the device reads 510.1 (both readings corrected for drift). (c) At what depth is the centre of a sphere of density 4 g cm3 and radius 5 km which is buried in the crust, given that the anomaly created at a point A, 12 km from the

78

Gravity

centre of the sphere, not in the same vertical, is 321 gu. Also calculate the horizontal distance from the centre to point A. Take, for compensation, cylinders of 10 km radius. g ¼ 9.8 m s2 (a) The isostatic anomaly is given by
g I ¼ gA  g þ 3:086h  0:419rC h þ C I

ð43:1Þ

To calculate the isostatic correction CI, assuming the Airy hypothesis, we must first calculate the root t that corresponds to the height h t¼

rC h ¼ 7800 m rM  rC

As in other problems the isostatic correction is calculated using qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 2pGðrM  rC Þ b þ a2 þ ðc  bÞ2  a2 þ c2 and substituting the values b ¼ t ¼ 7800 m; a ¼ 10 km;

c ¼ h þ H þ t ¼ 39900 m;
r ¼ rM  rC ¼ 700 kg m3

we obtain C I ¼ 84 gu Solving for gA in Equation (43.1) we obtain gA ¼
gI þ g  3:086h þ 0:419rC h  C I ¼ 9795 748 gu (b) For a Worden gravimeter the increment in gravity between two points (Dg) is proportional to the increment in the values given by the instrument (DL) corrected by the instrumental variations
g ¼ K
L gB  gA ¼ KðLB  LA Þ ) gB ¼ gA þ KðLB  LA Þ ¼ gA  364 gu ¼ 9795 384 gu where K is the constant of the gravimeter. (c) The anomaly produced by a sphere buried at depth d under sea level at a point at height h and at a horizontal distance x from the centre of the sphere is given by

C am


 4 G pa3 rsph  rC ðh þ dÞ ¼ 3
3=2 x2 þ ðh þ dÞ2

ð43:2Þ

79

Gravity anomalies. Isostasy

A

h

r d rC a x

x rsph

Fig. 43

For point A (Fig. 43) if r is the distance from the centre of the sphere to the point A, x2 þ (h þd)2 ¼ r2 and solving for d in Equation (43.2) gives d¼

C am r3
h 4 G pa3 rsph  rC 3

Substituting the values r ¼ 12 km, a ¼ 5 km, rsph – rC ¼ 1400 kg m3, h ¼ 2100 m, C am ¼ 321  106 m s2, we obtain: d ¼ 9245 m The horizontal distance is: x¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2  ðh þ dÞ2 ¼ 3910 m

44. In a gravity survey, two points A and B on the Earth’s surface gave the values 159 and 80 mGal for the free-air anomaly, and 51 and 25 mGal for the Bouguer anomaly, respectively. Given that B is at an altitude 1000 m lower than A, and that the density of the mantle is 25% greater than that of the crust, calculate: (a) The value of gravity at A and B, and the densities of the crust and mantle. (b) The isostatic anomaly according to the hypotheses of Airy (H ¼ 30 km) and Pratt (D ¼ 100 km, r0 the value determined in the previous part) at point A. Take, for compensation, cylinders of 10 km radius. g ¼ 980 Gal. (a) The free-air anomaly at point A is given by
gAFA ¼ gA  g þ CAFA ¼ gA  g þ 3:086hA

ð44:1Þ

80

Gravity

The Bouguer anomaly is
gAB ¼ gA  g þ CAFA  CAB ¼
gAFA  0:419rC hA Changing values from mGal to gu, we write for the Bouguer anomalies at points A and B
gAB ¼ 510 ¼ 1590  0:419rC hA ) 0:419rC hA ¼ 2100 gu
gBB ¼ 250 ¼ 800  0:419rC hB ) 0:419rC hB ¼ 1050 gu

ð44:2Þ

Dividing both equations, we find hA ¼2 hB Knowing that the difference in height between A and B is 1000 m, we obtain for the heights of both points, hB ¼ hA  1000 ) hA ¼ hB þ 1000 ¼ 2hB ) hB ¼ 1000 m ) hA ¼ 2000 m The density of the crust can be obtained from Equation (44.2): 0:419rC hB ¼ 105  105 m s2 ¼ 0:419rC  1000 ) rC ¼ 2:505 g cm3 The density of the mantle is 25% more than that of the crust, so rM ¼ rC ð1 þ 0:25Þ ¼ 1:2  2:505 ¼ 3:131g cm3 The gravity at A and B is obtained using Equation (44.1): gA ¼
gAFA  3:086hA þ g ¼ 1590  3:086  2000 þ 9800 000 ¼ 9 795 418 gu gB ¼ 800  3:086  1000 þ 9800 000 ¼ 9797 714 gu (b) For the isostatic anomaly at point A, according to the Airy hypothesis, we first calculate the value of the root t corresponding to its height: t¼

rC h A hA ¼ 8000 m ¼ 1:25rC  rC 0:25

For the isostatic correction we use, as in other problems, a cylinder qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

I C ¼ 2pGðrM  rC Þ b þ a2 þ ðc  bÞ2  a2 þ c2 Substituting the values b ¼ t ¼ 8000 m; a ¼ 10 km;

c ¼ h þ H þ t ¼ 40 000 m;
r ¼ rM  rC ¼ 626 kg m3

we obtain C I ¼ 77 gu

81

Gravity anomalies. Isostasy

The isostatic anomaly is
gAI ¼
gAB þ C I ¼ 510 þ 77 ¼ 433 gu If we use the Pratt hypothesis, we first calculate the density corresponding to the material under point A: r¼

Dr0 100  2:505 ¼ ¼ 2:456 g cm3 Dþh 100 þ 2

and the contrast of density
r ¼ r0  r ¼ 2:505  2:456 ¼ 0:049 g cm3 The isostatic correction is determined using a cylinder, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 2pGðrM  rC Þ b þ a2 þ ðc  bÞ2  a2 þ c2 where Dr ¼ 0.049 g cm3, b ¼ D ¼ 100 km, a ¼ 10 km, c ¼ D þ h ¼ 102 km. Then, we obtain C I ¼ 158 gu The isostatic anomaly will be the Bouguer anomaly plus the isostatic correction
gAI ¼
gAB þ C I ¼ 510 þ 158 ¼ 352 gu In both cases the anomaly is negative, but using the Airy model the value is greater than using the Pratt model. 45. At a point P on the Earth’s surface, the observed value of gravity is 9.795 636 m s2, and the Bouguer anomaly is 26 mGal. Assuming the Airy hypothesis (rC ¼ 2.7 g cm3, rM ¼ 3.3 g cm3, H ¼ 30 km), calculate: (a) The height of the point. (b) The isostatic anomaly. (c) The value of gravity that would be observed at the point if beneath it were a sphere at a depth of 10 km below sea level, with a density of 2.5 g cm3 and a radius of 5 km, such that the compensation was total. Compensation with cylinders of 5 km radius; g ¼ 9.8 m s2. (a) We calculate the height of point P from the Bouguer anomaly:
gB ¼ g P  g þ 3:086h  0:419rC h ) h ¼


gB  gP þ g 3:086  0:419rC

so h¼

260  9 795 636 þ 9 800 000 ¼ 2099:8 m ffi 2100 m 3:086  0:419  2:7

82

Gravity

(b) To calculate the isostatic anomaly, using the Airy hypothesis, we first calculate the value of the root t corresponding to the height of the point: t¼

rC 2:7 h¼  2100 ¼ 9450 m rM  rC 3:3  2:7

Using a cylinder, the isostatic correction is given by qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2

ð45:1Þ

where we substitute the values a ¼ 5 km; b ¼ t ¼ 9450 m; c ¼ h þ H þ t ¼ 2100 þ 30 000 þ 9450 ¼ 41 550 m;
r ¼ 3:3  2:7 ¼ 0:6 g cm3 and obtain C I ¼ 22 gu The isostatic anomaly is then
gI ¼
gB þ C I ¼ 260 þ 22 ¼ 238 gu (c) If the compensation is total then the isostatic anomaly must be zero. But now we have to include the gravitational effect Cam produced by the presence of the anomalous mass of the sphere.
g I ¼ 0 ¼ gP  g þ 3:086h  0:419rC h  C am þ C I Solving for gP: gP ¼ g  3:086h þ 0:419rC h þ C am  C I

ð45:2Þ

where the effect of the sphere is given by
 4 G pa3 rsph  rC C am ¼ 3 ¼ 48 gu ðh þ d Þ2 We calculate the isostatic correction according to the Airy hypothesis. First we calculate the value of root t, but now we add the effect of the sphere on point P (Fig. 45):
 4 rC pa2 h þ rC pa2 H þ pa3 rsph  rC þ rC pa2 t ¼ rC pa2 H þ rM pa2 t 3 Solving for t, we obtain

 4
rC h þ a rsph  rC 3 ¼ 7228 m t¼ rM  rC

83

Gravity anomalies. Isostasy

P h

d x a rsph

H

rC

rM

t

Fig. 45

We substitute this value of t in Equation (45.1) together with the other values a ¼ 5 km;

b ¼ 7 ¼ 7228 m;

c ¼ h þ H þ t ¼ 39 328 m;


r ¼ 3:3  2:7 ¼ 0:5 kg m3

and obtain C I ¼ 18 gu By substitution in (45.2) we find the value of the gravity at P under the given conditions: gP ¼ 9800000  3:086  2100 þ 0:419  2:7  2100  48  18 ¼ 9795830 gu 46. Consider a point on the surface of the Earth in an overcompensated region at which the values of the free-air and the Bouguer anomalies are 1300 gu and 1200 gu, respectively. (a) Is this a mountainous or an oceanic zone? Give reasons. (b) Calculate the altitude and the value of gravity at the point given that the density of the crust is 2.72 g cm3. (c) If the isostatic anomaly is -1062 gu calculate, according to the Airy hypothesis (rC ¼ 2.72 g cm3, rM ¼ 3.30 g cm3, H ¼ 30 km), the value of the root responsible for this anomaly. Compare it with the value that it would have if the region were in isostatic equilibrium.

84

Gravity Compensation with cylinders of 10 km radius; g ¼ 9.8 m s2. (a) Since the free-air anomaly is positive and the Bouguer anomaly is negative, this indicates that this is a mountainous region. (b) From the free-air and Bouguer anomalies we can easily calculate the height of the point:
g FA ¼ g P  g þ 3:086h
g B ¼ gP  g þ 3:086h  0:419rC h

ð46:1Þ

and solving for h, h¼


gFA 
g B ¼ 2193 m 0:419rC

The observed gravity at P can be obtained from either of the two equations (46.1): gP ¼ g  3:086h þ
gFA ¼ 9794 532 gu (c) The isostatic correction is found from the known Bouguer and isostatic anomalies:
g I ¼
gB þ C I ) C I ¼ 1062 þ 1200 ¼ 138 gu The isostatic correction, using the Airy hypothesis, is given, as in previous problems, as a function of the root t, by qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 138 ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2 where we substitute
r ¼ 3:3  2:72 ¼ 0:58 g cm3 ; a ¼ 10 km; b ¼ t; c ¼ h þ H þ t ¼ 2:193 þ 30 þ t km and obtain for t, t ¼ 19 984 m If the region is in equilibrium the root due to the height h would be t¼

rC h 2:72  2193 ¼ 10 284 m ¼ rM  rC 3:3  2:72

Since we have already obtained a larger value (t ¼ 19 984 m), this indicates that the region is overcompensated. 47. In an oceanic region, gravity is measured at a point on the surface of the sea, obtaining a value of 979.7950 Gal. Calculate, using the Airy hypothesis (H ¼ 30 km, rC ¼ 2.9 g cm3, rM ¼ 3.2 g cm3, rW ¼ 1.04 g cm3): (a) The isostatic anomaly if the thickness of the crust is 8.4 km.

85

Gravity anomalies. Isostasy

(b) The thickness that the water layer would have to have if 15 km vertically below the point there was centred a sphere of 10 km radius such that the anti-root is null and the compensation total. Also calculate the density of the sphere. For the compensation, take cylinders of 10 km radius. g ¼ 9.8 m s2. (a) We are in an oceanic region, therefore in the calculation of the root for the isostatic compensation according to the Airy hypothesis we have to consider the layer of water of density rW. The value of the root is now given by t0 ¼

rC  rW 0 2:9  1:04 0 h ¼ h ¼ 6:2h0 3:2  2:9 rM  r C

ð47:1Þ

According to Fig. 47a, we have the following relation: H  h0  t 0 ¼ e ) h0 ¼ H  t 0  e, where e is the thickness of the crust at the oceanic region, H the thickness of the normal (sea level) continental crust, h0 the thickness of the water layer, and t 0 the negative root. Substituting the values of t 0 from Equation (47.1) we obtain for h0 h0 ¼ 30  6:2h0  8:4 ) h0 ¼ 3 km ¼ 3000 m t 0 ¼ 6:2  3000 ¼ 18 600 m From the value of the root we calculate the isostatic correction using a cylinder of 10 km radius, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

I C ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2

where Dr ¼ 3.2  2.9 ¼ 0.3 g cm3 ¼ 300 kg m3, b ¼ t0 , c ¼ H ¼ 30 000 m, and so C I ¼ 269 gu

To calculate the isostatic anomaly we first have to apply the Bouguer correction which in this case consists of two terms: the first to eliminate the attraction of the water layer P h⬘

rW

rC

e

t⬘

Fig. 47a

rM

H

86

Gravity (2pGrWh0 ) and the second to replace this layer by one of density equal to the crustal density (þ2pGrCh0 ). Since the point is at sea level the free-air correction is null: DgI ¼ g  g þ C B  C I C B ¼ 0:419ðrC  rW Þh0 ¼ 2338 gu Dg I ¼ 19 gu (b) If the anti-root is null and there is total compensation, then we have t0 ¼ 0
g I ¼ 0 ¼ g  g  2pGrW h0 þ 2pGrC h0  C am

ð47:2Þ

Since the isostatic anomaly must be null, then the anomalous spherical mass and the water layer must compensate each other. The attraction of the anomalous mass is Cam ¼ GDM/d2 where d is the depth of its centre below sea level. Then we can write g  g þ 2pGðrC  rW Þh0 

G
M ¼0 d2

where the mass of the sphere is 4
M ¼ pa3
rsph 3 If the point P is totally isostatically compensated and the anti-root is null, then (Fig. 47b) 4 pa2 h0 rW þ pa2 ðH  h0 ÞrC þ pa3 ðrsph  rC Þ ¼ pa2 HrC 3 Solving for h0 gives h0 ¼


 4a rC  rsph

ð47:3Þ

3ðrW  rC Þ

P h⬘

rW d H

e

Fig. 47b

x a rsph

rC

87

Gravity anomalies. Isostasy

Substituting this value in (47.2) we obtain rsph ¼

gg

þ rC ¼ 3372 kg m3 4 a3 pG 2a  2 3 d

and putting this value in (47.3), h 0 ¼ 3369 m. 48. At a point on the Earth’s surface, 500 m below sea level, a gravity value is measured of 980.0991 Gal. If the region is in isostatic equilibrium calculate, using the Airy hypothesis (H ¼ 30 km, rC ¼ 2.7 g cm3, rM ¼ 3.2 g cm3): (a) The thickness of the crust. (b) The isostatic anomaly in gu, with reasons for the sign of each correction. Take compensating cylinders of 5 km radius; g ¼ 9.8 m s2. (a) We calculate first the root, according to the Airy hypothesis which corresponds to the depth of the point, applying the condition of isostatic equilibrium (Fig. 48) rC H ¼ rC ðH  h0  t 0 Þ þ rM t 0 so t0 ¼

rC h0 ¼ 2700 m rM  rC

The thickness of the crust e at that point is e ¼ H  h0  t 0 ¼ 30 000  2700  500 ¼ 26 800 m (b) The isostatic anomaly is given by
g I ¼ gP  g  3:086h0 þ 0:419rC h0  C I The free-air correction (3.086h0 ) is negative because the point is below sea level. For the same reason the Bouguer correction (0.419 rC h0 ) is positive. The isostatic correction, using the Airy hypothesis, is calculated as in previous problems using a cylinder under the point of 5 km radius and density contrast D r ¼ 3.2  2.7 ¼ 0.5 g cm3. The value of the anti-root t 0 has already been calculated, so qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

I C ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2 h⬘

P

rC

e

t⬘ rM

Fig. 48

H

88

Gravity Substituting c ¼ H – h0 and b ¼ t0 , we obtain C I ¼ 9 gu Then the isostatic anomaly is DgI ¼ 5 gu. 49. In an oceanic region where the density of the crust is 2.90 g cm3 and that of the mantle 3.27 g cm3, the value of gravity measured at a point P on the sea floor at depth 4000 m is 9.806 341 m s2. Calculate, according to the Airy hypothesis: (a) The thickness of the crust. (b) The isostatic anomaly in gravimetric units. Data: rw ¼ 1.04 g cm3, H ¼ 30 km, g ¼ 9.8 m s2. Take, for compensation, cylinders of 10 km radius. (a) First we calculate the value of the root according to the Airy hypothesis t0 ¼

rC  rW 0 h ¼ 20 108 m rM  rC

The thickness of the crust under the point is found by (Fig. 49) e ¼ H  h0  t 0 ¼ 30000  40000  2018 ¼ 5892 m (b) Because the point is located at the bottom of the sea, to reduce the observed value of gravity to the surface of the geoid (sea level) we eliminate first the attraction of the water layer. Then we apply the free-air and the Bouguer corrections, to take into account the attraction of a layer of crustal material which replaces the water. Finally we apply the isostatic correction:
g I ¼ gP  g þ 0:419rW h0  3:086h0 þ 0:419rC h0  C I The isostatic correction is calculated using a cylinder of 10 km radius, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2 h⬘

rW

rC

e

t⬘

Fig. 49

P

rM

H

89

Gravity anomalies. Isostasy where b ¼ t0 ¼ 20 108 m, c ¼ H  h0 ¼ 26 000 m, Dr ¼ 3.27  2.9 ¼ 0.37 g cm3 ¼ 370 kg m3 By substitution we obtain C I ¼ 598 gu The isostatic anomaly is:
gI ¼ 9 806 341  9 800 000  3:086  4000 þ 0:419  ð1:04 þ 2:9Þ  4000  598 ¼ 4 gu 50. At a point with coordinates 42.78º N, 0.5º E and height 1572 m, the observed value of gravity is 980.0317 Gal. (a) Calculate the free-air and Bouguer anomalies. (b) If cylinders of 10 km radius beneath that point are used for the isostatic compensation, calculate the gravimetric attraction of the mass defect corresponding to the altitude of the point according to the Airy and Pratt hypotheses. Take, for the crust, H ¼ 30 km, rC ¼ 2.67 g cm3, for the mantle, rM ¼ 3.27 g cm3, and D ¼ 100 km for the Pratt level of compensation. (c) How deep should the root of the Airy model be for the compensation to be total? (a) We calculate first the normal gravity at the point where gravity has been observed using the expression   g ¼ 9:780 32 1 þ 0:005 3025 sin2 ’ ¼ 9:804 243 m s2

where ’ is the latitude. The free-air anomaly is given by


g FA ¼ g P  g þ C FA ¼ 9800 317  9804 243 þ 3:086  1572 ¼ 925 gu and the Bouguer anomaly by
gB ¼ gP  g þ C FA þ C B ¼ 9800 317  9804 243 þ 1:967  1572 ¼ 834 gu (b) If we approximate the isostatic compensation by means of a cylinder of radius a under the point, we use the expression qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2 ð50:1Þ

where Dr is the density contrast, b the height of the cylinder, and c the distance from the base of the cylinder to the observation point. Airy: We calculate first the root given by the equation t ¼ 4:45  h ¼ 4:45  1572 ¼ 6995 km For the isostatic correction we substitute in (50.1) the values

90

Gravity

a ¼ 10 km;

b ¼ t ¼ 6995 km;

c ¼ t þ H þ h ¼ 38 567 km;


r ¼ rM  rC ¼ 3:27  2:67 ¼ 0:6 g cm3 and obtain C I ¼ 2  3:1416  6:67  1011 m3 s2 kg1  0:5  103 kg m3  A where: A¼



6:995 þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

102 þ ð38:567  6:995Þ2  102 þ 38:5672  103 ¼ 270 m

so C I ¼ 68 gu Pratt: The contrast of densities is now given by
r ¼

h 1575 r ¼  2:67 ¼ 0:04 g cm3 D þ h 0 100 000 þ 1572

and substituting in Equation (50.1) with the values a ¼ 10 km, b ¼ 100 km, c ¼ D + h ¼ 101 572 m, we have C I ¼ 135 gu (c) If the isostatic compensation is total (isostatic anomaly null) the isostatic correction, according to the Airy hypothesis, coincides with the value of the Bouguer anomaly (834 gu): qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi


g B ¼ 2pG
r t þ a2  ðH þ hÞ2  a2 þ ðt þ H þ hÞ2 so tþ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
gB a2  ðH þ hÞ2  a2 þ ðt þ H þ hÞ2 ¼ 2pG
r

In this expression we solve for the value of the root t: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
gB a2 þ ðt þ H þ hÞ2 ¼ t þ a2  ðH þ hÞ2  ¼tþN 2pG
r a2 þ ðt þ H þ hÞ2 ¼ t 2 þ N 2 þ 2tN

a2 þ t 2 þ ðH þ hÞ2 þ 2tðH þ hÞ ¼ t 2 þ N 2 þ 2tN so t¼

N 2  a2  ðH þ hÞ2 2ðH þ h  N Þ

91

Gravity anomalies. Isostasy

By substitution of the values of N, a, H, and h we obtain t ¼ 58 875 m Because this root has a negative value greater than the thickness of the crust, total compensation is not possible. 51. Calculate the free-air anomaly observed on a mountain of height 2000 m which is fully compensated by a root of depth t ¼ 10 km. The compensation is by a cylinder of radius 20 km, the density of the crust is 2.67g cm3, and that of the mantle is 3.27g cm3. The free-air anomaly is given by
gFA ¼ g  g þ 3:086h Since the point is isostatically compensated, we calculate the isostatic correction using a cylinder as in previous problems: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

C I ¼ 2pG
r b þ a2 þ ðc  bÞ2  a2 þ c2

where we substitute b ¼ t ¼ 10 km, c ¼ h þH þ t ¼ 42 km, and obtain

a ¼ 20 km,

Dr ¼ 600 kg m3

CI ¼ 306 gu Since the point is totally compensated the isostatic anomaly must be zero:
gI ¼ g  g þ C FA  C B þ C I ¼ 0 The free-air anomaly can now be written as
gFA ¼ g  g þ C FA ¼ C B  C I

ð51:1Þ

We can calculate the Bouguer correction: C B ¼ 1:119h ¼ 1:119  2000 ¼ 2238 gu and substituting in (51.1) we obtain, for the free-air anomaly,
gFA ¼ 2238  301:5 ¼ 1932 gu 52. Calculate for a point P at height 2100 m, latitude 40º N, and observed gravity 979.7166 Gal the refined Bouguer anomaly in gravimetric units. Consider a surplus mass compartment of 3000 m, mean height, 3520 m inner radius, 5240 m outer radius, with n ¼ 16. The density of the crust is 2.5 g cm3. The refined Bouguer anomaly is obtained using, besides the free-air and Bouguer corrections, the correction for the topography or topographic or terrain correction (T):
gB ¼ g  g þ C FA  C B þ T

92

Gravity

a2 P Px

a1

a1

a2 hm

h rC

Fig. 52

The normal gravity at latitude 40º N is given by   g ¼ 9:780 32 1 þ 0:0053025 sin2 ’ ¼ 9:801 747 m s2

the free-air correction by

C FA ¼ 3:086h ¼ 6481 gu and the Bouguer correction by C B ¼ 0:419rC h ¼ 2200 gu The topographic correction is introduced in order to correct for the topographic masses not included in the Bouguer correction, that is, in this case those above the height h (Fig. 52). Remember that the Bouguer correction corrects for an infinite layer or plate of thickness h and doesn’t consider the additional masses above h or the lack of masses below h. The topographic correction is always positive because the masses above height h produce on point P an attraction of negative sign which must be added and the lack of mass under h must also be taken into account with a positive sign, since it has been subtracted in the Bouguer correction. To calculate the attraction of the mass above height h we use the attraction of concentric cylinders (in our case two) with axis passing through point P and with height equal to the difference between the height h of the point P and the height of the mass of the topography hm above h. The cylinder is divided into n sectors with radius a1 and a2 to approximate the topography (Fig. 52). Then the topographic correction is given by qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2pGrC T¼ a22 þ ðc  bÞ2  a22 þ c2  a21 þ ðc  bÞ2 þ a21 þ c2 n In our case we substitute the values a2 ¼ 5240 m;

a1 ¼ 3520 m;

b ¼ hm  h ¼ 900 m;

c ¼ 0;

n ¼ 16

93

Gravity anomalies. Isostasy

and obtain T ¼ 2 gu The refined Bouguer anomaly is
gB ¼ g  g þ C FA  C B þ T ¼ 298 gu 53. Calculate the topographic correction for a terrestrial compartment of inner radius a1 ¼ 5240 m, outer radius a2 ¼ 8440 m, n ¼ 20, mean height 120 m, with 2000 m being the height of the point P. Take r ¼ 2.65 g cm3. In this problem we consider the topographic correction for the case of the lack of mass in the topography at heights below that of the point P. Since in the Bouguer correction we have subtracted an infinite layer of thickness h, we have to correct for the places where the mass was not present (Fig. 53). The topographic or terrain correction T in this case is calculated in the same way as in the previous problem. Thus we take n sectors of cylinders with axis at point P and height equal to the difference between h and hm (Fig. 52b). The correction is then given by qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2pGrC a22 þ ðc  bÞ2  a22 þ c2  a21 þ ðc  bÞ2 þ a21 þ c2 T¼ n where we substitute a2 ¼ 8440 m, a1 ¼ 5240 m, b ¼ c ¼ h – hm, n ¼ 20 to obtain T ¼ 0:67 mGal 54. Calculate the topographic correction for an oceanic sector or compartment of inner radius a1 ¼ 5240 m, outer radius a2 ¼ 8440 m, n ¼ 20, mean depth 525 m, with 600 m being the height of the point P. Take rC ¼ 2.67 g cm3, rW ¼ 1.03 g cm3. In this problem we have to correct for the lack of mass in the oceanic area near the point P, between the sea level and height h (column 1 in Fig. 54). Also we have to take into account the attraction produced by the water layer between sea level and the bottom of the sea (column 2 in Fig. 54). a2 a1

P

h

hm

Fig. 53

rC

94

Gravity

a2 a1

h

1

P

2

P

h

rw

Fig. 54

To calculate the necessary topographic correction we proceed as in Problems 52 and 53, using cylindrical sectors: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2pGr T¼ a22 þ ðc  bÞ2  a22 þ c2  a21 þ ðc  bÞ2 þ a21 þ c2 n where a2 ¼ 8440 m;

a1 ¼ 5240 m;

n ¼ 20

For the correction corresponding to the attraction of column 1, between height h and sea level, we substitute the values: b ¼ h ¼ c;

r ¼ rC

and obtain T1 ¼ 0:07 mGal For the correction of the attraction of column 2 between sea level and the bottom of the sea we use the difference between the densities of the crust and of water: b ¼ p ¼ 525 m c ¼ p þ hs ¼ 1125 m c  b ¼ hs r ¼ rC  rW T2 ¼ 0:11 mGal The total topographic correction is T ¼ T1 þ T2 ¼ 0:18 mGal

95

Tides

Tides 55. Two spherical planets A and B of radii 2a and a and masses 3m and m are separated by a distance (centre to centre) of 6a. The only forces acting are gravitational, and the system formed by the two planets rotates in the equatorial plane. (a) Calculate the value of the components of the acceleration of the tides at the Pole of each planet directly and using the tidal potential. On which planet are they greater? (b) If each planet spins on its axis with the same angular velocity as the system, what, for each planet, is the ratio between the centrifugal force and the maximum of the tidal force at the equator? On which planet is this ratio greater? (a) From Fig. 55a we can deduce that at the Pole of planet A, the radial component of the acceleration of the tides produced by planet B is grT ¼ 

Gm cos a q2

where q is the distance from the Pole of planet A to the centre of planet B and a the angle formed by q and the radius at the Pole of planet A. By substitution of the required values we obtain qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi q ¼ ð6aÞ2 þ ð2aÞ2 ¼ 40a 2a cos a ¼ pffiffiffiffiffi ) a ¼ 71:6 40a Gm Gm grT ¼  cosð71:6Þ ¼ 0:008 2 2 40a a

The tangential component gyT is given by (Fig. 55a) Gm Gm Gm Gm þ 2 sin a ¼ þ sinð71:6Þ 36a2 q 36a2 40a2 Gm Gm ¼ 2 ð0:028 þ 0:024Þ ¼ 0:004 2 a a

gyT ¼ 

A

2a

a

q

B

q⬘ b

6a 3m

Fig. 55a

a

m

96

Gravity

If we use the tidal potential,  GMr2  3cos2 #  1 3 2R

c¼

where R is the distance between the centres of planets A and B (Fig. 55a), and # is the angle the position vector r forms with the distance vector R (in this case it is equal to the colatitude, # ¼ y)  GMr2  3cos2 y  1 c¼ 3 2R The radial and tangential components of the acceleration are given by

  @c @ Gmr2  Gmr  ¼ 3cos2 y  1 ¼ 3cos2 y  1 3 3 @r @r 2ð6aÞ 216a 2   1 @c 1 @ Gmr Gmr ¼ 3cos2 y  1 ¼ 3ðcos y sin yÞ gyT ¼ r @y r @y 2ð6aÞ3 216a3

grT ¼

For planet A, at the Pole, r ¼ 2a and y ¼ 90º, so grT ¼ 

Gm Gm ¼ 0:009 2 2 108a a

gyT ¼ 0 For planet B, we proceed in a similar manner: q02 ¼ 36a2 þ a2 ¼ 37a2 a cos a ¼ pffiffiffiffiffi ) a ¼ 80:5 37a

Therefore,

G3m Gm cosð80:5Þ ¼ 0:013 2 37a2 a G3m G3m Gm T þ sinð80:5Þ ¼ 0:003 2 gy ¼  36a2 37a2 a

grT ¼ 

Using the tidal potential, we obtain the acceleration components c¼

G3mr2  3

3cos2 y  1



2ð6aÞ  @c Gmr  3cos2 y  1 ¼ grT ¼ 3 @r 72a 1 @c 9 Gmr gyT ¼  ¼ ðcos y sin yÞ r @y 216 ðaÞ3 Substituting at the Pole of planet B, r ¼ a and y ¼ 90º, we have Gm grT ¼ 0:014 2 a T gy ¼ 0

97

Tides

w

A

B

2a (3/2)a

(9/2)a

6a

m

3m

Fig. 55b

(b) First we calculate the centre of gravity of the system formed by the two planets measured from the centre of planet A (Fig. 55b): x¼

3m  0 þ m  6a 3 ¼ a 3m þ m 2

The rotation radius for planet A is 3/2a and for planet B 3 9 6a  a ¼ a 2 2 In the rotating system the centrifugal force equals the force of gravitational attraction, which at the equator (y ¼ 0º) is fg ¼

Gm3m ð6aÞ2

9 ¼ fC ¼ o2 r ¼ mo2 a 2

From this expression we obtain the value of the angular velocity o of the rotation system: 3Gm2 9 Gm ¼ mo2 a ) o2 ¼ 2 36a 2 54a3 Since the angular velocity of the spin of each planet is equal to that of the system, the spin centrifugal force at the equator of planet A is fC ¼ o 2 r ¼

Gm Gm 2a ¼ 3 54a 27a2

The tidal force is fT ¼ grT ¼

2Gmr 2Gm2a ¼ 3 3 R3 6 a

98

Gravity

and their ratio Gm fC 27a2 ¼ ¼2 Gm fT 54a2 If we repeat these calculations for planet B, we obtain Gm Gm a¼ 3 54a 54a2 2G3ma Gm fT ¼ 3 3 ¼ 6a 36a2

fC ¼ o 2 r ¼

and the ratio Gm fC 54a2 ¼ 0:666 ¼ Gm fT 36a2 The ratio is larger for planet A, as expected owing to its larger radius. 56. Two planets of mass M and radius a are separated by a centre-to-centre distance of 8a. The planets spin on their own axes with an angular velocity such that the value of the centrifugal force at the equator is equal to the maximum of the tidal force (the equatorial plane is the plane in which the system formed by the two planets rotates around an axis normal to that plane). (a) Calculate the value of the components of the vector g as multiples of GM/a2 for a point of l' ¼ 60º and w ¼ 45º (with l ¼ 0º, being the meridian in front of the other planet) including all the forces that act. (b) What is the relationship between the angular velocity of each planet and that of the system? (a) The tidal potential is given by c¼

 GMr2  3cos2 #  1 3 2R

where R is the centre-to-centre distance between the planets, and # the angle formed by the vector r to a point and R (Fig. 56a). From this potential we calculate the radial component of the tidal force:

  @c @ GMr2  GMr  2 ¼ 3cos #  1 frT ¼ ¼ 3 3cos2 #  1 3 @r @r 2R R At the equator of one planet # ¼ 0º, r ¼ a, and R ¼ 8a, so frT ¼

GMa ð8aÞ

3

2¼

GM 256a2

99

Tides

Ω w

w gr gλ θ

gθ a

x R M

M

Fig. 56a

The spin centrifugal force is fC ¼ o 2 a Equating these two expressions we find the value of the spin angular velocity, rffiffiffiffiffiffiffiffi GM 1 GM T 2 )o¼ fC ¼ fr ) o a ¼ 256a2 16 a3 At a point on the surface of one of the planets the total potential is the sum of the gravitational potential V, the spin potential F, and the tidal potential c: U ¼V þFþc¼

 GM 1 2 2 2 GMr2  þ o r cos ’ þ 3cos2 #  1 3 r 2 2R

For a point P at latitude ’ and longitude l (Fig. 56b)

cos # ¼ cos ’ cos l and the potential U is U¼

 GM 1 2 2 2 GMr2  þ o r cos ’ þ 3cos2 ’cos2 l  1 3 r 2 2R

The components of gravity including the three effects are,  @U GM GMr  ¼  2 þ o2 r cos2 ’ þ 3 3 cos2 ’ cos2 l  1 @r r R 1 @U GMr 2 ¼ o r cos ’ sin ’ þ 3 3 cos2 l cos ’ sin ’ g’ ¼  r @’ R 1 @U 1 GMr2 gl ¼ ¼ 6 cos2 ’ cos l sin l r cos ’ @l r cos ’ 2R3 GMr ¼  3 3 cos ’ cos l sin l R gr ¼

100

Gravity

P ϑ ϑ j j a

l

Fig. 56b

At the required point, r ¼ a;

’ ¼ 45 ;

l ¼ 60 ) cos # ¼ cos 45 cos 60 ¼ 0:35 ) # ¼ 69:3 so



GM GM 1 GMa 11 GM 3 gr ¼  2 þ a þ  1 ¼ 0:9993 2 3 3 a 256a 2 ð8aÞ 24 a GM 1 1 GMa 1 1 1 GM g’ ¼ a pffiffiffi pffiffiffi þ 3 pffiffiffi pffiffiffi ¼ 0:0027 2 256a3 a 2 2 ð8a3 Þ 4 2 2 pffiffiffi pffiffiffi GMa GM 21 3 3 gl ¼  ¼ 0:0018 2 a ð8aÞ3 2 2 2 (b) To obtain the angular velocity of the rotation of the system (O) we take into account that the centrifugal force due to the rotation of the system at the equatorial plane is equal to the gravitational attraction between the two planets: M 2 p ¼ G

MM ð8aÞ2

where p is the distance from the centre of one planet to the centre of gravity of the system. Then we find M 2 4a ¼ G and finally O ¼ o.

