Gravitation

CLASS

CBSE-i

XI

UNIT-6

PHYSICS

GRAVITATION

CENTRAL BOARD OF SECONDARY EDUCATION Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India

Contents Preface

v

Acknowledgements

viii

Teachers’ manual Learning outcomes

ix



Teaching Notes

xi

Lesson Plan Matrix

xv



Pre-requisites

xix



Weblinks/vediolinks/other references

xix

Students’ Manual

• Introduction



2



• Astronomy in Ancient India

4



• Kepler’s Laws of Planetary Motion

5



• Newton’s Law of Gravitation

9



• Important Characteristics of Gravitational force

10



• The Universal Gravitational Constant (G)

11



• Vector Form of Newton’s Law of Gravitation

13



• Principle of Superposition

13



• Acceleration Due to Gravity (G)

16



• Variation in G Due to Shape

19



• Variation in G with Height

19



• Variation in G with Depth

22



• Gravitational Field

25



• Gravitational Potential Energy

27



• Gravitational Potential (V)

32



• Escape Speed (Ve)

33



• Motion of Satellite

35





Post content student worksheet 1

40





Post content student worksheet 2

41





Post content student worksheet 3

42





Post content student worksheet 4

44





Post content student worksheet 5

45

Preface The Curriculum initiated by Central Board of Secondary Education -International (CBSE-i) is a progressive step in making the educational content and methodology more sensitive and responsive to global needs. It signifies the emergence of a fresh thought process in imparting a curriculum which would restore the independence of the learner to pursue the learning process in harmony with the existing personal, social and cultural ethos. The Central Board of Secondary Education has been providing support to the academic needs of the learners worldwide. It has about 11500 schools affiliated to it and over 158 schools situated in more than 23 countries. The Board has always been conscious of the varying needs of the learners and has been working towards contextualizing certain elements of the learning process to the physical, geographical, social and cultural environment in which they are engaged. The CBSE-i has been visualized and developed with these requirements in view. The nucleus of the entire process of constructing the curricular structure is the learner. The objective of the curriculum is to nurture the independence of the learner, given the fact that every learner is unique. The learner has to understand, appreciate, protect and build on knowledge, values, beliefs and traditional wisdom. Teachers need to facilitate the leaner to make the necessary modifications, improvisations and additions wherever and whenever necessary. The recent scientific and technological advances have thrown open the gateways of knowledge at an astonishing pace. The speed and methods of assimilating knowledge have put forth many challenges to the educators, forcing them to rethink their approaches for knowledge processing by their learners. In this context, it has become imperative for them to incorporate those skills which will enable the young learners to become ‘life long learners’. The ability to stay current, to upgrade skills with emerging technologies, to understand the nuances involved in change management and the relevant life skills have to be a part of the learning domains of the global learners. The CBSE-i curriculum has taken cognizance of these requirements.

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The CBSE-i aims to carry forward the basic strength of the Indian system of education while promoting critical and creative thinking skills, effective communication skills, interpersonal and collaborative skills along with information and media skills. There is an inbuilt flexibility in the curriculum, as it provides a foundation and an extension curriculum, in all subject areas to cater to the different pace of learners. The CBSE introduced the CBSE-i curriculum in schools affiliated to CBSE at the international level in 2010 at primary and secondary level in classes I and IX and subsequently in the session 2011-12 initiated the curriculum at Class II, VI and class X. The current session will take the curriculum forward to classes III, VII and XI. An important feature of the Senior Secondary Curriculum is its emphasis on the specialisation in different fields of study and preparing a student for higher professional life and career at the work place. The CBSE-i, keeping in mind, the demands of the present Global opportunities and challenges, is offering the new curriculum in the subject of English, Physics, Chemistry, Biology, Geography, Accountancy, Business Studies, Information and Communication Technology, and Mathematics at two levels, Mathematics-I for the students of pure sciences and Mathematics-II for the students of Commerce and other subjects. There are some non-evaluative components in the curriculum which would be commented upon by the teachers and the school. The objective of this part or the core of the curriculum is to scaffold the learning experiences and to relate tacit knowledge with formal knowledge. This would involve trans-disciplinary linkages that would form the core of the learning process. Perspectives, SEWA (Social Empowerment through Work and Action), Life Skills and Research would be the constituents of this ‘Core’. The Core skills are the most significant aspects of a learner's holistic growth and learning curve. The International Curriculum has been designed keeping in view the foundations of the National Curricular Framework (NCF 2005) NCERT and the experience gathered by the Board over the last seven decades in imparting effective learning to millions of learners, many of whom are now global citizens. The Board does not interpret this development as an alternative to other curricula existing at the international level, but as an exercise in providing the much needed Indian leadership for global education at the school level. The Curriculum envisages pedagogy which would involve building on learning experiences

