The Paradoxical Universe

The Paradoxical Universe 250 Problems in Astronomy

V. V. Ivanov, A. V. Krivov, P. A. Denissenkov Edited By- Udbhav Singh

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Saint-Petersburg State University

Explanations, councils and the manual This training management is formally addressed to the students of low-order it is course the astronomical departments of universities, that study its first astronomical course - the course of general astronomy. But the authors hope that on these electronic pages interesting for themselves they will find everything, coma road astronomy - from the qualified amateurs to the astronomers - professionals. As the basis of benefit is assumed the many-year experience of teaching by the authors of the course of general astronomy for the students - the astronomers of the first course of St. Petersburg university. Was included also certain additional, only more complex material. Tasks are anticipated by the list of values of the thoroughly selected most important astronomical and physical quantities. This list, named by us the "universe in the numbers", is planned as the "numerical portrait" of the universe at all its structural levels - from the atomic nucleus to the metagalaxy. Student, who attentively studied the numbers, their analyzed, comparable with each other, etc., will obtain excellent idea about how the peace surrounding us is arranged. We recommend to always have a copy of the "universe in the numbers" with itself - in the briefcase, on the working table, in the computer. But are still better the data by the "universe in the numbers" to learn by heart. Having a minimum of constants near at hand, and those more holding them in the memory, it is possible with the ease to make different estimations, and estimation - important tool of scientific research. In many instances, when it does not be required precise or special information, the "universe in the numbers" will completely replace thick reference book, for example, the known book of Allen "astrophysical values" [ 1 ] or the encyclopedia "physics of space" [ 6 ] , although, of course, and without them sometimes not to manage. Subsequently we assume the numbers, which entered the "universe in the numbers", known and we do not indicate them in the formulations of tasks. The tasks (sometimes, faster, questions for the reflections) and their detailed solutions compose major portion of this teaching aid. They are frequently accompanied by extensive commentaries and additions, which are the important component element of benefit and make its not only with the assignment book, but also with the book for reading - by something like the reader. However, the aspect of the tasks - standard (frequently somewhat reformulated), undertaken from the known benefits (for example , [ 3 ]), majorities belongs to the authors. Special support is made to the tasks of estimated nature, whose solution develops in students the skill to make simple and efficient estimators and thus rapidly and without the

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cumbersome calculations to obtain correct idea about the diverse astronomical objects and the phenomena, which then, if this is necessary, they can be investigated more accurately and fully. We tried to follow known motto the "purpose of calculations - not the number, but understanding". Even microcalculator is not necessary during the solution of almost all problems. The authors know according to the experience of teaching that to instruction in accurate calculations with the high accuracy both on the average and in the higher school is given considerably more attention, than to the ordinal estimations or to estimate calculations with one significant place. Many student- freshmen, using the calculator (and all the more computer), diligently extract from the indicator (or the display screen) the long tail of the significant places and propose this as the answer. Let us give a characteristic example. On one of first practical training to students it is proposed to estimate the mass of air in the audience. The question excitedly is posed from the places: and are such the dimensions of audience? Answer: you will estimate "to the view". And even after this frequently is given an answer of the type: 438.75 kg instead of of correct, matched with the accuracy of initial data, answer: near polutonny. Solution of many problems actually is reduced to combine several numbers of the "universe in the numbers" in one - two simple formulas, brought out at the lectures, and to obtain the required answer. The task can serve as typical illustration: to estimate the mass of the earth's atmosphere. The answer in an obvious manner is obtained, for example, from radius of the earth, the height of homogeneous atmosphere and air density on the earth's surface (all three numbers in the "universe in the numbers" are). It is possible, it goes without saying, to solve problem and differently (as?). Let us give three useful advice. The first - to produce calculations with the reasonable accuracy - we already discussed. Second council - in the solution of one problem consecutively using one and the same system of units. It is possible to consider cgs system de facto standard in astronomy; use, for example, it. In this case it is possible not to follow the dimensionality of the intermediate results: it is clear, in what units will come out the answer (if mass - that in the grams). The third council - is more ideological, than technical: do not forget to evaluate everythingthat you make and which in you is obtained, from the positions of the common sense. And intermediate results, and final answer compulsorily must be comprehended and estimated, they are plausible. Commonplace council? But students continue to pass the solved tasks with the following answers to instructors: a radius of the Moon = 2.54 parsecs; the mass of the Earth = 5 . 10-5 grams; the number of stars in our galaxy =732 (all examples real). The authors of assignment book, checking tests, see such answers each yr! Truly, the collection of such results could compose the alternative "universe in the numbers" - the "universe in the number1s from the point of view of student, which does not follow to commonplace councils"... We relate to the proposed tasks as to the means, which makes it possible to develop the habits of logical thinking, creative relation to the problem, the fantasy. For this purpose we frequently propose tasks with a question "why?", and also namerenno by the "illegible" ("neshkol'nymi") formulations, which contrast with the standard "it is given - to find". As an example let us give the following task: "to estimate the time, for which a change in the outlines of constellations will become noticeable by rule of thumb". It is possible to discuss 3