MM ð8aÞ2

) 2 ¼

GM GM ¼ 4  64a3 256a3

101

Tides

57. Consider two planets of equal mass m and radius a separated by a centre-to-centre distance of 8a. Only gravitational forces act. (a) Calculate the tidal force at the equator on one of the planets directly, using the formula of the tidal potential (do so at l ¼ 0º, i.e. the point in front of the other planet). Express the result in mGal given that Gm/a2 ¼ 980 000 mGal. (b) Compare and comment on the reason for the difference between the results of the direct calculation and using the tidal potential. (c) What relationship must there be between the angular velocities of the planet’s spin and of the system’s rotation for the centrifugal force due to the planet’s spin to be equal to the tidal force at the equator and l ¼ 0º? (a) The exact calculation of the tidal force at a point located at the equatorial plane in front of the other planet is (Fig. 57)

Gm Gm Gm 1 1 fT ¼   ¼ 2 ¼ 46 875 gu a 49 64 ð7aÞ2 ð8aÞ2 Using the tidal potential c¼

Gmr2 

2ð RÞ

3

3cos2 #  1



where R is the centre-to-centre distance between the planets (8a), r the radius to the point where the tide is evaluated (a), and # the angle between r and R (at the equator in front of the other planet # ¼ 0º), the radial component of the tidal force can be derived from the potential. At the equator this is the total tidal force !  @c @ Gmr2  2Gma 2 T ¼ 3cos #  1 ¼ 2 ¼ 38 281 gu fr ¼ @r @r 2ð8aÞ3 2ð512a3 Þ (b) The difference between the value obtained by the exact calculation and by using the tidal potential is 8594 gu, that is, 18%. This is explained because the tidal

Ω w

w

a x

x 8a m

Fig. 57

m

a

102

Gravity

potential is a first-order approximation corresponding to terms of the order of (r/R)2. For a relatively large value of r/R (1/8) this approximation is not very good. (c) Now we make the spin centrifugal force equal to the tidal force for a point at the equator of one of the planets which is given by exact calculation

Gm 1 1 fT ¼ 2  a 49 64 If the spin angular velocity is op, the spin centrifugal force at the equator is given by fC ¼ o2p a Equating these two expressions we obtain



Gm 1 1 Gm 1 1  ) o2p ¼ 3  fC ¼ fT ) o2p a ¼ 2 a 49 64 a 49 64

ð57:1Þ

The centrifugal force due the rotation of the system with angular velocity os is equal to the gravitational attraction between the two planets: mo2s 4a ¼

Gm Gm2 ) o2s ¼ 2 64a 256a3

ð57:2Þ

From (57.1) and (57.2) we obtain the relation between the two angular velocities:

Gm 1 1  o2p op a3 49 64 ¼ 1:22 ) ¼ ¼ 1:10 Gm o2s os 256a3 58. Two planets of equal mass m and radius a are separated by a distance R. The spin angular velocity of each planet is such that the centrifugal force at the equator is equal to the maximum of the tidal force. If the sum of the two forces at the equator cancels the gravitational force, what is the distance R? The tidal potential is given by c¼

 Gmr2 1  3cos2 #  1 3 R 2

where R is the centre-to-centre distance between the planets, r the radius to the point where the tide is evaluated, and # the angle between r and R. The maximum value is at a point at the equator in front of the other planet, # ¼ 0º and r ¼a. Then fT ¼

 2Gma @c Gmr  ¼ 3 3cos2 #  1 ¼ @r R R3

ð58:1Þ

The spin centrifugal force for a point at the equator is fC ¼ o 2 a

Equating (58.1) and (58.2) we obtain the spin angular velocity,

ð58:2Þ

103

Tides

2Gma 2Gm ¼ o2 a ) o2 ¼ 3 3 R R If, at the point considered, the sum of the spin centrifugal force and the tidal force cancel the gravitational force of the planet, then F þ fC þ fT ¼ 0 where F ¼  Gm/a2. The value of R must be ffiffiffiffiffi p 4Gma Gm 3 ¼ 2 ) R ¼ 4a 3 R a

59. Two spherical planets A and B of radii 2a and a and masses 5M and M spin on their axes with equal angular velocities. They are separated by a centre-to-centre distance of 8a, and form a system that rotates in the equatorial plane of both planets with an angular velocity that is equal to that of the spin angular velocity of each one. (a) Determine the total potential for points on planet A. (b) Determine the expression for the three components of the total gravity, including the tide, for a point on the surface of planet A at longitude 0º. (c) If the Love number h on planet A is 0.5, determine the height of the terrestrial tide as a multiple of a at the equator, at local noon with respect to planet B, in the case that the system’s rotational angular velocity is the same as that of the spin of the two planets about their axes. (a) We calculate the centre of gravity of the system, measured from the centre of planet A (Fig. 59a): X ¼

5M  0 þ M  8a 4 ¼ a 5M þ M 3

A B

2a (4/3)a

x

x 8a 5M

Fig. 59a

M

104

Gravity

P ϑ

ϑ

j

j 2a

l

Fig. 59b

The total potential U at a point on the surface of planet A at latitude ’ is given by the sum of the gravitational potential V, plus the spin potential Ф, plus the tidal potential c produced by planet B: U ¼V þFþc U¼

 5GM 1 2 2 2 GMr2  þ o r cos ’ þ 3cos2 #  1 3 r 2 ð8aÞ 2

ð59:1Þ

where according to Fig. 59b

cos # ¼ cos ’ cos t where t is the local time of planet B with respect to planet A (hour-angle), at a point of l ¼ 0º, the geographical longitude at planet A. For t ¼ 0, the planet B is in front of the point, so U¼

 5GM 1 2 2 2 GMr2  þ o r cos ’ þ 3cos2 ’cos2 t  1 3 r 2 2ð8aÞ

(b) The components of gravity are gr ¼

 @U 5GM GMr  ¼  2 þ o2 rcos2 ’ þ 3cos2 ’cos2 t  1 3 @r r ð8aÞ

gy ¼  gl ¼

1 @U GMr ¼ o2 r cos ’ sin ’ þ 3 cos ’ sin ’cos2 t r @’ ð8aÞ3

1 @U GMr ¼ 3 cos ’ cos t sin t r cos ’ @t ð8aÞ3

105

Tides

At a point on the surface of planet A, r ¼2a,  5GM GM 2a  þ o2 2acos2 ’ þ 3cos2 ’cos2 t  1 3 2 4a ð8aÞ GM 2 gy ¼ o2 2a cos ’ sin ’ þ 3 2 3 cos ’ sin ’cos2 t 8 a GM 2 gl ¼ 3 2 3 cos ’ cos t sin t 8 a gr ¼ 

(c) At the equator ’ ¼ 0º, at 12 h with respect to B, t ¼ 180º, h ¼ 1/2, and o ¼ O. The height of the equilibrium terrestrial tide is given by B¼h

c g

At the equator of planet A the tidal potential (59.1) is c¼

2GM ð2aÞ2 2ð8aÞ3

¼

GM 128a

If we approximate g by gr 5GM GM 2a 5GM GM þ o2 2a þ 2¼ þ þ o2 2a 3 2 2 4a2 4a 128a ð8aÞ 159GM ¼ þ 2o2 a 128a2

gr ¼ 

and the height of the equilibrium tide is GM 1 128a

B¼ 2 159MG 2 þ 2ao 128a2

ð59:2Þ

We know that the angular velocity of the rotation of the system is equal to the spin angular velocity of both planets, so the spin angular velocity is given by 4 G5M 2 3GM ) o2 ¼ 5M o2 a ¼ 2 3 256a2 ð8aÞ By substitution in (59.2) GM 1 a 128a

¼ B¼ 3GM 2 159MG 312 þ 2a 128a2 256a2 60. The Earth is formed by a sphere of radius a and density r, and a core of radius a/2 and density 2r, in the northern hemisphere, centred on the axis of rotation and

106

Gravity tangent to the equatorial plane. The Moon has mass M/4 (where M ¼ (4/3)pra3), is at a distance (centre-to-centre) of 4a, and orbits in the equatorial plane. Determine: (a) The total potential and the components of gravity including the tidal forces. (b) The total deviation of the vertical from the radial at lunar noon, and the deviation due to the tide at the same hour for latitude 45º N, with m ¼ 1/8. (a) The total potential U is equal to the gravitational potential of the planet V1 with uniform density plus that of the core V2 using the differential mass, the spin potential F, and the tidal potential c. The gravitational potentials are given by (Fig. 60): GM r GM 0 V2 ¼ q V1 ¼

where the differential mass of the core is 4 a3 M GM M 0 ¼ pð2r  rÞ ¼ ) V2 ¼ 3 8 8q 8 and q is the distance to the centre of the core. Its inverse can be approximated by  
a 2 1   1 1 a 2 ¼ 1 þ cos y þ 3cos y  1 q r 2r 2r 2

The spin potential is

1 F ¼ o2 r2 sin2 y 2 and the total potential is " ! #
r 3 m  GM 1 a a2 1  2 2 1þ 1 þ cos y þ 3cos y  1 þ sin y þ c U¼ r 8 2r a 2 ð2rÞ2 2 w

2r a/2

q

P ϑ

q r

L

j 4a t – 180° r

Fig. 60

107

Tides where m ¼ o2 a3/GM. The tidal potential c due to the Moon is given by c¼ According to Fig. 60

 GML r2  3cos2 #  1 2R3

ð60:1Þ

cos2 # ¼ cos2 ’cos2 ðt  180Þ By substitution in (60.1), since R (the centre-to-centre distance between the Earth and the Moon) is 4a, ML ¼ M/4, and ’ ¼ 90º  y, we obtain c¼ The potential U is U ¼ GM



 GMr2  3cos2 ’cos2 ðt  180Þ  1 3 512a

 5 a a2  þ sin ’ þ 3 sin2 ’  1 8r 16r2 64r3   r2 m 2 r2  2 2 þ 3 cos ’ þ 3 cos ’ cos t  1 a 2 512a3

The components of gravity are found by taking the derivatives of U with respect to r and ’:   5 2a 3a2  sin ’  3 sin2 ’  1 gr ¼ GM  2  3 4 8r 16r 64r    r 2r 2 2 3 cos ’ cos þ 3 m cos2 ’ þ t  1 a 512a3 1 @U r @’ GM a a2 ¼  cos ’  ð6 cos ’ sin ’Þ 2 r 16r 64r3

r2 m r2 2 6 sin ’ cos ’ cos t þ 3 2 sin ’ cos ’ þ a 2 512a3

gy ¼ 

By substituting r ¼ a, ’ ¼ 45º, m ¼ 1/8, and at 12 h lunar time, t ¼ 180º, we obtain gr ¼ 1:128

GM a2

gy ¼ 0:023

GM a2

(b) The deviation of the vertical with respect to the radial direction at 12 h lunar time is given by tan i ¼

gy ¼ 0:020 ) i ¼ 1:2 gr

108

Gravity

The part of the deviation due to the lunar tide is given by tan i0 ¼

gyM 1 1 @c 1 GMr2 ¼ 6 sin ’ cos ’cos2 t ¼ gr r @’ gr r512a3 gr

and by substitution of the same values i0 ¼ 0:3 The greater part of the deviation is due to the core. 61. Two spherical planets, planet A of radius 2a and mass 3m and planet B of radius a and mass m, are separated by a centre-to-centre distance of 6a. The system rotates in the equatorial plane and each planet spins on its axis with the same angular velocity. What, for each planet, is the ratio between the force of gravity and the maximum of the tidal force at the equator in front of the other planet? The centre of gravity of the system measured from the centre of planet A is (Fig. 61) 0  3m þ 6am 3 ¼ a 4m 2 The angular velocity of the system is given by 3mGm

3 Gm ¼ 3mo2 a ) o2 ¼ 3 2 54a ð6aÞ 2

The tidal force (radial component) can be calculated from the tidal potential c¼

  GMr2  @c GMr  ¼ 3 3cos2 #  1 3cos2 #  1 ) frT ¼ 3 @r R 2R

where R is the centre-to-centre distance between the planets. For a point on the equator in front of the other planet, # ¼ 0º, and frT ¼

2GMr

ð6aÞ3 2Gm2a Gm Planet A : frT ¼ ¼ 0:018 2 216a3 a 2G3ma Gm T ¼ 0:028 2 Planet B : fr ¼ 216a3 a Gravity without tides is the sum of the gravitational and centrifugal forces: g¼

Gm  o2 r r2

G3m 3Gm Gm Gm  o2 2a ¼  2a ¼ 0:71 2 4a2 4a2 54a3 a Gm Gm Gm Gm 2 a ¼ 0:98 2 Planet B : g ¼ 2  o a ¼ 2  a a 54a3 a

Planet A : g ¼

109

Tides

w

A

B

2a (3/2)a

(9/2)a

x

x 6a

a

m

3m

Fig. 61

The ratios between the gravity and tidal forces are: Planet A :

g 0:71 ¼ 39:44 ¼ T fr 0:018

Planet B :

g 0:98 ¼ ¼ 35:00 T fr 0:028

62. The Earth is of radius a and density r, with a core of radius a/2 and density 3r on the axis of rotation in the southern hemisphere tangent to the equatorial plane. The Moon has mass M/2 and its centre is at 4a from the centre of the Earth (M ¼ 4/3pra3). (a) Write down the total potential. (b) What is the value of the angular velocity of the Earth if at the point 30º N, 30º E at 06:00 lunar time the radial component of gravity is equal to GM/a2? (c) In this case, what is the ratio between the angular velocity of the Earth’s rotation and that of the system? (a) As in previous problems the total potential U is given by U ¼ V1 þ V2 þ F þ c The differential mass of the core is: M1 ¼

4p a3 M ð3r  rÞ ¼ 3 8 4

110

Gravity

w

P ϑ q

r

90° – q 4a

q

L

t a/2 x 3r

Fig. 62

Using the approximation for 1/q, where q is the distance from a point on the surface of the Earth to the centre of the core, the gravitational potential due to the core is    GM GM 1 a a2  V2 ¼ ¼  2 cos y þ 3 3cos2 y  1 4q 4 r 2r 8r The potential due to the spin of the Earth is given by

1 GM m F ¼ o2 r2 sin2 y ¼ 3 r2 sin2 y 2 a 2 where m¼

o2 a3 GM

and the tidal potential due to the Moon is c¼

 GMr2   GM r2 1  2 3cos #  1 ¼ 3sin2 ycos2 t  1 3 3 2 64a 2 256a

where t is the hour-angle of the Moon (Fig. 62) Then, the total potential U is   5 a a2   2 cos y þ 3 cos2 y  1 U ¼ GM 3 4r 8r 32r  3  r m r2  2 2 þ 3 sin2 y þ 3 sin y cos t  1 a 2 256a3 (b) The radial component of gravity is given by gr ¼

@U @r

  5 2a 3a2  ¼ GM  2 þ 3 cos y  3 cos2 y  1 4r 8r 32r4   rm sin2 y 2r  2 2 þ 3 sin y cos t  1 þ a3 256a3

111

Tides

Substituting y ¼ 60º and at 6 h, t ¼ 90, we have gr ¼

GM a2



141 3 1 GM þm   ¼ 2 128 4 128 a

Solving for m we obtain m ¼ 2.81, and the spin angular velocity is rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2:81GM rad s1 o¼ a3 (c) The centre of gravity r0 of the Earth–Moon system, measured from the centre of the Earth, is given by M 4a M r þ M r 8a T 1 L 2 r0 ¼ ¼ 2 ¼ 5 1 MT þ ML 7 þ M 4 2 where the mass of the Earth MT includes that of the core, MT ¼ M þ

M M and ML ¼ 4 2

and r1 ¼ 0 and r2 ¼ 4a. To calculate the angular velocity of the system, O, we put the centripetal force at the Moon equal to the gravitational force between the Earth and the Moon: ML ðr2  r0 Þ 2 ¼ G

MT ML r2

Substituting and solving for O, we obtain 5 2

M M 2 8 8 G ¼ O 4  a; 16a2 2 7

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi GM 7 so O ¼ a3 256

Then, the ratio between the angular velocities of the spin of the Earth and of the system is: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2:81GM o a3 ¼ rffiffiffiffiffiffiffiffiffiffiffiffi ¼ 10:14

7GM 256a3 63. Consider two planets of equal mass m and radius a separated by a centre-tocentre distance of 8a. The planets revolve around their centre of mass and spin around their own axes. Their spin angular velocity is such that the value of the centrifugal force is equal to the maximum tidal force of the two at the equator.

112

Gravity

(a) Calculate, for a point on the equator of one of the planets and longitude l ¼ 90º, the value of the vector g including all the forces acting at that point (l ¼ 0º corresponds to the point on the line joining the two centres of the planets) at t ¼ 0. (b) What is the deviation of the vertical from the radial at the point w ¼ 45º, l ¼ 0º? (a) If o is the spin angular velocity of the two planets, the centrifugal force at the equator y ¼ 90 , r ¼ a, only has radial component: frC ¼ o2 a

ð63:1Þ

The radial component of the tidal force can be obtained from the tidal potential c which, in the first-order approximation, is given by (Fig. 63) c¼

 Gmr2  3cos2 #  1 3 2R

where, if ’ is the latitude and l the longitude,

cos # ¼ cos ’ cosðt  lÞ On the equator f ¼ 0º, so c¼

 Gmr2  3cos2 ðt  lÞ  1 2R3

The radial component of the tidal force is     @c @ Gmr2  Gmr  2 frT ¼ ¼ 3cos ðt  lÞ  1 ¼ 3 3cos2 ðt  lÞ  1 3 @r @r 2R R Ω

w

w

a

8a x m

Fig. 63

90°

x m

a

113

Tides

The maximum value is for t ¼ l and putting R ¼ 8a and r ¼ a, we obtain frT ¼

Gma ð8aÞ

3

2¼

Gm 256a2

ð63:2Þ

Then, as the centrifugal force is equal to the tidal force, we put (63.1) equal to (63.2) and solve for o: o2 a ¼

Gm Gm ) o2 ¼ 256a2 256a3

ð63:3Þ

We know that m O2 4a ¼ Gm2/R2, so solving for O,

2 ¼

Gm 256a3

and then using (63.3), we obtain o/O ¼ 1. The total potential U is the sum of the gravitational, spin, and tidal potentials, which for t ¼ 0, is given by U ¼V þFþc¼

 Gm 1 2 2 2 Gmr2  þ o r sin y þ 3sin2 ycos2 l  1 3 r 2 2ð8aÞ

The components of gravity including the tidal forces are gr ¼ gy ¼ gl ¼

 @U Gm Gmr  ¼  2 þ o2 rsin2 y þ 3sin2 ycos2 l  1 3 @r r ð8aÞ 1 @U Gmr ¼ o2 r sin y cos y þ 3 sin y cos ycos2 l r @y ð8aÞ3

1 @U Gmr ¼ 3 sin y cos l sin l r sin y @l ð8aÞ3

By substitution of r ¼ a, l ¼ 90º, and y ¼ 90º we have gr ¼ 

Gm Gm þ o2 a  a2 512a2

gy ¼ 0 gl ¼ 0 (b) By substitution of r ¼ a, l ¼ 0, and y ¼ 45º in (63.4) we obtain

Gm 1 Gm 3 gr ¼  2 þ o2 a þ  1 a 2 512a2 2 Gm Gm 1 Gm 1 1021 Gm þ ¼ ¼ 2 þ 2 2 a 256a 2 512a 2 1024 a2 1 Gm 3 Gm 1 Gm 3 5 Gm ¼ þ ¼ gy ¼ o 2 a þ 2 2 2 2 512a 2 256a 2 512a 2 1024 a2 gl ¼ 0

ð63:4Þ

114

Gravity

The deviation of the vertical with respect to the radial direction is tan i ¼

gy ¼ 0:28 gr

64. Two spherical planets of radii 2a and a and masses 8M and M separated by a centre-to-centre distance of 4a spin on their own axes and rotate in the equatorial plane with the same angular velocity. (a) Determine all the forces acting at a point on the smaller planet at geocentric coordinates w ¼ 60º N, l ¼ 0º (00:00 h local time corresponds to passage of the other planet through the zero meridian). (b) For this same point, calculate the astronomical latitude and the tidal deviation of the vertical. (a) First we determine the centre of gravity of the system, putting the origin at the centre of the small planet (Fig. 64): x¼

0  M þ 4a  8M 32a ¼ 9M 9

Because the spin angular velocity of each planet is equal to the angular velocity of the system (o ¼ O), we can write, for the small planet, putting the gravitational attraction of the two planets equal to the centripetal force: G8MM ð4aÞ

2

¼ M o2

32a 9

and solving for o, o2 ¼

9GM 64a3 Ω w

w

q

q r

a x

x 4a

M

Fig. 64

8M

2a

115

Tides

On the small planet the gravitational force is gr ¼ 

GM r2

gy ¼ 0 and the force due to its spin is fr ¼ o2 rsin2 y fy ¼ o2 r cos y sin y For the point under consideration, r ¼ a, y ¼ 30º, we obtain Gm 9 GM 1 247GM þ ¼ 2 2 2 a 64 pffiffiffi a 4 pffiffi256a ffi 3 GM 9 3 9 GM ¼ ¼ 2 a 4 64 a3 a 256

grGC ¼  gyGC

To add the tidal force we use the tidal potential in the first-order approximation, c¼

 G8Mr2  3cos2 #  1 3 2R

where cos # ¼ sin y cos (t  l). The tidal force for the point considered, r ¼ a, y ¼ 30º, t ¼ l ¼ 0º, and R ¼ 4a, is given by @c GM 1 ¼ 2 @r a 32 pffiffiffi 1 @c GM 3 3 T ¼ 2 fy ¼ r @y a 32

frT ¼

The total force acting at the point is the sum of the three forces, gravitational, centrifugal, and tidal: 255GM 2 256a pffiffiffi GM 12 3 ¼ 2 a 256

grtotal ¼  gytotal

(b) The astronomical latitude is given by ’a ¼ ’ þ i, where i is the deviation of the vertical without considering the tide: pffiffiffi g total 9 3 ¼ 0:06 ) i ¼ 3:6 ) ’a ¼ 60 þ 3:6 ¼ 63:6 tan i ¼ ytotal ¼ 247 gr The maximum deviation of the vertical due to the tide at the point considered is i0 given by

116

Gravity

fyT grtotal

tan i0 ¼

pffiffiffi 3 3 ¼ 32 ¼ 0:163 ) i0 ¼ 9:3 255 256

Gravity observations 65. Determine the values of gravity at the following series of points belonging to a gravimetric survey with a Worden gravimeter, specifying the drift correction for each of them. Station

Time

Reading

A (base) B C D A

08:30 09:21 11:34 13:20 14:20

562.5 400.7 437.9 360.1 568.8

The gravity at the base is 980.139 82 Gal, and the gravimeter constant is 0.301 81 mGal/ru (ru: reading unit). The instrument drift is given by d¼

LAe  LAb tAe  tAb

where LAb and LAe are the readings at the base A at the beginning and end of the measurements taken at times tAb and tAe, respectively. By substitution we obtain d¼

568:8  562:5 ¼ 1:08 ru=hour 14:33  8:50

The corrected reading for station j is given by Ljc ¼ Lj – d (tj – tAb) where Lj is the reading taken at time tj. For a Worden gravimeter the increment in gravity between two points (D g) is proportional to the increment in the readings corrected by the instrument drift (D Lc ): D g ¼ K D Lc where K is the instrument constant. Thus, from the readings we obtain the following results.

117

Gravity observations

Station

Corrected reading

D g (mGal)

g (mGal)

A (base) B C D A

562.5 399.8 434.6 354.9 562.5

49.10 10.50 24.05 62.66

980 980 980 980 980

139.82 090.72 101.22 077.17 139.83

66. Point A is at a geopotential level of 97.437 43 gpu. Point B is at a difference of 15.213 m in height relative to A, and has a value of gravity of 9.712 611 m/s2. Calculate: (a) The value of gravity at point A, if the difference in readings of a Worden gravimeter between B and A is 17.8 ru, and the gravimeter constant is 0.308 21 mGal/ru. (b) The geopotential number, dynamic height, and Helmert height of the point B given that the normal gravity at a point of latitude 45º on the ellipsoid is 980 629.40 mGal. (a) Using a Worden gravimeter the increment of gravity between points A and B is given by DgAB ¼ KDLBA ¼ 5:5 mGal where K is the instrument constant and DL is the difference between the readings at points A and B. The gravity at point A is gA ¼ gB þ DgAB ¼ 971 266:6 mGal (b) The geopotential number at B can be calculated from the value at A in the form CB ¼ CA þ


g þ g  A B
hBA ¼ 82:661 59 gpu 2

where gravity is given in Gal and increments in height in km, because the geopotential units are, 1 gpu ¼ 1 kGal m ¼ 1 Gal km. The dynamic height is given by HDB ¼

CB ¼ 84:294 m g45

The Helmert height can be calculated from the dynamic height by H¼

C g þ 0:0424H

where C is given in gpu, g in Gal, and H in km. Solving for H, we obtain pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g  g2 þ 4  0:0424C H¼ 2  0:0424 Taking the positive solution because point B is above the geoid (CB > 0) we obtain HB ¼ 85:107 m

118

Gravity

67. In a geometric survey with measurements of gravity using a Worden gravimeter, the following values were obtained: Station

Gravimeter reading (ru)

Time

A (base) B C A

1520.23 1759.15 1583.11 1521.30

8h 9h 9h 9h

50 15 35 50

m m m m

Gravity (gu)

Height difference (m)

9 793 626.8 9 794 363.9 9 793 820.7

30.410 301.863

Calculate the gravimeter readings corrected for drift, and the gravimeter constant The instrument drift is given by d¼

LAe  LAb tAe  tAb

where LAb and LAe are the readings at the base A at the beginning and end of the measurements taken at times tAb and tAe, respectively. By substitution we obtain, d ¼ 1:07 ru=hour A reading corrected at station j is given by Lcj ¼ Lj  dðtj  tAb Þ where Lj is the reading at time tj. For a Worden gravimeter the increment in gravity between two points (D g) is proportional to the increment in the readings corrected by the instrument drift (D Lc): D g ¼ K DLc where K is the instrument constant. Thus, K can be calculated in the form K¼


g
Lc

From each pair of observations we obtain a value of K. Finally we take the arithmetic mean (Km) from all the values obtained. The results are given in the following table. Station

Corrected reading (ru)

Gravity (mGal)

K (mGal/ru)

A (base) B C A

1520.23 1758.70 1582.31 1520.23

979 362.68 979 436.39 979 382.07

0.3091 0.3079 Km ¼ 0.3085

68. The following table is obtained from observations with a Lacoste–Romberg gravimeter:

119

Gravity observations

Station

Gravimeter reading

Time

A B C A

3614.351 3650.242 3610.633 3614.414

10:10 10:25 10:37 11:02

The gravimeter scale factor is 1.000 65, and the equivalence between reading units and the relative value of gravity in mGal is given by

Reading

Value in mGal

Interval factor

3600 3700

3846.02 3953.15

1.071 25 1.071 40

Given that the value of gravity at point A is 9.794 6312 m s2, calculate the values at B and C. First we correct the readings by the instrument drift: d¼

LAe  LAb tAe  tAb

where LAe and LAb are the readings at the base A at the end and the beginning of the survey at times tAe and tAb. Then d ¼ 0:0727 ru=hour The corrected reading at each station j is given by Lcj ¼ Lj  dðtj  tAb Þ where Lj is the reading at time tj. The corrected readings are: LcA ¼ 3614:351 LcB ¼ 3650:224 LcC ¼ 3610:600 These readings are converted into relative gravity values Rj using the conversion table. The reading at station A is LcA ¼ 3600 þ 14:351 and the relative gravity value is RA ¼ (3846.02 þ (14.351  1.07125)) 1.00065 ¼ 3863.90 mGal For stations B and C,

120

Gravity

RB ¼ (3846.02 þ (50.224  1.07125)) 1.00065 ¼ 3902.36 mGal RC ¼ (3846.02 þ (10.600  1.07125)) 1.00065 ¼ 3859.88 mGal To convert the relative values into absolute values we need to know both values at one station, in our case in station A: gA ¼ 9.794 6312 m s2 ¼ 979 463.12 mGal gB ¼ gA – RA þ RB ¼ 979 501.58 mGal gC ¼ gA – RA þ RC ¼ 979 459.10 mGal

3

Geomagnetism

Main field 69. Assume that the geomagnetic field of the Earth is a geocentric dipole with a North Pole at 80 N, 45 E and a magnetic moment 8  1022 A m². Calculate for a point with geographical coordinates 45 N, 30 W the components NS, EW, and Z of the Earth’s magnetic field, the declination and inclination, and the geomagnetic longitude. Earth’s radius: 6370 km and the constant C ¼ 107 H m1 (this value is used in all problems). We calculate first the geomagnetic latitude and longitude (f  , l  ) from the geographical coordinates (f, l) of the point and the geographical coordinates of the geomagnetic North Pole (fB, lB) by the equation sin ’ ¼ sin ’B sin ’ þ cos ’B cos ’ cosðl  lB Þ sin l ¼

sinðl  lB Þ cos ’ cos ’

Substituting the values fB ¼ 80 N lB ¼ 45 E f ¼ 45 N l ¼ 30 W ¼ 330 we obtain f ¼ 46:70 l ¼ 84:82 In the geocentric magnetic dipole model, the vertical (Z  ) and horizontal (H  ) components of the magnetic field can be obtained from Z  ¼ 2B0 sin f H  ¼ B0 cos f Cm B0 ¼ 3 a 121

ð69:1Þ

122

Geomagnetism

In these equations B0 is the geomagnetic constant, m the magnetic moment of the dipole, a the Earth’s radius, and the constant C ¼107 H m1. In this case we are given that m ¼ 8  1022 A m2 a ¼ 6370 km ¼ 6.37  106 m By substitution in Equations (69.1) we obtain: B0 ¼ 30 951 nT Z  ¼ 45 051 nT H  ¼ 21 227 nT The geomagnetic declination is given by  cos fB sinðl  lB Þ cos f   D ¼ 14:16

sin D ¼

The NS (X  ) and EW (Y  ) components are X  ¼ H  cos D ¼ 20 582 nT Y  ¼ H  sin D ¼ 5193 nT Finally, the geomagnetic inclination or dip (I  ) at that point is given by tan I  ¼ 2 tan f ) I  ¼ 64:77 70. Assume that the geomagnetic field is produced by a geocentric dipole of magnetic moment 8  1022 Am², with North Pole at 80 N, 70 W, and that the Earth’s radius is 6370 km. Calculate for a point with geographical coordinates 60 N, 110 E: (a) Its geomagnetic coordinates, the components of the Earth’s magnetic field (X  , Y , Z  , H  ), the total field, the declination, and the inclination. (b) The equation of the line of force passing through it. (a) For this point the difference in longitude from the Geomagnetic North Pole (GMNP) is 180º (Fig. 70), so both are on the same great circle. Then, the geomagnetic coordinates are obtained from f ¼ f  ð90  fB Þ ¼ 50 l ¼ 180 The expressions for the geomagnetic vertical and horizontal components and for the total geomagnetic field are Z  ¼ 2B0 sin f H  ¼ B0 cos f pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F  ¼ H 2 þ Z 2 B0 ¼

Cm a3

Main field

123

GMNP

GNP P x f∗

fB

f

l=0

l l∗

Fig. 70

The values of the constants are m ¼ 8  1022 A m2 a ¼ 6370 km ¼ 6.37 106 m C ¼ 107 H m1 Substituting in the above equations we obtain B0 ¼ 30 951 nT Z  ¼ 47 420 nT H  ¼ 19 895 nT F  ¼ 51 424 nT For this point the geomagnetic declination D  ¼ 0. So the NS (X  ) and EW (Y  ) components are D ¼ 0 X  ¼ H  cos D ¼ H  ¼ 19 895 nT Y  ¼ H  sin D ¼ 0 The geomagnetic inclination (I  ) is given by tan I  ¼ 2 tan f ) I  ¼ 67:23 (b) The equation of the line of force passing through a point with geomagnetic co-latitude ỵ is r ¼ r0 sin2 y In this equation r0 is the distance from the Earth’s centre to a point on the line of force with y ¼ 90º. The distance r0 is different for each line of force.

124

Geomagnetism For the point with geomagnetic latitude f  ¼ 50º located in the Earth’s surface (r ¼ a), y ¼ 90º – f  ¼ 40º, so a r0 ¼ 2 ¼ 15 417km sin y 71. Assume that the geomagnetic field is produced by a geocentric dipole of magnetic moment 7.5  1022A m², with North Pole at 75 N, 65 W, and that the Earth’s radius is 6372 km. Calculate: (a) The NS and EW components for a point on the Earth’s surface at which the inclination is 67 and the geomagnetic longitude is 120 . (b) The geographical coordinates of that point. (c) The geomagnetic coordinates, field components, declination, and inclination of the point on the geographical equator of zero geomagnetic longitude. (a) The geomagnetic latitude f  is obtained from tan I  ¼ 2 tan f ) f ¼ 49:7 The horizontal component, H  , can be calculated from the geomagnetic constant, B0, and the geomagnetic latitude: Cm B0 ¼ 3 ¼ 28 989 nT a H  ¼ B0 cos f ¼ 18 761 nT To obtain the NS (X  ) and EW (Y  ) components it is necessary to calculate the declination (D  ) from the spherical triangle with vertices at the Geographical North Pole (GNP), Geomagnetic North Pole (GMNP), and the point P (Fig. 71a). But we need to calculate the geographic latitude firstly by solving the spherical triangle. Applying the cosine law to the angle (90º – f): cosð90  fÞ ¼ cosð90  fB Þ cosð90  f Þ þ sinð90  fB Þ sinð90  f Þ cosð180  l Þ sin f ¼ sin fB sin f  cos fB cos f cos l

ð71:1Þ

By substitution of the values, f ¼ 55:1 To obtain the geomagnetic declination we apply the sine law in the spherical triangle of Fig. 71a: sin D sinð180  l Þ cos fB sin l  ¼ ) sin D ¼  sinð90  fB Þ cos f sinð90  fÞ

ð71:2Þ

and substituting the values we find D ¼ 23:1 It is important to note that we have added a minus sign in the last equation in order for the declination be positive toward the east.

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125

GNP 90º– fB l – lB 180º– l∗

90º – f

GMNP q = 90º– f∗

D∗

P

Fig. 71a

The NS (X  ) and EW (Y  ) components are X  ¼ H  cos D ¼ 17 262 nT Y  ¼ H  sin D ¼ 7 349 nT (b) The calculated geographical latitude is f ¼ 55:1 The geographical longitude is obtained by applying the cosine law to the spherical triangle of Fig. 71a: cosð90  f Þ ¼ cosð90  fB Þ cosð90  fÞ þ sinð90  fB Þ sinð90  fÞ cosðl  lB Þ

ð71:3Þ

cos y ¼ sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ so cosðl  lB Þ ¼

sin f  sin fB sin f cos fB cos f

ð71:4Þ

Substituting the values, gives l  lB ¼ 101:6 To take the inverse cosine in the correct quadrant we bear in mind that l  < 0 implies that the point is to the west of the Geomagnetic North Pole, that is l – lB < 0. So we obtain l ¼ 166:6 W (c) If this point is on the geographical equator (f ¼ 90 ) and has zero geomagnetic longitude (l  ¼ 0º), it is on the same geographical meridian as the Geomagnetic North Pole. Then from Fig. 71b

126

Geomagnetism

GMNP

GNP

fB

P f∗ x

Fig. 71b

f ¼ 90  fB ¼ 15:0 l ¼ 0  Z  ¼ 2B0 sin f ¼ 15 006 nT H  ¼ B0 cos f ¼ 28 001 nT D ¼ 0 X  ¼ H Y  ¼ 0 nT tan I  ¼ 2 tan f ) I  ¼ 28:2 72. Assume that the geomagnetic field is that of a dipole with North Pole at 75 N, 0 E. What is the conjugate point of that of geographical coordinates 30 N, 30 E? First, we calculate the geomagnetic coordinates (f  , l  ) (Problem 71; Fig. 71a): sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ sin l ¼

ð72:1Þ

sinðl  lB Þ cos f cos f

The values of the geographical coordinates are fB ¼ 75 f ¼ 30

lB ¼ 0 l ¼ 30

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127

GMNP

GNP

fB

Px f∗

– f∗ P1 x

Fig. 72

By substitution in Equations (72.1) we obtain: f ¼ 42:6 l ¼ 36:0 A magnetic conjugate point is a point on the Earth’s surface that is located on the same line of force and in the opposite hemisphere (Fig. 72, P and P1). Then, its geomagnetic coordinates (f1 , l1 ) are f1 ¼ f ¼ 42:6 l1 ¼ l ¼ 36:0 To calculate the geographical coordinates for this point (f1, l1) we use the spherical triangle of Fig. 71a. We calculate the geographical latitude applying the cosine law: cosð90  f1 Þ ¼ cosð90  fB Þ cosð90  f1 Þ þ sinð90  fB Þ sinð90  f1 Þ cosð180  l1 Þ sin f1 ¼ sin fB sin f1  cos fB cos f1 cos l1 f1 ¼ 53:9 To calculate the geographical longitude we apply the cosine law again: cosð90  f1 Þ ¼ cosð90  fB Þ cosð90  f1 Þ þ sinð90  fB Þ sinð90  f1 Þ cosðl1  lB Þ

128

Geomagnetism

cosðl1  lB Þ ¼

sin f1  sin fB sin f cos fB cos f

l1  lB ¼ 47:3 and taking the solution in the correct quadrant l1 > 0 ) l1  lB > 0 l1 ¼ 47:3 73. Assume the centred dipole approximation, with the coordinates of the Geomagnetic North Pole being 65 N, 0 E, and the magnetic moment of the dipole 8  1022 A m². Calculate, for a point on the Earth’s surface at geographical coordinates 30 N, 30 E: (a) The geographical coordinates of the conjugate point. (b) The declination, inclination, and vertical and horizontal components of the field at both points. Compare and contrast the results. Earth’s radius: 6370 km. (a) First, we calculate the geomagnetic coordinates (f  , l  ) for point P with geographical coordinates f ¼ 30 N, l ¼ 30 E using the equations (Problem 71; Fig. 71a) sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ sin l ¼

ð73:1Þ

sinðl  lB Þ cos f cos f

The geographical coordinates of the Geomagnetic North Pole are fB ¼ 65 ;

lB ¼ 0 

Substituting in Equations (73.1) results in f ¼ 50:4 l ¼ 42:7 The geomagnetic coordinates (f1 , l1 ) of the magnetic conjugate point P1 satisfy (Problem 72): f1 ¼ f ¼ 50:4 l1 ¼ l ¼ 42:7 To calculate the geographical coordinates for this point (f1, l1) we use the spherical triangle of Fig. 71a. We calculate the geographical latitude applying the cosine law cosð90  f1 Þ ¼ cosð90  fB Þ cosð90  f1 Þ þ sinð90  fB Þ sinð90  f1 Þ cosð180  l1 Þ sin f1 ¼ sin fB sin f1  cos fB cos f1 cos l1 f1 ¼ 63:7

Main field

129

To calculate the geographical longitude we apply the cosine law again and, solving for l1, sin f1  sin fB sin f1 cos fB cos f1  l1  lB ¼ 77:1 l1 > 0 ) l1  lB > 0 cosðl  lB Þ ¼

l1 ¼ 77:1 (b) First we calculate the geomagnetic constant B0 B0 ¼

Cm 8  1022 ¼ ¼ 30 951 nT a3 ð6379Þ3  109

We calculate the declination D  , inclination I  , and vertical Z  and horizontal H  components of the field at both points. The results are shown in the table, where we notice that except for the declinations, which are very different, all other values are equal for both points except in sign. P

P1

 cos fB sinðl  lB Þ sin D ¼ cos f D  ¼ 19.33 tan I  ¼ 2 tan f  ) I  ¼ 67.5 Z  ¼ 2B0 sinf  ¼ 47 662 nT H  ¼ B0 cosf  ¼ 19 750 nT

 cos fB sinðl1  lB Þ sin D1 ¼ cos f1 D1 ¼ 40:2 tan I1 ¼ 2 tan f1 ) I1 ¼ I1 ¼ 67:5 Z1 ¼ 2B0 sin f1 ¼ Z  ¼ 47 662 nT H1 ¼ B0 cos f1 ¼ H  ¼ 19 750 nT

74. Assume the centred dipole approximation, with the coordinates of the Geomagnetic North Pole 78.5 N, 70.0 W, and the magnetic dipole moment being 8.25  1022 A m². Calculate, for a point on the surface with coordinates 60.0 S, 170.0 W: (a) Its geomagnetic coordinates, declination, inclination, and vertical and horizontal components of the field. (b) The potential at that point. (c) The declination and inclination at the point diametrically opposite to it. Earth’s radius: 6370 km. (a) We calculate the geomagnetic coordinates (f  , l  ) using the equations (Problem 71, Fig. 71a) sin f ¼ cos y ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ sin l ¼

ð74:1Þ

sinðl  lB Þ cos f cos f

The values of the geographical coordinates are fB ¼ 78:5 f ¼ 60:0

lB ¼ 70:0 l ¼ 170:0

130

Geomagnetism

Substitution in Equations (74.1) gives f ¼ 60:0 l ¼ 80:0 The geomagnetic declination is given by (Problem 71, Fig. 71a)  cos fB sinðl  lB Þ cos f   D ¼ 23:1 sin D ¼

The inclination (I  ) at that point is given by tan I  ¼ 2 tan f ) I  ¼ 73:9 The vertical (Z  ) and horizontal (H  ) components of the magnetic field can be obtained from Z  ¼ 2B0 sin f H  ¼ B0 cos f Cm B0 ¼ 3 a Substituting the values given we obtain: B0 ¼ 31 918 nT Z  ¼ 2B0 sin f ¼ 55 279 nT H  ¼ B0 cos f ¼ 15 963 nT (b) The potential at a point on the Earth’s surface (r ¼ a) at geomagnetic latitude f  is given by F¼

Cm cos y Cm sin f ¼ ¼ 176 T m a2 a2

(c) We can observe in Fig. 74 that at the point diametrically opposite the geographical and geomagnetic coordinates are f1 ¼ f ¼ 60:0 N l1 ¼ l þ 180 ¼ 10:0 E f1 ¼ f ¼ 60:0 l1 ¼ l þ 180 ¼ 100:0 The geomagnetic declination at that point satisfies (Fig. 71a)  cos fB sinðl1  lB Þ  cos fB sinðl þ 180  lB Þ ¼ cos f1 cos f cos fB sinðl  lB Þ ¼ ¼  sin D cos f

sin D1 ¼

Main field

131

GNP

GMNP

fB

P1 f1

f

f∗ l∗ P

Fig. 74

Then, D1 ¼ D . The geomagnetic inclination is given by tan I1 ¼ 2 tan f1 ¼ 2 tan f ) I1 ¼ I  So, we can notice that the two points P and P1 (Fig. 74) that are diametrically opposite are not magnetic conjugate points because the geomagnetic longitudes are different by 180º. 75. Consider a point P on the Earth’s surface at coordinates 30 S, 10 W at which the NS component of the geomagnetic field is 27 050 nT and the EW component is 5036 nT, with the geomagnetic inclination being negative. Assuming the centred dipole hypothesis with magnetic moment 7.8  1022 A m², calculate: (a) The geographical coordinates of the Geomagnetic North Pole. (b) The geomagnetic coordinates of P’s conjugate point. (a) We calculate first the geomagnetic constant B0: B0 ¼

Cm ¼ 30 177 nT a3

The geomagnetic declination D  is obtained from the NS (X  ) and EW (Y  ) components of the geomagnetic field: tan D ¼

Y  5036 ¼ ) D ¼ 10:5 X  27 050

132

Geomagnetism The geomagnetic latitude f  is calculated from the horizontal component H  : pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi H  ¼ X 2 þ Y 2 ¼ 270502 þ ð5036Þ2 ¼ 27 515 nT H  ¼ B0 cos f ) f ¼

H ¼ 24:3 B0

We then have two solutions for the geomagnetic latitude. To choose the correct one we bear in mind that a negative value of the geomagnetic inclination implies a negative value of the geomagnetic latitude: tan I  ¼ 2 tan f I  < 0 ) f < 0 f ¼ 24:3 With these results we calculate the geographical coordinates of the Geomagnetic North Pole (fB, lB) using the spherical triangle in Fig. 71a. Applying the cosine rule for the angle 90º – fB: cosð90  fB Þ ¼ cosð90  f Þ cosð90  fÞ þ sinð90  f Þ sinð90  fÞ cos D sin fB ¼ sin f sin f þ cos f cos f cos D fB ¼ 79:0 To calculate the longitude lB of the Geomagnetic North Pole, we apply the cosine law for the angle 90º – f  : cosð90  f Þ ¼ cosð90  fB Þ cosð90  fÞ þ sinð90  fB Þ sinð90  fÞ cosðl  lB Þ sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ cosðl  lB Þ ¼

sin f  sin fB sin f ) l  lB ¼ 61:0 cos fB cos f

To choose the correct sign for the longitude we notice that the declination is negative and then the point must be to the east of the Geomagnetic North Pole: D  < 0 ) l  lB > 0 lB ¼ 61  10 ¼ 71:0 W (b) The geomagnetic coordinates (f1 , l1 ) of P’s conjugate point verify that f1 ¼ f ¼ 24:3 l1 ¼ l We calculate the geomagnetic longitude l  by sinðl  lB Þ cos f cos f  l ¼ 56:2 ¼ l1 sin l ¼

Main field

133

76. At a point P on the Earth’s surface with coordinates 45 N, 30 W, the value of the total geomagnetic field is 49 801 nT, the horizontal component is 21 227 nT, and the EW component is 5171 nT, with the magnetic inclination being positive. Calculate: (a) The geographical coordinates of the Geomagnetic North Pole. (b) The value of the geomagnetic potential at P. (c) The distance from the Earth’s centre to the point at which the line of force passing through P intersects the geomagnetic equator. Earth’s radius: 6370 km. (a) We calculate first the geomagnetic inclination, latitude, and declination by cos I  ¼

H ) I  ¼ 64:8 F

tan I  ¼ 2 tan f ) f ¼ 46:7 sin D ¼

Y ) D ¼ 14:1 H

With these results we calculate the geographical coordinates of the Geomagnetic North Pole solving the spherical triangle (Fig. 71a) in the same way as in Problem 71: sin fB ¼ sin f sin f þ cos f cos f cos D fB ¼ 80:0 sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ sin f  sin fB sin f cos fB cos f l  lB ¼ 75:2 D > 0 ) l  lB < 0

cosðl  lB Þ ¼

lB ¼ 45:2 E (b) The geomagnetic potential at point P on the Earth’s surface (r ¼ a ¼ 6370 km) is given by F¼

Cm cos y ¼ B0 a sin f a2

ð76:1Þ

We calculate the geomagnetic constant B0 from the horizontal component H  : H  ¼ B0 cos f ) B0 ¼

21 227 ¼ 30 951 nT cosð46:7 Þ

Substituting in the potential equation (76.1) we obtain F ¼ 143 T m (c) The equation of the line of force passing through a point with geomagnetic colatitude y is r ¼ r0 sin2 y

134

Geomagnetism

In this equation r0 is the distance from the Earth’s centre to the point at which the line of force passing through P intersects the geomagnetic equator. Substituting r ¼ a ¼ 6370 km gives r0 ¼

a a ¼  ¼ 13 543 km 2 sin y cos2 f

77. Assume the centred dipole approximation, with the coordinates of the Geomagnetic North Pole being 75 N, 65 W, and the magnetic moment of the dipole 7.5  1022 A m2. For a point on the Earth’s surface at which the inclination is 67 and the geomagnetic longitude is 120 , calculate: (a) The NS and EW components. (b) Its geographical coordinates. The Earth’s radius: 6372 km. (a) We calculate first the geomagnetic constant B0, latitude f  , and horizontal H  component: B0 ¼

Cm ¼ 28 989 nT a3

tan I  ¼ 2 tan f ) f ¼ 49:7 H  ¼ B0 cos f ¼ 18 761 nT To calculate the NS (X  ) and EW (Y  ) components it is necessary to obtain first the geographic latitude (f  ) and the geomagnetic declination (D  ). Applying the cosine rule for the angle 90º – f (Fig. 71a), sin f ¼ sin fB sin f  cos fB cos f cos l f ¼ 55:1 The geomagnetic declination is given by sin D ¼

 cos fB sin l cos f

D ¼ 23:1 From this value we obtain the NS and EW components: X  ¼ H cos D ¼ 17 262 nT Y  ¼ H sin D ¼ 7349 nT (b) The geographic latitude was already obtained, f ¼ 55:1

Main field

135

To calculate the geographical longitude we apply the cosine law for the angle 90º – f  (Fig. 71a): sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ cosðl  lB Þ ¼

sin f  sin fB sin f cos fB cos f

l  lB ¼ 101:7 To choose the correct solution we notice that the declination is positive and then the point must be to the west of the Geomagnetic North Pole D  > 0 ) l  lB < 0 l ¼ 101:7  65 ¼ 166:7 W 78. Assume a spherical Earth of radius 6370 km, with magnetic field produced by a centred dipole whose northern magnetic pole is at 70 N, 60 W. Given that for a point on the surface with coordinates 50 S, 80 W the horizontal component is 24 890 nT, calculate: (a) The magnetic dipole moment. (b) The geographical coordinates of the conjugate point. (a) We calculate first the geomagnetic latitude by (Fig. 71a) sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ f ¼ 30:9 To obtain the magnetic dipole moment m we need the geomagnetic constant B0, which is related with the horizontal component H by B0 ¼

H ¼ 29 007 nT cos f

B0 ¼

Cm B0 a3 ¼ 7:5  1022 A m2 )m¼ 3 a C

(b) Let us obtain first the geomagnetic longitude by (Problem 71, Fig. 71a) sin l ¼

sinðl  lB Þ cos f cos f

l ¼ 14:8 The geomagnetic coordinates of the conjugate point (f1 , l1 ) are (Fig. 78) f1 ¼ f ¼ 30:9 l1 ¼ l ¼ 14:8

136

Geomagnetism

GNP

GMNP

fB

P1

f∗1

P

f∗

l∗

Fig. 78

Solving again the spherical triangle (Fig. 71a) we calculate the geographical coordinates (f1, l1): sin f1 ¼ sin fB sin f1   cos fB cos f1  cos l1  f1 ¼ 11:4 cosðl  lB Þ ¼

sin f1   sin fB sin f1 cos fB cos f1

l1  lB ¼ 13:0 l < 0 ) l1  lB < 0 l1 ¼ 73:0 79. If the Earth’s geomagnetic field is produced by a centred dipole, tilted 15 away from the axis of rotation, of magnetic moment 7.6  1022 A m2, and the Geomagnetic North Pole is at longitude 65 W, calculate: (a) The geomagnetic constant in nT. (b) The geographical coordinates of a point on the Earth’s surface at which the declination is D  ¼ 14 15.50 and the inclination is I  ¼ 65 23.50 . Discuss the possible solutions. (c) The geographical and geomagnetic longitude of the agonic line. Assume a spherical Earth of radius 6370 km.