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inside the classroom over a period of time. The Board while addressing the issues of empowerment and capacity building of teachers believes that all school must budget for and ensure teachers involved with CBSE-i are continuously updated. I appreciate the sincere effort put in by Dr. Sadhana Parashar, Director (Training) CBSE, Dr. Srijata Das, Education Officer, CBSE and the team of Officers involved in the development and implementation of this material. The CBSE-i website enables all stakeholders to participate in this initiative through the discussion forums provided on the portal. Any further suggestions are welcome. Vineet Joshi Chairman, CBSE

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Acknowledgements

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Teachers' manual Syllabus

Learning Outcomes At the end of this unit, students would be able to: • k  now the long history of mankind’s efforts to understand planetary motion • differentiate between the geocentric and the heliocentric model and the historical context under which these ideas prevailed • know about the early astronomical observations and their significance and contribution towards the development of the law of gravitation. • state and interpret Kepler’s laws of planetary motion • recognize how Kepler’s laws originated from the analysis and interpretation of Tycho Brahe’s astronomical data • realize how measurements on planetary motion were in agreement with Kepler’s law of ‘periods’. • State Newton’s law of Gravitation • understand the concept of central forces • define G the universal gravitational constant and know about the various experiments proposed for the measurement of G • draw the vector diagram for the gravitational force between two masses • understand how to compute the net gravitational force due to a collection of masses • recognize that gravitational force is one of the ‘basic forces of nature’ and it is a weak force that is always attractive in nature. • obtain Kepler’s laws of planetary motion from Newton’s law of Gravitation • recall the concept of acceleration and use it to obtain the expression for g (the acceleration due to gravity) from Newton’s law of Gravitation

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• recognize that the shape of the earth affects the value of g for different points on its surface • comprehend the meaning and significance of the popular statement “Cavendish weighed the earth” • realize that the acceleration due to the pull of the earth (i.e. ‘g’) is independent of the mass of the object experiencing this acceleration. • obtain the expression for variation of g with height h, (i.e. g(h)) above the surface of earth, and understand the limitations of this expression. • recognize why the expression for g(h) cannot be used for negative values of h (i.e., for points below the surface of the earth) • obtain the expression for variation of g with depth d , (i.e. g(d) below the surface of earth • recognize that work is done, by or against, the gravitational force, when a mass is moved from one point to another • recall the concept of potential energy and define Gravitational potential energy • obtain the general expression for the potential energy of a mass at a height from the surface of earth and get its usual (approximate) form for small values of this height. • derive the formula for the gravitational potential energy associated with two masses separated by a distance • differentiate between gravitational potential energy and gravitational potential • know the concept of escape speed • obtain the expression for the escape speed from the law of conservation of energy • appreciate the scientific efforts to place artificial earth satellites in orbits around the earth and know about their different uses. • obtain the expression for the time period of an orbiting earth satellite • list different types of artificial satellites and obtain the condition under which an artificial satellite would become a geo-stationary satellite • know the range of applications of a geo-stationary satellite.