as follows. It is clear that a change in the outlines of constellations is caused by the presence in the stars of proper motions. Star drift? ? (into ugl.sek./god) it is connected with its tangential velocity?Vt (in km/s) and with distance to star R in the parsecs by known formula?: Vt= 4.74 R ?. It is obvious that the stars, which determine the outlines of constellations, be among bright, and it means, on the average, the closest to us. Therefore it is possible to consider that the distances to the stars comprise order of tens of parsecs. What?Vt, then, after recalling the typical three-dimensional speeds of stars in the galaxy, it is possible to accept?Vt ~of 10 km/s. Substituting these values into the formula, we do find ?~0.2? ugl.sek./god. Further, completely it is possible to consider that the constellation by noticeable means does change when the determining its form stars do displace on?1o= 3600". We immediately conclude that the desired time interval is about 20000, and it is better to say, tens of thousands of years this is - correct result. See another, even simpler approach in the task. But even when the formulation of task is maximally clear and its solution, it would seem, does not require special creation, in student the possibility to appear initiative remains. Thus, in task about mass mentioned above of the Earth's atmosphere simple calculation does give the answer: ~ 5. 1021 g. but not are more interesting whether, without being limited to this (correct) answer, to compare the obtained number with some others? Natural to take the mass of Earth itself for the comparison: ~ 6. 1027 g. We can say as the answer: the "mass of the earth's atmosphere - one millionth of the mass of the Earth". What answer is better? Some tasks have the increased difficulty, and their solution will be hardly accessible to student- freshmen. They are noted by asterisk (*), and separately difficult - by two asterisks (**). All such tasks are original and nonstandard. Their start, as we hope, will make this benefit interesting to the wider circle of the readers, and the students of loworder it is course, after becoming acquainted with the solutions, they will enlarge their horizon. We already said that partly this benefit is planned not only as assignment book, but simultaneously and as its kind reader.

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Contents Problems 1. First acquaintance 2. Kinematics of the sky 3. Tools 4. Kinematics of the solar system 5. The universal gravitation 6. Earth, the Moon the planet 7. The stellar magnitude 8. Emission 9. Spectra of the sun and stars 10. Sun 11. Fundamental characteristics of the stars 12. Physics of the stars 13. Dual and variable stars 14. The interstellar medium 15. Galaxy 16. Beyond the limits of the galaxy 17. New tasks

Solutions 1. First acquaintance 2. Kinematics of the sky 3. Tools 4. Kinematics of the solar system 5. The universal gravitation 6. Earth, the Moon the planet 7. Solutions to chapter 7 not available in English 8. Emission 9. Spectra of the sun and stars 10. Sun 11. Fundamental characteristics of the stars 12. Physics of the stars 13. Dual and variable stars 14. The interstellar medium 15. Galaxy 16. Beyond the limits of the galaxy

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1. First acquaintance 1.1 of the substance of the Earth made a wire length from the Earth to a) of the sun; b) Cen; c) of the Andromeda nebula. To estimate the diameters of these wires ( V.V.Sobolev). 1.2 You will estimate how many stars in our galaxy and how many galaxies in the entire universe are fallen to one person. 1.3 As you do think what more - stars in the galaxy or gnats on the Earth? 1.4 universe are entirely small: stars in the entire universe as much of, atoms as in the drop of water! Prover'te. 1.5 Which is more: the Coulomb force of the attraction of electron to the proton in hydrogen atom or the Newtonian attracting force of two cosmonauts, who are exchanged handshake in open space? 1.6 Which the depth of that potential pit, in which we do live? 1.7 Which distance modulus Saint Petersburg - Moscow (in the stellar magnitude)? 1.8 is evident whether the sun by naked eye from the Pleiades? 1.9 That is greater - the angular dimension of the disk of the Moon or Andromeda nebula? 1.10 Why it is at night dark?

2. Kinematics of the sky 2.1 After awaking on 21 March, you unexpectedly revealed that they were transferred by some miracle in the uninhabited atoll. Shines the sun, on you - only bathing suit. You will be able in the hour - other to say, in what hemisphere you are located - northern or southern?

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2.2 After awaking from the lethargic sleep, you revealed that you are located in the uninhabited atoll, which lies accurately on the equator. Shines the sun, on you - only bathing suit. Will be able you in the hour - to other to say that now you have house in Petersburg - winter or summer? 2.3 After awaking, you unexpectedly revealed yourselves among the primitive people. How you would begin to convince them, that the Earth - this is sphere and that it does revolve around its axis? 2.4 In what direction - from left to right or from right to left - is moved because of the precession the spring equinox point? 2.5 are such minimum and maximum midday altitude of sun in your city? 2.6 In which hour will today visit the sun? 2.7 On 22 March the sun visited to in Arzamase?

later than the day before. Where you - in Armavire or

Indication: sin = sin 231/2o = 0.40 . 2.8 What fraction of the total number of stars on this latitude never does go? Stars to consider evenly distributed in the celestial sphere. 2.9 Is visible the Moon into the June plenilune at the North Pole? 2.10 Why in summer the Moon in the plenilune is always dim and yellowish and is visible low above the horizon (in our latitudes), and in winter, on the contrary, it is bright and does rise highly? 2.11 Which the greatest and smallest height of the upper culmination of the Moon in your city? 2.12 is it possible in Russia to somewhere see the Moon in the zenith? 2.13 Moon is visible in last fourth. To what on the average are equal the radial velocities of stars, find on the sky not far from it? On 2.14 7 February the Moon was in last fourth. To what did be equal its right ascension? 2.15 constellation of Orion has equatorial coordinates near

, ~ 0o, and the constellation

of southern crown , ~ -40o. Are such the visibility conditions of these constellations in Petersburg now? What they will be in 13 000 years? On 2.16 22 June on the solar disk from the right side is visible damage. What these are beginning or the end of the eclipse?