Main field

137

(a) We calculate the geomagnetic constant from the magnetic moment m, the Earth’s radius a, and the constant C¼107 H m1, by B0 ¼

Cm ¼ 29 403 nT a3

(b) Let us obtain first the geomagnetic latitude from the inclination I  by tan I  ¼ 2 tan f ) f ¼ 47:5 If the dipole is tilted 15 away from the axis of rotation the latitude of the Geomagnetic North Pole will be fB ¼ 90  15 ¼ 75:0 With these results we calculate the geographical latitude f solving the spherical triangle (Fig. 71a). Applying the cosine rule, cosð90  fB Þ ¼ cosð90  f Þ cosð90  fÞ þ sinð90  f Þ sinð90  fÞ cos D sin fB ¼ sin f sin f þ cos f cos f cos D

ð79:1Þ

To obtain the geographical latitude from this equation we can carry out a change of variables, introducing two new variables (m, N ) such that sin f ¼ m cos N cos f cos D ¼ m sin N

ð79:2Þ

From these equations we can calculate P and N: cos f cos D cos D ¼ ) N ¼ 41:6 sin f tan f cos N P¼ ¼ 1:02 sin f

tan N ¼

Substituting Equations (79.2) in Equations (79.1) we obtain sin fB ¼ P cos N sin f þ P cos N cos f ¼ P sinðf þ N Þ sin fB sin fB cos N ¼ sinðf þ N Þ ¼ m sin f  f þ N ¼ 78:4 ) f ¼ 36:8 But another solution is also possible: f þ N ¼ 180  ð78:4 Þ ¼ 258:4 ) f ¼ 60:0 The two solutions are correct and we don’t have any additional information to choose one or the other. (c) The agonic line is the line where the declination is zero and this implies that the point is on the great circle that contains the Geographic North Pole and the Geomagnetic North Pole. So the geomagnetic longitude l  is zero or 180º: l ¼ 0 ) l ¼ lB ¼ 65 W l ¼ 180 ) l ¼ lB þ 180 ¼ 115 E

138

Geomagnetism

80. The Earth’s magnetic field is produced by two dipoles of equal moment (M ¼ Cm ¼ 9.43  109 nT m3) and polarity, forming angles of 30 and 45 with the axis of rotation, and contained in the plane corresponding to the 0 meridian. Find the potential of the total field and the coordinates of the resulting magnetic North pole, taking the Earth’s radius to be 6000 km. The total potential at a point is the sum of the potentials of the two dipoles. If M is the magnetic moment (M ¼ Cm), r is the distance from the dipole’s centre, and y1 and y2 are the geomagnetic co-latitude relative to each dipole (Fig. 80), the total potential F is given by F ¼ F1 þ F2 ¼

M cos y1 M cos y2 M ðcos y1 þ cos y2 Þ  ¼ r2 r2 r2

ð80:1Þ

We calculate the angles y1 and y2 (Fig. 71a) by cos y1 ¼ sin fB1 sin f þ cos fB1 cos f cosðl  lB1 Þ cos y2 ¼ sin fB2 sin f þ cos fB2 cos f cosðl  lB2 Þ The geographical coordinates of the two Geomagnetic North Poles are given by fB1 ¼ 90  30 ¼ 60 ; lB1 ¼ 0 fB2 ¼ 90  45 ¼ 45 ; lB2 ¼ 180

45°

GNP 30° GMNP1

GMNP2

q2 q1

P

f

Fig. 80

ð80:2Þ

Main field

139

Substituting these values in Equations (80.2): cos y1 ¼ sin fB1 sin f þ cos fB1 cos f cos l cos y2 ¼ sin fB2 sin f  cos fB2 cos f cos l Adding the two equations gives cos y1 þ cos y2 ¼ ðsin fB1 þ sin fB2 Þ sin f þ ðcos fB1  cos fB2 Þ cos f cos l and substituting in the equation of the potential (80.1) M ½ðsin fB1 þ sin fB2 Þ sin f þ ðcos fB1  cos fB2 Þ cos f cos l r2 ffiffiffi p pffiffiffi pffiffiffi  

M 3 þ 2 sin f þ 1  2 cos f cos l F¼ 2r2 F¼

If we call # the geographic co-latitude, # ¼ 90 f, then pffiffiffi pffiffiffi pffiffiffi 

M 3 þ 2 cos # þ 1  2 sin # cos l F¼ 2r2 The resulting magnetic North Pole, the point where the inclination I ¼ 90º, due to the combined effect of the two dipoles is given by tan I ¼

Z H

and therefore at the magnetic Pole, H ¼ 0. We derive the component H by taking the gradient of the potential F pffiffiffi  pffiffiffi pffiffiffi   M  3 þ 2 sin # þ 1  2 cos # cos l 1 @F X ¼ B# ¼ ¼ r @# 2r3 pffiffiffi  1 @F M 1  2 sin l Y ¼ Bl ¼ ¼ r sin # @l 2r3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 H ¼ X þY

Since the magnetic North Pole is contained in the plane corresponding to the 0 geographical meridian, then its longitude is either 0º or 180º. If the longitude is 0º
i pffiffiffi M h
pffiffiffi pffiffiffi l ¼ 0 ) H ¼ 3  3 þ 2 sin # þ 1  2 cos # ¼ 0 2r # ¼ 8 ¼ 172 But this result doesn’t correspond to the north hemisphere. Then we must take the geographical longitude 180º:
i pffiffiffi M h
pffiffiffi pffiffiffi l ¼ 180 ) H ¼ 3  3 þ 2 sin #  1  2 cos # ¼ 0 2r # ¼ 8 ) fB ¼ 82

140

Geomagnetism

This is the correct result and the coordinates of the magnetic North Pole are fB ¼ 82 ; lB ¼ 180 81. The Earth’s magnetic field is produced by one dipole in the direction of the axis of rotation (negative pole in the northern hemisphere) and another with the same moment in the equatorial plane which rotates with differential angular velocity v with respect to the points on the surface of the Earth (consider that the Earth doesn’t rotate). Its negative pole passes through the 45 E meridian at time t ¼ 0 and completes a rotation with respect to that point in 24 hours. Consider a point of geographical coordinates 45 N, 45 E. (a) Calculate the magnetic field components (Br, Bu, Bl) at that point. (b) Illustrate graphically how each of them varies with local time. (a) The total potential at a point on the surface of the Earth is the sum of the potentials of the two dipoles (Problem 80, Equation 80.1): F ¼ F1 þ F2 ¼

M cos y1 M cos y2 M ðcos y1 þ cos y2 Þ  ¼ r2 r2 r2

Dipole 1 is in the direction of the axis of rotation and so the geomagnetic co-latitude of the point with respect to this dipole (Fig. 81a) is equal to the geographical co-latitude, y1 ¼ 90  f cos y1 ¼ sin f

GMNP1 GNP

q1

P q2 f GMNP2

Fig. 81a

Main field

141

Dipole 2 is on the equatorial plane (fB2 ¼ 0) and rotates with respect to the points of the surface. Owing to this rotation its geographical longitude lB2 changes with time t in the form lB2 ¼ ot þ 45 where o is the angular velocity, o ¼ 360 /T, T being the rotation period of 24 h. The co-latitude y2 is cos y2 ¼ sin fB2 sin f þ cos fB2 cos f cosðl  lB2 Þ Substituting the geographical coordinates of the negative geomagnetic equatorial Pole (fB2, lB2): cos y2 ¼ cos ’ cosðl  ot  45 Þ Substituting in the equation for the potential F ¼ F1 þ F2 ¼

M ½sin f þ cos f cosðl  ot  45 Þ r2

If we consider the geographical co-latitude # ¼ 90  f, the potential is given by F¼

M ½cos # þ sin # cosðl  ot  45 Þ r2

We obtain the magnetic field components (Br , By, Bl) at the point (#, l) by taking the gradient in spherical coordinates of the potential F: Br ¼ 

@F 2M ½cos # þ sin # cosðl  ot  45 Þ ¼ @r r3

B# ¼ 

1 @F M ½ sin # þ cos # cosðl  ot  45 Þ ¼ r@# r3

Bl ¼ 

1 @F M sinðl  ot  45 Þ ¼ r sin # @l r3

Substituting the values # ¼ 45 , l ¼ 45 , B0 ¼ M/a3, gives 2B0 Br ¼  pffiffiffi ð1 þ cos ot Þ 2

B0 B# ¼  pffiffiffi ð1  cos ot Þ 2 Bl ¼ B0 sin ot

(b) The variation of each component with local time is shown in Fig. 81b

142

Geomagnetism

1

Bl

0

Br,q,l B0

Br

–1 Bθ

–2

0

5

10

15

20

t (h)

Fig. 81b

Magnetic anomalies 82. Calculate the magnetic anomaly created by a magnetic dipole buried at depth d, arbitrarily oriented, at an angle to the vertical of a. The negative pole is upwards. Consider a point P with coordinates (x, z), where x is measured along the horizontal from the projection of the centre of the dipole and z is the vertical from the reference level (the Earth’s surface). The position vector r forms an angle b to the vertical (Fig. 82). The anomalous magnetic potential created by the dipole for this point is
F ¼

Cm cosða þ bÞ Cmðsin b cos a þ cos b sin aÞ ¼ r2 r2

where (Fig. 82) zþd cos b ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ ðz þ dÞ2 x sin b ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ ðz þ dÞ2

ð82:1Þ

143

Magnetic anomalies

P

x z

X

r

d b

– a

+ Z

Fig. 82

Substituting in Equation (82.1) we obtain
F ¼

Cm½ðz þ dÞ cos a  x sin a h i3=2 x2 þ ðz þ dÞ2

To calculate the magnetic anomaly DB:
B ¼ rð
FÞ The vertical component of the magnetic field anomaly, taking the z-coordinate positive downward, is h
 i 2 2 Cm x þ ðz þ dÞ cos a  3ðz þ dÞ ½ ðz þ dÞ cos a  x sin a  @ð
FÞ ¼
Z ¼ h i5=2 @z x2 þ ðz þ dÞ2 For points on the Earth’s surface (z ¼ 0)
Z ¼

Cm½ðx2  2d 2 Þ cos a þ 3dx sin a ½x2 þ d 2 5=2

ð82:2Þ

The component of the magnetic anomaly in an arbitrary horizontal direction x for the Earth’s surface points (z ¼ 0) is given by
X ¼ 

@ð
FÞ Cm½ð2x2  d 2 Þ sin a  3dx cos a ¼ @x ½x2 þ d 2 5=2

ð82:3Þ

144

Geomagnetism

83. Calculate the magnetic anomaly produced at a point with geographical coordinates 38 N, 30 W by a horizontal dipole buried at a depth of 10 m with Cm ¼ 5  105 T m3 which is in the vertical plane of geographical east, and with the negative pole to the west. Also calculate the total values of the field in the NS, EW, and vertical directions, and total field F, as well as the variations in the magnetic declination and inclination due to the existence of the dipole. Consider the Earth’s magnetic field to be produced by a centred dipole with North Pole at 72 N, 30 W, and with B0 ¼ 32 000 nT. We calculate the horizontal and vertical components of the magnetic anomaly from Equations (82.2) and (82.3), taking a ¼ 90º because the dipole is horizontal and x ¼ 0 because the dipole’s centre is beneath the point (Fig. 83a). We call DX and DY the horizontal components in the NS and EW directions, respectively. Because the dipole is on the vertical east–west plane, the north–south component DX ¼ 0,
Z ¼ 0
Y ¼

Cm½ð2x2  d 2 Þ sin a  3dx cos a ½ x2

þ

d 2 5=2

¼

Cm d3

Substituting Cm ¼ 5  105 T m3 d ¼ 10 m we obtain
Y ¼ 50 nT j
Bj ¼
Y ¼ 50 nT
X ¼ 0 P

E

d

a +

–

Z

Fig. 83a

145

Magnetic anomalies

GMNP

GNP

fB

P f f∗

Fig. 83b

To calculate the components of the magnetic anomaly in the direction of the Earth’s magnetic field, F, and their horizontal component, H, we need to determine the magnetic declination and inclination at the point:
H ¼
X cos D þ
Y sin D
F ¼
H cos I  þ
Z sin I 

ð83:1Þ

Because the point has the same longitude as the Geomagnetic North Pole (Fig. 83b), f ¼ 90  ðfB  fÞ ¼ 56:0 D ¼ 0 The inclination is given by tan I  ¼ 2 tan f ) I  ¼ 71:4 Substituting these values in Equations (83.1) we obtain
H ¼ 0
F ¼ 0 The total value of the field in the NS, EW, and vertical directions, and total field F are XT ¼ X  þ
X YT ¼ Y  þ
Y ZT ¼ Z  þ
Z FT ¼ F  þ
F

146

Geomagnetism The vertical Z  and horizontal H  components of the geomagnetic field are given by Z  ¼ 2B0 sin f ¼ 53 058 nT H  ¼ B0 cos f ¼ 17 894 nT The NS (X  ) and horizontal EW (Y  ) components of the geomagnetic field and its magnitude F  are given by X  ¼ H  cos D ¼ H  ¼ 17 894 nT Y  ¼ H  sin D ¼ 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F  ¼ H 2 þ Z 2 ¼ 55 994 nT

We finally obtain that the total field components are

XT ¼ 17 894 nT YT ¼ 50 nT ZT ¼ 53 058 nT FT ¼ 55 994 nT The variations in magnetic declination and inclination due to the presence of the buried dipole are YT ) D0 ¼ 0:02 tan D0 ¼ XT D  D0 ¼ 0:02 tan I 0 ¼

ZT ) I 0 71:4 ¼ I  HT

84. Buried at a point with magnetic latitude 30 N and a depth of 50 m is a horizontal magnetic dipole with Cm ¼ 107 nT m3 with the positive pole to the geographical north. (a) Calculate DF if B0 ¼ 30 000 nT and the declination at that point is 15 . Find the ratio DF /F. (b) How far from the dipole’s centre along the north–south line will the dipole field strength be in the same direction as that of the Earth (take D  ¼ 0  ). (a) The component of the magnetic anomaly DF in the direction of the Earth’s magnetic field (the total field anomaly) is given by
F ¼
H cos I  þ
Z sin I 

ð84:1Þ

We first calculate the components in the geographical directions of the magnetic anomaly produced by the buried dipole using Equations (82.2) and (82.3) of Problem 82, substituting a ¼ 90º because the dipole is horizontal, and x ¼ 0 because the dipole’s centre is beneath the point. In this problem DY ¼ 0 because the dipole is on the geographical north– south vertical plane (Fig. 84). Then

147

Magnetic anomalies

P

N

d

a +

–

Z

Fig. 84


X ¼ 

Cm d3


Y ¼ 0
Z ¼ 0 Substituting the values Cm ¼ 107 nT m3 d ¼ 50 m we obtain
X ¼ 80 nT Substituting D  ¼ 15º, the component of the magnetic anomaly in the direction of the horizontal component H of the Earth’s magnetic field is
H ¼
X cos D ¼ 77 nT At a point of magnetic latitude f  ¼ 30 the magnetic inclination is tan I  ¼ 2 tan f ) I  ¼ 49:1 Substituting in Equation (84.1), the total field anomaly is
F ¼ 50 nT To calculate the geomagnetic field F  we first obtain the components H  and Z  : Z  ¼ 2B0 sin f ¼ 30 000 nT H  ¼ B0 cos f ¼ 25 981 nT pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F  ¼ H 2 þ Z 2 ¼ 39 686 nT

148

Geomagnetism

The ratio of the total field anomaly and the Earth’s total magnetic field is
F ¼ 1:26  103 F (b) If the dipole field strength is in the same direction as that of the Earth then the inclination I 0 due to the dipole is equal to that of the Earth’s field I  , where
Z
H Z tan I  ¼  H tan I 0 ¼

Assuming D  ¼ 0º, then
H ¼
E If we substitute in Equations (82.2) and (82.3) of Problem 82, the angle a ¼ 90º because the dipole is horizontal, we obtain 3Cmdx
Z ¼

5=2 2 x þ d2

Cmð2x2  d 2 Þ
X ¼

5=2 x2 þ d 2

We have changed the sign of the vertical component because the negative pole is toward the south. Applying the condition, tan I 0 ¼ tan I  , we obtain
Z
Z Z ¼ ¼ 
H
X H 3Cmdx  5=2 2 x þ d2 3dx Z  2 ¼ ¼ Cm 2x  d 2 Þ 2x2  d 2 H   5=2 x2 þ d 2

2Z  x2  3dH  x  Z  d 2 ¼ 0 Substituting the values d ¼ 50 m Z  ¼ 30 000 nT H  ¼ 25 981 nT and solving the equation, we obtain x1 ¼ 80 m x2 ¼ 15 m

We have two solutions: a point 80 m to the north from the surface projection of the dipole’s centre and another 15 m to the south.

149

Magnetic anomalies 85. Located at a point with geocentric geographical coordinates 45 N, 30 W, at a depth of 100 m, is a dipole of magnetic moment Cm¼ 1 T m3, tilted 45 from the horizontal to true north, with the negative pole to the north and downwards. At this point on the surface, the following magnetic field values were observed (in nT): F ¼ 55 101; H ¼ 12 413; DF ¼ 1268; DH ¼ 547. Determine: (a) At the indicated point, the main field components X  , Y , Z  . (b) At the indicated point, the deviation of the compass with respect to geomagnetic north. (c) The geocentric geographical coordinates of the North Pole of the Earth’s dipole. Precision 1 nT. (a) We calculate first the magnetic anomaly produced by the dipole, applying Equations (82.2) and (82.3) of Problem 82, substituting a ¼ 225º and x ¼ 0. The horizontal component is in the NS direction (DX) (Fig. 85a) 2Cm cos a ¼ 1414 nT d3 Cm sin a
X ¼ ¼ 707 nT d3
Y ¼ 0
Z ¼

ð85:1Þ

To calculate the declination we use the equation
H ¼
X cos D ) cos D ¼


H
X

D ¼ 39:3

P

N

d + a

45°

– Z

Fig. 85a

150

Geomagnetism

To obtain the Earth’s main field we eliminate the buried dipole contribution from the observed values: F  ¼ F 
F ¼ 56 369 nT H  ¼ H 
H ¼ 11 866 nT qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z  ¼ ðF  Þ2  ðH  Þ2 ¼ 55 106 nT

and the X  and Y  components

X  ¼ H  cos D ¼ 9182 nT Y  ¼ H  sin D ¼ 7516 nT (b) The observed declination is given by tan D0 ¼

Y Y  þ
Y ¼  ) D0 ¼ 37:2 X X þ
X

The deviation of the compass due to the buried dipole with respect to geomagnetic north is D0  D ¼ 2:1 (c) We calculate first the geomagnetic latitude of the point from the vertical and horizontal components: Z  ¼ 2B0 sin f H  ¼ B0 cos f Z 2H   f ¼ 66:7

tan f ¼

With this value, the declination D  and the geographical coordinates of the point (f, l), we can solve the spherical triangle (Fig. 85b) and obtain the geographical coordinates of the Geomagnetic North Pole: GNP 90º – fB l – lB 90º – f

180º – l∗ GMNP q = 90º– f ∗

D∗

P

Fig. 85b

151

Magnetic anomalies

cosð90  fB Þ ¼ cosð90  f Þ cosð90  fÞ þ sinð90  f Þ sinð90  fÞ cos D sin fB ¼ sin f sin f þ cos f cos f cos D fB ¼ 60:0 cosð90  f Þ ¼ cosð90  fB Þ cosð90  fÞ þ sinð90  fB Þ sinð90  fÞ cosðl  lB Þ sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ sin f  sin fB sin f cos fB cos f l  lB ¼ 30:0

cosðl  lB Þ ¼

The correct solution is the negative one because a positive value of the declination implies that the point is to the west of the Geomagnetic North Pole: D  > 0 ) l  lB < 0 lB ¼ 0 86. Located at a point with geographical coordinates 45 N, 30 W, at a depth of 100 m, is a dipole of magnetic moment Cm ¼ 1 T m3, inclined 45 to the vertical towards the south, with the positive pole upwards, and in the geographical north–south vertical plane. The Earth’s dipole has its north pole at 60 N, 0 E and B0 ¼ 30 000 nT. Calculate: (a) The values of Z, H, F at the given point. (b) Where does the compass point to at that same point? (a) We calculate first the geomagnetic latitude corresponding to the point by sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ f ¼ 66:7 From this value we obtain the geomagnetic components Z  , H  and the total main field F  : Z  ¼ 2B0 sin f ¼ 55 107 nT H  ¼ B0 cos f ¼ 11 866 nT qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F  ¼ ðH  Þ2 þ ðZ  Þ2 ¼ 56 370 nT

The geomagnetic declination D  is given by

 cos fB sinðl  lB Þ cos f D ¼ 39:2 sin D ¼

and the geomagnetic inclination I  by tan I  ¼ 2 tan f ) I  ¼ 77:8

152

Geomagnetism

The magnetic anomaly created by the dipole buried at depth d is given by Equations (85.1) of Problem 85. Substituting Cm ¼ 1 T m3, d ¼ 100 m, and a ¼ 45º, we obtain 2Cm cos a ¼ 1414 nT d3 Cm sin a
X ¼ ¼ 707 nT d3
Z ¼


Y ¼ 0 The field anomalies DH and DF are given by
H ¼
X cos D ¼ 548 nT
F ¼
H cos I  þ
Z sin I  ¼ 1266 nT Finally, the observed values are Z ¼ Z  þ
Z ¼ 53 693 nT F ¼ F  þ
F ¼ 55 104 nT H ¼ H  þ
H ¼ 12 414 nT (b) To calculate in what direction the compass points we need the value of the observed declination D’ including the effects of the geomagnetic field and the buried dipole: tan D0 ¼

Y Y  þ
Y ¼  X X þ
X

Y  ¼ H  sin D ¼ 7500 nT X  ¼ H  cos D ¼ 9195 nT D0 ¼ 37:1 87. Located at a point on the Earth with geographical coordinates 45 N, 30 E, at a depth of 100 m, is a dipole of magnetic moment Cm ¼ 107 nT m3, tilted 45 to the vertical towards the south, with the positive pole downwards, and contained in the plane of true north. The Earth’s field is produced by a centred dipole tilted 30 from the axis of rotation in the plane of the 0 meridian, with B0 ¼ 30 000 nT. Calculate the total values of F, Z, and H observed at the point of the surface above the centre of the buried dipole. We first calculate the geographical coordinates of the Geomagnetic North Pole and the geomagnetic latitude fB ¼ 90  30 ¼ 60 lB ¼ 0 sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ f ¼ 66:7

153

Magnetic anomalies The geomagnetic field components Z  , H  and the total main field F  are given by Z  ¼ 2B0 sin f ¼ 55 107 nT H  ¼ B0 cos f ¼ 11 866 nT qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  F ¼ ðH  Þ2 þ ðZ  Þ2 ¼ 56 370 nT

We calculate the geomagnetic declination D  by

 cos fB sinðl  lB Þ cos f D ¼ 39:2

sin D ¼

The inclination I  is given by tan I  ¼ 2 tan f ) I  ¼ 77:8 We obtain the magnetic anomaly produced by the buried dipole applying Equations (82.2) and (82.3) of Problem 82, substituting a ¼ 45º and x ¼ 0. The horizontal component is in the NS direction (DX): 2Cm cos a ¼ 14 nT d3 Cm sin a ¼ 7 nT
X ¼ d3
Y ¼ 0
Z ¼

The field anomalies DH and DF are given by
H ¼
X cos D ¼ 5 nT
F ¼
H cos I  þ
Z sin I  ¼ 13 nT Finally the observed values are Z ¼ Z  þ
Z ¼ 55 121 nT F ¼ F  þ
F ¼ 56 383 nT H ¼ H  þ
H ¼ 11 861 nT 88. Buried at a point with geographical latitude 20 N and the same longitude as the geomagnetic pole, at a depth of 200 m, is a sphere of 50 m radius of material with magnetic susceptibility 0.01. The Earth’s field is produced by a centred dipole tilted 10 from the axis of rotation and magnetic moment M ¼ 1030 g cm3 (Earth’s radius: 6000 km). Calculate: (a) The anomaly produced by induced magnetization in the sphere at a point on the Earth’s surface above the centre of the sphere. Give the vertical and horizontal components in units of nT. (b) The total anomaly for a point on the Earth’s surface 100 m south of the above point.

154

Geomagnetism

GNP

GMNP q = 90° – f ∗

10°

P 20°

Fig. 88a

(a) We first calculate the geomagnetic co-latitude (y) and latitude (f  ) of the point, knowing that it is in the same meridian as the Geomagnetic North Pole (Fig. 88a): y ¼ 90  f ¼ 90  10  20 ¼ 60 f ¼ 30 The geomagnetic field is given by M ¼ 4630 nT a3 Z  ¼ 2B0 sin f ¼ 4630 nT B0 ¼

The inclination is given by

H  ¼ B0 cos f ¼ 4009 nT qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F  ¼ ðH  Þ2 þ ðZ  Þ2 ¼ 6124 nT tan I  ¼ 2 tan f ) I  ¼ 49

The magnetic anomaly created by a sphere is the same as the anomaly created by a magnetic dipole oriented in the same direction as the geomagnetic field, that is, tilted 90 – I  to the vertical and with the negative pole upwards (Fig. 88b). So we use Equations (82.2) and (82.3) taking a ¼ 41º and x ¼ 0, but we change the sign of Equation (82.3) because the negative pole is toward the north. The horizontal component is DX ¼ DH because the dipole is in the magnetic north-vertical plane and DY ¼0: 2Cm cos a d3 Cm sin a
H ¼ d3
Z ¼

155

Magnetic anomalies

F∗

49°

P

GN MN

41°

Z

Fig. 88b

To calculate Cm we use the magnetic susceptibility w and the volume V of the sphere: Cm ¼ wF  V ¼ 3:2  107 nT m3 Substituting this value in the equation of the components of the magnetic anomaly we obtain
Z ¼

2Cm cos a 2  3:2  107  cos 41 ¼ ¼ 6 nT 8  106 d3


X ¼

Cm sin a 3:2  107  sin 41 ¼ ¼ 3 nT d3 8  106

(b) The anomaly created by the sphere at a point at a distance x ¼100 m to the south of the above point is given by

Cm ðx2  2d 2 Þ cos a þ 3dx sin a ¼ 0:8 nT

5=2 x2 þ d 2

Cm ð2x2  d 2 Þ sin a  3dx cos a ¼ 3:3 nT
H ¼

5=2 x2 þ d 2
Z ¼

The total magnetic field anomaly is therefore


F ¼
H cos I  þ
Z sin I  ¼ 3:4 nT

156

Geomagnetism 89. Buried at a point on the Earth at magnetic latitude 45 N, at a depth d, is a vertical dipole of magnetic moment M (Cm) with negative pole upwards. If M/d3 ¼ 10B0 (B0 is the geomagnetic constant of the main field) calculate how far along the magnetic meridian the direction of the buried dipole’s field will coincide with that of the Earth (the terrestrial dipole field). First we calculate the geomagnetic field components and the inclination by pffiffiffi Z  ¼ 2B0 sin f ¼ 2B0 nT pffiffiffi 2 B0 nT H  ¼ B0 cos f ¼ 2 Z tan I  ¼  ¼ 2 ) I  ¼ 63:4 H The components of the magnetic field created by the dipole are given by Equations (82.2) and (82.3) of Problem 82, putting a ¼ 0º:   Cm x2  2d 2
Z ¼  5=2 x2 þ d 2 Cm3dx
H ¼  5=2 2 x þ d2

If the buried dipole’s field coincides with that of the Earth the magnetic inclinations due to both have to be equal and so
Z ¼ tan I  ¼ 2
H
pffiffiffiffiffi x2 þ 6xd  2d 2 ¼ 0 ) x ¼ d 3  28

Of the two solutions, x ¼ 2.3d and x ¼ 8.3d, only the positive corresponds to the equal direction of the two fields.

External magnetic field 90. The Earth’s magnetic field is produced by two dipoles of equal moment and polarity that are at an angle of 60 to each other, with the bisector being the axis of rotation. The dipoles are contained in the plane of the 0 geographical meridian. (a) Calculate the potential on points of the Earth’s surface as a function of geographical coordinates f and l. (b) At what points on the surface are the magnetic poles located? (c) What form would the external field have in order to annul the internal field at the magnetic equator?

External magnetic field

157

30° GMNP2

GNP

GMNP1 q2

q1

P

Fig. 90

(a) The total potential at a point is the sum of the potentials of the two dipoles (see Equation 80.1 of Problem 80): F ¼ F1 þ F2 ¼

M cos y1 M cos y2 Mðcos y1 þ cos y2 Þ  ¼ r2 r2 r2

We calculate the geomagnetic co-latitudes by Equation (71.3) of Problem 71: cos y1 ¼ sin fB1 sin f þ cos fB1 cos f cosðl  lB1 Þ cos y2 ¼ sin fB2 sin f þ cos fB2 cos f cosðl  lB2 Þ

ð90:1Þ

The geographical coordinates of the North Pole of each dipole are given by (Fig. 90) fB1 ¼ fB2 ¼ 90  30 ¼ 60 lB1 ¼ 0 lB2 ¼ 180 Substituting these values in Equation (90.1): pffiffiffi 3 1 sin f þ cos f cos l cos y1 ¼ 2 2 pffiffiffi 3 1 sin f  cos f cos l cos y2 ¼ 2 2 Adding the two equations: cos y1 þ cos y2 ¼

pffiffiffi 3 sin f

158

Geomagnetism

Substituting in the equation of the potential: pffiffiffi  3M sin f F¼ r2 We note that this p is ffiffiffithe same potential as that produced by only a centred dipole with magnetic moment 3M oriented in the direction of the rotation axis, because the geomagnetic co-latitude is 90º f. (b) Bearing in mind the last results, we calculate the inclination by tan I ¼ 2 cotð90  fÞ ¼ 2 tan f

ð90:2Þ

The magnetic poles are the points on the surface of the Earth where the value of the inclination is equal to  90º:  f ¼ 90 I ¼  90 ) f ¼ 90 Therefore the magnetic poles coincide with the geographical poles. (c) We derive the main field by taking the gradient of the potential F. The components are pffiffiffi @F 2 3M sin f ¼ @r r3 pffiffiffi 1 @F  3M cos f ¼ Bf ¼ r @f r3 1 @F Bl ¼  ¼0 r cos f @l Br ¼ 

At the magnetic equator the inclination is null (I ¼ 0) and according to Equation (90.2) the latitude is null too (f ¼ 0). Substituting in the last equations: Br ¼ 0

pffiffiffi  3M Bf ¼ r3 Bl ¼ 0 The external magnetic field to annul out the internal field is therefore pffiffiffi

3M Be ¼ 0; 3 ; 0 r 91. The Earth’s magnetic field is formed by a centred dipole with northern geomagnetic pole at 60 N, 0 E and B0 ¼ 32 000 nT and a uniform external field from the Sun of 10 000 nT parallel to the equatorial plane. (a) For a point at coordinates 60 N, 60 W, calculate the components X, Y, Z of the total field, and the values of D and I. (b) How do D and Z of the total field vary throughout the day with local time t?