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Teaching Notes The discovery of the concept of gravitational force, followed by the discovery of the quantitative law that determines the magnitude of this force between any two objects is one of the most significant discoveries in the history of Physics. It is useful to convey to the students the way Newton used astronomical data related to the moon to come to the conclusion that the gravitational force between two objects must vary as inverse of the square of distance between them. The ‘inverse square’ nature of this law gives the gravitational force several interesting properties notable among which are the (i) conservative nature of this force and (ii) zero value of the gravitational field inside a hollow object. The teacher may explain that the other important’inverse square’ force in nature—the electrical force—also has similar characteristics. One can therefore convey the ‘message’ that the same physical law often help us to understand a variety of apparently diverse physical phenomenon. It may also be explained that the ‘inverse square’ law for gravitational force, proposed by Newton, was also consistent with Kepler’s laws of planetary motion. These laws were based on the painstaking and thorough astronomical observations of the great Danish astronomer Tycho Brahe. Newton’s law of gravitation, therefore become the ‘core’ of our understanding of astronomical phenonmenon in addition to its use in understanding a variety of terrestrial phenomenon. The teacher may express this law in vector form and use ‘principle of superpositioin’ when the gravitational force on one object is due to a group of two or more other objects. It may also be stressed that unlike the electrical or the magnetic force, the gravitational force is always attractive in nature. The direction of the force between two (point) objects has to be along the line joining the two (point) objects. This is so because in a region of space where these two (point) objects alone may be present, the only direction that can be uniquely defined is the line joining these two points. It may also be pointed out that the direction of the gravitational force on object 2 due to object 1 has to be opposite to that of the position vector of object 2 with respect to object 1. The consistency of this statement with the ‘always attractive’ nature of the gravitational force has to be clearly brought out and emphasised.

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While discussing the universal nature of the law of gravitation, it needs to be stressed that we ordinarily do not perceive the effects of this force because of the very small value of the ‘universal constant’ of gravitation ‘G’ . The teacher may explain how it was thought that we would necessarily need to use very large, huge masses to measure the (very small) effects of the weak gravitational force. These ‘intial attempts’ need to be appreciated in their historical context. It would be interesting to point out that inspite of the relatively crude nature of these measurements. The value of G, obtained through these ‘experiments’ though not very accurate, was still of the correct’order of magnitude. While discussing the measurement of G, the teacher may point out why Cavendish’s experiment—the first ‘in-house’ or laboratory experiment—on the measurement of G, is regarded as one of the most important experiments in the history of Physics. His novel idea of measuring the (relatively) large effect of the moment of the couple due to a pair of (small) gravitational forces and his accurate and precise technique of carrying out his measurements needs to be ‘shown’ to the students. The concept of ‘acceleration due to gravity (g) --- a characteristic ‘constant’ for the earth (and other astronomical objects)- needs to be discussed carefully. The students need to be clarified that a common value of this ‘constant’ for all objects of different masses is a consequence of the equality of the ‘inertial’ and ‘gravitational’ mass on the other hand is the ratio of the (magnitudes) of the gravitational force on the object to its resulting (gravitational) acceleration. The equality of this (gravitational) mass to the inertial mass of the object and the nature of the gravitational force (proportional to the product of the mass of the object and that of the earth) then implies that the acceleration due to the gravity for the earth, would have the same value for objects of different masses. While discussing the variation of acceleration due to gravity with height above the earth’s surface, it needs to be pointed out that we make calculations by assuming the entire mass of the earth to be concentrated at its centre. It should also be pointed out that we can use the formula g

h/g

=

R2 2

(R + h)

=

1 h  1+  R 

2

in

1 − 2h  h m) The orbit parameters

is

characterized

(i) Semi major axis

by

Planet

Sun

B

A

two focl

Figure 4: Diagram showing elliptical path of a planet with Sun at one of the foci.

(ii) eccentricity (e) For circular orbits, eccentricity is zero and semi major axis is equal to the radius of the orbit. For most of the planets the eccentricity of their orbit is so small that we can assume them to be moving in circular orbits. For the sake of simplicity, we, therefore, generally assume all orbits to be circular unless mentioned otherwise.

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Kepler’s Second Law — Law of Areas The line joining a planet to the Sun sweeps out equal areas in equal interval of time in the orbital plane of the planet. i.e. the rate dA dt

at which it sweeps out area A (Areal velocity) is a constant. Kepler’s second law is the most important of the three laws.

dt A2

dt

A1=A2

A1 sun

planet

Figure 5: Diagram showing area covered by a planet at two different positions on its orbit.