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2.17 today occurred the solar eclipse. It means, the angular distance of the sun from the node of moon's orbit is less than 17o. You will explain, why. Indication: the angle of the slope of lunar orbit to the plane of ecliptic

.

2.18 v which to hour (Moscow time) today will begin true noon in that place, where you are located? 2.19 sun is located in the upper culmination in time to disregard).

universal time. To find length (by equation of

2.20 Earth - spheroid with compression by 1/300. A) To determine, on how much in the twentyfour hours will leave forward pendulum hours at the terrestrial pole in comparison with the analogous hours at the equator. b) On how much should be lengthened pendulum it is hour at the pole, so that the hours you dispatch so? Centrifugal force not to consider. 2.21 It is known that the duration of twenty-four hours increases by seconds in 1000 years. Therefore in 1000 years of indication it is hour, synchronized in the direction of the rotation of the Earth taking into account its age-long retarding, they will differ from the indications of atomic hour, synchronized with a constant initial velocity of the rotation of the Earth. To it is how much? (EXPLOSIVES. Vityazev)

3. Tools 3.1 to estimate the safe speed of the motion of the controlled from the Earth Mars rover, equipped with the television camera, which "sees" only on 10 m. 3.2 Where on the Earth it is necessary to place telescope so that its German mounting would be simultaneous and azimuthal? 3.3 will be sufficient one disk in order to write down the entire the information about the celestial sphere, which humanity was located up to the moment of the invention of telescope? (Sun, the Moon, planet and comet - not on account.) 3.4 Which smallest linear dimension of the formations on the Moon, which can be distinguished by the naked eye? 3.5 satellite does fly above the earth's surface at the height of 200 km which minimum size of the details on the Earth, which can be from it photographed?

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Indication: the size of the disk of the vibration of star because of the influence of the earth's atmosphere comprises

.

3.6 beyond the orbit of Neptune, at heliocentric distances from 30 to 50 AU, is located that opened in 1992. Kuiper's belt, which consists of the small bodies of the type of comets and asteroids. At the beginning of 1996 in the belt are discovered 32 objects; in 1996 it is opened 7 more. It is assumed that there there are tens of thousands of bodies of larger than 100 km they are such the approximately minimal sizes of the bodies of Kuiper's belts, which can be revealed with the aid of the khabblovskogo space telescope? 3.7 To estimate the photon flow (number of fotonov/(sm s)), that is necessary from: A) the suns; b) Cen; c) of star . Indication: flow from the source of zero stellar magnitude composes approximately fotonov/(sm s). 3.8 how many photons do fall in 1 second on the mirror of the largest in the world telescope of cake (D = of 10 m) from Vegi and from the star ?

4. Kinematics of the solar system 4.1 as in the task , you again among the primitive people. Continuing to sow reasonable, good, eternal, you explained by them, as the solar system was arranged, but they revealed to horror their that distance from the sun and the periods of revolution of Venus and Jupiter you forgot. What observations you should be produced in order to determine their values? 4.2 na how much approximately degrees was displaced Pluto along its orbit from the time of its discovery? 4.3 period of revolution of Neptune around the sun is 165 years. Will be in the sun visible disk, if we look at it from Neptune by the naked eye, or it will be point target? 4.4 to depict in the projection on the plane of ecliptic the orbit of celestial body, which has the elements: AU, e = 0 . For the comparison on the same drawing and on the same scale to depict the orbit of the Earth. 4.5 How angular distance of the Earth from the sun in the elongation with the observation from Jupiter? 4.6 At what greatest angular distance from the sun there is visible Jupiter with Cen?

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4.7 sidereal period of the axial rotation of the sun (at the equator) of --- of 25 days. To what is equal the synodic period of its rotation with the observation with A) mercury; b) the Earth; c) of Pluto? 4.8 You will estimate the characteristic transit time of Venus on the solar disk. How it is moved along the disk with the observation from middle latitudes of the northern hemisphere of --- from right to left or from left to right? 4.9 diameter of Pluto composes 2300 km, distances from the sun (to the end of the current millenium) of --- 30 AU. You will estimate zone width on the earth's surface, in which it is possible to observe coating star with Pluto, and its duration. 4.10 astronomers of future produce the radar of the asteroid, which moves along the circular orbit by radius 5/3 AU. By how many times more powerful pulse they must send in the connection, than in the opposition in order to register the signal of identical force reflected? To how many stellar magnitude this asteroid brighter in the opposition than in the connection?

5. The universal gravitation 5.1 which would become with the solar system, if the mass of the sun did instantly decrease doubly? 5.2 as would change the orbit of the Earth, if the mass of the sun suddenly was doubled? 5.3 why in all without the exception of the bodies of the solar system, from the sun to the asteroids and the satellites of planets inclusively, the minimum times of their circling do have one and the same order of magnitude? 5.4 speed of the motion of comet in the perihelion three times as great than in the aphelion. To what is equal eccentricity of its orbit? 5.5 as does depend the linear speed of the motion of planet along the circular orbit on a radius of orbit R? No matter how unexpectedly, a question refers straight to the famous problem of the "concealed mass" in the galaxies 5.6 to determine the semimajor axis, the revolution period, eccentricity and perihelion distance of comet, which at a distance of 1 AU from the sun has a speed, directed perpendicularly to radius-vector, also, along the value ten times less than the speed of the Earth.