External magnetic field

159

(a) We calculate the geomagnetic main field components Z  , H  obtaining the geomagnetic latitude given by (71.3): sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ f ¼ 61 Z  ¼ 2B0 sin f ¼ 55 976 nT H  ¼ B0 cos f ¼ 15 514 nT To calculate the NS (X  ) and EW (Y  ) components we need the geomagnetic declination D  (71.2):  cos fB sinðl  lB Þ cos f   D ¼ 63 X  ¼ H  cos D ¼ 7043 nT sin D ¼

Y  ¼ H  sin D ¼ 13 823 nT The external field is parallel to the equatorial plane and has a diurnal period (o ¼ 2p/24) because it comes from the Sun. We assume that at local time t ¼ 0 the Sun is at the point’s meridian. If we denote by N the modulus of the external field (N ¼ 10 000 nT) and bearing in mind Fig. 91a (representation of the plane parallel to the equator that contains the point) we have at time t 6¼ 0 Y ¼ Bel ¼ N sin ot The radial and tangential components (Fig. 91b) are

(a) P N coswt wt

N

Fig. 91a

B eλ

160

Geomagnetism

GNP

X = Bf

Ncosωt P

z = –Br

r

f

Fig. 91b

Z ¼ Br ¼ N cos ot cos f X ¼ Bf ¼ N cos ot sin f The total magnetic field is the sum of the two contributions: ZT ¼ ð55 976 þ 5000 cos otÞ nT XT ¼ ð7043 þ 8660 cos otÞ nT YT ¼ ð13 823 þ 10 000 sin otÞ nT The declination D and inclination I are given by tan D ¼ tan I ¼

YT 13 823 þ 10 000 sin ot ¼ XT 7043 þ 8660 cos ot ZT ZT ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi HT XT2 þ YT2

(b) To see how D and Z vary during the day with local time t we substitute several values for t, obtaining the values in the table: t (h)

Z (nT)

D (º)

0 6 12 18

60976 55976 50976 55976

41 28 83 73

External magnetic field

161

92. The Earth’s magnetic field is formed by a centred dipole of moment m and north pole 60 N, 0 E, and a uniform external field of magnitude N ¼ B0/4 (B0 is the geomagnetic constant of the internal field) parallel to the axis of rotation. Determine: (a) (b) (c) (d)

The total potential at any point. The coordinates of the boreal magnetic pole. The magnetic declination at the point 45 N, 45 E. The angle along the meridian between that point and the magnetic equator. (a) The total potential (F) is the sum of two contributions: the main (internal) field (Fi) and the external field (Fe): F ¼ Fi þ Fe

The main field is formed by a centred dipole of moment m so the potential is given by Fi ¼

Cm cos y r2

We calculate the geomagnetic co-latitude y ¼ 90 – f  at a point with geographical coordinates (f, l) by (71.3): pffiffiffi 1 3 cos y ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ ¼ sin f þ cos f cos l 2 2

GNP

P

ze

Xe f

Fig. 92a

Fe

162

Geomagnetism

The external field is parallel to the axis of rotation; therefore its components are in the vertical and NS directions (Fig. 92a). If we call # ¼ 90  f the geographical co-latitude, the components are given by Ber ¼ N cos # ¼ N sin f Be# ¼ N sin # ¼ N cos f Z e ¼ Ber ¼ N sin f X e ¼ Be# ¼ N cos f Bearing in mind that Be ¼ ∇Fe the potential for the external field is Fe ¼ Nr cos # ¼ Nr sin f Therefore the total potential is given by
pffiffi  Cm 23 sin f þ 12 cos f cos l þ Nr sin f F¼ r2 (b) At the magnetic boreal pole the inclination is I ¼ 90º and the horizontal field components H, X, Y are given by Z )H ¼0 H  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi X ¼0 2 2 H ¼ X þY ) Y ¼0 tan I ¼

We obtain the components X and Y by taking the gradient of the potential Be ¼ rFe We bear in mind the relations @F @F ¼ @y @f X ¼ By and obtain that the components are given by
pffiffi  3 1 Cm cos f  sin f cos l 2 2 1 @F ¼ X ¼  N cos f ¼ 0 r @f r3 1 @F Cm ¼  3 sin l ¼ 0 ) l ¼ 0 Y ¼ r cos f @l 2r Substituting in the equation of component X (92.1) the values l¼0 r¼a Cm a3 N ¼ B0 =4

B0 ¼

ð92:1Þ

External magnetic field

163

GNP

GMNP

fB P

f l feq

Fig. 92b

we obtain the geographical latitude of the boreal magnetic pole pffiffiffi 2 31 tan fBM ¼ ) fBM ¼ 51 2 Therefore the geographical coordinates of the magnetic boreal pole are 51 N, 0 . Notice that this is different from the geomagnetic pole. (c) The declination is given by Y tan D ¼ ¼ X



B0 sin l 2

pffiffiffi

3 1 B0 cos f  sin f cos l  cos f 2 2 4  sin l

¼ pffiffiffi 1 cos f  sin f cos l 3 2 B0

Substituting the geographical coordinates of the point (45 N, 45 E) we obtain D ¼ 62 (d) We call feq the angle between the geographical and magnetic equators at longitude l ¼ 45º (Fig. 92b). Then the angle to calculate will be feq þ 45 . To calculate feq we take into account that at the magnetic equator the vertical component is Z ¼ 0: pffiffiffi

@F 2CM 1 3 ¼ 3 Z ¼ Br ¼ sin feq þ cos feq cos l þ N sin feq ¼ 0 @r r 2 2 4 cos l tan feq ¼ pffiffiffi 4 3þ1

164

Geomagnetism

Substituting l ¼ 45º in the equation we obtain feq ¼ 20 Then the angle we are asked for is 45º – 20º ¼ 25º. 93. The internal magnetic field is formed by two orthogonal dipoles of equal moment M, one of them in the direction of the axis of rotation and the other contained in the 0 meridian on the equator. There is also an external field of constant intensity N ¼ B0/4 (B0 is the geomagnetic constant of the internal field) and lines of force parallel to the axis of rotation. (a) Calculate the potential of the total field for points on the surface. (b) At which latitude is Z maximum on the 0 meridian? (a) The potential of the total field is the sum of the two potentials of the internal dipoles and the potential of the external field: M cos y1 M cos y2  þ Fe r2 r2 Mðcos y1 þ cos y2 Þ ¼ þ Fe r2

F ¼ F1 þ F2 þ Fe ¼

In this equation r is the distance from the dipole’s centre (the Earth’s centre), y1 and y2 are the co-latitudes relative to each dipole, and M ¼ Cm. Dipole 1 is on the direction of the axis of rotation and so the geomagnetic co-latitude of the point with respect to this dipole is y1 ¼ 90  f cos y1 ¼ sin f Dipole 2 is in the equatorial plane (fB2 ¼ 0) and contained in the 0 meridian so the geomagnetic co-latitude y2 is given by (71.3) cos y2 ¼ sin fB2 sin f þ cos fB2 cos f cosðl  lB2 Þ fB2 ¼ 0 lB2 ¼ 0 cos y2 ¼ cos f cos l The equation for the potential of the external field is the same as that of Problem 92: Fe ¼ þNr sin f Therefore the potential of the total field is given by F¼

M ðsin f þ cos f cos lÞ þ Nr sin f r2

External magnetic field

165

(b) We obtain the component Z by taking the vertical component of the gradient of the potential (B ¼ ∇F) Z ¼ Br ¼

@F 2M ðsin f þ cos f cos lÞ ¼ þ N sin f @r r3

To calculate the maximum of Z on the 0 meridian we substitute l ¼ 0 and apply the condition that the first derivate with respect to the latitude is null: @Z 2M ¼ 3 ðcos f  sin fÞ þ N cos f ¼ 0 @f r At the Earth’s surface r ¼ a and we know that B0 ¼

M a3

Substituting this constant and solving the equation we determine the latitude at which the Z component is maximum: 2B0 ðcos f  sin fÞ þ

B0 cos f ¼ 0 ) f ¼ 48 4

94. The internal field has its northern geomagnetic pole at the coordinates 60 N, 0 E, and B0 ¼ 30 000 nT. At a point with coordinates 30 N, 45 W, one observes an increase of 7.7 in the value of the declination from 00:00 h to 09:00 h. There is known to be an external field parallel to the Earth’s axis of rotation in the direction from N to S which is null at 00:00 h and maximum at 12:00 h local times. Calculate: (a) The components of the internal and external fields. (b) The difference in the inclination at 00:00 h and 09:00 h. (c) The maximum value of the declination during the day. (a) To calculate the geomagnetic main field intensity components we obtain first the geomagnetic latitude (Equation 71.3): sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ f ¼ 47:7 The declination and inclination are given by (71.2):  cos fB sinðl  lB Þ ) D ¼ 31:7 cos f tan I  ¼ 2 tan f ) I  ¼ 65:5

sin D ¼

So the geomagnetic main field components are Z  ¼ 2B0 sin f ¼ 44 378 nT H  ¼ B0 cos f ¼ 20 190 nT X  ¼ H  cos D ¼ 17 178 nT Y  ¼ H  sin D ¼ 10 609 nT

166

Geomagnetism

The external field is parallel to the axis of rotation so its components are in the vertical and NS directions. This field is null at 00:00 h and maximum at 12:00 h local time (period T ¼ 24 h). Its components are given by N ð1  cos otÞ sin f 2 N X e ¼  ð1  cos otÞ cos f 2 Ye ¼ 0 Ze ¼

He ¼ X e o¼

2p 2p ¼ T 24

(b) To calculate the difference in the inclination we obtain first the value of N, bearing in mind the time variation of the declination. The observed declination as a function of time is given by tan D ¼

Y Y þ Ye ¼  ¼ X X þ Xe

Y X 

For t ¼ 0 h: tan D1 ¼

N ð1  cos otÞ cos f 2

Y ¼ tan D ) D1 ¼ D ¼ 31:7 X

Since we know the change in declination between 0 h and 9 h, we find the declination at 9 h, D2: D2  D1 ¼ 7:7 ) D2 ¼ 39:4 We know that at t ¼ 9 h Y

tan D2 ¼ X 

N ð1  cos otÞ cos f 2

¼ 0:82

Solving for N we obtain

2 Y  X  ¼ 5766 nT N¼ ð1  cos otÞ cos f tan D2 The magnetic inclination is given by N Z  þ ð1  cos otÞ sin f Z Z þ Ze 2 ¼ tan I ¼ ¼  N H H þ He  H  ð1  cos otÞ cos f 2 At t ¼ 0 h: I1 ¼ I  ¼ 65:5 At t ¼ 9 h: I2 ¼ 71:2

External magnetic field

167

The difference in the inclination is therefore I2  I1 ¼ 5:7 (c) The declination is given by Y

tan D ¼ X 

N ð1  cos otÞ cos f 2

ð94:1Þ

The maximum value is at 12 h because at that time the external field has the maximum value. Substituting ot ¼ p and the values obtained for X  , Y , N, and f: Dmax ¼ 41:0 95. At a point on the Earth with coordinates 45 N, 45 E, measurements are made of the magnetic field components at 00:00 h and 12:00 h in nT with a 2 nT precision: 0h X ¼ 20 732 Y ¼ 2500 Z ¼ 57 768 12 h X ¼ 24 267 Y ¼ 2500 Z ¼ 54 232 It is known that the modulus of the magnetic field intensity has a harmonic diurnal variation, and that the geomagnetic pole is on the zero meridian. Calculate: (a) The moment and coordinates of the main field dipole. (b) Expressions for the potential and components of the external field. (Earth’s radius a ¼ 6400 km, m0 ¼ 4p107 kg m s2 A2). (a) To calculate the moment and coordinates of the main field dipole we need to obtain the geomagnetic main field intensity components. The observed values are equal to the sum of the geomagnetic main field (X  , Y , Z  ) and the external field (X e, Y e, Z e): X ¼ X þ Xe Y ¼ Y þ Ye Z ¼ Z þ Ze The geomagnetic main field is constant but the external field changes with time. So if we denote by (X0, Y0, Z0) and (X12, Y12, Z12) the observed values at 0 h and at 12 h respectively then the differences are due to the variations of the external field: e  X0e X12  X0 ¼ 3535 nT ¼ X12 e Y12  Y0 ¼ 0 nT ¼ Y12  Y0e e Z12  Z0 ¼ 3536 nT ¼ Z12  Z0e

We notice that the Y component doesn’t vary, which implies that the component Y e is zero, and so the external field is parallel to the axis of rotation. We also notice that the NS component increases in the time interval between 0 h and 12 h, while the vertical component diminishes, which implies that the polarity of the external field is inverted with respect to that of the main field. The modulus of the magnetic field intensity has a harmonic diurnal variation and increases with time. Therefore the components of the external field are (Fig. 95a)

168

Geomagnetism

Fe

GNP

Xe

Ze P

f

Fig. 95a

X e ¼ N ð1  cos otÞ cos f Z e ¼ N ð1  cos otÞ sin f 2p 2p o¼ ¼ T 24h We notice that at time t ¼ 0 the external field is null and so e X12 ¼ 3535 nT e Y12 ¼ 0 nT e Z12 ¼ 3536 nT

Therefore we calculate the main field components by e X  ¼ X12  X12 ¼ 20 732 nT e Y  ¼ Y12  Y12 ¼ 2500 nT e ¼ 57 768 nT Z  ¼ Z12  Z12 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  H ¼ ðX  Þ2 þ ðY  Þ2 ¼ 20 882 nT

If we consider the centred magnetic dipole model, the vertical and horizontal field components are given by the equations Z  ¼ 2B0 sin f H  ¼ B0 cos f

External magnetic field

169

GNP 90º – fB l – lB 90º – f

180º – l ∗ GMNP q = 90º – f ∗

D∗

P

Fig. 95b

From these equations we obtain the geomagnetic latitude (f  ) and the geomagnetic constant B0: tan f ¼ B0 ¼

Z ) f ¼ 54:1 2H 

Z ¼ 35 657 nT 2 sin f

From B0 we calculate the magnetic moment: B0 ¼

m0 m ) m ¼ B0 107 a3 ¼ 9:3  1022 A m2 4p a3

The longitude of the Geomagnetic North Pole is lB ¼ 0 and we calculate the latitude fB from the spherical triangle (Fig. 95b), but obtaining first the declination from the X  and Y  components: tan D ¼

Y ) D ¼ 6:9 X

Applying the cosine rule: cosð90  fB Þ ¼ cosð90  f Þ cosð90  fÞ þ sinð90  f Þ sinð90  fÞ cos D sin fB ¼ sin f sin f þ cos f cos f cos D fB ¼ 79:9 (b) We obtain the radial and transverse components of the external field from the vertical and NS components: @Fe @r 1 @Fe e e Bf ¼ X ¼ N ð1  cos otÞ cos f ¼ r @f Ber ¼ Z e ¼ N ð1  cos otÞ sin f ¼

170

Geomagnetism

Therefore the potential of the external field is given by Fe ¼ Nrð1  cos otÞ sin f where N¼

e Z12 ¼ 2500 nT 2 sin f

96. The Earth’s main magnetic field is that of a centred dipole of moment M (M ¼ Cm) in the direction of the axis of rotation, and the external field is produced by electric currents of intensity J circulating in a clockwise sense in a ring in the plane of the ecliptic at a distance of 10 Earth radii around the Earth. (a) Express the potential of the total field and the components Br and Bø on the Earth’s surface for l ¼ 0 . (b) If the inclination of the ecliptic is 30 , and the external field strength is N ¼ M / 4R3, what is the latitude of the northern magnetic pole? (a) The potential of the total field is the sum of the potentials of the dipole and of the external field: FT ¼ F þ Fe ¼

M cos y þ Fe r2

where r is the distance from the dipole’s centre (Earth’s centre) and y is the geomagnetic co-latitude. The dipole is in the direction of the axis of rotation and so the geomagnetic colatitude is equal to the geographic co-latitude: y ¼ 90  f cos y ¼ sin f To calculate the potential of the external field we know that it is produced by electric currents of intensity J circulating in a clockwise sense at a distance of 10 Earth radii around the Earth. These electric currents produce, at remote points, a magnetic dipolar field whose modulus is m0J/2R, J being the current intensity and R the radius of the circular currents; the dipole is oriented in the direction of the ecliptic’s axes with the negative pole in the southern hemisphere, because the currents are clockwise. So the potential of the external field is given by Fe ¼

Cme cos y2 r2

In this equation me ¼ Jp100a2, a is the Earth’s radius, and y2 is the angle between the axes of the circular currents and the direction of the point from the negative pole (Fig. 96) Therefore y2 ¼ 180  ðy þ eÞ cos y2 ¼  cosðy þ eÞ where e is the angle between the axes of the circular currents and the axes of rotation of the Earth.

External magnetic field

171

GNP

+

P

q e

q2

e

–

Fig. 96

The potential of the external field is Fe ¼

Cme cosðy þ eÞ r2

and the total potential is given by FT ¼ 

M cos y Cme cosðy þ eÞ þ r2 r2

We calculate the components of the main field intensity by taking the gradient of the potential: @F 2M cos y ¼ @r r3 1 @F M sin y By ¼  ¼ r @y r3 Br ¼ 

The magnitude of the external field is given by Be ¼

m0 J ¼N 4 10a

The radial and tangential components are Ber ¼ N cosðy þ eÞ Bey ¼ N sinðy þ eÞ

172

Geomagnetism

The total field is the sum of the two contributions: 2M cos y þ N cosðy þ eÞ r3 M sin y By ¼  þ N sinðy þ eÞ r3 Br ¼ 

(b) We know that the Earth’s dipole is in the direction of the axis of rotation and y is the geographical co-latitude (Fig. 96). At the magnetic North Pole the total field is vertical and the tangential component is By ¼ 0: By ¼ 

M sin y þ N sinðy þ eÞ ¼ 0 r3

The values of N and e are known: M M ) 3 ¼ 4N 3 4r r e ¼ 30

N¼

From these values we calculate y, the geographical co-latitude of the magnetic North Pole, By ¼ 4N sin y þ N sinðy þ 30 Þ ¼ 0 pffiffiffi 1 3 sin y þ cos y ¼ 0 ) y ¼ 9  4 sin y þ 2 2 97. Two spherical planets of radius a and separated by a centre-to-centre distance of 4a orbit around each other and spin in the equatorial plane. Each has a magnetic field produced by a centred dipole in the direction of the axis of rotation, with the positive pole in the northern hemisphere and B0 ¼ 10 000 nT. Determine the components X, Y, Z, D, and I of the total magnetic field at the North Pole of one of the planets (precision 1 nT). The total magnetic field in each planet is the sum of its main field and the external field created by the other planet. To determine the main field of either of the planets we need the geomagnetic latitude, which is positive toward the negative pole, in this case, the South Pole. Therefore the geomagnetic latitude and the components of the main field at the North Pole are f ¼ 90 Z  ¼ 2B0 sin f ¼ 20 000 nT H  ¼ B0 cos f ¼ 0 X ¼ Y ¼ 0 The external field at one of the planets is created by the main field of the other planet and corresponds to that of a magnetic dipole. Its components are (Fig. 97) 2Cm cos y r3 Cm sin y Bey ¼ r3 Ber ¼

External magnetic field

173

B er q – 90°

+ r

B eθ q – 90° a

a 4a

θ

–

Fig. 97

At the North Pole r ¼ (Fig. 97)

pffiffiffiffiffi 17a ¼ 4:12a and we calculate the geomagnetic co-latitude y by a 1 ¼ 4a 4 y ¼ 14 þ 90 ¼ 104 tanðy  90 Þ ¼

Substituting these values in the equations for the radial and tangential components with respect to the planet producing the external field: 2Cm cos y Cm cos y B0 cos y ¼ ¼ ¼ 69 nT r3 35a3 35 Cm sin y Cm sin y B0 sin y Bey ¼ ¼ 139 nT ¼ ¼ r3 70a3 70 Ber ¼

From this value we calculate the vertical and horizontal components (Fig. 97): Z e ¼ Ber cosð180  yÞ þ Bey cosðy  90 Þ ¼ Ber cos y þ Bey sin y ¼ 118 nT H e ¼ Ber sinð180  yÞ þ Bey sinðy  90 Þ ¼ Ber sin y  Bey cos y ¼ 33 nT X e ¼ H e ¼ 33 nT Ye ¼ 0 The components of the total field are finally Z ¼ Z  þ Z e ¼ 19882 nT H ¼ H  þ H e ¼ 33 nT X ¼ X  þ X e ¼ 33 nT Y ¼ Y þ Ye ¼ 0 Y tan D ¼ ) D ¼ 0 X Z tan I ¼ ) I ¼ 89:9 H

174

Geomagnetism

Main (internal), external, and anomalous magnetic fields 98. At a point with geographical coordinates 40 N, 45 E, measurements are made of the magnetic field components, obtaining the values (in nT): At 06:00 h: X ¼ 19 204; Y ¼ 0; Z ¼ 38 195 At 12:00 h: X ¼ 11 544; Y ¼ 0; Z ¼ 44 623 Buried at a depth of 20 m below this point is a dipole of magnetic moment Cm ¼ 0.01 T m3, oriented in the NS plane at an angle of 45 with the vertical towards the south, and the positive pole upwards. Given that the external field at 12:00 h is twice that at 06:00 h, determine: (a) The geomagnetic constant Bo and the coordinates of the northern geomagnetic pole. (b) The magnitude and direction of the external field. How does the magnitude of the external field vary with time? (a) The observed magnetic field is composed of three parts: the geomagnetic main (internal) field, the anomalous field (magnetic anomaly) created by the buried dipole, and the external field. We determine first the magnetic anomaly produced by the buried dipole, applying Equations (82.2) and (82.3) in Problem 82, substituting a ¼ 225º and x ¼ 0. The horizontal component is in the NS direction (DX), because the buried dipole is in the NS-vertical plane (Fig. 98a): 2Cm cos a ¼ 1768 nT d3 Cm sin a
X ¼ ¼ 884 nT d3
Y ¼ 0
Z ¼

P

N

d

+ 45°

α – Z

Fig. 98a

175

Main (internal), external, and anomalous magnetic fields

The components of the total magnetic field at 06:00 h are given by X1 ¼ X  þ
X þ X e ¼ 19 204 nT Y1 ¼ Y  þ
Y þ Y e ¼ 0 Z1 ¼ Z  þ
Z þ Z e ¼ 38 195 nT At 12:00 h given that the external field is twice that at 06:00 h: X2 ¼ X  þ
X þ 2X e ¼ 11 544 nT Y2 ¼ Y  þ
Y þ 2Y e ¼ 0 Z2 ¼ Z  þ
Z þ 2Z e ¼ 44 623 nT If we subtract both sets of equations and obtain X2  X1 ¼ X e ¼ 7660 nT Y2  Y1 ¼ Y e ¼ 0 Z2  Z1 ¼ Z e ¼ 6428 nT Now we can calculate the elements of the main field X  ¼ X1 
X  X e ¼ 25 980 nT Y  ¼ Y1 
Y  Y e ¼ 0 Z  ¼ Z1 
Z  Z e ¼ 33 535 nT qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi H  ¼ ðX  Þ2 þ ðY  Þ2 ¼ X  ¼ 25 980 nT tan D ¼

Y ) D ¼ 0 H

We calculate the geomagnetic latitude of the point f and the geomagnetic constant B0 from the vertical and horizontal geomagnetic main field components by Z  ¼ 2B0 sin f H  ¼ B0 cos f Z ) f ¼ 32:8 tan f ¼ 2H  Z B0 ¼ ¼ 30 953 nT 2 sin f We obtain the coordinates of the Geomagnetic North Pole by (Fig. 98b) D ¼ 0 ) lB ¼ 180 þ l ¼ 225 E ¼ 135 W 90  fB ¼ f  f ) fB ¼ 82:8 (b) The magnitude of the external field at 06:00 h is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi B6e ¼ Xe2 þ Ye2 þ Ze2 ¼ 10 000 nT

176

Geomagnetism

GMNP

GNP

P

f∗ f

90° – fB

Fig. 98b

At 12:00 h the magnitude is 6 B12 e ¼ 2Be ¼ 20 000 nT

The direction of the external field is in the NS-vertical plane because the EW component is null, forming with the horizontal an angle Ie (Fig. 98c). This direction is the same at 06:00 h and at 12:00 h. We calculate the angle Ie by Ze Xe Ie ¼ 40

tan Ie ¼

Xe

N

Ie Ze Be

Z

Fig. 98c

Main (internal), external, and anomalous magnetic fields

177

Because at 06 : 00 h ðt ¼ p=2Þ ! N ¼ 10 000 12 : 00 hðt ¼ pÞ ! N ¼ 20 000 the variation of the magnitude of the external field with time is given by N ¼ 10 000ð1  cos tÞ 99. Buried at a depth of 100 m at a point with geographical coordinates 45 N, 45 W is a dipole anomaly of Cm ¼ 0.1 T m3, inclined 45 from the horizontal northwards in the vertical plane with the negative pole downwards. Measurements gave the following results (in nT): 09:00 h X ¼ 27 759; Y ¼ 0; Z ¼ 30 141 12:00 h X ¼ 28 052; Y ¼ 0; Z ¼ 30 141 Find: (a) The coordinates of the magnetic dipole’s North Pole. (b) The value of B0. (c) An expression for the variation Sq knowing that it is zero at 00:00 h and maximum at 12:00 h. (a) As in Problem 98, the observed field is the result of three parts: the main (internal) field, the buried dipole field, and the external field. To calculate the coordinates of the magnetic dipole’s North Pole we need to obtain the components of the geomagnetic main field from the components of the total field. With this aim we begin by calculating the magnetic anomaly created by the buried dipole, applying Equations (82.2) and (82.3), and substituting a ¼ 225º and x ¼ 0. The horizontal component is in the NS direction (DX) given that the dipole is on the NS-vertical plane (Fig. 99a). 2Cm cos a ¼ 141 nT d3 Cm sin a
X ¼ ¼ 71 nT d3
Y ¼ 0
Z ¼

The total observed field at 09:00 h is X1 ¼ X  þ
X þ X1e Y1 ¼ Y  þ
Y þ Y1e ¼ 0 ) Y  ¼ Y1e Z1 ¼ Z  þ
Z þ Z1e The total field at 12:00 h is X2 ¼ X  þ
X þ X2e Y2 ¼ Y  þ
Y þ Y2e Z2 ¼ Z  þ
Z þ Z2e

178

Geomagnetism

∆X P

N ∆Z d

+

a

45°

– Z

Fig. 99a

Subtracting both sets of equations we obtain X2  X1 ¼ X2e  X1e ¼ 293 nT Y2  Y1 ¼ Y2e  Y1e ¼ 0 ) Y2e ¼ Y1e ¼ Y  Z2  Z1 ¼

Z2e



Z1e

¼0)

Z2e

¼

ð99:1Þ

Z1e

We assume that the time variation of the observations is due to the diurnal Sq variation which is zero at 00:00 h and maximum at 12:00 h. Therefore the only possible values for the components Ye and Ze are zero because these components have the same values at 09:00 h and at 12:00 h: Y2e ¼ Y1e ¼ Y  ¼ 0 Z2e ¼ Z1e ¼ 0 Then, the intensity of the external field is given by X e ¼ N ð1  cos otÞ o¼

2p 24

The NS components of this field at 09:00 h (X1) and at 12:00 h (X2) are pffiffiffi



3p 2 e ¼N 1þ X1 ¼ N 1  cos 2 4 X2e ¼ N ð1  cos pÞ ¼ 2N

Subtracting the two values and using Equation (99.1) we obtain pffiffiffi

pffiffiffi e 2 2 X2 e e X2  X1 ¼ 1  ¼ 293 nT ) X2e ¼ 2001 nT N ¼ 1 2 2 2 X1e ¼ 1708 nT

179

Main (internal), external, and anomalous magnetic fields

GMNP

GNP

P

f∗ f

90° – fB

Fig. 99b

The components of the geomagnetic main field intensity are given by X  ¼ X1 
X  X1e ¼ 26 122 nT Y  ¼ Y1 
Y  Y1e ¼ 0 Z  ¼ Z1 
Z  Z e ¼ 30 000 nT qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  H ¼ ðX  Þ2 þ ðY  Þ2 ¼ X  ¼ 26 122 nT tan D ¼

Y ) D ¼ 0 H

The geomagnetic latitude of the point f  is determined from the vertical and horizontal geomagnetic main field components: Z  ¼ 2B0 sin f H  ¼ B0 cos f tan f ¼

Z ) f ¼ 29:9 2H 

We obtain the coordinates of the Geomagnetic North Pole by (Fig. 99b) D ¼ 0 ) lB ¼ 180 þ l ¼ 135 E 90  fB ¼ f  f ) fB ¼ 74:9

180

Geomagnetism

(b) The geomagnetic constant B0 is given by H  ¼ B0 cos f H B0 ¼ ¼ 30 133 nT cos f (c) We have obtained above that F e ¼ X e ¼ N ð1  cos otÞ To calculate N we take into account that the Sq variation is maximum at 12:00 h: X2e ¼ 2N ) N ¼ 1000 nT 100. At a point with geographical coordinates 60 N, 45 E, a magnetic observation results in the following values: FT ¼ 48 277 nT, DT ¼ 2.9 , IT ¼ 63.7 . It is known that at 20 m below this point there is a horizontal magnetic dipole of moment Cm ¼ 0.01 T m3, with the positive pole oriented in the direction N 60 E. There is also an external field parallel to the axis of rotation directed southwards of magnitude 1000 nT. Determine the main (internal) field components, the constant B0, and the geocentric geographical coordinates of the northern and southern geomagnetic poles (precision 1 nT). We first calculate the components of the total field intensity from FT, DT, and IT (Fig. 100a) by HT ¼ FT cos IT ¼ 21 390 nT ZT ¼ FT sin IT ¼ 43 280 nT XT ¼ HT cos DT ¼ 21 363 nT YT ¼ HT sin DT ¼ 1082 nT

XT DT

YT

X

HT Y

ZT

IT

FT

Z

Fig. 100a

181

Main (internal), external, and anomalous magnetic fields

P X60

d a +

–

Z

Fig. 100b

The observed field is the result of three parts: the main field (X  , Y , Z  ), the buried dipole field (DX, DY, DZ), and the external field (Xe, Ye, Ze): XT ¼ X  þ
X þ X e YT ¼ Y  þ
Y þ Y e ZT ¼ Z  þ
Z þ Z e We determine the magnetic anomaly created by the buried dipole from Equations (82.2) and (82.3) substituting a ¼ 90º and x ¼ 0. If we call X60 the direction N 60º E the horizontal component of this anomaly is DX60 (Fig. 100b): 2Cm cos a ¼ 0 nT d3 Cm sin a
X60 ¼ ¼ 1250 nT d3 The NS and EW components will be given by (Fig. 100c)
Z ¼


X ¼
X60 cos 60 ¼ 625 nT
Y ¼
X60 sin 60 ¼ 1083 nT The external field is parallel to the Earth’s axis of rotation in the southwards direction so its components are in the vertical and NS direction and are given by (Fig. 100d) Z e ¼ 1000 sin f ¼ 866 nT X e ¼ 1000 cos f ¼ 500 nT We calculate the main field components from these values: X  ¼ XT 
X  X e ¼ 21 238 nT Y  ¼ YT 
Y  Y e ¼ 1082 nT Z  ¼ ZT 
Z  Z e ¼ 42 414 nT qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi H  ¼ ðX  Þ2 þ ðY  Þ2 ¼ 21 266 nT Y tan D ¼  ) D ¼ 2:9 X

182

Geomagnetism

N

60°

∆Y E

∆X ∆X60

Fig. 100c GNP

P Ze Xe

f

Fe

Fig. 100d

The geomagnetic latitude of the point f  and the geomagnetic constant B0 are found from the vertical and horizontal geomagnetic main field components by Z  ¼ 2B0 sin f H  ¼ B0 cos f Z tan f ¼ ) f ¼ 44:9 2H  Z ¼ 30 953 nT ¼ 30 044 nT B0 ¼ 2 sin f

Main (internal), external, and anomalous magnetic fields

183

GNP 90º – fB l – lB 90º – f

180º – l∗ GMNP q = 90º – f∗

D∗

P

Fig. 100e

To calculate the geographical coordinates of the Geomagnetic North Pole we use the corresponding spherical triangle (Fig. 100e). We obtain the latitude fB by applying the cosine law for the angle 90º – fB: cosð90  fB Þ ¼ cosð90  f Þ cosð90  fÞ þ sinð90  f Þ sinð90  fÞ cos D sin fB ¼ sin f sin f þ cos f cos f cos D fB ¼ 75:0 We obtain the longitude lB by applying the cosine rule for the angle 90º – f  : cosð90  f Þ ¼ cosð90  fB Þ cosð90  fÞ þ sinð90  fB Þ sinð90  fÞ cosðl  lB Þ sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ sin f  sin fB sin f cos fB cos f  l  lB ¼ 180:0 cosðl  lB Þ ¼

lB ¼ 135:0 W Therefore the coordinates of the Geomagnetic North Pole are fB ¼ 75 N

lB ¼ 135 W

The coordinates of the Geomagnetic South Pole (the antipodal point) are: fA ¼ fB ¼ 75 S

lA ¼ 180 þ lB ¼ 45 E

101. At a point with geographical coordinates 30 N, 30 E, the observed geomagnetic field components are (in nT): X ¼ 15 364, Y ¼ 7660, Z ¼ 48 980. The northern geomagnetic pole is at 60 N, 0 E, and B0 ¼ 30 000 nT. There is also a constant external magnetic field normal to the equatorial plane, with a southwards direction, of 1000 nT intensity. Buried 10 m below the observation point is a magnetic dipole. Calculate: (a) The magnetic anomalies DX, DY, DZ, DH, DF.

184

Geomagnetism (b) The orientation and magnetic moment (Cm, in nT m3) of the buried dipole. (a) The observed values are the sum of the geomagnetic main field, the magnetic field due to the buried dipole, and the external field: X ¼ X  þ
X þ X e Y ¼ Y  þ
Y þ Y e Z ¼ Z  þ
Z þ Z e To obtain the magnetic anomalies from these equations we first calculate the geomagnetic latitude and the vertical and horizontal components of the geomagnetic main field by (71.3): sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ f ¼ 54 Z  ¼ 2B0 sin f ¼ 48 479 nT H  ¼ B0 cos f ¼ 17 676 nT The declination and inclination are given by  cos fB sinðl  lB Þ ) D ¼ 25 cos f tan I  ¼ 2 tan f ) I  ¼ 70

sin D ¼

The NS and EW components are X  ¼ H  cos D ¼ 16 007 nT Y  ¼ H  sin D ¼ 7498 nT The external field is parallel to the axis of rotation directed southwards so its components are in the vertical and NS direction and are given by (Fig. 101a) Z e ¼ 1000 sin f ¼ 500 nT X e ¼ 1000 cos f ¼ 866 nT Ye ¼ 0 From these values we calculate the magnetic anomalies DX, DY, DZ, DH, DF:
X ¼ X  X   X e ¼ 223 nT
Y ¼ Y  Y   Y e ¼ 162 nT
Z ¼ Z  Z   Z e ¼ 1 nT
H ¼
X cos D þ
Y sin D ¼ 271 nT
F ¼
H cos I  þ
Z sin I  ¼ 94 nT (b) We call b the angle between the geographical north and the buried dipole directions. Then using Equations (82.2) and (82.3) we obtain
Z ¼

2Cm cos a d3

ð101:1Þ

185

Main (internal), external, and anomalous magnetic fields

GNP

P Ze Xe

f

Fe

Fig. 101a


X ¼

Cm sin a cos b d3

ð101:2Þ


Y ¼

Cm sin a sin b d3

ð101:3Þ

To solve this system of three equations in three unknowns (Cm, a, b) we divide Equation (101.3) by (101.2): tan b ¼


Y ) b ¼ 36 þ 180 ¼ 144
X

This value of the angle b implies that the dipole is oriented in the N 144º E direction. To calculate the angle a between the buried dipole and the vertical we divide Equation (101.2) by (101.1):
X 1 ¼  tan a cos b
Z 2 2
X ) a 90 tan a ¼  cos b
Z Therefore the dipole is practically horizontal (Fig. 101b) and this explains the small value of the vertical component DZ. Finally we calculate the magnetic moment of the buried dipole from Equation (101.2): Cm ¼ 

d3
X ¼ 2:8  105 nT m3 sin a cos b

186

Geomagnetism

P N 144° E

d a +

–

Fig. 101b

102. At a point with geographical coordinates 30 N, 30 E, magnetic measurements give the following values: F ¼ 52 355 nT, I ¼ 70.5 , and D ¼ -26.0 . The terrestrial dipole has a north pole at 60 N, 0 E and B0 ¼ 30 000 nT. There is also a constant external field of 1000 nT with lines of force contained in the plane of the 30 E meridian at an angle of 60 with the equatorial plane and directed southwards. (a) Calculate the anomalies DX, DY, DZ. (b) If the anomalies are produced by a dipole buried at 10 m below the point, what is its orientation and its magnetic moment, Cm? Neglect values less than 10 nT. (a) We first calculate the components of the total field from F, D, and I by H ¼ F cos I ¼ 17 476 nT Z ¼ F sin I ¼ 49 352 nT X ¼ H cos D ¼ 15 707 nT Y ¼ H sin D ¼ 7661 nT The magnetic anomalies are found by subtracting from the observed values the main and external field contributions:
X ¼ X  X   X e
Y ¼ Y  Y   Y e
Z ¼ Z  Z   Z e First we determine the geomagnetic latitude and declination using (71.3) and (71.2): sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ f ¼ 53:9 sin D ¼

 cos fB sinðl  lB Þ ) D ¼ 25:1 cos f

The vertical and horizontal components are given by Z  ¼ 2B0 sin f ¼ 48 479 nT H  ¼ B0 cos f ¼ 17 676 nT X  ¼ H  cos D ¼ 16 007 nT Y  ¼ H  sin D ¼ 7498 nT

Main (internal), external, and anomalous magnetic fields

187

GNP

P

60° Xe

Ze f

Fe

Fig. 102a

N

60° E f Xe Ze

Fe

Fig. 102b

The external field is on the plane containing the vertical and NS directions (Figs 102a and 102b): Z e ¼ 1000 cosð60  fÞ ¼ 866 nT X e ¼ 1000 sinð60  fÞ ¼ 500 nT Ye ¼ 0

188

Geomagnetism

Finally, the magnetic anomalies are given by
X ¼ X  X   X e ¼ 200 nT
Y ¼ Y  Y   Y e ¼ 163 nT
Z ¼ Z  Z   Z e ¼ 7 nT (b) We call b the angle between the positive pole of the buried dipole and the geographical north. Then applying Equations (82.2) and (82.3) we obtain 2Cm cos a d3

ð102:1Þ


X ¼

Cm sin a cos b d3

ð102:2Þ


Y ¼

Cm sin a sin b d3

ð102:3Þ


Z ¼

We divide Equation (102.3) by Equation (102.2) and obtain tan b ¼


Y ) b ¼ 39:2 þ 180 ¼ 140:8
X

Therefore the dipole is oriented in the N 140.8º E direction. To calculate the inclination of the dipole from the vertical we divide Equation (102.2) by Equation (102.1):
X 1 ¼  tan a cos b
Z 2 2
X tan a ¼  ) a ¼ 89:2 cos b
Z This value implies that the dipole is nearly horizontal. Finally we calculate Cm from the total anomalous field DB pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2 þ
Y 2 þ
Z 2 ¼ 258 nT Cm
B ¼ 3 ) Cm ¼ 258 000 nT m3 d


B ¼

103. The Earth’s magnetic field is formed by a centred dipole with a geomagnetic pole at 60 N, 0 E, and B0 ¼ 30 000 nT, and an external field of 10 000 nT parallel to the equatorial plane and to the zero meridian. (a) Calculate the components X, Y, Z of the total field at a point P with geographical coordinates 60 N, 30 W. (b) If at 30 m in the direction of the compass needle from P there is a vertical dipole of moment Cm ¼ 4000 nT m3 buried 40 m deep, what would be the anomaly DZ produced at P?

189

Main (internal), external, and anomalous magnetic fields

(a) The components of the total field are the sum of the geomagnetic main field and the external field: X ¼ X þ Xe Y ¼ Y þ Ye Z ¼ Z þ Ze Let us first calculate the geomagnetic latitude, declination, and inclination using (71.3) and (71.2): sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ f ¼ 75:1  cos fB sinðl  lB Þ ) D ¼ 76:5 cos f tan I  ¼ 2 tan f ) I  ¼ 82:4

sin D ¼

The vertical and horizontal components are given by Z  ¼ 2B0 sin f ¼ 57 983 nT H  ¼ B0 cos f ¼ 7714 nT X  ¼ H  cos D ¼ 1801 nT Y  ¼ H  sin D ¼ 7501 nT The external field is parallel to the equatorial plane and to the zero meridian. If its magnitude is N ¼ 10 000 nT, its components, from Fig. 103a (plane through the point parallel to the equator) and Fig. 103b (plane through the geographical meridian of the point), are given by Z e ¼ Ber ¼ N cos l cos ’ ¼ 4330 nT X e ¼ Bef ¼ N cos l sin ’ ¼ 7500 nT Y e ¼ Bel ¼ N sin l ¼ 5000 nT Therefore the components of the total field are Z ¼ 62 313 nT X ¼ 9301 nT Y ¼ 12 501 nT (b) The buried vertical dipole is in the direction of the compass, that is, in the direction of the magnetic north (Fig. 103c). We calculate the anomaly DZ produced at P using Equations (82.2) and (82.3) substituting a ¼ 0º and x ¼ 30:
Z ¼

Cm½ðx2  2d 2 Þ cos a þ 3dx sin a

½x2 þ d 2 5=2 Cmðx2  2d 2 Þ
Z ¼ ¼ 0:029 nT ½x2 þ d 2 5=2

190

Geomagnetism

B eλ = Nsinl

P

Be

Ncosλ l

Plane parallel to the equator

Fig. 103a

GNP B ef Ncosλ

P

Ber f

Fig. 103b

191

Main (internal), external, and anomalous magnetic fields

P

x NM – d

+

z

Fig. 103c

104. The Geomagnetic North Pole is at 60 N, 150 W, with B0 ¼ 30 000 nT, and there is an external magnetic field of intensity 3000 nT parallel to the axis of rotation pointing away from the North Pole. Buried 10 m below a point with coordinates 30 N, 30 E there is a horizontal dipole with Cm ¼ 40 000 nT m3 and the negative pole pointing in the direction N 45 E. (a) What are the components X, Y, Z of the total field? (b) Calculate the total field anomaly DF. (c) What is the angle between the direction of the compass and geographic north? (a) The components of the total field are the sum of the geomagnetic main field, the magnetic field due to the buried dipole, and the external field: X ¼ X  þ
X þ X e Y ¼ Y  þ
Y þ Y e Z ¼ Z  þ
Z þ Z e We determine first the geomagnetic latitude, declination, and inclination using (71.3) and (71.2): sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ f ¼ 0 sin D ¼

 cos fB sinðl  lB Þ ) D  ¼ 0 cos f

tan I  ¼ 2 tan f ) I  ¼ 0

192

Geomagnetism

P

X45

d

a –

+

z

Fig. 104

The vertical and horizontal components are given by Z  ¼ 2B0 sin f ¼ 0 nT H  ¼ B0 cos f ¼ 30 000 nT X  ¼ H  cos D ¼ 30 000 nT Y  ¼ H  sin D ¼ 0 nT We calculate the magnetic anomaly created by the buried dipole from Equations (82.2) and (82.3), substituting a ¼ 270º and x ¼ 0. The vertical component DZ and the horizontal component DX45 of the anomaly in the direction N 45 E are (Fig. 104) 2Cm cos a ¼ 0 nT d3 Cm sin a ¼ 40 nT ¼ d3


Z ¼
X45

The NS and EW components are given by
X ¼ 40 sin 45 ¼ 28:3 nT
Y ¼ 40 cos 45 ¼ 28:3 nT The external field is parallel to the axis of rotation directed southwards, so its components are in the vertical and NS direction and are given by Nr ¼ Z e ¼ N cos 60 ¼ 1500 nT N# ¼ X e ¼ N sin 60 ¼ 2598 nT The components of the observed total magnetic field are X ¼ 27 430 nT Y ¼ 28 nT Z ¼ 1500 nT

193

Main (internal), external, and anomalous magnetic fields

(b) The total field anomaly DF is given by
F ¼
X ¼ 28 nT (c) The direction of the compass is affected by the three fields. The angle D between the direction of the compass and the geographic north is obtained from the horizontal components of the total observed field tan D ¼

Y 28 ¼ ) D ¼ 0:06 X 27430

105. The geomagnetic field is that of a dipole in the direction of the axis of rotation and B0 ¼ 30 000 nT. There is also a constant external field of 2500 nT normal to the equatorial plane in the direction of the South Pole. (a) Calculate the value of the inclination observed at a point P with coordinates 45 N, 45 E given that, at 10 m below it, there is a vertical dipole with the negative pole upwards and moment Cm ¼ 40 000 nT m3. (b) For a point 20 m north of P, calculate the observed inclination and declination, and the total field anomaly DF. (a) The components of the total observed field are the sum of the geomagnetic main field, the magnetic field due to the buried dipole, and the external field: X ¼ X  þ
X þ X e Y ¼ Y  þ
Y þ Y e Z ¼ Z  þ
Z þ Z e The magnetic dipole is oriented in the direction of the axis of rotation and therefore f ¼ f ¼ 45 D ¼ 0 Then the components of the geomagnetic main field are Z  ¼ 2B0 sin f ¼ 42 426 nT H  ¼ B0 cos f ¼ 21 213 nT X  ¼ H  cos D ¼ 21 213 nT Y  ¼ H  sin D ¼ 0 The magnetic anomaly created by the dipole is obtained from Equations (82.2) and (82.3) substituting a ¼ 0º and x ¼ 0 (Fig. 105a): 2Cm cos a ¼ 80 nT d3 Cm sin a
X ¼ ¼0 d3
Y ¼ 0
Z ¼

194

Geomagnetism

P X

– d

+

z

Fig. 105a GNP

P Ze Xe f

Fe

Fig. 105b

The external field is parallel to the axis of rotation directed southwards so its components are in the vertical and NS direction (Fig. 105b) and are given by Z e ¼ 2500 sin f ¼ 1768 nT X e ¼ 2500 cos f ¼ 1768 nT Ye ¼ 0

Main (internal), external, and anomalous magnetic fields

195

Therefore the components of the observed field are X ¼ X  þ
X þ X e ¼ 19 445 nT Y ¼ Y  þ
Y þ Y e ¼ 0 H ¼X Z ¼ Z  þ
Z þ Z e ¼ 44 274 nT From these values we calculate the observed inclination tan I ¼

Z ) I ¼ 66:3 H

(b) For a point Q located 20 m to the north of P we can assume that the main and external fields have the same value as at point P and only the magnetic anomaly created by the buried dipole is different. We calculate this anomaly from Equations (82.2) and (82.3) substituting a ¼ 0º and x ¼ 20:
Z ¼
X ¼

Cm½ðx2  2d 2 Þ cos a þ 3dx sin a ½x2 þ d 2 5=2 Cm½ð2x2  d 2 Þ sin a  3dx cos a ½x2

þ

d 2 5=2

¼

¼

Cmðx2  2d 2 Þ ½x2 þ d 2 5=2 Cm3dx

½x2 þ d 2 5=2

¼ 1 nT

¼ 4 nT


Y ¼ 0 Therefore, the components of the observed field at that point are X ¼ X  þ
X þ X e ¼ 19 441 nT Y ¼ Y  þ
Y þ Y e ¼ 0 H ¼X Z ¼ Z  þ
Z þ Z e ¼ 44 193 nT From these values we calculate the observed inclination and declination by the expressions Z ) I ¼ 66:2 H Y tan D ¼ ) D ¼ 0 X

tan I ¼

The total field anomaly DF is given by
F ¼
X cos I þ
Z sin I ¼ 4 cos 66  1 sin 66 ¼ 3 nT 106. Consider a point with coordinates 30 N, 30 E under which is buried at a depth of 100 m a horizontal dipole of moment m0 m/4p ¼ 1 T m3, with the positive pole in the direction N 60 E. The terrestrial field is formed by a centred dipole in the direction of the axis of rotation and a constant external field of 10 000 nT from the Sun, B0 ¼ 30 000 nT. (a) Calculate F, D, and I at that point on December 21 at 12 noon.