The area swept by the planet dA in time dt can be expressed as

dA =

1 2 r 2

dq

(Area of triangle =

1 2

rdq

base × height)

where r is the distance between Sun and planet, dθ is the angle swept by the (radial line of the) planet in time dt, as shown in the accompanying figure. Hence where

dA dt

dq

r

Figure 6:

= 1 r 2 dq = 1 r 2 ω 2

dt

2

ω = angular speed of the planet

The magnitude of the angular momentum of the planet, about an axis passing through the Sun, is L = rpn →

where pn is the component of its linear momentum p along the direction normal to r L = r(mv) = r (mwr) = mr2w From these equations, we get

dA dt

= L

2m

Since, the gravitational force is a central force, it acts along the line joining the

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→

Sun to planet i.e. parallel to r . Hence the torque, exerted by the gravitational force, on the planet, is zero. →

In the absence of torque, the angular momentum L remains conserved in accordance with the law of conservation of angular momentum. Hence the areal velocity

dA = dt

a constant

One simple consequence of Kepler’s second law is that a planet moves faster when it is closer to Sun and slower when it is far from it. Can you think of an explanation?

Kepler’s Third Law — Law of Periods The square of the time period of revolution of a planet around the Sun, is directly proportional to the cube of the semi – major axis of the orbit. (radius of the orbit when the orbit can be assumed to be circular.) Thus

T2 ∝ r3

where T = time period of revolution of the planet and r = radius of its orbit. Alternately, we can also write this law as

T12 T22

=

r13 r23

For an interactive animation visit:

http://www.drennon.org/science/kepler.htm

   Concept Probe 1. C  onsult a table of planetary data. Calculate T2/r3 for each planet. Verify that this quantity is almost constant for all planets. 2. Y ou are given that the period of rotation of the moon around the earth is approximately 30 days and its distance from the earth is approximately 64 earth radii. Can you calculate the height of geosynchronous satellite?

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Brahe was the master of quantitative observations but was deficient in their theoretical interpretation. On the other hand, Johannes Kepler (1571-1630), a German, who went to Prague to become Brahe’s assistant, had a strong theoretical intuition. Kepler and Brahe did not get along well. Brahe apparently mistrusted Kepler; He thus allowed only limited access to Kepler to his voluminous data.

Figure 7: Tycho Brahe

He set Kepler the task of understanding the orbit of the planet Mars, which was particularly troublesome. It is believed that part of the motivation for giving the Mars problem to Kepler was that it was difficult, and Brahe hoped it would keep Kepler occupied while Brahe worked on his theory of the Solar System. In a supreme irony, it was precisely the Martian data that allowed Kepler to formulate the correct laws of planetary motion, thus eventually achieving a place in the development of astronomy that far surpassed that of Brahe. It was left to to Kepler to provide an answer to the final piece of the puzzle. After a long struggle, in which he tried mightily to avoid his eventual conclusion, Kepler was finally forced to come to the conclusion that the orbits of the planets were not the circles (demanded by Aristotle and assumed implicitly by Copernicus), but were instead the “flattened circles” that geometers call ellipses. The irony, noted above, lies in the realization that the difficulties with the Martian orbit originate precisely from the fact that the orbit of Mars was the most elliptical of all the planets for which Brahe had extensive data. Thus, Brahe had unwittingly given Kepler the very part of his data that allowed Kepler to eventually formulate the theory of the Solar System that surpassed Brahe’s own theory!

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5.4 Newton’s Law of Gravitation

We all are familiar with the famous legend of Newton observing the free fall of an apple and formulating the law of gravitation. However, it will be appropriate to mention that Newton gave the law of gravitation after almost 20 years of his first thought about it which is quite often associated with the episode of a falling apple.

Figure 8: Sir Issac Newton

Newton’s law for gravitational force between two bodies is one of the most far – reaching laws in the history of human scientific endeavor. Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the two bodies.

If must be noted that the law as stated above applies to point bodies. For extended bodies, the distance must be taken between their centres of mass (geometrical centres if bodies are regular and homogeneous) and the direction of the force is along the line joining their centres of mass.

F ∝ m1m2



F ∝



F ∝

m2

m1

r

Figure 9: Two masses m1 and m2 separated by distance r.

1 r2 m1m2 r2

where m1 and m2 are the masses of the two bodies and r is the distance between their centers.