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5.7 satellite "Molniya"; in the perigee it is separated from the center of the earth by 7000 km, in the apogee - on 46000 km you will calculate circling time of satellite around the Earth and eccentricity of its orbit. 5.8 you will estimate the time of the overflight of automatic spacecraft from the Earth to Mars on the semi-ellipse, in the perihelion which is concerned the orbit of the Earth, and in the aphelion - the orbit of Mars (Hohmann ellipse). Semimajor axis of the orbit of Mars to accept equal to 1.5 AU. 5.9 Geliostatsionarnoy is called the circular orbit, which lies at the equatorial plane of the sun, with the revolution period, equal to the sidereal period of the axial rotation of the sun. To find its semimajor axis. 5.10 which the minimum speed of a drop in the meteorite on the Moon? 5.11 what must be the rate of accretion to Jupiter (in the masses of Jupiter per year) so that its bolometric luminosity would grow doubly? 5.12 when people intended to land on the Moon, actively was discussed a question, it was not covered with the thick layer of dust, in which it is possible to sink. Then in students this question arose: how long must on the Earth work vacuum cleaner, so that the isolated energy would be sufficient for its delivery to the Moon? 5.13 which limit of size of the asteroid, from which still it is possible to jump into space? 5.14 it is improbable, but the fact: from the friction against air the satellite, which flies in the rarefied layers of the atmosphere, is not impeded, but it is accelerated! To understand, as so it can be, after proving that energy, which the satellite expends on overcoming of air resistance, exactly is equal to an increase (positive) in its kinetic energy. 5.15 how must be distance to the Moon, so that in its orbit in the solar system would be points of inflection? To consider that the Moon dvizhetsya strictly in the plane of ecliptic, and its orbit relative to the Earth - circular. 5.16 spacecraft investigates neutron star. At what approximately distance from it tidal forces yet will not create the danger in the health of cosmonaut? 5.17 several years ago in the solar system is opened the new class of the objects of --- dual asteroids. You will roughly estimate the greatest possible distance from the 100-kilometer asteroid to its satellite - the asteroid of smaller size. 5.18 surface, on which the attracting forces to the Earth and to the sun are equal, is called the sphere of the attraction of the Earth relative to the sun. Is the sphere of attraction actually sphere? With what mass of the Earth the sphere of attraction would be plane?

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5.19 raised stone to height R above the surface of the planet of radius R and mass M and let go there with the zero speed. From dimensional considerations to show that the time of its free fall on the surface of planet exists

where --- dimensionless coefficient (order of one). Without attempting to find obtain the following estimation:

accurately, to

5.20 Task- parody to the fantastic subjects about the starry wars. Reconnaissance reported, that the nuclear-powered submarine of enemy must float at the north pole for 10 minutes exactly at noon on our time. DAN order it to destroy. For this to the pole in the morning arrived our submarine, from onboard of which was vertically upward neglected the combat missile. In the very short time (count - instantly) rocket collected the speed, equal to the first space, after which its engines were opened. Our submarine here was immersed and swam away. After a certain time - exactly at noon as prevented the reconnaissance, on the surface appeared enemy submarine. But it did not pass even five minutes after its emersion, as it was destroyed by rocket fallen to it from space. How much did show the hours of the commander of our valiant submarine, when it did return command "launching!"?

6. Earth, the Moon the planet 6.1 to what is equal the temperature of the surface of the Moon at sunflower point? 6.2 with the building of lunar observatory muddler- astronaut at hot noon spilled to the lain on the soil thick iron sheet the bucket of remarkable white- white whiting (albedo A=0.99). It was revealed after a certain time that to work, costing on this sheet, is completely impossible. How you do think, why? 6.3 let us assume that around the star with the temperature and the average density dvizhetsya the planet. Let it not have an atmosphere, it is inverted to the star by one side and has the same albedo A, as on the Moon. To what is equal the period of revolution of this planet along the orbit, if temperature on its surface is the same as on the Moon? How does depend

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temperature on the planetary disc on the zenith distance of star? It is considered that also the star, and planet emit as blackbody. 6.4 is there sufficient whether the kinetic kinetic energy of the Earth along the orbit in order to ionize the mass of hydrogen, equal to the mass of the Earth? 6.5 what fraction of the mass of the Earth is fallen to its atmosphere? 6.6 with a good approximation (by what?) it is possible to consider that the sun illuminates the Earth by parallel light beams. As then to explain the picture, which you for sure repeatedly observed: from under the cloud, which closes the sun, do go the clearly divergent rays? 6.7 the earth's atmosphere has optical thickness along the normal (at the wavelength ). Visualize that the mass of the atmosphere grew 5. With what will see the cloudless sky of peterburzhets at noon and in midnight? But that if the mass of the atmosphere does decrease 5? 6.8 with a smooth increase in the mass of the atmosphere from zero skies would be in the daytime first black, then everything would become brighter, then again by less bright and finally would begin the dark. At what optical thickness is reached the maximum of the brightness of sky? 6.9 sky from the morning to the evening is tightened by the overcast, but the disciplined astronomer honestly works --- on its solar telescope it it attempts to be occupied by the determination of the chemical composition of the sun. How you do think, can this something leave? 6.10 sun arises from behind the sea horizon. One person admires rise from deck of ship, and another of --- from the mountain peak with a height of 4 km on the located in ocean volcanic island (let us say, from the observatory of Mauna Kea in Hawaii). In which of the observers the solar disk will be more brightly, also, into how much approximately once? 6.11 everyone knows that on the rise and with the sunset is red. To show that this in a sense illusion, in fact it infrared. More precisely saying, to show that the maximum in the energy distribution in the spectrum of the direct radiation of the sun, recorded at the moments of rise and sunset, lies at the IR range. To consider that 1) in the exoatmospheric spectrum of the sun the energy distribution blackbody with the maximum at the wavelength extinction of light in the atmosphere is caused by molecular scattering.

and 2)

Note. The calculator will be required for the solution of this problem. The solution is sufficiently long.