196

Geomagnetism

(b) How do D and I vary throughout the year? (a) The components of the total field intensity are the sum of the geomagnetic main field, the magnetic field due to the buried dipole, and the external field: X ¼ X  þ
X þ X e Y ¼ Y  þ
Y þ Y e Z ¼ Z  þ
Z þ Z e We first calculate the geomagnetic latitude and the declination and inclination. The magnetic dipole is oriented in the direction of the axis of rotation and therefore f ¼ f ¼ 30 D  ¼ 0 Then the components of the geomagnetic main field are Z  ¼ 2B0 sin f ¼ 30 000 nT H  ¼ B0 cos f ¼ 25 981 nT X  ¼ H  cos D ¼ 25 981 nT Y  ¼ H  sin D ¼ 0 We calculate the magnetic anomaly produced by the buried dipole from Equations (82.2) and (82.3) substituting a ¼ 90º and x ¼ 0. We call X60 the direction N 60º E and DX60 the horizontal component of the anomaly (Fig. 106a): 2Cm cos a ¼ 0 nT d3 Cm sin a ¼ 1000 nT ¼ d3


Z ¼
X60

P X 60

d a +

–

z Fig. 106a

197

Main (internal), external, and anomalous magnetic fields

N

60°

∆Y

E ∆X

∆X60

Fig. 106b GNP N = Be P f +e f e

Sun

Fig. 106c

The NS and EW components are given by (Fig. 106b)
X ¼
X60 cos 60 ¼ 500 nT
Y ¼
X60 sin 60 ¼ 866 nT The external field has a diurnal period (o ¼ 2p / 24 h) and we know that the Sun is at the meridian point at 12:00 h (solar time). This field changes through the year because the Sun moves on the ecliptic plane (apparent motion) which is tilted with respect to the equatorial plane by an angle e ¼ 23º. Therefore the solar declination (d), the angle from the Sun to the celestial equator, changes through the year. On December 21 (winter solstice) this angle is d ¼ e ¼ 23º (Fig. 106c). If we call N the magnitude of the external field (N ¼ 10 000nT) its components on December 21 at 12:00 are (Fig. 106d) Z e ¼ Br ¼ N cosðf þ eÞ ¼ 6018 nT X e ¼ Bf ¼ N sinðf þ eÞ ¼ 7986 nT Ye ¼ 0

198

Geomagnetism

GNP X = Bf

N

P f+e Z = – Br

r

f

Fig. 106d

The components of the observed field are X ¼ X  þ
X þ X e ¼ 33 467 nT Y ¼ Y  þ
Y þ Y e ¼ 866 nT Z ¼ Z  þ
Z þ Z e ¼ 36 018 nT Therefore the total field F and the declination D are pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F ¼ X 2 þ Y 2 þ Z 2 ¼ 49 166 nT Y tan D ¼ ) D ¼ 1:5 X Z sin I ¼ ) I ¼ 47:1 F (b) Any other day at 12:00 h X ¼ X  þ
X þ X e ¼ 25 481 nT þ N sinðf  dÞ Y ¼ Y  þ
Y þ Y e ¼ 866 nT Z ¼ Z  þ
Z þ Z e ¼ 30 000 nT þ N cosðf  dÞ 107. The internal field of the Earth corresponds to a centred dipole with the negative pole in the northern hemisphere at coordinates 80 N, 130 W and B0 ¼ 30 000 nT. There is a uniform external field from the Sun of 1000 nT. Buried at 500 m depth under a point P with geocentric coordinates 40 N, 50 E there is a positive magnetic pole of strength CP ¼ 0.5 T m². Calculate: (a) The total field components X, Y, and Z, and the magnetic and geomagnetic declination at P on March 21 at 12:00 h.

199

Main (internal), external, and anomalous magnetic fields

(b) The same parameters for a point 200 m north of P, assuming that neither the internal nor the external fields change (precision 1 nT). (a) The components of the intensity of the total field are the sum of the geomagnetic main field, the external field, and the magnetic field due to the buried pole. To calculate the main field we determine first the geomagnetic latitude and the declination by (71.3) and (71.2): sin f ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ f ¼ 30 sin D ¼

 cos fB sinðl  lB Þ ) D ¼ 0 cos f

From these values we obtain the vertical and horizontal components of the geomagnetic main field: Z  ¼ 2B0 sin f ¼ 30 000 nT H  ¼ B0 cos f ¼ 25 981 nT X  ¼ H  cos D ¼ 25 981 nT Y  ¼ H  sin D ¼ 0 To calculate the external field we notice that it comes from the Sun which on March 21 (spring equinox) is on the equatorial plane so that the external field is parallel to this plane. In addition this field changes during the day as a function of local time t with a diurnal period (o ¼ 2p/24 h). At t ¼ 12 h, the external field is maximum given that at this time the Sun is at the meridian point (Fig. 107a). Calling N its magnitude (N ¼ 1000 nT), the components of the external field are given by GNP Xe P

N Ze f

Fig. 107a

200

Geomagnetism

Z e ¼ Br ¼ N cos f ¼ 766 nT X e ¼ Bf ¼ N sin f ¼ 643 nT Y e ¼ Bel ¼ 0 The magnetic field anomaly created by the buried pole is derived from its potential DF, given by CP
F ¼ r Applying the gradient, we obtain
B ¼ 
F but the only component is the vertical:
Z ¼ 
Br ¼

@ð
FÞ CP ¼ 2 @r r

Substituting r ¼ d ¼ 500 m in this equation we obtain
Z ¼ 2000 nT Therefore the components of the total field and the observed declination are given by X ¼ X  þ X e þ
X ¼ 26 624 nT Y ¼ Y  þ Y e þ
Y ¼ 0 Z ¼ Z  þ Z e þ
Z ¼ 28 766 nT tan D ¼

Y ) D ¼ 0 X

(b) The radial component of the magnetic field anomaly created by the buried pole for a point 200 m north of P (x ¼ 200 m) is given by
Br ¼ 

@ð
FÞ CP ¼ 2 @r r

From Fig. 107b the vertical and NS components are CP d CPd ¼ r2 r ðx2 þ d 2 Þ3=2 CP x CPx
X ¼ 
Br sin a ¼ 2 ¼ 2 r r ðx þ d 2 Þ3=2
Z ¼ 
Br cos a ¼


Y ¼ 0 Substituting the values given (d ¼ 500 m, x ¼ 200 m, CP ¼ 0.5 Tm2), we obtain
Z ¼ 1601 nT
X ¼ 640 nT
Y ¼ 0

201

Paleomagnetism

P

x

X

r

a

d α +

z

Fig. 107b

The components of the observed total field and the declination are the sum of the three contributions: X ¼ X  þ X e þ
X ¼ 27 264 nT Y ¼ Y  þ Y e þ
Y ¼ 0 Z ¼ Z  þ Z e þ
Z ¼ 29 165 nT tan D ¼

Y ) D ¼ 0 X

Paleomagnetism 108. At a point with geographical coordinates 60 N, 60 W a 1 cm3 sample was taken of a rock with remanent magnetism, age 10 000 years, specific susceptibility 0.01 cm3. The magnetization components of the rock were: X ¼ 40, Y ¼ 30, Z ¼ 50 nT (N, E, nadir). The current field is B0 ¼ 30 000 nT and the geomagnetic pole coincides with the geographical pole. Calculate: (a) The coordinates of the virtual geomagnetic pole which corresponds to the sample. (b) The magnetic moment of the terrestrial dipole 10 000 years ago. (c) The secular variation of F, D, and I in nT and minutes per year assuming that the variation since that time has been constant.

202

Geomagnetism

GNP 90º – fB l – lB

90º – f

180º – l∗

VP

90º – f ∗

D∗

P

Fig. 108

(a) First we determine the declination D and the geomagnetic co-latitude y, corresponding to the virtual pole, from the magnetization components of the rock X, Y, and Z: Y tan D ¼ ) D ¼ 36:9 X pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi H ¼ X 2 þ Y 2 ¼ 50 nT Z tan I ¼ ) I ¼ 45 H tan I ¼ 2 cot y ) y ¼ 63:4 ; fvirtual ¼ 26:6 Since at present the geomagnetic pole coincides with the geographical pole, the geographical latitude of the point coincides with the present geomagnetic latitude: f ¼ fpresent ¼ 60 N To determine the coordinates of the virtual Geomagnetic North Pole (VP), corresponding to the magnetization of the rock, we solve the spherical triangle of Fig. 108 for ’B and lB using the obtained values of y and D. The latitude fB applying the cosine rule is given by cosð90  fB Þ ¼ cos y cosð90  fÞ þ sin y sinð90  fÞ cos D sin fB ¼ cos y sin f þ sin y cos f cos D fB ¼ 48:2 To obtain the longitude lB we again apply the cosine rule: cos y ¼ sinð90  f Þ ¼ cosð90  fB Þ cosð90  fÞ þ sinð90  fB Þ sinð90  fÞ cosðl  lB Þ cos y ¼ sin fB sin f þ cos fB cos f cosðl  lB Þ

203

Paleomagnetism

cos y  sin fB sin f cos fB cos f  l  lB ¼ 126:4 cosðl  lB Þ ¼

To choose between the positive and negative solution we bear in mind that the declination is negative and so the point is to the east of the virtual magnetic North Pole: D < 0 ) l  lB > 0 lB ¼ 173:6 E (b) To obtain the magnetic moment we first calculate the constant B0. The susceptibility w relates the magnetization and the magnetic field. If we call F the magnitude of the paleomagnetic field and F0 the remanent magnetization, the relation between them is F 0 ¼ wF

ð108:1Þ

w ¼ 0:01 We calculate F0 from its components pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F 0 ¼ X 2 þ Y 2 þ Z 2 ¼ 71 nT The field F of the virtual pole is given by pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F ¼ B0 1 þ 3 cos2 y

Substituting in Equation (108.1) we obtain pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F 0 ¼ wB0 1 þ 3 cos2 y F0 B0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 5610 nT w 1 þ 3 cos2 y

From this value we calculate the magnetic moment of the virtual pole taking a ¼ 6370 km for the Earth’s radius B0 ¼

Cm ) m ¼ 1:45  1022 A m2 a3

(c) The magnetic field, the declination, and inclination 10 000 years ago were F0 ¼ 7100 nT w D ¼ 36:9 F¼

I ¼ 45 At present the values of these parameters are F p ¼ Ba0 D p ¼ 0

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 3 sin2 fpresent ¼ 54 083 nT

tan I p ¼ 2 tan f ) I p ¼ 73:9

204

Geomagnetism

The secular variation of F, D, and I, for this period of time, is Fp  F ¼ 4:7 nT=yr 10 000 p D D ¼ 0:220 =yr 10 000 Ip  I ¼ 0:170 =yr 10 000 109. The following table gives the demagnetization data for a sample that was subjected to stepwise thermal demagnetization of its natural remanent magnetization (NRM). Demagnetization temperature (°C)

Declination (D, °E)

Inclination (I, °)

NRM Intensity (J, mA/m1)

20 100 200 300 400 500 600 650 700

32 36 38 39 41 41 41 41 300

33 22 12 4 5 5 5 5 55

0.056 0.056 0.057 0.058 0.058 0.050 0.016 0.009 0.000

Calculate the direction of each stable component identified by the demagnetization curve. First, construct a vector component diagram of the demagnetization data. Decompose each observation into its north (X), east (Y), and vertical (Z) components: H ¼ J cos I X ¼ H cos D Y ¼ H sin D Z ¼ J sin I In Fig. 109 plotting X versus Y gives the projection of the demagnetization vector onto the horizontal plane, while plotting X versus Z gives the projection onto the vertical plane. A stable component of NRM is represented by collinear points on the vector component diagrams, so that two stable components can be identified in the range 20–300  C and in the range 400–700  C. The declination of a stable component is determined by measuring or by calculating the angle between the north axis and the trajectory of the stable component in the horizontal plane.

205

Paleomagnetism

0.06

0.06

0.05

0.05 0.04

0.03

0.03

X

X

a 0.04

0.02 0.01

γ

0.02 d

b

0.01

0.00 0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035

0.00 –0.01

0.00

0.01 Z

Y

0.02

0.03

Fig. 109

For the 20–300  C component:



X300  X20 0:0051 ¼ 23:9 a ¼ tan1 ¼ tan1 0:0115 Y300  Y20 D ¼ 180 þ 23:9 ¼ 203:9

For the 400–700  C component: D ¼ b ¼ tan

1



Y400 X400





0:0379 ¼ tan ¼ 41:0 0:0436 1

Note: this value can be obtained directly from the declination of the observations between 400 and 650  C. The apparent inclination, Iap, of a stable component is determined by measuring or by calculating the angle between the north axis and the trajectory of the stable component in the vertical plane. Iap is related to the true inclination, I, by: tan I ¼ tan Iap j cos Dj For the 20–300  C component: Iap ¼ g ¼ tan

1



Z20  Z300 X300  X20



¼ tan

1



0:0265 0:0051

I ¼ tan1 ðtanð79:1Þj cosð203:9ÞjÞ ¼ 78:1



¼ 79:1

For the 400–700  C component: Iap ¼ d ¼ tan1



Z400 X400



¼ tan1



0:0051 ¼ 6:7 0:0436

I ¼ tan1 ðtanð6:7Þj cosð41ÞjÞ ¼ 5:0

Note: this value can be obtained directly from the inclination of the observations between 400 and 650  C. Therefore the stable component isolated in the range 20–300  C has D ¼ 203.9 and I ¼ 78.1 and the stable component isolated in the range 400–700  C has D ¼ 41.0 and I ¼ 5.0 .

206

Geomagnetism

110. A palaeomagnetic study of a late Jurassic limestone outcrop near Alhama de Granada (37 N, 4 W) in southern Spain yielded a well-defined primary remanent magnetization whose directions are given in the table below. Calculate the mean direction of the primary remanence of the seven samples. Compare this direction with that defined by the reference late Jurassic palaeomagnetic pole for the stable Iberian tectonic plate (252 E, 58 N). How much vertical axis rotation has the studied outcrop suffered with respect to stable Iberia?

Declination (D, °E)

Inclination (I, °)

30 28 34 25 32 35 26

43 39 44 45 38 44 40

Use unit vector addition to calculate the mean direction of the primary remanence. Calculate the direction cosines of each direction, the resultant total field vector, F, and then the mean direction using: X ¼ cos I cos D Y ¼ cos I sin D Z ¼ sin I vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !2ffi !2 u N !2 N N u X X X F¼t Z ¼ 6:98542 Y þ X þ Xmean ¼

N P

i

i

i

i¼1

i¼1

i¼1

Xi

i¼1

F N P Yi

¼ 0:6446;

Ymean ¼ i¼1 ¼ 0:37186; F N P Zi Zmean ¼ i¼1 ¼ 0:66799 F

Ymean Dmean ¼ tan1 ¼ 30:0 Xmean Imean ¼ sin1 ðZmean Þ ¼ 41:9

The mean direction of the primary remanence has a declination of 30.0 and an inclination of 41.9 .

207

Paleomagnetism

Next, calculate the expected field direction at the site using the reference palaeomagnetic pole. The first step is to determine y ¼ 90  f  , (Equation 71.3), from the pole (fp, lp) to the site (fs, ls) using spherical triangles: sin f ¼ sin ’p sin ’s þ cos ’p cos ’s cosðls  lp Þ ) f ¼ 24:1 The expected inclination can then be calculated using: tan Iexp ¼ 2 tan f ) Iexp ¼ 41:8 The expected declination can be calculated by (71.2): sin Dexp ¼

 cos ’p sinðls  lp Þ ) Dexp ¼ 34:5 W ¼ 325:5 cos f

rotation about the vertical axis should give rise to a difference between the observed and expected declinations, defined as positive for an observed declination clockwise from the expected declination. Therefore the outcrop has suffered 64.5 of clockwise rotation with respect to stable Iberia.

4

Seismology

Elasticity 111. Determine the principal stresses and principal axes of the stress tensor: 0 1 2 1 1 @ 1 0 1 A 1 1 2

Find the invariants I1, I2, I3, the deviator tensor, its eigenvalues, and the invariants J2 and J3. To calculate the principal stresses (s1, s2, s3) and principal axes (n1i, n2i, n3i), we calculate the eigenvalues and eigenvectors of the matrix. They are found through the equation ðtij  sdij Þni ¼ 0

ð111:1Þ

The eigenvalues are the roots of the cubic equation for s resulting from putting the determinant of the matrix in (111.1) equal to zero:    2  s 1 1     1 ¼0  1 s    1 1 2  s ) ð2  sÞðsÞð2  sÞ  1  1 þ s  ð2  sÞ  ð2  sÞ ¼ 0

s3  4s2 þ s þ 6 ¼ 0

The three roots of the equation are the principal stresses s1 ¼ 1 s2 ¼ 2 s3 ¼ 3 From these values we obtain s0 ¼ 13 ðs1 þ s2 þ s3 Þ ¼ 43. The invariants of the matrix are the coefficients of the characteristic equation s3  I1 s2 þ I2 s  I3 ¼ 0

208

209

Elasticity

which, in terms of the roots of the equation, are I1 ¼ 4 ¼ s1 þ s2 þ s3 I2 ¼ 1 ¼ s1 s2 þ s1 s3 þ s2 s3 I3 ¼ 6 ¼ s1 s2 s3 The principal axes of stress are the eigenvectors ni associated with the three eigenvalues. For s1 ¼ 1 0 10 1 3 1 1 n1   @ 1 1 1 A@ n2 A ¼ 0 ) n1 ; n1 ; n1 ¼ ð1; 2; 1Þ 1 2 3 1 1 3 n3 For s2 ¼ 2

For s3 ¼ 3

0

0 1 @ 1 2 1 1

10 1 1 n1   1 A@ n2 A ¼ 0 ) n21 ; n22 ; n23 ¼ ð1; 1; 1Þ 0 n3

0

10 1 1 n1   1 A@ n2 A ¼ 0 ) n31 ; n32 ; n33 ¼ ð1; 0; 1Þ 1 n3

1 1 @ 1 3 1 1

The deviatoric stress tensor is defined as

t0ij ¼ tij  s0 dij where in our problem, s0 ¼ 4/3. The three components of the principal diagonal of the deviatoric tensor are t011 ¼

2 3

t022 ¼  t033 ¼

4 3

2 3

To calculate its eigenvalues we proceed as we did before: 0 1 2 1 1 C B3 s C B 13 38 4 C B ¼0 B 1   s 1 C ¼ 0 ) s3  s þ B C 3 27 3 A @ 2 s 1 1 3

Comparing with the characteristic equation

s3  J1 s2 þ J2 s  J3 ¼ 0

210

Seismology

the invariants are 13 3 38 J3 ¼  27

J2 ¼ 

Solving the cubic equation we obtain s1 ¼ 2.22, s2 ¼ 0.33, s3 ¼ 1.89. 112. Given the stress tensor 0

1 4

B B B B B B 25 B 4 B B 3 @ 1 2 2

5 2 4 1 4 3 1 2 2 2

3 1 1 2 C C 2 C 3 C 2 1 C C 2 2 C C C A 3 2

calculate: (a) The principal stresses. (b) The angles formed by the greatest of these stresses with the axes 1, 2, 3. (a) As in the previous problem to find the principal stresses we calculate the eigenvalues of the stress matrix 1 0 3=2 1 5 1 B C  C B 4s 4 2 B C 3=2 C B C B 5 1 1 B  C ¼ 0 ) s3  2s2  s þ 2 ¼ 0 s C B 4 4 2 B C 3=2 C B 3=2 A @ 1 1 3 s  2 2 2

Solving the cubic equation, its three roots are

s1 ¼ 2 s2 ¼ 1 s3 ¼ 1 The largest is s1. The associated eigenvector corresponds to the axis of greatest stress whose direction cosines are (n1, n2, n3). They are found by solving the equation 0 3=2 1 7 5 1 B C   B C0 1 4 4 C n1 B 2

B C 5 7 1 C@ n2 A ¼ 0 B   3=2  B C 4 4 2 C n3 B C B 3=2 3=2 A @ 1 1 1    2 2 2

211

Wave propagation. Potentials and displacements

Solving this equation with the condition that n21 þ n22 þ n23 ¼ 1, we obtain, n1 ¼ n2 ¼ 1 n3 ¼ pffiffiffi 2

1 2

(b) From these values we obtain #, the angle with the vertical axis (x3) and ’, the angle which forms its projection on the horizontal plane with x1: 9 1 > > n1 ¼ sin # cos ’ ¼ > 2 > > 1= n2 ¼ sin # sin ’ ¼  ) ’ ¼ 315 ; # ¼ 45 2> > > 1 > > n3 ¼ cos # ¼ pffiffiffi ; 2 113. The stress tensor t ij in a continuous medium is 0 1 3x1 x2 5x22 0 @ 5x2 0 2x3 A 2 0 2x3 0 Determine the stress vector Tin acting at the point (2, 1, √3) through the plane tangential to the cylindrical surface x22 1 x23 at that point. First we calculate the value of the stress tensor at the given point: 0 1 6 5 0 ffiffiffi pffiffiffi p 0 ffiffiffi 2 3 A tij ð2; 1; 3Þ ¼ @ 5 p 0 2 3 0

A unit vector normal to the surface f ¼ x22 þ x23  4 ¼ 0 at the given point is 0 1 @f @f @f pffiffiffi

B @x1 @x2 @x3 C grad f C ¼ 0; 1 ; 3       ¼B ni ¼ ; ; 2 2 jgrad f j @ @f   @f   @f A @x  @x  @x  1

2

3

Then, the stress vector acting at the point through that surface is given by 0 1 0 1 0

1 C 6 5 0 ffiffiffi B p B C pffiffiffi AB 2 C ¼ 5 ; 3; 3 0 2 3 Tin ¼ tij nj ¼ @ 5 B pffiffiffi C pffiffiffi 2 @ 3A 0 2 3 0 2

Wave propagation. Potentials and displacements 114. The amplitudes of P- and S-waves of frequency 4/2p Hz are

4 4 4 pffiffiffi ; pffiffiffi ; pffiffiffi uP ¼ 3 2 3 2 3
pffiffiffi pffiffiffi pffiffiffi S u ¼  3;  3; 2

212

Seismology and their speeds of propagation are 6 km s1 and 4 km s1, respectively. Find the scalar and vector potentials. Displacements are always given in µm. The displacements of P-waves can be deduced from a scalar potential function ’ such that uPi ¼ ðr’Þi . The general form of the potential function for P-waves for harmonic motion is ’ ¼ A exp ika ðni xi  atÞ

ð114:1Þ

where A is the amplitude, ni the direction cosines of the ray or propagation direction, a the velocity of propagation, and ka the wavenumber. If uj is given in µm and ka in km1, then A is given in 103 m2. Taking the derivatives in (114.1) we obtain for the components of the displacement uPj ¼

@’ ¼ Aika nj exp ika ðnk xk  at Þ @xj

Their amplitude is uPj ¼ Aka nj

ð114:2Þ

and the wavenumber is 4 o 2p 2p 2 ¼ km1 ka ¼ ¼ 6 a 3 By substitution in (114.2), we obtain for the two horizontal components 4 uP1 ¼ Aka n1 ¼ pffiffiffi mm 3 2 4 uP2 ¼ Aka n2 ¼ pffiffiffi mm 3 2

Dividing these two expressions and writing the direction cosines of the ray in terms of the incident angle i and azimuth az, n1 ¼ sin i cos az v2 ¼ sin i sin az n3 ¼ cos i

ð114:3Þ

we have uP1 n1 sin i cos az ¼1¼ ¼ ) a ¼ 45 P n2 sin i sin az u2 Using the uP3 and uP1 components, pffiffiffi uP3 Aka n3 cos i ¼ ¼ ¼ 3 ) i ¼ 30 uP1 Aka n1 sin i cos az

From the values of the direction cosines and the amplitude of the displacement we calculate the amplitude of the potential A: A¼

uP1 ) A ¼ 4  103 m2 ka n1

213

Wave propagation. Potentials and displacements

Finally, the expression for the scalar potential of P-waves is pffiffiffi pffiffiffi pffiffiffi

2 2 2 3 ’ ¼ 4 exp i x1 þ x2 þ x3  6t 3 4 4 2 Displacements of the S-wave are obtained from a vector potential ci of null divergence, whose general form is   ci ¼ Bi exp ikb nj xj  bt where b is the velocity of propagation and kb the wavenumber. The displacements are given by uS ¼ r  c

ð114:4Þ

The wavenumber is kb ¼ o/b ¼ 1 km1. According to (114.4) the relation between the components of the displacement and of the amplitude of the potential is pffiffiffi pffiffiffi pffiffiffi 2 3 S  B2 ¼  3 mm u1 ¼ B3 4 2 pffiffiffi pffiffiffi pffiffiffi 3 2 uS2 ¼ B1  B3 ¼  3 mm 2 4 pffiffiffi pffiffiffi 2 2 pffiffiffi uS3 ¼ B2  B1 ¼ 2 mm 4 4 The potential must have null divergence, pffiffiffi pffiffiffi pffiffiffi 3 2 2 rc¼ B3 ¼ 0 B1 þ B2 þ 2 4 4 From these equations we obtain, in units of 103 m2, B1 ¼ 2 B2 ¼ 2 B3 ¼ 0 The S-wave vector potential is pffiffiffi pffiffiffi pffiffiffi

2 2 3 cj ¼ ð2; 2; 0Þ exp i x1 þ x2 þ x3  4t 4 4 2 Note: These units will be used for all problems but not explicitly given. 115. The components of the S-wave with respect to the axes (x1, x2, x3) are (6, 3.25, 3) where x2 is the vertical axis, the azimuth is 60 , and the angle of incidence is 30 . Determine the amplitude and direction cosines of the SV and SH components. From the azimuth and incident angles we calculate the direction cosines (x2 is the vertical axis)

214

Seismology

X2

SV

r 30° SVH

X1 60° SH

90° X3

R

Fig. 115a

1 n1 ¼ sin i cos az ¼ 4 pffiffiffi 3 n2 ¼ cos i ¼ 2 pffiffiffi 3 n3 ¼ sin i sin az ¼ 4 Since the SV and SH components are on a plane normal to the direction of the ray r (Fig. 115a) unit vectors in the direction of SV (a1, a2, a3) and of SH (b1, 0, b3) must satisfy the equations pffiffiffi pffiffiffi a1 a2 3 a3 3 þ ¼0 þ 4 2 4 pffiffiffi b1 b3 3 ¼0 SH  r ¼ 0 ) þ 4 4

SV  r ¼ 0 )

ð115:1Þ

SH  SV ¼ 0 ) a1 b1 þ a3 b3 ¼ 0 The projections on the horizontal plane R of the ray r and SH are perpendicular (Fig. 115b). Then SH forms an angle of 180 – 30 with the x1 axis. The direction cosines of SH are pffiffiffi 3 b1 ¼  2 1 b3 ¼ 2

215

Wave propagation. Potentials and displacements

SVH

30° X1

30° 60°

SH X3

R

Fig. 115b

SV forms an angle of 60 with the vertical axis x2 (Fig. 115a). Then a2 ¼  sin i ¼  12 From a2 using Equations (115.1) and a21 þ a22 þ a23 ¼ 1, we calculate a1 and a2: pffiffiffi 3 a1 ¼ 4 3 a3 ¼ 4 pffiffiffi pffiffiffi pffiffiffi



 7 5 7 7 116. Given the potential ci ¼ pffiffiffi ; pffiffiffi ; 6 exp 4i p1ffiffiffi x1 þ p1ffiffiffi x2 þ pffiffiffiffiffi x3  4t , 5 3 5 5 15 calculate the polarization angle. First we calculate the amplitudes of the components of the displacement of the S-wave from the vector potential 8 13 > > uS1 ¼ c3;2  c2;3 ¼ 4 pffiffiffi ¼ 30:02 > > > 3 > >

> < 7 6 S S p ffiffi ffi p ffiffi ffi u ¼ c  c ¼ 4 ¼ 13:97 þ ui ¼ r  c i ) 1;3 3;1 2 > 5 5 3 > > pffiffiffi

> > > pffiffiffi 7 > S > : u3 ¼ c2;1  c1;2 ¼ 4 7  pffiffiffiffiffi ¼ 7:85 15 The modulus is

uS ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u21 þ u22 þ u23 ¼ 34:03

From the vertical component uS3 we calculate the SV component (Fig. 116) knowing that rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 8 7 7 n3 ¼ cos i ¼ pffiffiffiffiffi ) sin i ¼ 1  ¼ pffiffiffiffiffi 15 15 15

Then, uS3 ¼ uSV cosð90  iÞ

) uSV ¼ 10:75.

216

Seismology

X3 r i

SV

90° – i R

Fig. 116

To find the SH component we use the relation qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi uS ¼ ðuSV Þ2 þ ðuSH Þ2 ) uSH ¼ 32:29

From the SV and SH components we obtain the polarization angle e: tan e ¼

uSH ) e ¼ 71:6 uSV

117. Given the amplitudes of the P- and S-waves (ka ¼ 1): uP1 ¼ 4

uS1 ¼ 8

pffiffiffi uS2 ¼ 2 2
pffiffiffi uS3 ¼  4 þ 2

uP2 ¼ 4 uP3 ¼ 8

determine the angle of incidence i, azimuth az, polarization angle « of the S-wave, and apparent polarization angle g. Given that the displacements of the P-wave are on the incident plane, in the direction of the ray r, the angle of incidence i can be obtained from the modulus and the vertical component: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi uP ¼ 42 þ 42 þ 82 ¼ 4 6 cos i ¼

uP3 8 ¼ pffiffiffi ) i ¼ 35:3 P u 4 6

The azimuth, the angle between the horizontal projection of the ray and the north (x1), is obtained from the horizontal components, uP1 and uP2 : tan az ¼

uP2 ) az ¼ 45 uP1

217

Wave propagation. Potentials and displacements

90° – az

uS 1

az uS H

u S2 az

X1

γ

SH R

X2

Fig. 117

We calculate the SV component from the vertical component uS3 , as in the previous problem: uSV ¼

uS3 uS ¼ 3 ¼ 9:37 cosð90  i0 Þ sin i0

The SH component is found from the horizontal components (Fig. 117) uSH ¼ uS2 cos az  uS1 sin az ¼ 3:66 From SV and SH we find the polarization angle e (Fig. 117): tan e ¼

uSH 3:66 ¼ uSV 9:37

) e ¼ 21:3

To calculate the apparent polarization angle g, the angle between the horizontal component of S, (uSH), and the radial direction R (Fig. 117), we use the relation tan e ¼ cos i0 tan g

) g ¼ 25:5

  118. Given the values az ¼ 60 , « ¼ 30 , g ¼ 45 , and ub  ¼ 5, calculate the amplitudes of the components 1, 2, and 3 of the S-wave. From the modulus of the displacement of S-waves and the polarization angle, we calculate the SV and SH components (Fig. 118a): uSV uSH

pffiffiffi 5 3 ¼ u cos e ¼ 2 5 S ¼ u sin e ¼ 2 S

218

Seismology

SV

e

SH

Fig. 118a

X3 r

SV uS 3 i i R SV uH

Fig. 118b

From the angles e and g we obtain the incidence angle i: tan e ¼ tan g cos i

1 ) cos i ¼ pffiffiffi 3

rffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 2 1 ) sin i ¼ 1  ¼ pffiffiffi 3 3

The vertical component u3 of the S-wave is obtained from the value of SV (Fig. 118b): pffiffiffi pffiffiffi 5 3 2 S SV pffiffiffi ¼ 3:53 u3 ¼ u cosð90  iÞ ¼ 2 3

To calculate the horizontal components we have to take into account the horizontal component of SV (Fig. 118b): SV uSV cos i ¼ H ¼ u

5 2

219

Wave propagation. Potentials and displacements

u SV H

az X1 uS H

az u SH

az

SH R

X2

Fig. 118c

From SH and the horizontal component of SV we find the horizontal components of S (Fig. 118c): pffiffiffi

5
uS1 ¼  uSH sin az þ uSV 1 þ 3 ¼ 3:42 H cos az ¼  4
pffiffiffi 5 1  3 ¼ 0:92 uS2 ¼ uSH cos az  uSV H sin az ¼ 4 119. For a scalar potential w and a vector potential c i, it is known that A ¼ 3, Bi ¼ (2, 2, 0), ka ¼ 2/3, T ¼ p/2, and Poisson’s ratio is s ¼ 1/4. Calculate the amplitudes u1, u2, and u3 of the P- and SV-waves. on the free surface (x3 ¼ 0) of the components pffiffiffi

3 1 1 The direction cosines are pffiffiffi ; pffiffiffi ; . 2 2 2 2 2

The general expressions for the scalar and vector potentials are

’ ¼ A exp ika ðn1 x1 þ n2 x2 þ n3 x3  at Þ ci ¼ Bi exp ikb ðn1 x1 þ n2 x2 þ n3 x3  bt Þ Since ka ¼ o/a then a ¼ o/ka ¼ 6 km s1. If Poisson’s ratio is 0.25 then 1 l s¼ ¼ 4 2 ð l þ mÞ

)l¼m

Substituting this condition in the equation for the P-wave velocity a, we find for the velocity b of S-waves is given by sffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffi pffiffiffi pffiffiffi l þ 2m 3m a ¼ ¼ b 3 ) b ¼ pffiffiffi ¼ 2 3 km s1 a¼ r r 3 Then the wavenumber of S-waves is

kb ¼

o 2 ¼ pffiffiffi b 3

220

Seismology

Then we can write the complete expressions for the potentials: pffiffiffi

2 1 1 3 p ffiffi ffi p ffiffi ffi ’ ¼ 3 exp i x1 þ x2 þ x3  6t 3 2 2 2 2 2 pffiffiffi

pffiffiffi 3 2 1 1 x 3  2 3t ci ¼ ð2; 2; 0Þ exp i pffiffiffi pffiffiffi x1 þ pffiffiffi x2 þ 2 3 2 2 2 2

ð119:1Þ

To determine the displacements of the P- and S-waves we use the relations uP ¼ r’ uS ¼ r  c obtaining 1 uP1 ¼ pffiffiffi 2 1 uP2 ¼ pffiffiffi 2 p ffiffi ffi uP3 ¼ 3

uS1 ¼ 2 uS2 ¼ 2 pffiffiffi 2 2 uS3 ¼ pffiffiffi 3

From the components of the displacement of the S-wave we can obtain the SV component. The values of the angles of incidence and azimuth are found from the direction cosines, 1 n1 ¼ sin i cos az ¼ pffiffiffi 2 2 1 n2 ¼ sin i sin az ¼ pffiffiffi 2 2 pffiffiffi 3 n3 ¼ cos i ¼ 2

) i ¼ 30 ;

az ¼ 45

and from the angle i we obtain the SV component from uS3: pffiffiffi 4 2 S SV SV u3 ¼ u cosð90  iÞ ) u ¼ pffiffiffi ¼ 3:27 3

From the horizontal components of the S-waves and the azimuth we calculate the SH component: uSH ¼ uS2 cos az  uS1 sin az ¼ 0 120. In an elastic medium of density r ¼ 3 g cm3 and Poisson ratio 1/3 there

propagate P- and S-waves of frequency 1 Hz in the direction



pffiffiffi

7 1 1 ; pffiffiffi ; pffiffiffi . Given 3 2 3 2

that the pressure of the P-wave is 5000 dyn cm2, the magnitude of its displacement, 10 mm, is twice that of the S-wave, and the angle g ¼ 45 , find all the parameters involved in the expression of the potentials ’ and ci. (It is not necessary to solve the equations to obtain the coefficients Bi)

221

Wave propagation. Potentials and displacements

Given that Poisson’s ratio is 1/3 the relation between the elastic coefficients l and m is, s¼

l 1 ¼ ) l ¼ 2m 2ðl þ mÞ 3

and between the velocities of P (a) and S (b)-waves is sffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffi l þ 2m 4m ) a ¼ 2b ¼ a¼ r r The bulk modulus K is 2 8 K ¼lþ m¼ m 3 3 Expressing m in terms of b and a, we get for K: rffiffiffi m 8 8 a2 2 b¼ ) m ¼ rb2 ) K ¼ rb2 ¼ r ¼ ra2 r 3 3 4 3 Taking into account that the bulk modulus K is defined as the applied pressure divided by the change in volume per unit volume y, 2 P ¼ Ky ¼ ra2 y 3 2 y ¼ r ’ ¼ Aka2 y¼

ð120:1Þ

3P 3P ¼ Aka2 ) a2 ¼ 2ra2 2rAka2

The expression for the scalar potential is ’ ¼ A exp ika ðn1 x1 þ n2 x2 þ n3 x3  at Þ where the wavenumber for P-waves is ka ¼

o 2pf ¼ a a

P P  P u  ¼ jr’j ¼ ka A ) A ¼ ju j ¼ ju ja ka 2pf

By substitution in (120.1) we obtain the values of a and b: a2 ¼ where P ¼ 5000 dyn cm2 r ¼ 3 g cm3 f ¼ 1 Hz uP ¼ 10 mm

3P 3Pa 3P ¼ )a¼ 2rAka2 2rjuP j2pf 2rjuP j2pf

ð120:2Þ

222

Seismology We obtain a ¼ 3.98 km s1 and b ¼ 1.99 km s1. Since we know A, a, and ka we can write the complete expression for the scalar potential pffiffiffi

7 1 1 ’ ¼ 10 exp i1:58 x1 þ pffiffiffi x2 þ pffiffiffi x3  3:98t 3 3 2 2

The vector potential of the S-waves is given by

ci ¼ Bi exp ikb ðn1 x1 þ n2 x2 þ n3 x3  bt Þ

ð120:3Þ

We calculate kb: kb ¼

o ¼ 3:16 km1 b

The displacements are given by uS ¼ r  c and uS1 ¼

@c3 @c2  ¼ kb ðB3 n2  B2 n3 Þ @x2 @x3

uS2 ¼

@c1 @c3  ¼ kb ðB1 n3  B3 n1 Þ @x3 @x1

uS3 ¼

@c2 @c1  ¼ kb ðB2 n1  B1 n2 Þ @x1 @x2

ð120:4Þ

The incidence angle i is found from n3 and, using tane ¼ cosi tang, we find the polarization angle e: pffiffiffi 7 n3 ¼ cos i ¼ pffiffiffi ) i ¼ 31:95 ) e ¼ 31:95 3 2 The azimuth is

az ¼ tan1

n2 ¼ 64:76 n1

Since the amplitude of the S-wave displacement is 5 mm, knowing the value of e we can find the values of the SV and SH components: uSV ¼ uS cos e ¼ 4:24 mm uSH ¼ uS sin e ¼ 2:65 mm From uSV we calculate its vertical and horizontal components: uS3 ¼ uSV cosð90  iÞ ¼ 2:25 mm SV uSV cos i ¼ 3:61 mm H ¼ u

223

Wave propagation. Potentials and displacements Using uHSV and uSH we find the two horizontal components:

uS1 ¼  uSH sin az þ uSV H cos az ¼ 3:94 mm uS2 ¼ uSH cos az  uSV H sin az ¼ 2:14 mm

Using the values found for the displacements and Equations (120.4) and ∇ · c ¼ 0 (n1 B1 þ n2 B2 þ n3 B3 ¼ 0) we find the values of B1, B2, B3. Substituting all the values in (120.3) we obtain for the vector potential pffiffiffi

7 1 1 p ffiffi ffi p ffiffi ffi ci ¼ ð1:17; 2:5; 3:44Þ exp 3:16i x1 þ x2 þ x3  1:99t 3 2 3 2

121. At the origin in an infinite medium in which s, Poisson’s ratio, is 0.25, and the density is 3 g cm3, there is an emitter of elastic plane waves of frequency 0.5 cps. Calculate: (a) The equation of the P- and S-waves in exponential form and with arbitrary amplitudes for the wave arriving at the point A(500, 300, 141) km. (b) The arrival time. (a) First we calculate the distance to point A and the direction cosines of the direction of the ray (r) (Fig. 121): pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5002 þ 3002 þ 1412 ¼ 599:90 600 km 500 5 ¼ n1 ¼ 600 6 300 1 ¼ n2 ¼ 600 2pffiffiffi 2 141 n3 ¼ ¼ 6 600 r¼

X3

r 141 X1 300 500

X2

Fig. 121

224

Seismology

The S-wave velocity is b¼ Given that Poisson’s ratio is 0.25,

rffiffiffi pffiffiffiffiffi m ¼ 10 km s1 r

pffiffiffi pffiffiffiffiffi s ¼ 0:25 ) l ¼ m ) a ¼ b 3 ¼ 30 km s1

The wavenumbers of the P- and S-waves are

2pf 2p  0:5 p ¼ ¼ pffiffiffiffiffi km1 a 5:5 30 2pf 2p  0:5 p ¼ pffiffiffiffiffi ¼ pffiffiffiffiffi km1 kb ¼ b 10 10 ka ¼

The general expressions for the scalar and vector potentials are ’ ¼ A exp ika ðn1 x1 þ n2 x2 þ n3 x3  at Þ ci ¼ Bi exp ikb ðn1 x1 þ n2 x2 þ n3 x3  bt Þ Leaving the amplitudes A and Bi in arbitrary form and substituting the obtained values we have for the potentials, pffiffiffi

pffiffiffiffiffi p 5 1 2 x1 þ x2 þ x3  30t ’ ¼ A exp i pffiffiffiffiffi 2 6 30 6 pffiffiffi

pffiffiffiffiffi p 5 1 2 x1 þ x2 þ x3  10t ci ¼ Bi exp i pffiffiffiffiffi 2 6 10 6 (b) The travel times for P- and S-waves from the origin to the given point are r 600 ¼ pffiffiffiffiffi ¼ 109:5 s a 30 r 600 t b ¼ ¼ pffiffiffiffiffi ¼ 189:7 s b 10

ta ¼

Reflection and refraction 122. A P-wave represented by the potential

x1 x2 x3 p ffiffi ffi p ffiffi ffi p ffiffi ffi þ þ  4t w ¼ 4 exp 0:25i 6 3 2

is incident on the surface x3 ¼ 0 separating two liquids of densities 3 g cm3 and 4 g cm3. If the speed of propagation in the second medium is 2 km s1, write the expressions for the potentials of the reflected and transmitted waves.