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Note that as the masses of the bodies increase, the force between them increases. Also, as the distance between their centres increases, the force decreases.

6.5 important characteristics of gravitational force

The force of gravitation is associated with the following characteristics: (a) The gravitational force is a central force. It acts along the line joining the centres of two bodies. (b) It is a conservative force. This means that the work done by the gravitational force in displacing a body from one point to another is only dependent on the initial and final positions of the body and is independent of the path followed.

1 2

3

Figure 10: Work done by gravitation force along path 1 is the same as that along path 2 or path 3.

(c) It is a long range force. The gravitational force is effective even at large distances. (d) Unlike electrostatic and magnetic forces, the gravitational force is always attractive. The entire Universe is held together by the gravitational force. It is, therefore, the most important force in nature. However, it is the weakest of all the fundamental forces in nature. Note that earlier mass was regarded only as a measure of the inertia of the body. This led to the concept of inertial mass. Newton’s law of gravitation, however, bestows mass with another property. Mass is also a measure of the gravitational force. This leads us to the concept of the gravitational mass of an object. However, all the precise experiments done so far, show that both inertial and gravitational masses are equal.

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   Concept Probe 1. W  hat happens to the gravitational force between two bodies if the mass of one of the bodies is halved and the distance between them is doubled? 2. T he gravitational force acts on all bodies. Why, then does an apple fall towards earth but the earth does not move towards the apple?

The formulation of Newton’s law of gravitation is a story of human determination and quest for scientific enquiry. In the first instance, Newton‘s calculations did not work and he put aside his papers, in a drawer, for almost 20 years. It was during the advent of a comet in 1680, and at the prodding of Sir Edmund Halley, his friend, that he again worked on his calculations and, subsequently, obtained excellent results.

For an interesting explanation, visit the link:

http://www.physicsclassroom.com/class/circles/u6l3c.cfm

6.6 The Universal Gravitational Constant(G)

The proportionality sign in the Newton’s law of gravitation can be eliminated by putting a constant of proportionality, denoted by G. The equation then becomes

F = G

m1m2 r2

Here G is called the gravitational constant. It is a universal Figure 11: Henry Cavendish constant because the gravitational force between two bodies placed at a certain distance remains the same, wherever these bodies may be placed in the universe. The constant G is also independent of the medium in which the interacting bodies are placed. The value of G, in SI units, was later on, found to be equal to 6.67 × 10–11 Nm2/kg2. The units of G are obtained by using the fact that the force has to be expressed in newtons.

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The numerical value of G would change with a change in the units of measurement for mass and distance. The very small value of G points to the fact that the gravitational force is an extremely weak force. In fact, it is the weakest of all the fundamental forces (Can you find out, which are the other fundamental forces?) and it becomes important only when the masses of the bodies involved are very large. That is why the gravitational force plays such an important role in the case of heavenly bodies. http://www.physicsclassroom.com

   Concept Probe 1. If there is a gravitational attractive force between all objects, why do we not feel ourselves attracted towards massive structures in our surroundings? 2. W  hy is the gravitational force an important force for heavenly (or astronomical) objects?

The value of G was first measured by an English Physicist, Henry Cavendish in the eighteenth century. He achieved this by measuring the small force between lead masses with an extremely sensitive torsion balance. A better method was later developed by Philipp von Jolly. As long as the sizes of the objects are small compared to the distance between them they can be treated as point objects which simplifies considerably the mathematics of their gravitational interaction. The Sun and Saturn are far enough (in comparison to their sizes) for them to be treated as point particles. If the distance between two objects is very large (in comparison to their sizes), we can take them as point objects. What about the case of bodies on the earth? For such bodies, the earth does not seem like a point object. The answer to this dilemma lies in Newton’s shell theorem: A uniform spherical shell of matter attracts a body that is outside the shell as if all the mass of the shell were concentrated at its center.

According to this theorem, Earth can be regarded as a ‘point mass’, located at the center of Earth and with mass equal to that of Earth. We usually follow

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this theorem in all our calculations.

5.7 VECTOR FORM OF NEWTON’S LAW OF GRAVITATION

Since force is a vector quantity, it must be expressed in a vector form. The gravitational force can also be expressed in a vector form by attaching a unit vector to the expression for this force.