7. The stellar magnitude 13

7.1 to largest ground-based telescopes (in particular, to two largest in the world 10- meter telescopes of cake) are accessible stars . By how many times they are weaker than stars, hardly distinguished by the naked eye? 7.2 the variable o cet (peace of whale) in the maximum has visual luster , in the minimum of --. In how much approximately once do change its luminosity in the visible region? 7.3 difference in the stellar magnitude of two stars of identical luminosity is equal many times one of them is further another?

. By how

7.4 where it is brighter than --- in the daytime on Pluto or on the lunar night on the Earth? 7.5 of the substance of the Moon into the plenilune made millions of identical spherical satellites, after leaving them approximately in the same place, but so that they would not overshade each other. Which the stellar magnitude of the received cluster? Stellar magnitude of full moon is known to you. 7.6 binary star has components

and

. To find the summary stellar magnitude of dual.

7.7 Zatmenno - binary system has identical components. On how many stellar magnitude do change the luster of system at the moment of the total eclipse of one component of another? 7.8 v star cluster N of the stars of stellar magnitude m each. To find the summary stellar magnitude of accumulation. 7.9 before the sky of approximately 6000 stars is brighter than the 6th stellar magnitude. Considering, that in all stars identical luminosity and that they are distributed in the space evenly, to estimate, it is how much before the sky of stars of brighter than mof -1 stellar magnitude. 7.10 * on what approximately distance it is necessary to place 100 - watt lamp so that it would look like the star of 0-1 stellar magnitude? 7.11 * is it possible from the Moon with the naked eye to see cities on the Earth? 7.12 * one of four Galilean satellites of Jupiter --- Europe --- has a radius of 1600 km and it is completely covered with ice. You will estimate the stellar magnitude of Europe at that moment, when Jupiter nearest of all to the Earth. 7.13 to show that with a small change in the distance to the luminous object ( visible stellar magnitude changes on

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) its

7.14 the stellar magnitude of planet in the opposition to 3.43m is less than in the connection. What this after planet? 7.15 how long did pass from the connection to the opposition of planet, if its stellar magnitude in this time did decrease by 0.85m ? 7.16 you will estimate the maximum distance (in PK), from which the sun by the also evidently naked eye. 7.17 as you do think, there will be the sun in the number of one - two ten brightest stars for the observer, who lives in environments Cen? 7.18 are accessible to the largest telescopes of the stars, similar to the sun, in the Andromeda nebula? 7.19 to estimate the absolute bolometric stellar magnitude of the working vacuum cleaner.

8. Emission

8.1 * the very good estimation of blackbody intensity in the maximum of Planckian curve is obtained, if we use ourselves the approximation of wine. You will explain, why this so, and give estimation of error. How far beyond maximum, i.e., in the region possible to use the approximation of wine? How

, still it is

in the maximum it does grow with T?

8.2 * you will examine the family of Planckian curves in the scale of the wavelengths, which correspond to the different values of the temperature (see any course of general physics). They do not intersect, those, which correspond to higher temperature, lying above (prove!), and maximum in them is located to the left (Wein's displacement law). In this case maximum in these Planckian curves the "sharper", the higher the temperature. Why this thus? But as does change with an increase in the temperature the form of Planckian curves in the frequency scale? 8.3 ** to obtain the following exponential approximation of the dependence of blackbody intensity

on the temperature in the environment

where

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:

It is hence clearly evident that upon transfer into the short-wave part of the spectrum the sensitivity to variations in the temperature becomes stronger of --- the very important property of the Planck function. 8.4 * why the medium energy of one blackbody photon is not equal 8.5 * to show that photon concentration T is equal

? To find it.

in the equilibrium field of emission with temperature

.

8.6 upon what transfer of electron in hydrogen atom is formed the lying at the red part of the spectrum line

)?

Indication: to consider known the maximum wavelength of the photon, even capable of ionizing hydrogen atom from the ground state:

.

8.7 are possible the ground observations of the line of interstellar hydrogen H electron from the 11th to the 10th level)? 8.8 to estimate the wavelength of the recombination radio link of hydrogen H 8.9 * to prove that the recombination radio links of hydrogen frequency. 8.10 is capable whether emission second level?

,

,

(passage of

.

are equidistant in the

of ionizing hydrogen atoms, which are located on the

8.11 * why in the spectrum of the solar chromosphere the evidently much larger number of lines of a Balmer series, than in the spectra of white dwarfs?

9. Spectra of the sun and stars 9.1 to estimate the minimum width of Fraunhofer lines. 9.2 star of class B0V have a period of axial rotation . To find the characteristic widths of lines in the spectrum of this star in the visible region of the spectrum (in the angstroms).