225

Reflection and refraction

The potentials of the reflected and transmitted waves are given by ’refl ¼ A exp ik ð tan e x3 þ x1  ct Þ ’trans ¼ A0 exp ik ðtan e0 x3 þ x1  ct Þ

ð122:1Þ

where e ¼ 90 – i is the emergence angle and i the incidence angle, and k ¼ ka cos e is the wavenumber corresponding to the apparent horizontal velocity, c ¼ a /cos e. These expressions are written for rays contained on the incidence plane (x1, x3). Then, we have to rotate the given potential to refer it to the incidence plane. First, from the direction cosines we calculate the incidence angle i and the azimuth az: 1 n1 ¼ sin i cos az ¼ pffiffiffi 6 1 n2 ¼ sin i sin az ¼ pffiffiffi 3 1 n3 ¼ cos i ¼ pffiffiffi ) i ¼ 45 ¼ e 2 pffiffiffi 1 2 cos az ¼ pffiffiffi ) sin az ¼ pffiffiffi 3 3 Using the rotation matrix we obtain the direction cosines on the plane of incidence (x1, x3): 1 0 1 10 1 0 01 0 p ffiffi ffi n1 n1 cos az sin az 0 B 2C C @ n02 A ¼ @  sin az cos az 0 A@ n2 A ) B B 0 C @ 1 A n03 1 n3 0 0 pffiffiffi 2 The values of c and k are pffiffiffi a a0 8 p ffiffi ffi c¼ ¼ ¼ 4 2 km s1 ¼ cos e cos e0 2 pffiffiffi 2 km1 k ¼ ka cos e ¼ ka0 cos e0 ¼ 8 Then the potential of the incident wave is now given by pffiffiffi  1
’inc ¼ 4 exp i pffiffiffi x3 þ x1  4 2 t 4 2 0 The angle i of the transmitted or refracted ray is found from Snell’s law: pffiffiffi sin i sin i0 2 0 ¼ 0 ) sin i ¼ ¼ cos e0 a 4 a pffiffiffiffiffi pffiffiffi 14 0 0 ) tan e0 ¼ 7 ) cos i ¼ sin e ¼ 4 Using the expressions for the reflection and refraction coefficients, V and W, we can calculate the amplitude of the reflected and refracted potentials: pffiffiffi pffiffiffi A r0 tan e  r tan e0 4  3 7 16  12 7 p ffiffi ffi p ffiffiffi ¼ 1:07 V ¼ ¼ 0 ¼ ) A ¼ A0 r tan e þ r tan e0 4 þ 3 7 4þ3 7 A0 2rtge0 6 24 pffiffiffi ) A0 ¼ pffiffiffi ¼ 2:01 ¼ 0 ¼ W ¼ A0 r tan e þ r tan e0 4 þ 3 7 4þ3 7

226

Seismology

By substitution in (122.1) we obtain pffiffiffi

8 2 ’refl ¼ 1:07 exp i x3 þ x1  pffiffiffi t 8 2 pffiffiffi

p ffiffi ffi 2 8 i 7x3 þ x1  pffiffiffi t ’trans ¼ 2:01 exp 8 2  pffiffiffi pffiffiffi pffiffiffi 123. A P-wave of amplitude 5 2; 5 6; 10 2 and frequency v ¼ 12 rad s1 in a semi-infinite medium of speed of propagation a ¼ 6 km s1 and Poisson’s ratio 0.25 is incident on the free surface. Calculate: (a) The potential of the incident P-wave. (b) The potential of the reflected S-wave. (c) The components u1, u2, u3 of the reflected S-wave. (a) The displacements of the P-wave can be deduced from its scalar potential: ’ ¼ A exp ika ðn1 x1 þ n2 x2 þ n3 x3  at Þ ð123:1Þ uP ¼ r’ where A is the amplitude, ka is the wavenumber (P), ni are the direction cosines, and a is the P-wave velocity. The wavenumber is found from the given angular frequency and velocity: ka ¼

o 12 ¼ ¼ 2 km1 a 6

Since we know the amplitudes of the components of the displacements we can find the incidence angle i and the azimuth az: pffiffiffi @’ ¼ Aka n1 ¼ A2 sin i cos az ¼ 5 2 @x1 pffiffiffi @’ ¼ Aka n2 ¼ A2 sin i sin az ¼ 5 6 uP2 ¼ @x2 pffiffiffi @’ uP3 ¼ ¼ Aka n3 ¼ A2 cos i ¼ 10 2 @x3 pffiffiffi Dividing the two first equations we obtain 3 ¼ tan az ) az ¼ 60 , and dividing the last two, pffiffiffi 5 2 1 pffiffiffi ¼ tan i cos az ¼ tan i ) i ¼ 45 2 10 2 uP1 ¼

The amplitude A is given by

 P  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  2  2 j uP j u  ¼ ¼ 102 m2 uP1 þ uP2 þ uP3 ¼ Aka ) A ¼ ka

Finally the potential is given by

’inc ¼ 102 exp i2

pffiffiffi pffiffiffi pffiffiffi

2 6 2 x1 þ x3  6t m2 x2 þ 4 4 2

227

Reflection and refraction

If we express the potential referred to the plane (x1, x3) as the incidence plane, as we did in Problem 122, then ’ ¼ A exp ik ðx1 þ tan e x3  ct Þ where pffiffiffi 1 k ¼ ka cos e ¼ 2 pffiffiffi ¼ 2 2 pffiffiffi a 6 c¼ ¼ ¼ 6 2 km s1 1 cos e pffiffiffi 2

e ¼ 90  i ¼ 45 ;

and the potential is

pffiffiffi
pffiffiffi  ’inc ¼ 102 exp i 2 x1 þ x3  6 2t m2

ð123:2Þ

(b) Since Poisson’s ratio is 1/4, l ¼ m, and

pffiffiffi a b ¼ pffiffiffi ¼ 2 3 km s1 3

The angle f of the reflected S-wave is obtained by Snell’s law: rffiffiffi pffiffiffi cos e cos f b 1 5 2 ¼ ) cos f ¼ cos e ¼ pffiffiffi ¼ pffiffiffi ) sin f ¼ a b a 6 6 2 3

From the values of e and f we calculate the P-to-S reflection coefficient VPS, using equation VPS ¼

4að1 þ 3a2 Þ 4ab þ ð1 þ 3a2 Þ2

where we substitute a ¼ tan e ¼ 1

and

b ¼ tan f ¼

so

pffiffiffi 5

4ð1 þ 3Þ VPS ¼ pffiffiffi ¼ 0:64 4 5 þ ð 1 þ 3Þ 2

From this coefficient we calculate the proportion of the incident P-wave which is reflected as an S-wave (only with SV component; the negative sign indicates the opposite sense of the reflected ray): B ¼ AVPS ¼ 10  0:64 ¼ 6:4  103 m2 When the ray is contained in the (x1, x3) plane we use a scalar potential for the S-wave which in this case is given by c ¼ B exp ik ðx1  tan f x3  ct Þ pffiffiffi
pffiffiffi  pffiffiffi ¼ 6:4  103 exp i 2 x1  5x3  6 2t m2

228

Seismology

(c) To calculate the amplitudes of the total displacements in terms of the two scalar potentials, we remember that for this orientation of the axes the displacements are given by @’ @c u1 ¼  ¼ uP1 þ uSV 1 @x1 @x3 @’ @c u3 ¼ þ ¼ uP3 þ uSV 3 @x3 @x1 The displacements of the SV reflected wave in this case are pffiffiffi @c ¼ 6:4 5 uSV 1 ¼  @x3 pffiffiffi @c ¼ 6:4 2 uSV 3 ¼ @x1

If we want to determine the components 1 and 2, referred to the original system of axes, we SV  project uSV 1 ¼ uR using the azimuth 60 : pffiffiffi SV uSV 1 ¼ uR cos az ¼ 3:2 5 pffiffiffiffiffi SV uSV 2 ¼ u2 sin az ¼ 3:2 15

124. An S-wave of vector potential pffiffiffi pffiffiffi


 pffiffiffi x1 3 3 þ x2 þ x3  4t ci ¼ 10 3; 2; 4 exp 5i 4 4 2

is incident on the free surface x3 ¼ 0. Find the SV and SH components of the reflected S-wave referred to the same coordinate system as the incident wave, and the coefficient of reflection. Poisson’s ratio is 3/8. According to the value of Poisson’s ratio the relation between l and m is s¼

l 3 ¼ ) l ¼ 3m 2 ð l þ mÞ 8

and the relation between the velocities of the P-waves and S-waves is sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisffiffiffiffiffiffi l þ 2m 5m pffiffiffi ¼ ¼ 5b a¼ r r

The incidence angle i and the azimuth az are obtained from the direction cosines: pffiffiffi 3 n3 ¼ cos i ¼ ) i ¼ 30 2 1 n1 ¼ sin i cos az ¼ ¼ sin 30 cos az ) az ¼ 60 4 Using Snell’s law we find the value of the critical angle sin ic 1 b 1 ¼ ) sin ic ¼ ¼ pffiffiffi ) ic ¼ 26:5 b a a 5

229

Reflection and refraction

X3

r i u SV 60°

R

Fig. 124a

Since i > ic, there is no reflected P-wave. The components of the incident S-wave are obtained from the potential 8 < uS1 ¼ 0 S ui ¼ r  c ) uS2 ¼ 80 : S u3 ¼ 40

  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi The modulus of the displacement is uS  ¼ ð80Þ2 þð40Þ2 ¼ 40 5: The SV component is given by (Fig. 124a) uSV ¼

u3 ¼ 80 cosð90  i0 Þ

and the SH component is uSH ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðuS Þ2 ðuSV Þ2 ¼ 40

The minus sign corresponds to that of uS2 The amplitude of the reflected SH wave is equal to that of the incident SH wave. We find the 1 and 2 components using the azimuth 60 (Fig. 124b): pffiffiffi pffiffiffi pffiffiffi



3 1 3 3 1 uSH ¼ 40  ; ; 0 exp 5i x x x   4t þ 2 3 1 i 2 2 4 2 4 For total reflection the amplitude of the reflected SV is equal to that of the incident one, but with a phase shift d. The components are given by (Figs 124a and 124b) SV cos i cos az uSV 1 ¼ u SV uSV cos i sin az 2 ¼ u SV uSV sin i 3 ¼ u

230

Seismology

X1

R

u SH

30°

X2

60° SV

uH

Fig. 124b

The components of the reflected SV wave are pffiffiffi pffiffiffi 

 pffiffiffi

3 3 3 3 1 1 SV ; ; x2  x3  4t þ id exp 5i x1 þ ui ¼ 80  4 4 2 4 2 4 To determine the phase shift d we have to determine the reflection coefficients for a free surface. Textbooks usually give those for Poisson’s ratio s = 1/4, but since in this problem s ¼ 3/8 we have to calculate them. On a free surface the boundary conditions are that stresses are null, which in terms of the scalar potentials ’ and c, are given by   t31 ¼ 0 ¼ m u3;1 þ u1;3 ¼ 2’;31  c;11 þ c;33   ð124:1Þ t33 ¼ 0 ¼ l u1;1 þ u3;3 þ 2mu3;3 ¼ 3’;11  5’;33  2c;13 where, using (x1, x3) as the plane of incidence, the scalar potential ’ of the reflected P-waves is ’ ¼ A exp ik ðax3 þ x1  ct Þ and the scalar potential of the incident and reflected S-wave is c ¼ B0 exp ik ðbx3 þ x1  ct Þ þ B exp ik ðbx3 þ x1  ct Þ Substituting in (124.1) we obtain for the coefficient of the reflected S-waves, VSS ¼

B i4^ab  ð1  b2 Þð3 þ 5^a2 Þ ¼ B0 i4ab þ ð1  b2 Þð3 þ 5a2 Þ

The phase shift is given by d ¼ tan

1



4^ab ð1  b2 Þð3  5^a2 Þ



231

Reflection and refraction

By substitution of the values of the problem, sffiffiffiffiffiffiffiffiffiffiffiffiffi c2 1 b ¼ tan f ¼  1 ¼ pffiffiffi 2 b 3 rffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi c2 22 ^ a¼ 1 2 ¼ a 30 and finally we obtain d ¼ 77.33 . 125. An S-wave represented by the potential pffiffiffi

pffiffiffi 3 1 x3  4 2 t c ¼ 10 exp i3 x1 þ 2 2

is incident from an elastic medium with l ¼ 0 onto a liquid with velocity a0 ¼ 4 km s1 (the two media have the same density). Derive the equations relating the amplitudes of the potentials of the incident, reflected, and transmitted waves. Given that the wave is incident from an elastic medium onto a liquid medium, there are reflected S- and P-waves in the elastic medium and transmitted P-waves in the liquid (Fig. 125). If l ¼ 0, the P-wave velocity in the elastic medium is sffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffi pffiffiffi pffiffiffipffiffiffi l þ 2m 2m ¼ ¼ b 2 ¼ 4 2 2 ¼ 8 km s1 a¼ r r Assuming (x1, x3) is the incidence plane, we use the scalar S-wave potential which, for the incident and reflected waves in the solid medium, is given by c ¼B0 exp ikb ðx1 cos f þ x3 sin f  bt Þ þ B exp ikb ðx1 cos f  x3 sin f  bt Þ X3 P

e⬘ a⬘= 4 b⬘= 0

M⬘ f

M

f

X1 a=8 b = 4√2

e

P

S

Fig. 125

S

232

Seismology

From the potential of the incident wave given in the problem, cos f ¼

1 ) f ¼ 60 ; 2

B0 ¼ 10;

kb ¼ 3

The potential can also be written in the form (Problem 122) c ¼ B0 exp ik ðx1 þ tan f x3  ct Þ þ B exp ik ðx1  tan f x3  ct Þ where 3 km1 2 pffiffiffi b ¼ 8 2 km s1 c¼ cos f

k ¼ kb cos f ¼

and the potential is given by pffiffiffi  pffiffiffi  pffiffiffi pffiffiffi 3
3
c ¼ 10 exp i x1 þ x3 3  8 2t þ B exp i x1  x3 3  8 2t 2 2

Applying Snell’s law we determine the angle e of the reflected P-wave in the solid medium and the angle e0 of the transmitted P-wave onto the liquid (Fig. 125): cos f cos e cos 60 cos e ¼ ) pffiffiffi ¼ ) e ¼ 45 b a 8 4 2 cos e0 cos f 1 ) cos e0 ¼ pffiffiffi ) e0 ¼ 69 ¼ 0 a b 2 2 rffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 7 1 sin e0 ¼ 1  ¼ pffiffiffi ) tan e0 ¼ 7 8 2 2

The potential of the reflected P-wave in the solid medium (M) is pffiffiffi  3
’ ¼ A exp ik ðx1  x3 tan e  ct Þ ¼ A exp i x1  x3  8 2t 2

and that of the transmitted P-wave in the liquid (M0 ) is ’0 ¼ A0 exp ik ðx1 þ x3 tan e0  ct Þ ¼ A0 exp i

pffiffiffi  pffiffiffi 3
x 1 þ x 3 7  8 2t 2

The relation between the amplitude of the potential of the incident S-wave (B0 = 10) and those of the reflected and refracted P-waves A, A0 and the reflected S-wave B can be obtained from the conditions at the boundary between the two media (x3 ¼ 0), that is, continuity of the normal component of the displacements (u3) and of the stress (t33) and null tangential stress (t31): u3 ¼ u03 ) ’;3 þ c;1 ¼ ’0;3 t31 ¼ 0 ) 2’;13  c;33 þ c;11 ¼ 0
   t33 ¼ t033 ) l0 ’0;33 þ ’0;11 ¼ 2m ’;33 þ c;13

233

Reflection and refraction By substitution of the potentials we obtain the equations (in units of 103 m2) pffiffiffi  A þ 10 þ B ¼ A0 7 A þ B þ 10 ¼ 0 pffiffiffi pffiffiffi 2A0 ¼ A þ 10 3  3B

Solving the system of equations we obtain pffiffiffi 40 3 0 pffiffiffi pffiffiffiffiffi ¼ 6:2 A ¼ 4 þ 7 þ 21 pffiffiffiffiffi 20 21 pffiffiffiffiffi ¼ 8:2 p ffiffi ffi A¼ 4 þ 7 þ 21 pffiffiffi pffiffiffiffiffi  10 4  7 þ 21 pffiffiffi pffiffiffiffiffi B¼ ¼ 1:8 4 þ 7 þ 21

126. An S-wave incident on the free surface of a semi-infinite medium with s ¼ 0.25 is given by (in units of 103 m2) pffiffiffi pffiffiffi


  1 pffiffiffi 3 3 ci ¼ 10 3; 2; 4 exp 5i x1 þ x2 þ x3  4t 4 4 2 Calculate:

(a) The amplitude of the components of the reflected P-wave. (b) The components of the reflected S-wave. (a) From the direction cosines we calculate the incidence i and emergence f angles and azimuth az of the incident S-wave: 1 ¼ sin i cos az 4 pffiffiffi 3 ¼ sin i sin az n2 ¼ 4 pffiffiffi 3 ¼ cos i ) i ¼ 30 ) f ¼ 60 n3 ¼ 2 1 1 ¼ cos az ) az ¼ 60 4 2 pffiffiffi pffiffiffi Since Poisson’s ratio is 0.25 then l ¼ m ) a ¼ 3b ¼ 4 3, and n1 ¼

sin ic 1 1 ¼ ) sin ic ¼ pffiffiffi ) ic ¼ 35 b a 3

Since i < ic we have a reflected P-wave. The reflection coefficient at a free surface for a reflected P-wave from an incident S-wave is given by VSP ¼

4bð1 þ 3a2 Þ 4ab þ ð1 þ 3a2 Þ2

ð126:1Þ

234

Seismology

X1 f

f

e

P S

S

X3

Fig. 126

where a ¼ tan e and b ¼ tan f, and f is the emergence angles of the incident S-wave and e is that of the reflected P-wave (Fig. 126). The relation between f and e according to Snell’s law is pffiffiffi 1 cos f cos e a ¼ ) cos e ¼ cos f ¼ 3 ) e ¼ 30 b a b 2

Substituting in Equation (126.1):

VSP ¼

pffiffiffi 4 3ð 1 þ 1Þ

4 þ ð 1 þ 1Þ

2

¼

pffiffiffi 3

We can write the potential of the incident S-wave referred to the incidence plane (x1, x3) by means of the rotation matrix 0 1 cos az sin az 0 @  sin az cos az 0 A 1 0 0

and substituting 0

1 B 2 B pffiffiffi B @ 3 2 0

pffiffiffi 1 3 8 pffiffiffi 0 C0 10pffiffi3ffi 1 0 B1 1 > < u1 ¼ 20 3 uP ¼ r’ ) uP2 ¼ 60 > > : P u3 ¼ 40

(b) For the reflected S-wave we have to separate the SV and SH components. The SV component can be deduced from the scalar potential pffiffiffi

1 3 r cSV ¼ B exp ikb ðcos f x1  sin f x3  4tÞ ¼ B exp 5i x1  x3  4t 2 2

We obtain B by means of the reflection Vss: VSS ¼

4ab  ð1 þ 3a2 Þ 4ab þ ð1 þ

2

3a2 Þ2

¼0¼

The reflected S-wave doesn’t have an SV component.

B )B¼0 B0

236

Seismology

For the reflected SH component we use the displacement of the u2 instead of the potential. The displacements of the SH component of the incident wave are obtained from (126.2): uiSH ¼ c01;3  c03;1 ¼ 40 Referred to the reference of the plane of incidence, the displacement is given by pffiffiffi

3 1 i x3  4t uSH ¼ 40 exp 5i x1 þ 2 2 The amplitude of the reflected SH wave is equal to that of the incident SH wave. Referred to the incidence plane system of reference, pffiffiffi

3 1 r uSH ¼ 40 exp 5i x1  x3  4t 2 2 The displacement of the reflected S-wave referred to the original system of axes is pffiffiffi pffiffiffi

1 3 3 uri ¼ ðBr1 ; Br2 ; Br3 Þ exp i5 x1 þ x2  x3  4t 4 4 2 Since the SV component is zero, Br3 ¼ 0; Br1 and Br2 are found using the equations jur j ¼ urSH ) ðBr1 Þ2 þ ðBr2 Þ2 ¼ 1600 pffiffiffi 1 r 3 r r B ¼0 ui n i ¼ 0 ) B 1 þ 4 4 2 resulting in pffiffiffi Br1 ¼ 20 3

Br2 ¼ 20

127. A wave represented by the potential

x1 x2 x3 w ¼ 4 exp 0:25i pffiffiffi þ pffiffiffi þ pffiffiffi  4t 6 3 2

is incident on the surface x3 ¼ 0 of separation between two liquids. If the speed of propagation in the second medium is 2 km s1, the pressure exerted by the incident wave on the surface of separation is 5  109 Pa, and the transmitted energy is four times greater than the reflected energy, calculate: (a) The energy transmitted to the second medium. (b) The potentials of the transmitted and reflected waves referred to the same coordinate system as the incident potential. (a) The intensity or energy per unit surface area of the wavefront of an incident P-wave is given in units of J m2 by Iinc ¼ A20 o2 ka2 ar

ð127:1Þ

237

Reflection and refraction

From the given potential we have A0 ¼ 4  103 m2 ka ¼ 0:25 km1 a ¼ 4 km s1 o ka ¼ ) o ¼ 1 s1 a We need to know the value of the density r. Since the medium is liquid l ¼ K (bulk modulus) and then l ¼ P/y, where P is the pressure and y the cubic dilatation (change of volume per unit volume). For liquids the shear modulus m is zero and from the velocity of P-waves we obtain, sffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffi l þ 2m l ¼ ) l ¼ a2 r a¼ r r The cubic dilatation is obtained from the potential ’: y ¼ r2 ’ ¼ ka2 A ¼

1 1 4¼ 16 4

Then, we obtain l ¼ a2 r ¼

P 5  109 Pa 5 ¼ ) r ¼ g cm3 2 2 1 y m s 4  16  106 4

By substitution in (127.1) the incident energy is Iinc ¼ 16  1 

1 5  4  ¼ 5 J m2 16 4

The energy transmitted into the second medium is Itras ¼ A02 o2 ka02 a0 r0 ¼ W 2 A20 ka02 a0 r0

ð127:2Þ

Irefl ¼ Ao2 ka02 a0 r0 ¼ V 2 A20 ka02 a0 r0

ð127:3Þ

and the energy reflected is

where W and V are the transmission and reflection coefficients, respectively: A0 2r tan e ¼ A0 r0 tan e þ r tan e0 A r0 tan e  r tan e0 ¼ 0 V ¼ A0 r tan e þ r tan e0

W ¼

The emergence angle e of the incident wave is 1 n3 ¼ pffiffiffi ¼ cos i ¼ sin e ) i ¼ e ¼ 45 2

ð127:4Þ

238

Seismology and from Snell’s law the emergence angle of the transmitted wave e0 is rffiffiffiffiffiffiffiffiffiffiffi pffiffiffi cos e cos e0 a0 2 1 1 1 7 0 0 ¼ 0 ) cos e ¼ cos e ¼ pffiffiffi ¼ pffiffiffi ) sin e ¼ 1  ¼ pffiffiffi a a a 4 2 2 2 8 2 2 Given that the transmitted energy is four times the reflected energy, Itras Iref

W 2 A20 r0 o4 V 2 2r0 a 0 ¼ 4 ¼ 2 a2 4 ) 2 ¼ W 5ra0 V A0 ro a

If we substitute in (127.4) 5 pffiffiffi 7 4 V ¼ 5 pffiffiffi 7 r0 þ 4 5 2 4 W ¼ pffiffiffi 5 7 r0 þ 4 r0 

We have three equations for r’, V, and W. The solution for positive values of the variables is  W ¼ 0:23 r0 ¼ 7:7 ) V ¼ 0:40 (b) The potential of the reflected P-wave is ’ref ¼ VA0 exp ik a ðn1 x1 þ n2 x2  n3 x3  at Þ

1 1 1 1 pffiffiffi x1 þ pffiffiffi x2  pffiffiffi x3  4t ¼ 1:6 exp i 4 6 3 2

To determine the potential of the transmitted wave we have to calculate the direction cosines of the transmitted ray. The azimuth is the same as that of the incident wave which can be deduced from the direction cosines and the value of i: 1 1 n1 ¼ sin i cos az ¼ pffiffiffi ) cos az ¼ pffiffiffi 6 3 pffiffiffi 1 2 n2 ¼ sin i sin az ¼ pffiffiffi ) sin az ¼ pffiffiffi 3 3

The direction cosines of the transmitted ray in the medium M0 are 1 1 1 n01 ¼ sin i0 cos az ¼ pffiffiffi pffiffiffi ¼ pffiffiffi 2 2 3 2 6 1 n02 ¼ sin i0 sin az ¼ pffiffiffi 2 3 pffiffiffi 7 n03 ¼ cos i0 ¼ pffiffiffi 2 2

239

Reflection and refraction

and the potential of the transmitted wave is pffiffiffi

1 1 1 7 pffiffiffi x1 þ pffiffiffi x2 þ pffiffiffi x3  2t ’tras ¼ 0:92 exp i 2 2 6 2 3 2 2

128. Two liquid media are separated at x3 ¼ 0, the first of volumetric coefficient 1 K ¼  109 Pa and density 1 g cm3. The amplitudes of the components of an pffiffiffi 2  incident wave of frequency 3 Hz pffiffiare ffi
ui ¼ 18p 1;1; 6 mm and those of the wave 63 2p pffiffiffiffipffiffiffi pffiffiffi transmitted to medium 2 are 2; 2; 3 mm. Given that the amplitude of 7 the transmitted potential is twice that of the reflected potential, find expressions for the incident, reflected, and transmitted potentials. In liquids only P-waves are propagated and their displacements can be deduced from the scalar potential ’ ¼ A0 exp ika ðn1 x1 þ n2 x2 þ n3 x3  at Þ ) uP ¼ r’ Then in our case the components of the displacement in mm are uP1 ¼

@’ ¼ A0 ka n1 ¼ A0 ka sin i cos az ¼ 18p @x1

uP2 ¼

@’ ¼ A0 ka n2 ¼ A0 ka sin i sin az ¼ 18p @x2

uP3 ¼

pffiffiffi @’ ¼ A0 ka n3 ¼ A0 ka cos i ¼ 18p 6 @x3

and we find az ¼ 45 ; i ¼ 30 ; e ¼ 60 , A0 ¼ 6  103 m2. The emergence angle of the refracted wave, e0 , can be found from its displacements, pffiffiffi pffiffiffi pffiffiffi pffiffiffi

pffiffiffi 2 2 3 3 63p 2 2 0 0 P pffiffiffi ; pffiffiffi ; pffiffiffi ) n3 ¼ sin e ¼ pffiffiffi ) cos e0 ¼ pffiffiffi utras ¼ pffiffiffi 7 7 7 7 7 7 The P-wave velocity in the medium of the incident wave is sffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffi l þ 2m K 1 a¼ ¼ ¼ pffiffiffi km s1 r r 2

Using Snell’s law we find the velocity of the medium of the refracted wave, pffiffiffi cos e a 2 2 0 ¼ ) a ¼ pffiffiffi cos e0 a0 7

From the values of the velocities in the two media we calculate their densities: 9 1 0 > > r 2 0 > a 7 Kr > 2 > = ¼ ¼ ¼ 02 0 0 3 3 a 16 K r K ) r0 ¼ r ¼ g cm3 p ffiffi ffi > 2 2 qffiffiffiffi 2 2 > 8 0> > 0 K0 0 > a ¼ r0 ¼ pffiffiffi ) K ¼ r ; 7 7

240

Seismology

The reflection V and transmission W coefficients are found using their expressions and from them we get the relation between the amplitude A0 of the incident wave potential and those of the reflected A and refracted A0 waves, and substituting the value for A0 ¼ 6, we obtain A0 ¼ 6 ) A ¼ 3  103 m2 From these values we can write the potentials of the incident, reflected, and transmitted waves: pffiffiffi

pffiffiffi 1 3 1 1 x3  pffiffiffi t ’inc ¼ 6 exp i6p 2 pffiffiffi x1 þ pffiffiffi x2 þ 2 2 2 2 2 2 pffiffiffi

pffiffiffi 1 3 1 1 x3  pffiffiffi t ’ref ¼ 3 exp i6p 2 pffiffiffi x1 þ pffiffiffi x2  2 2 2 2 2 2 pffiffiffi pffiffiffi

pffiffiffi pffiffiffi pffiffiffi 2 2 2 3 7 ’tras ¼ 6 exp i3p pffiffiffi pffiffiffi x1 þ pffiffiffi x2 þ pffiffiffi x3  2 pffiffiffi t 7 7 7 7 2

129. Two liquids in contact have speeds of propagation of 4 and 6 km s1. The density of the first is 2 g cm3 and is less than that of the second. For waves of normal incidence, the reflected and transmitted energies are equal. A wave of v ¼ 1 s1 and with a potential of amplitude A0 ¼ 2103 cm2 is incident from the first onto the second at an angle of 30 . Calculate:

(a) The transmitted and reflected energies. (b) An expression for the transmitted potential. (a) For normal incidence, the reflection and transmission coefficients in terms of the refractive index m ¼ a/a0 and the density contrast m ¼ r0 /r are given by mn mþn 2 Wn ¼ mþn Vn ¼

If the reflected energy is equal to the transmitted energy, then Vn2 ¼ mnWn2 )

ðm  nÞ2 ðm þ nÞ2

¼

4mn ðm þ nÞ2

Substituting n ¼ a/a0 ¼ 2/3, from (129.1) we obtain the value of m:  4 m ¼ 3:9 4 2 m  4m þ ¼ 0 ) m ¼ 0:1 9 Trying both values we obtain for r0 r0 ¼ 0:1 ) r0 ¼ 0:2 g cm3 r m ¼ 4 ) r0 ¼ 8 g cm3 m¼

ð129:1Þ

241

Reflection and refraction

i = 30°

r = 2 g cm–3

M e

a = 4 km s–1

e⬘

a ⬘ = 6 km s–1 M⬘

Fig. 129

But the problem states that r ¼ 4 < r0 , so r0 ¼ 8 g cm3. For a wave with incidence angle 30 (e ¼ 60 ) the emergence angle of the transmitted wave e0 is, according to Snell’s law (Fig. 129), given by rffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi cos e cos e0 3 9 7 0 0 ¼ 0 ) cos e ¼ ) sin e ¼ 1  ¼ a 4 16 a 4 The partition of energy between the reflected and refracted waves is given by sin e0 2 W þ V2 ¼ 1 sin e The reflection V and transmission W coefficients are pffiffiffi pffiffiffi 3 2 7 m sin e  n sin e0 4 2  3 4 pffiffiffi ¼ 0:8 ¼ pffiffiffi V ¼ m sin e þ n sin e0 3 2 7 4 þ 2 pffiffi3ffi 4 3 2 sin e pffiffiffi ¼ 0:46 W ¼ ¼ pffiffiffi 0 m sin e þ n sin e 3 2 7 4 þ 2 3 4 mn

The incident, reflected, and transmitted energies per unit time and surface area are (Problem 127) ro4 2 A sin e ¼ 17:3 erg cm2 s a 0 ro4 2 ro4 2 2 Eref ¼ A sin e ¼ A V sin e ¼ 11:1 erg cm2 s a a 0 r0 o4 r0 o4 Etrans ¼ 0 A02 sin e0 ¼ 0 A20 W 2 sin e0 ¼ 7:5 erg cm2 s a a Einc ¼

242

Seismology

(b) The potential of the transmitted wave is ’tras ¼ A0 exp ik ðx3 tan e0 þ x1  ct Þ where A0 ¼ A0 W ¼ 2  103  0:46 ¼ 920 cm2 o 1 k ¼ ka0 0 cos e0 ¼ 0 cos e0 ¼ km1 a 8 a0 c¼ ¼ 8 km s1 cos e0 pffiffiffi

7 1 x3 þ x1  8t ’tras ¼ 920 exp i 8 3 130. An SV wave is incident on the free surface of an elastic medium of Poisson ratio 0.25. If the potential of the wave is (in units of 103m2) pffiffiffi



3 5 5 1 1 c i ¼ pffiffiffi ;  pffiffiffi ; 0 exp i pffiffiffi x1 þ pffiffiffi x2 þ x3  4t 2 2 2 2 2 2 2

find the components of the amplitude of the reflected P-wave referred to this set of axes. From the direction cosines we find the incidence angle i, the emergence angle f, and the azimuth az of the incident SV wave (Fig. 130): pffiffiffi 3 n3 ¼ cos i ¼ sin f ¼ ) i ¼ 30 and f ¼ 60 2 1 1 n1 ¼ sin i cos az ¼ cos az ¼ pffiffiffi ) az ¼ 45 2 2 2

Bearing in mind that Poisson’s ratio is 0.25, from Snell’s law we find the emergence angle e of the reflected P-wave: pffiffiffi pffiffiffi s ¼ 0:25 ) l ¼ m ) a ¼ 3b ¼ 4 3 km s1 pffiffiffi cos f cos e 3 ¼ ) cos e ¼ ) e ¼ 30 b a 2 X3

e

f f

P

i

SV

Fig. 130

S

243

Ray theory. Constant and variable velocity

The reflection coefficient for the reflected P-wave gives us the relation between the amplitude of the potential, B0, of the incident SV wave and A, that of the reflected P wave: VSP ¼

pffiffiffi A 4 tan f ð1 þ 3 tan2 eÞ ¼ ¼ 3 2 B0 4 tan e tan f þ ð1 þ 3 tan2 eÞ

To find B0 we write the potential of the incident SV wave referred to the (x1, x3) plane of incidence using the rotation matrix 0 5 1 0 1 0 1 pffiffiffi 0 cos az sin az 0 B 2 C B C C C B C B @  sin az cos az 0 AB B  p5ffiffiffi C ¼ @ 5 A @ 2A 0 0 0 1 0 pffiffiffi B0 ¼ 5  103 m2 ) A ¼ 5 3  103 m2 Referred to this system of axes the potential of the reflected P-wave is given by ’ ¼ A exp ik ðx3 tan e þ x1  a t Þ kb 1 ¼ 1 ¼ 2 km1 cos f 2

pffiffiffi pffiffiffi 1 ’ ¼ 5 3 exp i2 x3 pffiffiffi þ x1  4 3t 3 k¼

The amplitudes of the displacements of the reflected P-wave referred to this set of axes are pffiffiffi @’ ¼ 10 3 mm @x1 @’ u3 ¼ ¼ 10 mm @x3

u1 ¼

Referred to the original set of axes the horizontal components are pffiffiffi 10 3 0 p ffiffiffi mm u1 ¼ u1 cos az ¼ 2 pffiffiffi 10 3 u20 ¼ u1 sin az ¼ pffiffiffi mm 2

Ray theory. Constant and variable velocity 131. Assume that the Earth’s crust consists of a single layer of thickness H and a constant speed of propagation of seismic waves of v1 on top of a mantle of velocity of propagation 20% greater than the crust. Given that a focus on the surface produces a reflected wave that takes 17.2 s to reach a distance of 99 km, and that this is the

244

Seismology

xc F

S

H ic

ic

v1

v2 Fig. 131a

critical distance, calculate the values of H, v1, and v2. Plot the travel-time curve (t, x) for this specific case with numerical values. The critical distance xc is the distance at which a ray that is reflected with the critical angle at the top of the mantle arrives at the surface and is given by the equation (Fig. 131a) xc ¼ 2H tan ic ¼ 99 km

ð131:1Þ

where H is the thickness of the crust. Since we know the relation between the velocities in the crust and the mantle, we can calculate the critical angle v2 ¼ 1:2v1 )

sin ic 1 1 ) ic ¼ 56:44 ¼ ) sin ic ¼ v1 v2 1:2

If we substitute in Equation (131.1) we obtain the thickness of the crust, H: 99 ¼ 2H tan 56:44 ) H ¼ 32:8 km The travel time of the critically reflected ray is H H ¼2 ¼ 17:2 v1 cos ic v1 cos 56:44 32:84 ) v1 ¼ 2 ¼ 6:9 km s1 ) v2 ¼ 8:3 km s1 17:2 cos 56:44 t¼2

To draw the travel-time curve for different distances of the direct, reflected, and critically refracted waves we use the equations x t1 ¼ v1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 x2 þ H2 t2 ¼ v1 4 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2H v22  v21 t3 ¼ þ v2 v1 v2

245

Ray theory. Constant and variable velocity

We obtain the following values x(km)

t1 (s)

t2 (s)

t3 (s)

0 30 60 90 99 120 150

0 4.3 8.7 13.0 14.3 17.4 21.7

9.5 10.4 12.9 16.1 17.2 19.8 23.7

– – – – 17.2 19.7 23.4

The travel-time curves are drawn in Fig. 131b.