1

r^

A

2 B

Figure 12: Diagram showing direction of the gravitational force acting along the line joining the center of two bodies.

By convention, the direction of unit vector is always taken as directed from the body experiencing the force (body 1) towards the body exerting the force (body 2). Therefore,



F = G

m1 m2 r2

rˆ

If we are calculating the force on body A due to body B, then rˆ will be the unit vector drawn from A towards B. It will be imperative to mention here that this vector notation is consistent with the basic fact that the gravitational force is always an attractive force. This implies that the gravitational force is a central force, and hence the direction of this force has to be along the line joining the center of the two bodies.

5.8 PRINCIPLE OF SUPERPOSITION

Newton‘s law has been stated for two point bodies. How do we calculate the force on a body if there are more than two bodies interacting with one other?

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The solution to this situation lies in what is called the principle of superposition. In a group of objects, the net gravitational force, on any one of the objects, is the vector sum of the forces due to all the other objects. The principle implies that we first calculate the gravitational force that acts on an object due to each of the other objects as if all other objects are absent. After doing this for all possible pairs, the net force on the object under consideration is calculated by the vector sum of all the forces acting on it. →

→

→

→

→

F 1 = F 12 + F 13 + F 14 + F 15 + ...

→

Here F 1 is the net force on object 1 due to all the other objects 2, 3, 4, 5, … The Principle of superposition is based on the fact that the gravitational interaction between two bodies is independent of the presence of other bodies in the neighborhood. The same concept is applicable to electrostatic force which will be studied in class XII.

Illustration. Three objects of masses 5 kg, 3 kg and 3 kg are placed at the corners

of an equilateral triangle of side 20 cm. Calculate the net gravitational force on the object of 5 kg.

Solution. The magnitude of the force on object A

A

due to object B is

FAB = G

mA mB r2



It is in the direction of AB



Putting the values we get,



B

C

FAB  5 × 10–9 N

 imilarly, the magnitude of the force on S A due to body C is



FAC  5 × 10–9 N



It is the direction of AC.



As per the principle of superposition, the net force on the body A is the →

→

vector sum of forces F AB and F AC . Applying the laws of vector addition,

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we get the magnitude of the resultant as

FA =

2 2 FAB + FAC + 2 FABFAC cos 60°



FA =

1 (5 × 10 −9 N )2 + (5 × 10 −9 N )2 + 2 (5 × 10 −9 N )2   2

= 5 3 × 10–9 N









The resultant force is directed along the bisector of the angle between the →

→

two forces F AB and F AC .

Practice problem Four equal masses are placed at the corners of a square of side 2cm as shown in the figure. Another mass is placed at the centre of the square. Find the magnitude and direction of the net force on the body kept at the center of the square due to all the other masses.

?

Did You Know?

There are two high tides per day in oceans due to the gravitational pull of the moon on the earth. Surprisingly, although the gravitational pull of Sun is 180 times greater than the pull of the Moon, the effect of Sun is much less on ocean tides. Search for the possible reason for this?

Derivation of Kepler’s law of time periods from Newton’s law of gravitation For simplicity let us assume that the orbit of the planet is circular. The gravitational force exerted by the sun on the planets provides the necessary centripetal force for the planet to remain in its orbit. So, applying the force equation,

Centripetal force = Gravitational force

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we get

M p v2 r

v =

putting



=

M p 4 p2 r T2

\

=

GMs M p r2 2 pr T GMs M p r2  4 p2  GMs

T2 = 

 3 r  

The quantity in the parentheses is a constant that depends only on the mass Ms of the central massive body which is the Sun in this case. It can also be the Earth if we are talking of the motion of an artificial satellite. \

T2 ∝ r3

This, as you have learnt above, is Kepler’s third law for planetary motion.

6.9 Acceleration due to Gravity

We assume that the Earth is a uniform sphere of mass M and radius R. We can assume then that the mass of earth is concentrated at its center. The gravitational force of Earth, on a particle of mass m, at a distance r from the center of Earth, is given by Newton’s Law as follows:

F =

GMm r2



…(i)

where M is the mass of the earth. This force gives rise to an acceleration in the particle which is called the acceleration due to gravity.