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9.3 * lines H and K of ionizovannogo calcium of --- the strongest in the visible part of the spectrum suns. (their equivalent widths of --- 14 and 19 , respectively.) The strongest of the hydrogen lines of --- these are Balmer lines and (their equivalent widths they are close they are about 4 ). Why lines H and K are considerably stronger than hydrogen lines, although hydrogen of --- fundamental component of solar atmosphere, and calcium of --- the minor constituent? 9.4 ** the prevailing detail of the spectra of the stars of class A0 in the optical part of the spectrum of --- large Balmer jump on . Is great whether the darkening of the disks of these stars to the edge na and on ? but 9.5 ** energy distribution in the spectrum of sun (G2V) is close to the blackbody with k. why in the spectrum Of vegi (A0V) it in no way similarly to the Planckian s k? 9.6 ** the atmosphere of the sun can be considered gray. This means that the emission of all wavelengths (in the visible part of the spectrum) is weakened by it equally. Why then the darkening of solar disk to the edge with the decrease of wavelength does become more? 9.7 ** as is known, the fundamental component the atmosphere suns of --- these are hydrogen and helium. The number of atoms of all other elements, together, --- so-called "metals" of --- is from the number of hydrogen atoms. How will change the mass of solar atmosphere, i.e., those it is layer, from where emission does come to us directly, if the content of "metals" decreased 10 times?

10. Sun 10.1 ** as does be born daylight? It is more precise: which that concrete elementary atomic process, proceeding in the atmosphere of the sun, with which are emitted the photons, received by us as sunlight? 10.2 visualize improbable: the gigantic spot, which shut entire solar disk, developed in the sun. It will become in the daytime darker than on the lunar night? 10.3 prove that the gravitational pole of the Sun not is capable of holding the electrons of the solar corona. 10.4 ** the brightness of crown from the brightness of the sun. It shines due to the scattering of light of the sun on the free electrons. To estimate the mass of crown.

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10.5 approximately days after the chromospheric flare appear different geophysical disturbances. Which kinetic energy of their causing protons (in eV)?

11. Fundamental characteristics of the stars

11.1 stars of the main sequence of the spectral class B0V ( K) have a mass After using mass-ratio of --- luminosity, to estimate their average density.

.

11.2 along luminosity and radius of the sun to calculate flow with 1 cm2 of its surface, and on it -- effective temperature. 11.3 as will change the luminosity of star, if its radius slightly decreased (let us say, to 2%), and the effective temperature of --- on so many increased? 11.4 to find a difference in the absolute stellar magnitude of two stars of identical size, whose effective temperatures differ to 11%. 11.5 luminosity of one of two identical with respect to size stars to 4% is more than another. Considering that emission of both stars blackbody, to find, on are how much characterized by A) temperatures of stars; b) the wavelengths, which correspond to maximum in the energy distribution in the spectrum; c) of the intensity of emission in the maximum of spectrum; g) the intensity of emission at one wavelength in Rayleigh -djinsovsko1 the region of the spectrum. 11.6 * to find height of the uniform hydrogen atmosphere for A) the suns and b) white dwarf s k, 11.7 white dwarf has magnitude.

and

. to and

. To find his absolute bolometric stellar

11.8 40 Eri B of --- one of the first open white dwarfs (and one of most studied at the present time). It has effective temperature 17000 k and absolute stellar magnitude . To find its radius. 11.9 * to estimate the thickness of foil from shokoladki, sowing to which in the environment of the star of class O5, gnat could fly away on it as on the photon sail to other stars. At the moment, when gnat sits down itself on the foil, it rests.

12. Physics of the stars 18

12.1 rate of energy release per unit of mass in the human tele- several orders higher than in the sun. Why we are much colder? 12.2 molar mass of the substance of solar depths is close to 0.6. Why it is less than one? You will estimate the total number of particles, which compose the sun. 12.3 density and temperature in the center of the sun are equal, correspondingly, 150 g/sm3 and K. What it is there more than --- of photons or particles? 12.4 * average solar density (1.4 g/sm3) is more than water density. Why it it is possible to consider gas sphere? 12.5 mass - particle approximately to 1 % is lower than summary mass of four protons. You will ascertain that the thermonuclear synthesis reactions - particles are capable of supporting the present luminosity of the sun at duration of billions years. 12.6 to estimate the relation of the numbers of photons and neutrino, second-by-second radiated by the sun. With synthesis of one - particle is separated the energy 26.7 MEV, the neutrino taking away only the small part of this energy (of about 2%). 12.7 * you will determine the rate of accretion (in the masses of the sun per year), which could ensure the observed luminosity of the sun. How this accretion it would affect the duration of year? 12.8 temperatures in the centers of two stars are equal TO K and To estimate the ratio of the corresponding velocities of energy release. 12.9 which smallest possible period of the axial rotation of the red giant s radius km?

K.

, which has a

12.10 in the quite short-period pulsar PSR 1937---214 period is one-and-a-half milliseconds. To estimate its density. 12.11 * visualize impossible: in interior of the suns gas pressure ceased to exist. In what time it will collapse into the point? But how long does depart to the compression of interstellar cloud with an initial density of g/smof 3?

12.12 ** when began interstellar overflights, about one of the stars was opened the planet, which consists of the uncommon substance. A study of the models of this substance showed that it is compressed with the pressing; moreover pressure P and density of substance are connected

19

with the relationship laboratory.

, where K of --- certain constant, which was measured in the

Many other properties of this substance made with its truly priceless. Therefore, in spite of the distance of planet and the high cost of transport, was begun the output of this substance on enormous, truly space scales. But here was revealed something improbable: how much not took away substances from that surprising planet, its size from this did not change! Prove that this "miracle" is possible, and find a radius of this planet. (just in case is immediately it prevented, that relativistic effects here not with what; gravity - Newtonian). Indication. To use the method of dimensional analysis. For obtaining the precise value of a radius use an equation of hydrostatic equilibrium.