40 3

t (S)

30

20 1

2

10

ti

0 0

Fig. 131b

50

xc

150 x (km)

200

250

300

246

Seismology

x P

h

F

Fig. 132

132. In a seismogram recorded at a regional distance, the S-P time lag is 5.5 s, and the focus is at a depth x/2, where x is the epicentral distance. The pffiffiffi model1Earth has a single layer of Poisson ratio 0.25 and constant S-wave velocity 3 km s . Calculate:

(a) The depth of the focus. (b) The epicentral distance.

(a) For a direct wave from point F to point P (Fig. 132) the difference of the arrival times of the P- and S-waves (the S-P interval) is t S-P ¼ 5:5 ¼

FP FP  b a

The distance FP can be expressed in terms of x as (Fig. 132) rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi
x 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 FP ¼ x2 þ h2 ¼ x2 þ ¼x 2 2 The S-P interval is given by

pffiffiffi 5ab 5:5 ¼ x 2 ab Since Poisson’s ratio is 0.25 and knowing the S-wave velocity we obtain pffiffiffi pffiffiffi pffiffiffi 5 31 pffiffiffipffiffiffi s ¼ 0:25 ) a ¼ b 3 ) 5:5 ¼ x 2 3 3 x ¼ 21 km x h ¼ ¼ 10:5 km 2 133. The Earth consists of a layer of thickness 20 km and seismic wave velocity 6 km s1 on top of a medium of speed of propagation 8 km s1. A seismic focus is

247

Ray theory. Constant and variable velocity

S

x h H

F ic

ic

ic i

v1 v2

Fig. 133

located at a depth of 10 km. Calculate the difference in travel times between the reflected and the critical refracted waves observed on the surface at a distance of 150 km from the epicentre. This problem is similar to Problem 131, but now the focus is at depth h ¼10 km. The critical distance in this case is given by (Fig. 133) xc ¼ ð2H  hÞ tan ic

v1 1 6 sin ic ¼ ) ic ¼ sin ¼ 48:6 v2 8 ) xc ¼ ð2  20  10Þ tanð48:6Þ ¼ 34:0 km Since the distance 150 km is greater than the critical distance there arrive critically refracted rays. The travel times of the reflected (t2), and critically refracted (t3) rays at that distance are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ ð2H  hÞ2 1502 þ ð2  20  10Þ2 ¼ 25:5 s t2 ¼ ¼ v1 6 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi x ð2H  hÞ v22  v21 150 ð2  20  10Þ 82  62 t3 ¼ þ þ ¼ ¼ 22:1 s v2 8 v1 v2 86 The time difference between the travel times of the two rays is t3  t2 ¼ 22:06  24:49 ¼ 3:4 s 134. Consider a crust of thickness H and constant speed of propagation v1 on a mantle of constant speed of propagation v2. A seismic focus is located at depth H/2, the critical distance is 51.09 km, the delay time is 4.96 s, and the critical angle is 48.59 . Calculate the values of H, v1 and v2, and the depth of the focus. For a focus at depth h ¼ H/2, the travel times of the critically refracted (t3) rays and the critical distance are given by the expressions (Fig. 133).

248

Seismology

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi x ð2H  hÞ v22  v21 t3 ¼ þ v2 v1 v2 xc ¼ ð2H  hÞ tan ic ) 51:09 ¼



H 2H  tanð48:59Þ 2

) H ¼ 30 km; h ¼ 15km

Knowing the depth and thickness of the crust, using Snell’s law, the value of the critical angle, and the delay time ti, we find the velocities v1 and v2: sin ic 1 ¼ ) v1 ¼ 0:75v2 v1 v2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð 2  30  15 Þ v22  ð0:75v2 Þ2 ð2H  hÞ v2  v1 ti ¼ ) 4:96 ¼ v1 v2 0:75v22 ) v2 ¼ 8 km s1

v1 ¼ 6 km s1 135. In a seismogram, the S-P time difference is equal to 5.31 s, and corresponds to a regional earthquake that occurred at a depth h ¼ 2H, where H is the thickness of the crust. Given that the crust is formed by a layer of constant P-wave velocity of 3 km s1, that below it there is a semi-infinite mantle of double that speed of propagation, and that Poisson’s ratio is 0.25, determine: (a) An expression for the travel-time of the P- and S-waves. (b) The epicentral distance for an emerging P-wave with a take-off angle of 30 at the focus. (a) The travel time corresponding to the ray given in Fig. 135 is given by t¼

FA AS þ 2v v

F⬘

S

x

H

i0

i0

v

2H A ih

F

Fig. 135

S⬘

2v

249

Ray theory. Constant and variable velocity

where v ¼ a for P-waves and v ¼ b for S-waves. Using Snell’s law we find the relation between the incidence angle at the focus ih and at the station i0: sin ih sin i0 1 ¼ ) sin i0 ¼ sin ih 2v v 2 From Fig. 135 we obtain H H ) FA ¼ cos ih FA H H cos i0 ¼ ) AS ¼ cos i0 AS

cos ih ¼

From these equations we deduce the expression for the travel time: t¼

H H þ 2v cos ih v cos i0

ð135:1Þ

For the epicentral distance we obtain x ¼ F0 A þ AS0 ¼ H tan ih þ H tan i0

ð135:2Þ

Using Equations (135.1) and (135.2) and putting v ¼ a we obtain the travel time and epicentral distance for P-waves and putting v ¼ b for S-waves. (b) For a P-wave with take-off angle at the focus (ih) of 30 , we first find the value of the angle at the station i0, sin ih sin i0 ¼ ) i0 ¼ 14:47 2a a Substituting in (135.1) we obtain tP ¼

H H H pffiffiffi þ ¼ 1:61 a0:97 a 3 2a 2

The travel time of the S-wave with take-off angle jh ¼30 can also be calculated using (135.1). Since Poisson’s ratio is 0.25, the velocity of the S-wave is pffiffiffi pffiffiffi a s ¼ 0:25 ) a ¼ 3b ) b ¼ pffiffiffi ¼ 3 km s1 3

The incidence angle at the station, j0, using Snell’s law, is given by sin jh sin j0 ¼ ) j0 ¼ 14:47 2b b and the travel time is tS ¼

H H H þ ¼ 2:79 2b cos jh b cos j0 a

Since we know the S-P time interval we can obtain the value of h: t S-P ¼ 5:31 ¼ 2:79

H H  1:61 ) h ¼ 13:61 km 3 3

250

Seismology

The epicentral distance is found using (135.2): x ¼ 13:61 tan 30 þ 13:61 tan 14:47 ¼ 11:41 km 136. Consider a crust composed of two layers of thickness 12 and 18 km, and constant P-wave speeds of propagation of 7 and 6 km s1, respectively, on top of a semi-infinite mantle of constant speed of propagation 8 km s1. There is a seismic focus at a depth of 6 km below the surface. For a station located at 100 km epicentral distance, calculate the travel time of the direct, reflected, and critical refracted waves (neglecting waves with more than a single reflection or critical refraction). The travel times of the direct ray t1 and the ray reflected on the bottom of the first layer t2 are given by (Fig.136) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h2 þ x 2 62 þ 1002 ¼ 14:3 s ¼ t1 ¼ v1 7 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2H1  hÞ2 þx2 ð2  12  6Þ2 þ1002 t2 ¼ ¼ ¼ 14:5 s v1 7 As the velocity of the second layer is less than that of the first layer there is no critical refraction at that boundary. There is critical refraction at the boundary between the second layer and the mantle where the velocity is greater. Using Snell’s law, we can calculate the

E

S

x

h F∗ 2 (H1 – h)

H1

n1 i1

i1

F⬘ H2

n2

ic

ic

n3

Fig. 136

251

Ray theory. Constant and variable velocity

critical angle ic and from this value the incidence angle at the focus i1 for the critically refracted ray: sin i1 sin ic 1 6 ¼ ¼ ) sin ic ¼ ) ic ¼ 48:6 v1 v2 v3 8 7 sin i1 ¼ ) i1 ¼ 61:0 8 The travel time t3 of the critically refracted ray at the bottom of the second layer is given by FA AB BC CD DS þ þ þ þ v1 v2 v3 v2 v1 If the epicentral distance x is 100 km, the different segments of (136.1) are t3 ¼

ð136:1Þ

H1  h H1 ¼ ) FA ¼ 12:4 km; DS ¼ 24:8 km FA DS H2 cos ic ¼ ) AB ¼ CD ¼ 27:2 km AB BC ¼ x  2H2 tan ic  ðH1  hÞ tan i1  H1 tan i1 ¼ 26:8 km

cos i1 ¼

Finally, by substitution in (136.1) we obtain, t3 ¼ 17:7 s 137. Consider a two-layered structure of thickness H and speed of propagation v and 3v on top of a half-space medium of speed of propagation 2v. At a depth 3H below the surface there is a seismic focus. Write the expressions (as functions of H, v, and ih) for the travel times of waves that reach the surface without being reflected. Give the range of values of ih. In this problem the focus is located at the half-space medium at depth h¼3H under its boundary. Applying Snell’s law we can find the relation between the velocities, the incidence angles at the focus and at the bottom of each layer, and the critical angle at the boundary between the second layer and the half-space (Fig. 137): sin ih sin i2 sin i1 ¼ ¼ 2v 3v v sin ic 1 ¼ ) ic ¼ 41:8 2v 3v

ð137:1Þ

The rays which leave the focus and arrive at the surface at a distance x are only those with angles less than the critical angle (Fig. 137). The travel time for these rays is t¼

FA AB BS H H H þ þ ¼ þ þ 2v 3v v 2v cos ih 3v cos i2 v cos i1

ð137:2Þ

According to Equation (137.1) we have the relation between the incidence angles: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 1 4  9 sin2 ih sin i2 ¼ sin ih ) cos i2 ¼ 1  sin2 ih ¼ 2 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 sin i1 ¼ sin ih ) cos i1 ¼ 1  sin2 ih ¼ 4  sin2 ih 2 2

252

Seismology

E

S

x

i1 H

i0

v

B

H i2

3v

A ic ih

H

2v

F

Fig. 137

Substituting in (137.2) we write the travel time as function of the take-off angle ih: ! H 1 2 2 t¼ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v 2 cos ih 3 4  9 sin2 ih 4  sin2 ih We find a similar expression for the epicentral distance x: x ¼ H tan ih þ H tan i2 þ H tan i1 ! sin ih 3 sin ih x ¼ H tan ih þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4  sin2 ih 4  9 sin2 ih

The range of values of the take-off angle for rays which arrive at the surface is 0 < ih a.

253

Ray theory. Constant and variable velocity

E

S x

i

P

i

a

A

i

F

a

a 3v

v

Fig. 138a S

E x

ic

B

a A

ic ic

F

a v

3v

Fig. 138b

(a) For 0 < x a, we have the rays refracted at the boundary between the two media when the incidence angle is less than the critical angle (Fig. 138d), that is, i 0 and leaving the focus with take-off angles 0 < ih < 45 , the travel-times are given by FA AS a a þ ¼ þ 2v v 2v cos ih v cos r a cos ih ¼ FA a cos r ¼ AS t¼

ν a 2a

2v

Focus

Fig. 139

a

ð139:1Þ

257

Ray theory. Constant and variable velocity

X E

A⬘

S

r 2a

A

F⬘

v

2v a ih F a

Fig. 139a

Applying Snell’s law, sin ih sin r 1 ¼ ) sin r ¼ sin ih v 2 2v

ð139:2Þ

Substituting in (139.1): a 1 2 t¼ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v 2 cos ih 4  sin2 ih

!

ð139:3Þ

The relation between the epicentral distance x and the incidence angle ih is x ¼ F0 A þ A0 S F0 A ¼ a tan ih A0 S ¼ a tan r sin ih x ¼ aðtan i þ tan rÞ ¼ a tan i þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4  sin2 ih

!

From Fig. 139a we deduce:

tan ih ¼

F0 A a

tan r ¼

x  F0 A x ¼  tan ih a a

Using Equation (139.2), x sin ih 2 sin r x 1 2 a 2 ¼ ¼  tan ih ) ¼  cos r cos r a cos r sin ih cos ih

258

Seismology

x E

S

A⬘

n 2a r 2n

A

a ih

ih F

F⬘ a

Fig. 139b

By substitution in (139.3), t¼

2x 3a  v sin ih 2v cos ih

For ih ¼ 45 the epicentral distance is

1 x ¼ a 1 þ pffiffiffi 7



This is the limit of the epicentral distance at which these rays arrive. The corresponding time limit is t¼



a 1 4 pffiffiffi þ pffiffiffiffiffi v 2 14

For angles ih > 45 (Fig. 139b), the travel time and epicentral distance, as a function of the take-off angle ih, are t¼

FA AS a xa a xa þ ¼ ¼ þ þ sin ih 2v v 2v sin ih v sin r 2v sin ih v 2

x ¼ a þ A0 S A0 S 2a  F0 A a F0 A ¼ tan ih

a sin ih pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x ¼ a þ 2a  tan ih 4  sin2 ih tan r ¼

259

Ray theory. Constant and variable velocity

Using cos ih cos r 1 ¼ ) cos r ¼ cos ih 2v v 2 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4  cos2 ih sin r ¼ 2pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4  cos2 ih tan r ¼ cos ih we obtain for x and t,

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 4  cos2 ih tan ih cos ih a 2ðx  aÞ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi t¼ 2v sin ih v 4  cos2 ih x ¼ a þ 2a 

For i ¼ 90 , as expected the ray doesn’t arrive at the free surface. 140. For the structure in Fig. 140a, write the equations of the travel times of the direct, reflected, and transmitted waves (neglecting waves with more than a single reflection) as a function of the epicentral distance. Determine the times of intersection, and the minimum and maximum distances in each case in terms of a/v and v. Plot the travel-time curves. The travel-time of the direct wave for distance 0 < x < 1 is (Fig. 140a) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4x2 þ a2 t¼ 2v For the ray reflected on the horizontal surface at depth a the travel time is sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 3a pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 2 4x2 þ 9a2 ¼ t¼ 2v v x S a /2 F

∗

a ν a

2ν

Fig. 140a

260

Seismology

x S a/2

v

F

a a

2v

Fig. 140b

The range of distances for this ray is xmin ¼ 0 a xmax xmax ) a ¼ ) xmax ¼ 3a 3a 2 2 For the reflected ray on the surface at depth 2a the travel time is (Fig. 140b) sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 7a 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x þ 2 4x2 þ 49a2 t¼ ¼ 2v v The range of distances is 3a 3 a 2

xmin

¼ 7 2

a

) xmin ¼ 7a ) tmin

pffiffiffi 7 5a ¼ 2 v

xmax ¼ 1 The critically refracted ray on the surface at depth a, using the general expression, is given by pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi x ð2H  hÞ v22  v21 x a3 3 ¼ þ t¼ þ v2 v 1 v2 2v v 4 The minimum distance for this ray corresponds to the critical distance: sin ic 1 ) ic ¼ 30 ¼ v 2v pffiffiffi 3 a a xc ¼ tan ic þ a tan ic ¼ 2 2 and the maximum distance is xmax

pffiffiffi  a 3þ 3 a ¼ a þ pffiffiffi ¼ 3 3

261

Ray theory. Constant and variable velocity

3

10

t (a/v)

8

6

4

1

2 2 4

0

2

4

6 X (a)

8

10

12

Fig. 140c

On the surface at depth 2a there is no critically refracted ray, since the minimum take-off angle ih at that surface is a tan ih ¼ a ¼ 2 ) ih ¼ 63:4 2  greater than the critical angle 30 . The travel-time curves are drawn after rewriting the equations in units of a/v and a, and are represented in Fig. 140c
tv2
x 2 x 1. Direct ray:  ¼ 14 ; 0 < 1 (a hyperbola) a a a
2
 tv x 2 9 x 2. Reflected ray on the surface at depth a:  ¼ ; 0 3 (a hyperbola) a a 4 a
tv2
x2 49 x  ¼ ; 7 1 (a hyperbola) 3. Reflected ray on the surface at depth 2a: a a 4 pffiffiffi a tv 1 x 3 3 x 4. Critically refracted ray on surface at depth a: ¼ þ ; 0:87 1:58 (due a 2a 4 a to the short range of distances this is not noticeable in the figure)

262

Seismology

141. For the structure in the diagram, assume a seismic focus at the surface, and calculate the travel time of the direct, reflected, and critical refracted waves for epicentral distances between 0 and a. Calculate the critical distance, and the expression for the transmitted wave. Since the focus is at the free surface, the travel time of the direct ray is simply given by (Fig. 141a) t¼

x v

0 ic are totally reflected and don’t penetrate into the core. According to Snell’s law the critical angle is given by sin ic 1 ¼ ) ic ¼ 30:0 v 2v We calculate, using Snell’s law, the angle of incidence i corresponding to the take-off angle of 15 (Fig. 156b): 8 6 R sin ih R sin i 10 10 ¼ ) i ¼ 20:2 v v Since the incidence angle i (20.2 ) is less than the critical angle (30 ) and less than the angle corresponding to the maximum distance (48.6 ), the ray with take-off angle of 15 penetrates into the core.

294

Seismology

(b) Applying Snell’s law we find the angle of the transmitted ray in the core i2 (Fig. 156b): 8 6 R sin ih R sin i2 10 ¼ 10 v 2v

)

i2 ¼ 43:7

By consideration of triangles FOA and AOB, we determine D1 and D2 (Fig. 156b): ih þ
1 þ 180  i ¼ 180 )
1 ¼ 5:2 i2 þ
2 þ i2 ¼ 180 )
2 ¼ 92:6 and using Snell’s law we determine i0 the incidence angle at the surface and D3: 6 6 R sin i2 R sin i R sin i o 10 10 ¼ ¼ 2v v v
3 ¼ i  io ¼ 8:3

) io ¼ 11:9

The epicentral distance of the ray is
¼
1 þ
2 þ
3 ¼ 106 (c) The travel time is t¼

FA AB BS þ þ v 2v v

where ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 8R 6R 8 6 FA ¼ þ  2 R R cos
1 ¼ 0:21 R 10 10 10 10 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2

2 6 6 6 6 2 AB ¼ R þ R 2 R cos
2 ¼ 0:87 R 10 10 10 10 s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi

2 6 6 R þ R2  2 RR cos
3 ¼ 0:42 R BS ¼ 10 10 so t ¼ 1:07

R v

157. Assume a spherical Earth of radius R ¼ 6000 km and constant S-wave speed of propagation 4.17 km s1. Poisson's ratio is 1/4. At a station at epicentral distance 60 an earthquake is recorded with a time interval t S-P ¼ 554 s. Calculate the depth of the earthquake. Given that Poisson’s ratio is 0.25, the P-wave velocity is s¼

pffiffiffi 1 l ¼ ) l ¼ m ) a ¼ 3b ¼ 7:22 km s1 4 2ðl þ mÞ

295

Ray theory. Spherical media

S

F

R R–h Δ

Fig. 157

From the time interval S-P we can calculate the length of the ray FS (Fig. 157): FS FS ab  ¼ FS b a ab S-P t ab FS ¼ ¼ 5469 km ab

t S-P ¼

ð157:1Þ

The distance along the ray in terms of the angular epicentral distance D, using the cosine law, is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi FS ¼ R2 þ ðR  hÞ2  2RðR  hÞ cos
Substituting FS from (157.1):

5370 ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R2 þ ðR  hÞ2  2RðR  hÞ cos


ð157:2Þ

We substitute in (157.2) the values R ¼ 6000 km and D ¼ 60 and solve for h, finding two possible solutions: h1 ¼ 4706 km h2 ¼ 1294 km 158. Assume a spherical Earth of radius R and P-wave velocity which can be expressed by the equation v(r) ¼ a  br2. The speed of propagation at the surface of the Earth is v0 and at the centre of the Earth it is 2v0. What angular distance D does a wave reach which penetrates to a depth equal to half the Earth’s radius? If the velocity distribution inside the Earth is v(r) ¼ a  br2, the ray paths are circular with radius given by (Fig. 158)

296

Seismology

7/4 R 7/4 R

S

F

Δ/2 Δ/2

R/2

O

R

Fig. 158

r¼

r dv p dr

ð158:1Þ

From the conditions of the problem r ¼ R ) v ¼ v0 ¼ a  bR2 r ¼ 0 ) v ¼ 2v0 ¼ a and

v0 ) r2 2 R ) v ¼ v0 2  2 R a ¼ 2v0 b¼

The radius of curvature of the ray which penetrates to r ¼ R/2 is that corresponding to the ray parameter r0 R p0 ¼ 0 ¼ 0 ð158:2Þ 2v v The velocity at depth R/2 is 0 0 12 1 R B 7 B 0 2C C C v ¼ v0 B @2  @RA A ¼ v0 4

Substituting the ray parameter in (158.2): p0 ¼

R 2R ¼ 7 7v0 v0 2

297

Ray theory. Spherical media

The derivative of the velocity is  dv d v0 ¼ a  br2 ¼ 2br ¼ 2 2 r dr dr R

Substituting in (158.1) we obtain, for the radius of curvature (Fig. 158), r¼

r 7R ¼ pð2brÞ 4

The epicentral (angular) distance corresponding to this ray is found by applying the cosine law to the triangle POS: 2 2 7 9 9R
R ¼ R þ R2  2 R cos )
¼ 96:4 4 4 4 2 pffiffi 159. The Earth consists of a mantle of radius R and speed of propagation v ¼ a= r, and a core of radius R/2 and speed of propagation 4v0, where v0 is the speed of propagation at the Earth’s surface. Calculate:

(a) The maximum epicentral distance corresponding to a wave that travels only through the mantle. (b) The critical angle of the wave reflected at the core, and the angle at which it leaves the surface. (c) The epicentral distance Dc corresponding to the critical angle. (d) Plot the travel-time curve, specifying Dc and Dm. (a) The value of a in the velocity distribution is found from the value of velocity at the surface: 1 1 r ¼ R ! v ¼ v ¼ aR 2 ) a ¼ v R 2 0

0

The velocity distribution is a 12 R R v ¼ v0 ¼ v0 r r For this general type of distribution of velocity with depth a < 1, the angular distance for a surface focus (Fig. 159a) is given by

2 1 p
¼ ð159:1Þ cos 1þa 0 where r R  ¼ ) 0 ¼ v v0 The maximum distance for a ray which travels only through the mantle, that is that reaches depth R/2, can be calculated from the velocity at that depth, vm: 0 11 R R 2 pffiffiffi 1 R BRC vm ¼ v0 @ A ¼ v0 2 ) p ¼ 2 ¼ 2pffiffiffi ¼ pffiffiffi ¼ p R vm v0 2 2 2 v0 2

298

Seismology

Δc F

S

∗ i0

ic

Δ/2

Δ/2

O

R/2

R

Fig. 159a

By substitution in (159.1) we obtain, for the maximum distance, 1 1 R pffiffiffi B2 2 v0 C 2  C cos1 B
m ¼ @ R A ¼ 92:4 1 1þ v0 2 0

(b) The critical angle of a reflected ray at the mantle-core boundary applying Snell’s law is sin ic 1 pffiffiffi ¼ ) ic ¼ 20:7 v0 2 4v0 The take-off angle at the surface i0 for this ray is found by again applying Snell’s law: R  R sin i0 2 sinð20:7 Þ pffiffiffi ¼ ) i0 ¼ 7:2 v0 v0 2

(c) To find the critical distance we use the expression ð r0 p dr pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c ¼ 2 2 r   p2 rp where R 2 r0 ¼ R r0 sin i0 R ¼ p¼ v0 8v0 r r r3=2  ¼ ¼  1 ¼ pffiffiffi v v R 2 v0 R 0 rp ¼

r

Substituting the values of the problem in (159.2):
c ¼

1 4

ðR

dr rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R=2 r R3 3 r 64 R3=2

ð159:2Þ

299

Ray theory. Spherical media

This integral is of the type ð so we can write the solution
c ¼

dx 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffi cos1 n n x x a n an

rffiffiffiffiffi an xn

 

8 1 1 ¼ 18 cos1  cos1 pffiffiffi 6 8 8

(d) The travel time of the rays in the mantle is given by   20
R 4 3
sin ð1 þ aÞ sin for 0
92:4 ¼ t¼ 1þa 2 v0 3 4 By substitution of values of D we find the travel time curve given in Fig. 159b. 1.2

1.0

t (R/n)

0.8

0.6

0.4

0.2

0 0

Δc 20

Δm 40

60

80

Δ (⬚)

Fig. 159b

160. A spherical medium of radius R has a constant speed of propagation v0 from the surface down R/2, and from R/2 to the centre a core of variable speed of propaga to

1=2 R tion v ¼ v0 . r (a) What value should i0 have for the waves to penetrate into the core? (b) Calculate the epicentral distance reached by a wave leaving a focus at the surface at angle i0. (a) The velocity at the top of the core (r ¼ R/2) is

300

Seismology

S F

∗

l i0

P

Δ1

Δ1

Δ2

n0 R/2

O

R

Fig. 160

0 11 2

pffiffiffi BRC v1 ¼ v0 @ A ¼ v0 2 R 2

Applying Snell’s law we find the critical angle ic for incident rays at the core (Fig. 160): R R sin ic ¼ pffiffi2ffi v0 2v 0

) ic ¼ 45

The take-off angle i0, for a focus at the surface corresponding to the critical angle, using Snell’s law, is R R sin i0 2 sin ic ¼ v0 v0

) i0 ¼ 20:7

The rays that penetrate into the core must leave the focus with take-off angles less than 20.7 . (b) For a ray with take-off angle i0 which penetrates the core the epicentral distance is the sum of that corresponding to the part that has travelled through the mantle, D1, plus the part that has travelled through the core, D2:
¼ 2
1 þ
2 Since in the core the velocity varies with depth with the law given in the problem, the epicentral distance is given by

2 1 p cos
2 ¼ 1þa 1 where a is the exponent of the velocity distribution v ¼ v0 ra ) a ¼

1 2

301

Ray theory. Spherical media

and p is the ray parameter, which can be obtained using Snell’s law:

and

R R sin i0 2 sin i1 p¼ ¼ pffiffiffi v0 2v 0

Then, we find

R r  ¼ ) 1 ¼ pffiffi2ffi v 2v 0 0

1 R sin i0  B C 4 1
pffiffiffi 2 1 B v0 C
2 ¼ cos 2 ¼ cos 2 sin i 0 @ R A 3 1 þ 12 pffiffiffi 2 2v 0

The distance D1 can be determined using the sine and cosine laws for the triangle FOP (Fig. 160): R

2 ¼ l )
¼ sin1 2l sin i0 1 R sin i0 sin
1 l 2 þ R2  2Rl cos i0 ¼

R2 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ) l ¼ R cos i0  4R2 cos2 i0  3R2 4 2

and we find the expression in terms of i0:

 
pffiffiffi  ffi

1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4
¼ 2sin1 2 cos i0  4cos2 i0  3 sin i0 þ cos1 2 2 sin i0 2 3

161. Consider a spherical Earth of radius 6000 km and surface velocity of 6 km s1, with pffiffi a velocity distribution of the type vðrÞ ¼ a= r. At the distance reached by a wave emerging at a take-off angle of 45 from a focus on the surface, calculate the interval between the arrival times of the direct P- and reflected PP-waves (the PP-wave is one that is reflected at the surface at the midpoint between focus and the point of observation). the a R The velocity distribution is of the type v ¼ v0 where r pffiffiffi a r ¼ R ) v ¼ v0 ¼ pffiffiffi ) a ¼ v0 R R 12 R v ¼ v0 r and the ray parameter p for a ray with take-off angle of 45 is 1 6000  pffiffiffi R sin i0 2 p¼ ¼ 707 s ¼ 6 v0

302

Seismology

For this type of velocity distribution the relation between the ray parameter and the epicentral distance is 

 1þa p ¼ 0 cos
ð161:1Þ 2 In this problem the value of 0 is r R  ¼ ) 0 ¼ ¼ 1000 s v v0 Substituting the values in (161.1) we obtain the distance for the ray with take-off angle of 45 :

3 707 ¼ 1000 cos
)
¼ 60 4 The PP-wave which arrives at D ¼ 60 travels twice the distance which a P-wave does for a distance of 30 . For this type of velocity distribution the travel time for a distance D is given by   20 ð1 þ aÞ
sin t¼ 1þa 2 In our case for the P- and PP-waves at distance 60 we substitute the values of the problem and find tP ð60 Þ ¼ 943 s tPP ð60 Þ ¼ 2tP ð30 Þ ¼ 1021 s The time interval between the PP- and P-waves 60 is tPP  tP ¼ 1021  943 ¼ 78s 162. In an elastic spherical medium of radius r0, the velocity increases with depth according to v ¼ arb. If v0 ¼ 6 km s1, r0 ¼ 6000 km, and, at a point at distance D ¼ 90 , the slope of the travel-time curve is 500 s, determine: (a) The value of b. (b) The value of rp and of vp of the wave reaching an epicentral distance of 90 . (a) For this type of velocity distribution the travel time in terms of the epicentral distance is given by   20
t¼ sin ð1 þ bÞ ð162:1Þ 1þb 2 As we know the velocity at the surface, r r0  ¼ ) 0 ¼ ¼ 1000 s v v0

303

Ray theory. Spherical media

The ray parameter p is known, because it is equal to the slope of the travel-time curve which for D ¼ 90 is given as 500 s. Using the relation between p and D for this type of velocity distribution,   dt
¼ 0 cos ð1 þ bÞ p¼ d
2 h h pi pi 1 p p 1 500 ¼ 1000 cos ð1 þ bÞ ) cos ð1 þ bÞ ¼ ) ð1 þ bÞ ¼ ) b ¼ 4 4 2 4 3 3 (b) At the point of greatest penetration rp for D ¼ 90 , we have the relation p¼

rp sin 90 rp ¼ ) rp ¼ pvp vp vp

and also 1 r0 3 vp ¼ v0 rp From these two equations we obtain rp and vp: 1



1

3 1 3

rp ¼ v0 pr03 rp 3 ) rp ¼ v40 r04 p4 ¼ 3564 km rp 3564 vp ¼ ¼ ¼ 7:1 km s1 p 500 163. Consider a spherical Earth of radius R, the northern hemisphere with a constant speed of propagation v0, and the southern hemisphere with a speed of propagation of 1 R 2 . v ¼ v0 r (a) Calculate the travel time of seismic waves for a focus on the equator and stations on the same meridian. (b) In which hemisphere does the wave at a distance of 60 arrive first? (a) In the northern hemisphere the velocity is constant and the rays have straight paths and their travel time is (Fig. 163) 2R
sin 0 <
< 90 ð163:1Þ v0 2 In the southern hemisphere the velocity increases with depth and the rays have curved paths. Their travel time is given by tN ¼

tS ¼

20
sinð1 þ bÞ 2 1þb

where b R R 1 0 ¼ and v ¼ v0 )b¼ v0 r 2 Substituting in (163.2) we obtain for the travel time in the southern hemisphere

ð163:2Þ

304

Seismology

S1

Δ F

n0

∗ R

S2

Fig. 163

tS ¼

4R 3
sin 3 v0 4

0 <
< 90

ð163:3Þ

(b) The travel times for waves in the northern and southern hemisphere are given by Equations (163.1) and (163.3). By substitution of D ¼ 60 we obtain R v0 pffiffiffi R2 2 R S ¼ 0:47 t ¼ v0 6 v0 The waves arrive first in the southern hemisphere. tN ¼

164. Consider a spherical medium of radius R consisting of two concentric regions (mantle and core), the core of radius R/2. The speeds of propagation are v ¼ ar1=2 for the mantle and v ¼ aR1=6 r1=3 for the core. The surface velocity is v0. For a wave leaving a focus with angle of incidence 14.5 , calculate the angular distance D at which it reaches the surface. We calculate a by applying the boundary conditions 1

1

r ¼ R ) v ¼ v0 ) v0 ¼ aR 2 ) a ¼ v0 R2 1 R 2 Mantle : v ¼ v0 r 1 R 3 Core : v ¼ v0 r We determine the ray parameter corresponding to the ray with take-off angle i0 ¼ 14.5 , using Snell’s law (Fig. 164a): p¼

r sin i R sin 14:5 R ¼ ¼ v v0 4v0

305

Ray theory. Spherical media

S F

i0

∗ i0

i

Δ1

R/2

i2

Δ1

Δ2

O

R

Fig. 164a

At the bottom of the mantle at depth R/2 the velocity is 0 11 2

1 BRC v1 ¼ v0 @ A ¼ v0 22 R

2

The incident angle i of this ray on the mantle–core boundary, applying Snell’s law, is given by R sin i R sin i 0 2 ) i ¼ 45 ¼ 1 v 0 v 22 0

On the top of the core the velocity is 0 11 3

1 BRC v2 ¼ v0 @ A ¼ v0 23 R

2

which is less than at the bottom of the mantle and there is no critical angle. Applying Snell’s law again we obtain the angle i2 of the refracted ray in the core: sin i sin i2 ¼ ) i2 ¼ 39 v1 v2 The take-off angle of the last ray which travels only in the mantle is given by R R sin i0 ¼ 2 1 ) i0 ¼ 20:7 v0 v 0 22 In our case the angle 14.5 is less and the ray penetrates into the core. The epicentral distance is the sum of the distances corresponding to the paths in the mantle and in the core:
¼ 2
1 þ
2

306

Seismology

S F

Δ3

∗ Δ1

Δ4

Δ1

O

R

Fig. 164b

The distance corresponding to the path in the core is given by

2 p
2 ¼ cos1 1þb 0 where p¼

R 4v0

R 0 ¼ 2 1 v0 23 and where p is the ray parameter and  ¼ r/v. Substituting the values we obtain
2 ¼ 76:4 To calculate D1 we suppose that there is no core and a ray with take-off angle i0 ¼ 14.5 would arrive at distance D3 which is related with D1 by (Fig. 164b) 2
1 ¼
3 
4 The distances D3 and D4 can be determined using the equation

2 1 p cos
¼ 1þb 0 where for D3 0 ð RÞ ¼

R ¼  2

R v0 R 3 R 2 2 2 ¼ 0 11 v 0 2 BR C v0 @ A R 2

307

Surface waves

and we obtain 0

1 R B4v0 C 2  C
3 ¼ cos1 B @ R A ¼ 100:6 1 1þ 2

v

and by similar substitutions for D4

0

1 R C B B C 2 1 B 4v0 C  cos B
4 ¼ C ¼ 60:0 1 3 @ A 1þ R  2 2 2 v0

Then, 2D1 ¼ 100.6 – 60 ¼ 40.6 and the epicentral distance is
¼ 40:6 þ 76:4 ¼ 117:0

Surface waves 165. A Rayleigh wave in a semi-infinite medium has a 20 s period. If the P-wave velocity is 6 km s1 and Poisson’s ratio is 0.25, calculate the depth at which u1 ¼ 0, and at which depth the particle movement becomes prograde. Since Poisson’s ratio is 0.25, we find the relation between P- and S-waves: sffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffi 1 l l þ 2m 3m pffiffiffi s¼ ¼ )l¼m)a¼ ¼ ¼ 3b 4 2ð l þ m Þ r r 6 a ¼ 6 ) b ¼ pffiffiffi ¼ 3:4 km s1 3

For a half-space the velocity of Rayleigh waves is cR ¼ 0:919b ¼ 3:2 km s1 The displacement u1 is given by u1 ¼

@’ @c  @x1 @x3

where the potentials are given by ’ ¼ A expðikrx3 þ ik ðx1  cR t ÞÞ c ¼ B expðiksx3 þ ik ðx1  cR t ÞÞ

1=2 c2 r ¼ i 1  R2 ¼ 0:85i a

1=2 c2 ¼ 0:39i s ¼ i 1  R2 b

308

Seismology

Then, u1 ¼ 0 ) ikA expð0:85kx3 Þ  0:39ikB expð0:39kx3 Þ ¼ 0

ð165:1Þ

so k¼

2p 2p ¼ ffi 0:1 km1 l TcR

We can write B in terms of A using the boundary condition of zero stress at the free surface:   t31 ¼ 0jx3 ¼0 ) 2rA  1  s2 B ¼ 0 so

B ¼ 1:47iA Substituting in (165.1) we obtain the value of x3: expð0:85kx3 Þ ¼ 0:39  1:47 expð0:39kx3 Þ x3 ¼ 12 km At 12 km depth u1 is null and for greater values of depth the particle motion is prograde while for lesser values of depth it is retrograde. 166. Given a layer of thickness H and shear modulus m ¼ 0 on top of a half-space or semi-infinite medium in which l ¼ 0, study (without expanding the determinant) whether there exist surface waves that propagate in the x1-direction. Are they dispersive waves? In the liquid layer ( m ¼ 0) the P- and S-velocities are sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffi sffiffiffiffi 0 m l0 þ 2m0 l0 b0 ¼ 0 ¼ ) a0 ¼ ¼ r r r and in the solid half-space sffiffiffiffiffiffi 2m pffiffiffi ¼ 2b l¼0)a¼ r

The relation between the stress and strain is

tij ¼ lydij þ 2meij where eij ¼

1 2





ui; j þ uj;i .

t0ii ¼ l0 ðe011 þ e022 þ e033 Þ t0ij ¼ 0 In the half-space: l ¼ 0 ) tij ¼ 2 meij If there are surface waves propagating in the x1-direction, their displacements in terms of the potentials are given by (Fig. 166) In the layer: m0 ¼ 0 )



u1 ¼ ’;1  c;3 u2 ¼ u 2 u3 ¼ ’;3 þ c;1

309

Surface waves

x3 x1

H

μ⬘ = 0 0 λ=0

Fig. 166

The boundary conditions at the free surface are null normal stresses:  0 t33 ¼ 0 x3 ¼ H ) t031 ¼ 0 and at the boundary between the liquid layer and the solid half-space continuity of the normal component of the displacement and stress and zero tangential stresses, 8 u3 ¼ u03 > > > < t ¼ t0 33 33 x3 ¼ 0 ) 0 > t ¼ t > 32 ¼ 0 > 32 : 0 t31 ¼ 0

In the liquid layer there is only the P-wave potential ’. Taking (x1, x3) as the incidence plane ’0 ¼ A expðikr0 x3 þ ik ðx1  ct ÞÞ þ B expðikr0 x3 þ ik ðx1  ct ÞÞ rffiffiffiffiffiffiffiffiffiffiffiffiffiffi c2 0 r ¼ 1 a02 where c is the velocity of wave propagation in the x1-direction In the half-space c ¼ C expðiksx3 þ ik ðx1  ct ÞÞ u2 ¼ E expðiksx3 þ ik ðx1  ct ÞÞ ’ ¼ D expðikrx3 þ ik ðx1  ct ÞÞ rffiffiffiffiffiffiffiffiffiffiffiffiffi c2 1 r¼ a2 sffiffiffiffiffiffiffiffiffiffiffiffiffi c2 s¼ 1 b2 In the layer we have only guided P-waves and r0 is real, while in the half-space for surface waves, r and s must be imaginary. Then a > b > c > a0 must be satisfied.