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Applying Newton’s second law, force on a body can be expressed as

F = ma

…(ii)

Both the expressions (i) and (ii) are measuring the same force in two different ways. m

Hence, from the two equations (i) and (ii), we get ma =

\

a =

GMm r2

r

GM r2

We use the special symbol ‘g’ for this acceleration and call it as the acceleration due to gravity. Thus, the acceleration due to gravity (g) is the acceleration produced on a body, on, or near, the earth’s surface, due to the gravitational pull of earth. We have

g =

GM R2

Aristotle taught that heavy objects fall faster than light objects. Galileo argued that all objects, irrespective of their mass, should take the same time to fall to the ground from a given height. Do you agree with Galileo? If so, why? Search the internet for Galileo’s Pisa experiment. It is said that Galileo went to the top of the tower of Pisa and dropped bodies of various masses and showed that they all take the same time to fall to the ground.

For points lying very close to the surface of earth, we put r = R, the radius of the earth. We can, therefore, express acceleration due to gravity (g) as

g =

GM R2

For earth, taking M = 6 × 1024 kg and R = 6400 km, we get g = 9.8 m/s2. However, it must be noted that the value of 9.8 m/s2 is an average value and the value of g varies on earth from one place to another because of various factors. On a given planet, the average value of acceleration due to gravity is the same for all objects and is independent of the mass of the object. In fact, it is

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a characteristic constant for that planet as it depends only on the mass and radius of that planet.

Illustration. Find the value of g on a planet whose mass is half that of earth and radius twice that of earth. Given g on earth = 9.8 m/s2.

Soltuion. In any such question, the best way is to write two equations:

Value of g on earth



gp =

Re2

GM p Rp2

Inserting given values,



GM e

Value of g on planet



ge =

We get

2

R  1 1 1 = e  × = = M e  Rp  2 4 8 ge 1 gp = × 9.8 ms–2 = 1.225 ms–2 8

gp

Mp

≈ 1.2 ms–2 At the time when Cavendish determined the value of G, there was lot of excitement the world over. In fact, Cavendish’s determination of G was publicized by the popular statement “Cavendish weighed the earth”. Can you explain how knowing the value of G, it is possible to calculate the mass of the Earth?

Variation in the value of g The value 9.8 m/s2 for the acceleration due to gravity on Earth is the average value and is not the same at all places on the earth. It varies due to the following reasons: (i) Density of earth is not uniform at all places (ii) Earth is not a perfect sphere (iii) The earth has a rotational motion about its own axis.

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Unit 6 : G  ravitation

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Can you explain how variation of gravity can be caused by variation of density in the earth’s interior? In fact, variation of gravity inside the earth is a tool for exploration of mineral and other deposits inside the earth. Search for ‘gravity measurement and exploration of mineral deposits’ on the internet for more information on this topic.

It is easy to realize that ‘g’ would also vary as we go up above the surface of earth and as we go deep down inside the earth. Let us now study the variation in the value of g in some detail.

6.9.1 Due to the shape of the earth The earth is not a perfect sphere. In fact, the shape of earth is more like an oblate spheroid. It is bulging out at the equators and a little compressed at the poles.

Polar diameter 12,714 km

North Pole

Thus, since

Re > Rp,

we would have

South Pole

ge < gp



Equatorial diameter 12,756 km

Figure 13: Picture showing difference in the radius of Earth at equator and the poles.

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Did You Know?

A person weighs (a little) more at the poles than at the equator.

6.9.2 With height (h) The value of g at a point P on the surface of earth

g =

GM R2

Unit 6 : G  ravitation

19

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If we go up above the surface of earth, the value of g at a point P′ at a height ‘h’ from the surface of earth, would be

g′ =

\

g′ g

=

( R + h )2

( R + h )2

h

P

GM

R2

P′

R



…(i)

We can also write the formula as Figure 14: Diagram showing point P on earth and P’ at a height h from the surface of earth.

This formula is valid for all values of ‘h’. However, it is usually used only when h is comparable to R.

R2



g′ = g



g′ = g  1 + h 

h  R2  1 +  R  

2

−2

R

−2

For h