13. Dual and variable stars 13.1 * let --- the radial velocity of component spectral- dual with period P. Understand the physical sense of the values

and

In the second case we count the orbit of circular even we assume that the center of the masses of dual rests. 13.2 * dependence the period of --- absolute stellar magnitude in strip V for the cepheids takes the form < Mv> = -1.3m -3.0m lg P, where P of --- a day and < Mv> of --- the average value of absolute stellar magnitude. To estimate distance to Cep (period 5.3 d). 13.3 * in cepheid with the average period are observed its sine flutters with an amplitude of 43 seconds, proceeding strictly with the annual period. The period of cepheid is

20

maximum on 20 December. Why it is possible to assert that this star is located near the spring equinox point? 13.4 * the effective temperature of cepheid changes between 5500 and 7000 k, and its bolometric stellar magnitude of --- on . By how many times does change a radius of star in the course of pulsations? Phase lag in the curve of radial velocities to disregard. 13.5 Malomassivnyy component of binary star s , and by period is X-ray pulsar with the true period of 0.1 s. find, within what limits the observed period of pulsations because of the Doppler effect changes. Orbit circular, slope angle , i.e., line of sight lies in the orbital plane of orbit. 13.6 to make, which prove, that, live we in system Cen, to astronomers would be difficult to see planets near the sun. See, however, task . 13.7 which must be the accuracy of the determination of A) radial velocities and b) proper motions, so that the astronomer with Cen could suspect existence of planets in the sun?

14. The interstellar medium 14.1 * in the star of class A0 interstellar absorption in the visible range is equal observed color index of this star?

. Which

14.2 there is a planetary nebula, whose nucleus --- very hot star (T = 80000 K). The large part of the emission of star is absorbed in the fogginess. Why through the fogginess are well visible distant galaxies? 14.3 * how many interstellar atoms of hydrogen does ionize second-by-second star with WITH T = 100000 k and

?

14.4 * star with T = 16000 K are submerged in the interstellar medium. You will estimate, what portion of the energy emitted by it proceeds with the ionization of interstellar hydrogen. To consider that the star shines as blackbody. 14.5 * to show that a radius of zone HII depends on the concentration of hydrogen n as 14.6 to estimate the mass of the molecular cloud, which consists in essence of H2, if its size PK, and concentration of particles cm-3.

21

.

14.7 * you will estimate the mass of neutral hydrogen on the line of sight (into g/sm2), which generates noticeable ( %) absorption in the center of line L . Absorption coefficient in the center of line taking into account one atom approximately 100 k).

(at a temperature of gas of

15. Galaxy 15.1 za how many years the Earth in its annual motion around the sun does pass the way, equal to distance to Cen? This is - the important characteristic time: the form of the celestial sphere on the Earth substantially changes in this time interval. Why? 15.2 to estimate eccentricity of the parallactic ellipses of Spica ( Vir) and by polar. 15.3 "flying Barnard's star" have a record large proper motion 10" V and a parallax 0.5". To estimate its tangential velocity. 15.4 star, which is been located at a distance of 10 PK, dvizhetsya with a speed of about 50 km/s perpendicular to line of sight. To what is approximately equal to its proper motion? 15.5 Vegas - this is the star of the spectral class A0V. Its visible stellar magnitude estimate parallax.

. To

15.6 the Crab Nebula - this a remainder of the supernova, which flared up in 1054 it it has, speaking in general terms, the form of the luminous ellipsis with the size of ''.kh 6'. The measurement of the radial velocities of gas of fogginess showed that it is enlarged with a speed of about 1200 km/s. To estimate distance to the fogginess. 15.7 two cepheids exactly of one and the same period are located in the Andromeda nebula of -- one on the nearest, another on outermost from us the edges of its disk. You will estimate a difference in the stellar magnitude of these cepheids. 15.8 * five of seven stars of the ladle of Ursa Major (besides "extreme") have practically identical proper motions and nearly equal radial velocities. After recalling, as appears ladle before the sky in the comparison with other stars, to estimate distance to the Ursa Major (more precise, to these five stars of ladle). 15.9 astronomers for the first time carried out observations in the ultra-unknown (UN) and infranevedomom (IT WAS OTHER) ranges. In UN they found discrete sources, moreover it turned out that they were distributed throughout entire celestial sphere evenly. In they were also other discovered previously unknown objects, moreover it turned out that these sources are

22

concentrated to the galactic plane (but not to it to center). What it is possible to speak about object distances, opened in UN range? But about the distances to the Other- sources? 15.10 on the distance to the center of galaxy and velocity of the motion of the sun along its galactic orbit to estimate the mass of galaxy.

16. Beyond the limits of the galaxy 16.1 you, of course, remember, as appear the usual photographs of the Andromeda nebula (Mey). How you it does seem, which approximately angle of the slope of the galactic plane Of mey to the line of sight? After you, anywhere without glancing, made an estimation, prover'te yourselves, after taking the necessary measurements of the photograph of this fogginess in the answer. 16.2 as would appear the Andromeda nebula for the naked eye, if we did see it from the edge? 16.3 you will estimate the absolute stellar magnitude of the supernova, which flared up in 1987 in the large Magellan cloud. In the maximum of luster it had the visible stellar magnitude of approximately 3m. 16.4 width of line in the spectrum of the nucleus of Seyfert galaxy are about 30 . Which spread of characteristic velocities of the motion of the clouds of the radiating gas in the nucleus of this galaxy?