310

Seismology

From the boundary conditions we obtain the following equations: x3 ¼ H   0   0 t033 ¼ 0 ) A 1 þ r02 eikr H þ B 1 þ r02 eikr H ¼ 0 x3 ¼ 0 t31 ¼ 0 ) 2Dr  C þ Cs2 ¼ 0 u03 ¼ u3 ) Ar0  Br0 ¼ Dr þ C     t033 ¼ t33 ) l0 1 þ r02 ðA þ BÞ ¼ 2m Dr2 þ Cs

For a solution of the system the determinant must  0 0  eikr H eikr H   0 0  0 0  r r   l0 ð1 þ r02 Þ l0 ð1 þ r02 Þ

be zero: 0 s2  1 1 2ms

 0  2r  ¼0 r  2mr2 

Expanding the determinant and working r0, r, and s in terms of the variable c, we obtain c (k), the velocity of waves in the x1-direction. They have the form of guided waves in the liquid layer and surface waves in the half-space. Since the velocity c(k) is a function of the wavenumber the waves are dispersive. 167. There is a liquid layer of density r and speed of propagation a on top of a rigid medium (half-space). Derive the dispersion equation of waves in the layer by boundary conditions and by constructive interference in terms of v. Plot the dispersion curve for the different modes. Given that the layer is liquid the only potential is ’: ’ ¼ ðA exp ikrx3 þ B expðikrx3 ÞÞ exp ik ðx1  ct Þ

ð167:1Þ

The boundary conditions at the free surface are zero normal stress and at the boundary between the liquid layer and the rigid half-space zero normal component of displacement (Fig. 167a): x3 ¼ H ) t33 ¼ 0 x 3 ¼ 0 ) u3 ¼ 0 where t33 ¼ ly ¼ r2 ’ ¼ ro2 ’ ¼ 0 u3 ¼ ’;3 rffiffiffiffiffiffiffiffiffiffiffiffiffi c2 1 r¼ a2 By substitution of (167.1) we obtain AeikrH þ BeikrH ¼ 0 AB¼0)A¼B

311

Surface waves

x3

H

liquid layer x1

0 rigid half-space

Fig. 167a x3 O

H

B A

x1

0 P

Fig. 167b

Then, 2A cos krH ¼ 0. For waves propagating in the layer, r must be real and c > a. The solution is given by

1 krH ¼ n þ p; n ¼ 0; 1; 2; . . . ð167:2Þ 2 The solution can also be found by the method of constructive interference. The condition of constructive interference implies that waves coinciding at a given wavefront (AB) are in phase, that is, the distance along the ray must be an integer multiple of the wavelength, taking into account possible phase shifts (Fig. 167b). In our case on the free surface, x3 ¼ H, there is a phase shift of p (l/2) and we write the condition as (Fig. 167b) AP þ PQ þ QB 

la ¼ nla 2

or 2p ðA P þ P Q þ Q BÞ  p ¼ 2pn la

312

Seismology

Substituting A P þ P Q þ Q B ¼ 2H cos i we obtain 2p 2H cos i  p ¼ 2pðn þ 1Þ la

ð167:3Þ

According to Snell’s law, a sin i ¼ ) cos i ¼ c

rffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi a2 a c 2 a 1 2 ¼ 1¼ r 2 c c a c

a and ka ¼ k. c Substituting in (167.3), we obtain the same solution obtained in (167.2):

a 1 ka Hr ¼ n þ p c 2 This expression can also be written as " #12 rffiffiffiffiffiffiffiffiffiffiffiffi





1 oH c2 1 1 1 2 p2 ð167:4Þ 1 ¼ nþ p)c¼ 2  nþ krH ¼ n þ p ) 2 c 2 a 2 H 2 o2 a2 The fundamental mode (FM) corresponds to n ¼ 0, and n 1 to the higher modes (HM). In the FM and the higher modes, the frequency oc corresponding to the zero in the denominator in (167.4) is called the cut-off frequency, as there are no values of c for o < oc. For a mode of order n the cut-off frequency is given by

1 p nþ a 2 oc ¼ H The dispersion curve is shown in Fig. 167c. c FM

1 HM

2 HM

α

πα 2H

Fig. 167c

3πα 2H

5πα 2H

ω

313

Surface waves

168. Consider an elastic layer of coefficients l and m, thickness H, and density r on a rigid semi-infinite medium. Derive the dispersion equation c(v) for P-SV and SH-type channelled waves for the fundamental mode (FM) and the first higher mode (1HM). Plot the dispersion curve for the SH motion. For a SH-wave which propagates in the x1-direction its displacement is given by u2 ¼ ðE expðiksx3 Þ þ F expðiksx3 ÞÞ exp ik ðx1  ct Þ The P- and SV-waves are given by their scalar potentials ’ and c: ’ ¼ ðA expðikrx3 Þ þ B expðikrx3 ÞÞ exp ik ðx1  ct Þ c ¼ ðC expðiksx3 Þ þ D expðiksx3 ÞÞ exp ik ðx1  ct Þ where r and s were defined in Problem 166. The boundary conditions for SH-waves are null stress at the free surface and null displacement at the boundary with the rigid medium (Fig 168a): x3 ¼ H ) t32 ¼ 0 ¼ m

@u2 @x3

x 3 ¼ 0 ) u2 ¼ 0 By substitution we have EeiksH  FeiksH ¼ 0 E þ F ¼ 0 ) F ¼ E Eðe

iksH

þe

iksH

Þ ¼ 0 ) cosðksHÞ ¼ 0 ) ksh ¼

Substituting s and putting k ¼ o/c: Ho c

sffiffiffiffiffiffiffiffiffiffiffiffiffi

c2 1  1 ¼ n þ p)c¼ 2 b2



1 nþ p 2

! 1

2 1 1 2 p2  n þ 2 H 2 o2 b2

ð168:1Þ

This equation give us, for the SH component, the frequency dependence of the velocity c(o). The boundary conditions for P and SV are similarly

x3

x1

H

α, β, µ

0 rigid medium

Fig. 168a

314

Seismology

x3 ¼ H ) t31 ¼ 0; t33 ¼ 0 x3 ¼ 0 ) u1 ¼ 0; u3 ¼ 0 where t33 t31 u1 u3

¼ lðe11 þ e33 Þ þ 2me33 ¼ 2me31 ¼ ’;1  c;3 ¼ ’;3 þ c;1

Substituting the expression for the potentials we obtain ðl þ 2mÞðr2 AeikrH þ r2 BeikrH Þ þ lðAeikrH þ BeiksH Þ þ 2msðCeiksH  DeiksH Þ ¼ 0 2rðAeikrH  BeikrH Þ þ ð1  s2 ÞðCeiksH þ DeiksH Þ ¼ 0 ðA þ BÞ  sðC  DÞ ¼ 0 rðA  BÞ þ C þ D ¼ 0 For a solution we put the determinant of the system of equations equal to zero:     1 1 s s     r r 1 1   ikrH ikrH 2 iksH 2 iksH  ¼ 0 ð168:2Þ  2re 2re ð1  s Þe ð1  s Þe    ½l þ r2 ðl þ 2mÞeikrH ½l þ r2 ðl þ 2mÞeikrH 2mseiksH 2mseiksH 

Expanding the determinant and putting it equal to zero, we obtain the dependence with frequency of the velocity c(o) which gives us the dispersion curve. For the wave with SH component the dispersion curve is given in Fig. 168b: !12

1 1 2 p2 c¼  nþ 2 H 2 o2 b2 c FM

1 HM

2 HM

β

πβ 2H

Fig. 168b

3πβ 2H

ω

5πβ 2H

315

Surface waves

For n ¼ 0 the curves correspond to the fundamental mode and for 1 n to the higher modes. For all modes, including the fundamental mode, there is a cut-off frequency oc ¼ (nþ1)pb/2H, with n ¼ 0 for the fundamental mode and n 1, for higher-order modes. 169. For a liquid layer of thickness H with a rigid medium above and below, derive the dispersion equation c(v) of the fundamental and higher modes. For the FM, at what height above the layer is the motion circular? Given that the medium is a liquid, motion is represented only by the scalar potential f: ð169:1Þ ’ ¼ ðA exp ikrx3 þ B expðikrx3 ÞÞ exp ik ðx1  ct Þ rffiffiffiffiffiffiffiffiffiffiffiffiffi c2  1. where r ¼ a2 The boundary condition at the two boundaries between the liquid and rigid solid is that the normal component of the displacement is null (Fig. 169): x 3 ¼ 0 ) u3 ¼ 0 x 3 ¼ H ) u3 ¼ 0 Substituting u3 ¼ ’;3 we have AB¼0

ð169:2Þ

AeikrH  BeikrH ¼ 0 which leads to the equation

A eikrH  eikrH ¼ 0

ð169:3Þ

Consider first that r is real, that is, c >a. Then, from (169.1) 2iA sin krH ¼ 0 ) krH ¼ np; n ¼ 0; 1; 2; ::: with n ¼ 0, fundamental mode (FM), and n 1 for higher modes. For the FM, n ¼ 0 and r ¼ 0, and then rffiffiffiffiffiffiffiffiffiffiffiffiffi c2 1¼0)c¼a Hk a2 The displacements from (169.1) and (169.2) are x3 rigid medium H

liquid layer 0

x1 rigid medium

Fig. 169

316

Seismology

@’ ¼ Aikrðexp ikrx3  exp ikrx3 Þ exp ikðx1  ctÞ @x3 @’ ¼ Aik ðexp ikrx3 þ exp ikrx3 Þ exp ikðx1  ctÞ u1 ¼ @x1 For the FM r ¼ 0, then u3 ¼ 0 and this is a P-wave, with only a u1 component, which propagates in the x1-direction. For all HM the displacements have both components For the first higher mode (1HM), n ¼ 1: rffiffiffiffiffiffiffiffiffiffiffiffiffi o c2 H 1¼p c a2 1 c2 ¼ 1 p2  a2 o2 H 2 If 1 p2 ap )c!1  ¼0 then o¼ a2 o 2 H 2 H u3 ¼

The cut-off frequency is oc > ap/H. For each higher mode there is a cut-off frequency onc > nap/H. _ _ If r ¼ ir is imaginary, then c < a and this implies that 2 sinhðkr HÞ ¼ 0 which is impossible (1< sinhx = T2 ¼ 0:11 ¼ sin YT sin FT ) T ðYT ¼ 42:27 ; FT ¼ 9:32 Þ > ; T3 ¼ 0:74 ¼ cos YT

In the same way for the axis P 1 0 l1 P1 @ P2 A ¼ @ l2 P3 l3 0

n1 n2 n3

0 1 1 pffiffiffi 1 Z1 B 2 C   1 C Z2 AB @  pffiffiffi A ! PðYP ¼ 80:02 ; FP ¼ 268:03 Þ 2 Z3 0

178. The seismic moment tensor relative to the geographical axes (X1, X2, X3) (north, east, nadir) is 0 1 2 1 1 Mij ¼ @ 1 0 1 A 1 1 2 Find the values of the principal stresses, and the orientation of the tension and pressure stress axes.

First we calculate the eigenvalues of Mij. Since Mij is a symmetric tensor its eigenvalues are real and the corresponding eigenvectors mutually orthogonal (Problem 111):   0 1  2  s 1 3 1    1 0s 1  ¼ 0 ) s ¼ @ 2 A   1 1 1 2  s

Ordered by magnitude, the three eigenvalues are

s1 ¼ 3; s2 ¼ 2; s3 ¼ 1

332

Seismology

The diagonalized matrix is 0

3 Mij ¼ @ 0 0

0 2 0

1 0 0 A 1

In this form Mij is referred to the coordinate system formed by the eigenvectors or principal axes. Given that the sum of the elements of the principal diagonal is not zero, the source has net volume changes. Then, we can separate Mij into two parts: an isotropic part with volume changes (ISO) and a deviatoric part without volume changes. The second part can be separated into two parts: a part corresponding to a double-couple or shear fracture (DC) and a part corresponding to a non-double-couple source usually expressed as a compensated linear vector dipole (CLVD). Thus the moment tensor is separated into three parts, namely M ¼ M ISO þ M DC þ M CLVD The isotropic part is given by 1 4 M ISO ¼ s0 ¼ ðs1 þ s2 þ s3 Þ ¼ 3 3 The deviatoric part (DCþCLVD) is given by Mij0 ¼ Mij  dij so and in our case

Mij0

0

s1 ¼@ 0 0

05

0 B3 0 B 2 0 A¼B B0 3 @ s3 0 0 1

0 s2 0

Now we separate this part into two parts, DC and CLVD:

0

1

C C 0 C C A 7  3

M 0ij ¼ M DC þ M CLVD 1 ðs  s3 Þ B2 1 B 0 Mij ¼ B 0 @ 0 0

Mij0

0

2 B ¼ @0 0

0 0 0

0

1

0 0 0

s2 C B 2 C B CþB 0 0 A @ 1 0  ðs1  s3 Þ 2 1

0

0



1 3

B B C B 0 AþB 0 B @ 2 0

0

0

1

C C C 0 C C A 1 0  3 2 3

0



0 s2 0

0

1

C C 0 C s2 A  2

333

Focal parameters The orientation of the P and T axes is calculated from the double-couple part MDC: 0 1 1 0 0 MijDC ¼ 2@ 0 0 0 A 0 0 1

We can find the n and l axes:

MijDC ¼ M0 ðli nj  lj ni Þ

1 1 ~ ) nðYn ¼ 45 ; Fn ¼ 0 Þ n : pffiffiffi ; 0;  pffiffiffi 2 2

1 1 ~ p ffiffi ffi p ffiffi ffi ; 0; ) lðYl ¼ 45 ; Fl ¼ 0 Þ l: 2 2

In the same way as in Problems 176 and 177, we determine T and P from n and l, finding first Z ¼ n  l: 0 1 0 1 0 1 p1ffiffiffi T1 l1 n1 Z1 B 2 C C @ T2 A ¼ @ l2 n2 Z2 AB B 1 C @ pffiffiffi A T3 l3 n3 Z3 2 0 T1 ¼ 1 ¼ sin YT cos FT T2 ¼ 0 ¼ sin YT sin FT ! T ðYT ¼ 90 ; FT ¼ 0 Þ T3 ¼ 0 ¼ cos YT For the P-axis, 0

1

0

P1 l1 @ P2 A ¼ @ l2 P3 l3

n1 n2 n3

1 1 pffiffiffi Z1 B 2 C B 1 C Z 2 AB CPðYP ¼ 0 ; FP ¼ 0 Þ @  pffiffiffi A Z3 2 0 1

0

179. The magnitude Ms of an earthquake as calculated for surface waves of period 20 s is 6.13. (a) Calculate the amplitude of these waves at a station 3000 km away. If the instrument’s amplification is 1500, what will be the amplitude of the seismogram’s waves and the seismic energy? (b) If Ms ¼ Mw, and the area of the fault is 12 km  8 km with m ¼ 4.4104 MPa, find the fault slip Du. (a) The surface wave magnitude Ms is given by Ms ¼ log

A þ 1:66 log
þ 3:3 T

334

Seismology

where A is the ground motion amplitude, T is the period of the wave, and D is epicentral distance in degrees. Knowing the magnitude and period of the waves we can calculate the wave amplitude: A 3000 þ 1:66 log þ 3:3 T 111:11 A ) log ¼ 0:454 ) A ¼ 2:84  20  1500 ¼ 8:5 cm T

6:13 ¼ log

We have reduced the ground motion to the amplitude of the seismogram using the amplification of the instrument (1500). Knowing the magnitude we can calculate the seismic energy: log Es ¼ 11:8 þ 1:5Ms ) Es ¼ 1021 ergs ¼ 1014 J (b) Mw ¼ 6.13 ¼ 2/3 log M0 – 6.1; M0 ¼ 2.191018 Nm If M0 ¼ mDu S, with S ¼ 12 8 ¼ 9.6107 m2, then
u ¼

M0 2:19  1018 Nm ¼ ¼ 0:52 m mS 4:4  1010 N m2  9:6  107 m2

5

Heat flow and geochronology

Heat flow 180. Assume that the temperature variation within the Earth is caused by gravitational forces under adiabatic conditions. Knowing that the coefficient of thermal expansion at constant pressure is aP ¼ 2105 K1 and the specific heat at constant pressure is cP ¼ 1.3 kJ kg1 K1, determine an expression for the gradient of the temperature with depth. Compare it with the value observed at the surface which is 30 K km1, knowing that, at 200 km depth, T ¼ 1600 K. Under adiabatic conditions, there is no heat flow and the variation of pressure with depth z is a function of gravity g and density r: dP ¼ rgdz

ð180:1Þ

Using the first and second laws of thermodynamics dU ¼ dQ  PdV dQ ¼ TdS where Q is the heat, U is the internal energy, S is the entropy, T is the absolute temperature, P is the pressure, and V is the volume. If we use the specific variables (variables divided by mass) we can write du ¼ dq  pdv dq ¼ Tds Considering that dS ¼



@S @T



dT þ

P



@S @P



dP

T

we can write

@s dq ¼ Tds ¼ T @T





@s dT þ T @P P



dP

T

According to the definition of specific heat at constant pressure, cP and the increase in heat dq are given by 335

Heat flow and geochronology

336

cP ¼ T



@S @T



P

dq ¼ cP dT þ T



@s @P



ð180:2Þ dP

T

The Gibbs function G is defined as G ¼ u  Ts þ pv Taking the differential in this expression, and taking into account the second law of thermodynamics du ¼ Tds  pdv we obtain dG ¼ vdp  sdT If we compare this expression to the differential of the Gibbs function



@G @G dG ¼ dp þ dT @p T @T p we differentiate again and using the Schwartz theorem we obtain



@v @s ¼ @T P @P T But the coefficient of thermal expansion is defined as

1 @v ap ¼ v @T p In consequence we can write Equation (180.2) as dq ¼ cp dT  Tvap dp In our case the process is adiabatic and in consequence using this equation and Equation (180.2) we obtain dT T ap g ¼ cp dz

ð180:3Þ

where we have taken into account that the variables are by unit mass so rv ¼ 1, and substituting the values we obtain: dT 1600  2  105  10 K K1 ms2 ¼ ¼ 0:25 K km1 dz 1:3  103 J kg1 K1 We observe that this result is two orders of magnitude lower than the observed values. This shows that observations correspond to heat flow at the lithosphere and are not satisfied by purely adiabatic conditions.

Heat flow

337

181. If the Earth’s temperature gradient is 1  C/30 m, calculate the heat loss per second due to conduction from its core. Compare this with the average power received from the Sun. Data: Thermal conductivity K ¼ 4 W m1  C1. Earth’s radius R ¼ 6370 km. Solar constant: 1.35 kW m2. The heat flow is given by q_ ¼ KA

dT dr

where A is the area of the Earth’s surface A ¼ 4pR2 ¼ 5:10  1014 m2 and dT 1 C ¼ dr 30 m Substituting the values q_ ¼ 6:80  1013 J s1 ¼ 1:63  1013 cal s1 the average power received on the Earth’s surface by the radiation from the Sun is 1:35  103

W 5:10  1014 m2 ¼ 6:89  1017 J s1 ¼ 1:65  1017 cal s1 m2

In consequence, from these values we can see that on the Earth’s surface the average solar power is much larger than that due to the heat flow from inside the Earth. 182. At the Earth’s surface, the heat flow is 60 mW m2 and T0 ¼ 0  C. If all the heat is generated by the crust at whose base the thermal conductivity is K ¼ 4 W m1  C1, and T is 1000  C, determine the thickness of the crust and the heat production per unit volume If we assume that all the heat is generated at the crust and there is no heat flow from the mantle at the crust base, then we can write _ z¼H ¼ 0 qj Using the temperature equation for a flat Earth for one-dimensional heat-flow and the stationary case we can write T ¼ e¼

e 2 q_ 0 z þ z þ T0 K 2K

q_ 0 H

For z ¼ H: TH ¼

q_ 0 H ðTH  T0 Þ2K þ T0 ) H ¼ 2K q_ 0

Heat flow and geochronology

338

Substituting the values given in the problem we obtain H ¼ 133:3 km and the heat production by unit of volume is e¼

q_ 0 ¼ 4:5  104 mW m3 H

183. Consider the crust to be H ¼ 30 km thick and the heat flow at the surface to be 60 mW m2. (a) If all the heat is generated in the crust, what is the value of the heat generated per unit volume? (Take K ¼ 3 W m1 K1) (b) If all the heat is generated in the mantle with a distribution Aez=H mW m3 , what is the value of A? What is the temperature at 100 km depth? (a) We solve the heat equation for a stationary one-dimensional case, assuming a flat Earth with one-dimensional flow in the z-direction (vertical) positive downward. In this case the solution of the heat equation is given by T ¼

e 2 q_ 0 z þ z þ T0 2K K

where e is the heat generated by unit volume and time, K is the thermal conductivity, q0 and T0 are the heat flow and temperature at the surface of the Earth, respectively If all the heat is generated at the crust we can write (Fig. 183) z ¼ H ! q_ 0 ¼ 0 The heat generated by unit volume is 

 dT  e2z q_ 0  _ z¼H ¼ 0 ¼ K  ¼ K  þ qj z¼H ¼ 0 dz z¼H 2K K  e¼

q_ 0 60  103 ¼ ¼ 2  106 W m3 H 30  103

(b) If all heat is generated in the mantle with distribution e ¼ Aez=H the heat equation is d2 T e A ¼  ¼  ez=H dz2 K K and the solution is given by T ¼

A 2 z=H H e þ Cz þ D K

ð183:1Þ

where C and D are constants of integration. They may be estimated from the boundary conditions at the surface

Heat flow

339

• q0 T0

H

• qH TH

Fig. 183

A 2 A H þ D ! D ¼ T0 þ H 2 K K A z ¼ 0 ! q_ ¼ q_ 0 ¼ 0 ! C ¼  H K

z ¼ 0 ! T ¼ T0 ¼ 

Substituting in (183.1) we obtain T ¼

 A 2 z=H AH A AH
H e z þ T0 þ H 2 ¼ Hez=H  z þ H þ T0  K K K K

ð183:2Þ

If the heat has its origin in the mantle, the flow at the base of the crust is  dT  q_ 0 _ z¼H ¼ q_ o ¼ K  )A¼ qj dz z¼H H ðe1  1Þ

The temperature at z = 100 km may be estimated from (183.2) assuming that T0 = 0:
 q_ 0 z=H He  z þ H T ðzÞ ¼ K ðe1  1Þ T ðz ¼ 100Þ ¼

 60  103
3 100=30 3 3 30  10 e  100  10 þ 30  10 ¼ 2249 K 3ðe1  1Þ

184. Calculate the thickness of the continental lithosphere if its boundary coincides with the 1350  C geothermal, knowing that the surface temperature is 15  C, the heat flow at the surface is q_ 0 ¼ 46 mW m2 , the lithospheric mantle’s thermal conductivity is K = 3.35 W m1 K1, and the radiogenic heat production is P = 0.01  103 mW m3 The geothermal equation at depth z is given by: Tz ¼ T0 þ

q_ 0 P0 ðz  z0 Þ  ðz  z0 Þ2 K 2K

ð184:1Þ

Heat flow and geochronology

340

where K is the thermal conductivity, T0 is the temperature at the surface of the Earth (in K), q_ 0 is the heat flow at the surface, and P0 is the radiogenic heat production at the Earth’s surface. At the Earth’s surface z0 = 0, and Equation (184.1) becomes: Tz ¼ T0 þ

q_ 0 P0 2 z z K 2K

P0 2 q_ 0 z  z þ ðTz  T0 Þ ¼ 0 2K K So

z¼

q_ 0  K

ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 q_ 0 P0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 ðTz  T0 Þ K 2K q_ 0  q_ 20  2P0 KðTz  T0 Þ ¼ P0 P0 2 2K

Substituting the data given in the problem: K ¼ 3.35 W m1 K1 q_ 0 ¼ 46  103 W m2 P0 ¼ 0.01106 mW m3 Tz ¼ 1623 K T0 ¼ 288 K we obtain two solutions, but only z = 98.27 km is realistic (the second one gives a depth larger than the Earth’s radius). 185. On the surface of an Earth of radius 6000 km, the temperature is 300 K, the heat flow is 6.7 mW m2, and the thermal conductivity is 3 W m1 K1. If the heat production per unit volume inside the Earth is homogeneously distributed, what is the temperature at the centre of the planet? We begin solving the problem of heat conduction inside a sphere with constant internal heat generation per unit volume e and conductivity K. The differential equation for heat conduction with spherical symmetry is

1 d e 2 dT r ð185:1Þ ¼ 2 r dr dr K Integrating twice and using the boundary conditions: Surface: r = R ! T = T0 Center: r = 0 ! T finite we obtain the solution  e  2 T ¼ T0 þ R  r2 6K

The heat flow is given by dT q_ ¼ K ) q_ 0 ðr ¼ RÞ ¼ 6:7 mW m2 dr  e r d
e  2 T0 þ R  r2 ¼ ¼ K dr 6K 3

ð185:2Þ

Heat flow

341

Solving for the heat production e e ¼ 3:35  109 Wm3 From Equation (185.1) the temperature at the Earth’s centre is Tr¼0 ¼ 300 þ

3:35  109  62  1012 ¼ 7000 K 63

186. Consider a spherical Earth of radius R = 6000 km and a core at R/2, in which there is a uniform and stationary distribution of ε heat sources per unit volume. The heat flow at the surface is 5 mW m2, the thermal conductivity is 3 W m1 K1, and the temperature at the core–mantle boundary is 4000 K. Calculate the temperature at the Earth’s surface. We consider the problem as one of heat conduction inside a sphere with conductivity K and constant heat generation per unit volume e inside the core (radius R/2). We begin with Equation (185.1)

1 d e 2 dT r ¼ r2 dr dr K The boundary conditions at the core–mantle boundary and its centre are r ¼ R=2 ! T ¼ TN r ¼ 0 ! T finite where TN is the temperature at the core–mantle boundary Integrating twice we obtain e T ðrÞ ¼ TN þ 6K

! 2 R  r2 2

The heat flow is given by  dT  eR q_  3 ¼ )e¼ 0 q_ 0 ðr ¼ RÞ ¼ K  ¼ 2:5  109 W m3 dr r¼R 3 R

Then, the temperature at the Earth’s surface is 2:5  109 T ðr ¼ RÞ ¼ 4000 þ 63



6000  103 2

2

2

 6000  10

6

!

¼ 250 K

187. Consider the Earth of radius R0 = 6000 km formed by a spherical crust with its base at 500 km and constant thermal conductivity K. If the temperature at the base of the crust is T1 and at the surface of the Earth is T0 = 0  C, determine: (a) An expression for the heat flow through the crust.

Heat flow and geochronology

342

(b) An expression for the temperature distribution within the Earth. (c) The temperature at the base of the crust if, at that depth, q_ ¼ 5:5  1013 W and K = 4 W m1  C1. (a) We assume a spherical Earth where the temperature varies only in the radial direction. Then we can solve the problem as one of spherical unidirectional flow. For the stationary case, when the conductivity and heat generation are constant, the Fourier law may be written as q_ ¼ KA

dT dr

ð187:1Þ

where A ¼ 4pr2 is the area in the normal direction to the heat flow. Integrating this equation: q_ 4p

ð R0 r1

dr ¼ K r2

ð T0

dT

T1

where the conditions at the Earth’s surface are, r = R0 !T = T0 and at the base of the crust, r = r1 !T = T1. Solving Equation (187.1), assuming that K is constant, we obtain 4pK 4pKr1 R0 ðT0  T1 Þ ðT0  T1 Þ ¼ q_ ¼  1 1 R0  r1  r1 R0

ð187:2Þ

(b) The temperature distribution inside of the Earth may be obtained by integration of Equation (187.1): ð ðT q_ r dr ¼ K dT 4p r1 r2 T1 T ð r Þ ¼ T1 þ

R0 ðr  r1 Þ ðT0  T1 Þ ðR0  r1 Þr

(c) The radial distance to the base of the crust is r2 = 5500 km, so, using expression (187.2), we obtain T1 ¼ T0 þ

_ 0  r1 Þ qðR 5:5  1013  500  103 ¼0þ ¼ 16579  C 4pKr1 R0 4p  4  5500  103  6000  103

This result implies a constant increase of temperature from the Earth’s surface of 1 ºC each 33.2 m similar to the observed gradient in the real Earth of 1 ºC per 30 m 188. Assume that the heat flow inside the Earth is due to solar heating of the Earth’s surface. Calculate the maximum penetration of this flow in the diurnal and annual cycles. Take as typical values for the Earth K = 3 Wm1 K1, r = 5.5 g cm3, Cv = 1 kJ kg1 K1.

Heat flow

343

We assume the heat propagation inside the Earth coming from the solar radiation on its surface as unidirectional flow thermal diffusion (inside the Earth) with periodic variation of surface temperature. The diffusivity equation is k

@ 2 T @T ¼ @z2 @t

ð188:1Þ

K , K is the thermal conductivity, r is the density, rCv and Cv is the specific heat at constant volume. We solve Equation (188.1) using the separation of variables where the thermal diffusivity is k ¼

T ðz; t Þ ¼ Z ð zÞyðt Þ Substituting in (188.1) we obtain the solution Z ð zÞ ¼ Aeaz þ Beaz yðt Þ ¼ Ceka

2

t

where a is the constant of separation of variables. Using the boundary condition of periodic flow and the temperature T0 at the Earth’s surface, z ¼ 0 ) T ¼ T0 eiot and as Z(z) exists only inside the Earth, B = 0. At the surface, z = 0, so 2

ACeka t ¼ T0 eiot Then AC ¼ T0 ka2 ¼ io But putting, i ¼ 12 ð1 þ iÞ2 , we have rffiffiffiffiffiffi o a ¼ ð1 þ iÞ 2k Then, we can write the temperature variation inside of the Earth as:  rffiffiffiffiffiffi rffiffiffiffiffiffiffiffi

 o o zþi  z þ ot T ðz; t Þ ¼ T0 exp  2k 2k This equation corresponds to a periodic wave, with angular frequency o propagating for positive z values (to the Earth’s interior) and with the amplitude decreasing with depth. The propagation velocity and wavelength are given by rffiffiffiffiffiffi 2k v¼ o rffiffiffiffiffiffi 8k l ¼ 2pv ¼ p o

Heat flow and geochronology

344

The values of l corresponding to the daily and annual cycles give their maximum penetration: Daily cycle: 2p ¼ 7:2  105 s1 24  60  60 K 3 W m1 K1 ¼ ¼ 0:5  106 m2 s1 k¼ rCv 5:5  103 K gm3  103 J Kg1 K1

o¼

Then rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8  0:5  106 ¼ 0:74 m l¼p 7:2  105 Annual cycle: 2p ¼ 2  107 s1 365  24  60  60 l ¼ 14 m

o¼

The penetration of the solar radiation as periodic heat conduction inside the Earth is very shallow due to the poor heat conduction. 189. Consider a lithospheric plate of 100 km thickness created from asthenospheric material originating from a ridge in the asthenosphere with constant temperature Ta and in which no heat is generated. Given that k = 106 m² s1, that the temperature at the base of the lithosphere is 1100  C, and in the asthenosphere is 1300  C, calculate the age of the plate, and, if the velocity of drift is 2 cm yr1, how far it has moved away from the ridge. The heat propagation inside the plate is given by:

K @2T @2T @T þ 2 ¼u rcv @x2 @z @x

ð189:1Þ

where T is the temperature, r is the density, cv is the specific heat at constant volume, and u is the horizontal velocity of the plate in the x-direction (normal to the plate front). If we assume that the horizontal conduction of heat is insignificant in comparison with the horizontal advection and vertical conduction, we can write, using the following change of variable t = x/u, K @ 2 T @T ¼ rcv @z2 @t Integrating this equation and using the boundary conditions at the ridge and surface: x ¼ 0 ! T ¼ Ta z¼0!T ¼0 we obtain for the temperature distribution T ðz; t Þ ¼ Ta erf



z pffiffiffiffiffi 2 kt



345

Geochronology

where K rcv ð 2 x 2 erf ð xÞ ¼ pffiffiffi ey dy p 0 k¼

Substituting the data of the problem,



L L 1100 ¼ 1300erf pffiffiffiffiffi ) erf pffiffiffiffiffi ¼ 0:846 2 kt 2 kt

Values of the error function, erf(x), may be obtained from tables. If erf(x) = 0.846, x = 1.008, then

L L2 1010 pffiffiffiffiffi ¼ 1:008 ) t ¼ ¼ 4k  1:0082 4  106  1:0082 2 kt ¼ 2:5  1015 s ¼ 79 Myr If the displacement velocity is 2 cm yr1, the plate has moved 1580 km. 190. If the concentrations of 235U and 235Th in granite are 4 ppm and 17 ppm, respectively, and the respective values of heat production are 5.7  104 W kg1 and 2.7  105 W kg1, respectively, calculate the heat flow at the base of a granite column of 1 m² cross-section and 30 km height (the density of granite is 2.65 g cm3). We estimate first the mass of the granite column: M ¼ rV ¼ 2:65  103  1  30  103 ¼ 7:95  107 kg If the concentration of

235

U in the granite is 4 ppm, its quantity in the column is 235

Then the heat flow due to the

U:

235

4  7:95  107 ¼ 318 kg 106

U is

q_ ¼ 5:7  104  318 ¼ 181:26 mW m2 For 235Th, the heat flow is 235

Th : 7:95  107  17  106 ¼ 1351:5 kg

q_ ¼ 1351:5  2:7  105 ¼ 36:49 mW m2

Geochronology 191. The mass of 1 millicurie of 214Pb is 3  1014 kg. Calculate the value of the decay constant of 214Pb. The mass of the sample is

Heat flow and geochronology

346

3  1014 kg ¼ 3  1017 g ¼

N M N0

where N is the number of atoms in the sample, N0 is Avogadro’s number ¼ 6.02  1023, and M is atomic number ¼ 214. Solving for N we obtain N¼

3  1017  6:02  1023 ¼ 8:44  1038 atoms 214

The correspondence of a curie is 1 curie ¼

dN ¼ lN ¼ 3:7  1010 disintegrations s1 dt

where l is the decay constant and t is the time. Then l¼

3:7  107 ¼ 0:44  1031 s1 8:44  1038

192. The isotope 40K decays by emission of b particle with a half-life of 1.83  109 years. How many b decays occur per second in one gram of pure 40K? The average life t of a radioactive material is a function of the decay constant l: t ¼

1 1 !l¼ ¼ 0:55  109 yr1 ¼ 1:73  102 s1 l 1:83  109

The number of atoms N contained in 1 g of 40K may be estimated from Avogadro’s number N0 and the atomic number M: 1¼

N 1  6:02  1023 M !N ¼ ¼ 0:15  1023 atoms N0 40

The rate of disintegration is given by dN ¼ lN ¼ 0:26  1021 Bq dt 193. The half-life of 238U is 4468  106 yr and of 235U is 704  106 yr. The ratio 235 238 U/ U in a sample is 0.007257. Given that the ratio was 0.4 at the time of formation, calculate the sample’s age. From the half-life T1/2 we can obtain the decay constant l: 0:693 l 0:693 ¼ ¼ 1:5510  1010 yr1 4468  106 0:693 ¼ 9:8434  1010 yr1 ¼ 704  106

T1=2 ¼ l238 l235

The number N of disintegrating atoms at time t is given by N ¼ N0 elt

ð193:1Þ

347

Geochronology

where N0 is the number of atoms at time t ¼ 0. Then N235 ¼ N0235 el235 t N238 ¼ N0238 el238 t If we divide these equations: N235 N0235 ðl238 l235 Þt ¼ e N238 N0238 Substituting the values given in the problem we obtain  4:0095 ¼ 8:2924  1010 t ! t ¼ 4:8  109 yr 194. Date a meteorite which contains potassium knowing that its content of 40K is 1.19  1014 atoms g1, of 40Ar is 4.14  1017 atoms g1, and that the half-life of 40 K!40Ar is 1.19  109 years. We obtain the decay constant of the

40

K from its half-life T1/2:

0:693 T1=2 0:693 l¼ ¼ 0:58  109 yr1 1:19  109

l¼

We can solve the problem considering it as a case of radioactive ‘parent’ atom disintegrating to a ‘daughter’ stable atom. At time t ¼ 0 we have n0 ‘parent’ atoms at the sample, and at time t there remain NR radioactive atoms in the sample and NE daughter atoms, from the disintegration of the n0 parent atoms: n0 ¼ NR þ NE But n0 NR þ NE NE ¼1þ ¼ nt NR NR From Equation (193.1) we obtain the age of the sample: nt ¼ n0 elt !

n0 NR ¼ elt ¼ 1 þ nt NR

ð194:1Þ

so

1 NE t ¼ ln 1 þ l NR NE, the number of atoms of 40K in the sample, can be estimated from the number of atoms contained in 1 g of potassium: 1¼

N 40 ! N ¼ 1:506  1022 atoms g1 6:023  1023

Heat flow and geochronology

348

so NE ¼ 1:506  1022  1:19  1014 ¼ 1:792  1036 atoms g1 Then the age of the meteorite from (194.1) is

1 1:792  1036 t ¼ ln 1 þ ¼ 7:4  1010 yr l 4:41  1017 195. At an archaeological site, human remains were found and assigned an age of 2000 years. One wants to confirm this with 14C dating whose half-life is 5730 yr. If the proportion of 14C/12C in the remains is 6  1013, calculate their age. (Assume that at the initial time the 14C/12C ratio was 1.2  1012.) The decay constant l may be obtained from the half-life: T1=2 ¼

0:693 0:693 !l¼ ¼ 1:2094  104 yr1 l 5730

The activity in a sample is given by R ¼ R0 elt where R0 ¼ (14C/12C)t ¼ 0 and R ¼ (14C/12C). Then the age of the remains is



1 R0 1 1:2  1012 ¼ ln ¼ 2379 yr t ¼ ln l R 1:2094  104 9  1013 196. Mass spectrometry of the different minerals in an igneous rock yielded the following table of values for the concentrations of 87Sr originating from the radioactive decay of 87Rb and of 87Rb, with the concentration expressed relative to the concentrations of 86Sr of non-radioactive origin. Mineral

87

Sr /86Sr

A B C D E F

0.709 0.715 0.732 0.755 0.756 0.762

87

Rb /86Sr

0.125 0.418 1.216 2.000 2.115 2.247

Express on a 87Sr/86Sr–87Rb/86Sr diagram the isochron corresponding to the formation of the rock, and calculate the age of the rock. Take l ¼ 1.42  1011 yr1. For the decay of

87

Rb 



87 Sr now ¼ 87 Sr 0 þ 87 Rb now elt  1

ð196:1Þ

349

Geochronology

0.80

0.78

86Sr

87Sr

0.76

0.74

0.72

0.70 0.0

0.5

1.0

1.5

2.0

2.5

87Rb 86Sr

Fig. 196

where [87Sr]now and [87Rb]now are the number of atoms of each isotope at time t, [87Sr]0 is the amount of original number of atoms of the isotope 87Sr [87Sr]now, and l is the decay constant. Equation (196.1) may be written as    87  87   Sr Sr 87 Rb  lt ¼ 86  þ 86  e 1 ð196:2Þ 86 Sr Sr 0 Sr now now  87  Sr This equation corresponds to a line (y ¼ a þ bx) with intercept 86  and slope (elt  1), Sr 0 which is called an isochron. If we plot the values given in the problem (Fig. 196) we can obtain the equation of the line by least-squares fitting: y ¼ 0:025x þ 0:705 The age of the sample can be obtained from the slope b ¼ 0.025: el t  1 ¼ b lnð1 þ bÞ t¼ l Substituting the values of b and l: t ¼ 1:72  109 yr:

Heat flow and geochronology

350

197. Magma with a material proportion of 87Sr/86Sr equal to 0.709 crystallizes producing a series of rocks with different concentrations of 87Rb with respect to the content of 86Sr: Sample

87

Rb/86Sr

A B C D

1.195 2.638 4.892 5.671

(a) Calculate the proportions of 87Sr/86Sr and 87Rb/86Sr that these rocks will have after 500 Myr. Take l ¼ 1.421011 yr1. (b) Express in a 87Sr/86Sr–87Rb/86Sr diagram the isochrons corresponding to t ¼ 0 and t ¼ 500 Myr. (a) Using the same method as in the previous problem, we can write   87 87  Sr 87 Sr 87 Rb  lt Rb
1:421011 5108 ¼ þ 86 e  1 ¼ 0:709 þ 86 e 1 86 Sr 86 Sr  Sr Sr 0 0.78

0.76

86

87

Sr Sr

t = 500 Ma

0.74

0.72 t=0

0.70

0

1

2

3

4 87Rb 86Sr

Fig. 197

5

6

7

351

Geochronology

The results for each rock are given in the following table Sample

87

Sr/86Sr

A B C D

0.717 0.728 0.744 0.749

87

Rb/86Sr

1.187 2.619 4.857 5.631

(b) For t ¼ 500 Myr, we carry out a least-squares fitting to obtain the isochron, which results in y = 0.007x þ 0.709 In Fig. 197 the isochrones corresponding to t ¼ 0 and t ¼ 500 Myr are shown.

Bibliography

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Gravimetry Blakely, R. J. (1995). Potential Theory in Gravity and Magnetic Applications. Cambridge University Press, Cambridge. Bomford, G. (1980). Geodesy (4th edn). Clarendon Press, Oxford. Heiskanen, W.A. and H. Moritz (1985). Physical Geodesy. Freeman, San Francisco. Hofmann-Wellenhof, B. and H. Moritz. (2006). Physical Geodesy. Springer, New York. Lambeck, K. (1988). Geophysical Geodesy. Clarendon Press. Oxford. 352

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Seismology Aki, K. and P.G. Richards (2002). Quantitative Seismology. (2nd edn). University Science Books, Sausalito, California. Bullen, K.E. and B.A. Bolt (1985). An Introduction to the Theory of Seismology (4th edn). Cambridge University Press, Cambridge. Chapman, C. (2004). Fundamentals of Seismic Wave Propagation. Cambridge University Press, Cambridge. Dahlen, F.A. and J. Tromp (1998). Theoretical Global Seismology. Princeton University Press, Princeton Gubbins, D. (1990). Seismology and Plate Tectonics. Cambridge University Press, Cambridge. Lay, T. and T.C. Wallace (1995). Modern Global Seismology. Academic Press, San Diego. Kennnett, B.L.N. (2001). The Seismic Wavefield. Cambridge University Press, Cambridge. Pujol, J. (2003). Elastic Wave Propagation and Generation in Seismology. Cambridge University Press, Cambridge. Shearer, P.M. (1999). Introduction to Seismology. Cambridge University Press, Cambridge. Stein, S. and M. Wyssesion (2003). An Introduction to Seismology, Earthquakes and Earth Structure. Blackwell Publishers, Oxford. Udías, A. (1999). Principles of Seismology. Cambridge University Press, Cambridge.

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