16.5 radio source in the nucleus of active galaxy has angular dimension of 0.001", the amount of red shift z = 0.5. You will estimate the linear dimensions of source in PK. 16.6 galaxy are moved away from us at a rate of 5000 km/s. It is visible as object with size of 1'. To estimate its linear dimension. 16.7 * what more in the universe - protons or relict photons? Average density of substance in the universe to accept equal to 10-30 g/sm3. The temperature of relict emission is equal to 2.7 K.

New tasks 1.

23

2.

Surprising fact: The Earth obtains from the sun more heat, than Venus despite the fact that is located 1.4 times further from it! And the matter here completely not in the fact that the Earth only is more than Venus - the discussion deals with the heat, which is fallen per the unit surface area, and not to the planet as a whole. So in what the matter, as so it can be? And why then on Venus nevertheless much to zharche, than on the Earth?

3.

If the sun was wooden and oxygen it would be in the abundance, then in it not nuclear, but usual fire would blaze. Through how many days the firewood will burn down and the sun will go out? To consider that the solar firewood bonfire blazes with such force, which ensures the same energy release as in the present sun. The mass of firewood is equal to the mass of the sun. 4. After arriving in the machine of time to the ancient Greeks, you met in Rhodes with Hipparchus and told to it, as the solar system was arranged in reality. Some of its parameters, in particular, the orbit eccentricity of the Earth (e= 0.0167), you, unfortunately, forgot. However, to take with yourselves into the past at least reference book Allen [1], to say nothing of the computer connected to the Internet, you did not find time. To find in the antiquity it was for long cannot, without risking to remain there forever, your desheven'kaya machine of time it was entirely primitive. Therefore to describe to Hipparchus about the telescopes did not be sufficient time. It was necessary to urgently return to present, hardly after having time in the last minute to present to shaken Hipparchus its wrist quartz hours. Hipparchus, after believing in Kepler's laws, decided to determine itself from the observations of the sun, to what it is equal to e. What it could dostich', if soon it did not die? 5. 17 June midday altitude of sun to less than on 22 June. On how much approximately it does grow from 22 December until 1 January? 6. We will indicate that the sun in the zenith, if it shuts zenith with its disk. Where such it is possible to see more frequently - in Quito or into San -Paulu? 7. For several days before the new year the fashionable hotel has on famous oceanic Kapakabana on the bank, that in the center of Rio de Janeiro, appeared this advertising panel: "nowhere in the world you will be able never to tan after sea bathing more rapidly than on Kapakabane on the following week!" To prove that this is truth. Latitude of Rio de Janeiro

.

8. Russian astronomer arrived to work on the contract to Rio de Janeiro. Manifesting enviable diligence, whereas during the first workday it observed on its astrometric tool the sun from the rise to the sunset. In this case with the consternation (but on the truth speaking, even with a certain fear - "already I do not descend 4 from the mind?") it revealed that from the morning the sun was moved on the sky from left to right, and in the after-dinner time - from right to left. What was days 10 prior to far - Easter or Christmas? But will not be able you to say, what precisely number our suffering from the incredible heat Russian did approach the work in

24

Brazil? The latitude of the university observatory, on which it observed, was equal . 9.

In 1997 the declination of the Moon varied each lunar month approximately from

to . In what year the declination of the Moon will vary from to ? But here is a question potrudneye: in what constellation find in 1997 the ascending node of moon's orbit? 10. Are such the smallest and greatest duration of sunrise, i.e., the time interval, during which the solar disk "does come out" from behind the horizon? Refraction not to consider. 11. To estimate the duration of sunrise on the Moon on the average lunar latitudes. 12. How many radians one hour? 13. Twenty-four hours on Mars are nearly equal terrestrial - in all to 2.5% is more prolonged than our, and the period of its revolution around the sun - about two years. To how much approximately on Mars the sideral days are shorter than average solar? 14. That that yr - this is the period of revolution of the earth around the sun, any child knows. This assertion is inaccurate in reality. Why? To estimate a relative error in this well-known truth. 15. The so-called equation of time

is the difference between the "present", i.e.,

evenly current time t and by apparent solar time moment t: With an accuracy to

, i.e., by the hour angle of the sun at

it is given by the following expression:

where

How the duration of apparent solar days it does change during the year? When it greatest and when - smallest? To the how much longest solar days of more prolonged than the short? 16. Student- astronomers decided to finish celebrating the astronomical beginning of the third millenium. What number and into how much Moscow time they had to raise glasses with champagne, if it is known that astronomically 1997 it began on 31 December, 1996, v on Greenwich. 17. In one and the same date, on 27 February, 1900 , at noon local time one steamship left from Vladivostok to San Francisco and, etc. - from San Francisco to Vladivostok. Both find in the way exactly on 10 days. Immediately on the arrival into the ports of designation the captains of each of the steamships sent on the telegram, which notified their shipowners about the satisfactory completion of voyage. What date was written on the telegram, which left from Vladivostok? But on that that it was sent from San Francisco?

25

18. Live Hipparchus on Mars, that from that opened by them from Rhodes it, observing sky from Mars, would not reveal? 19. In Durlandii from 1 January is produced the denomination of the speed of light. Its value is reduced 1000 , with 300000 km/s to 300 km/s. How from this did change the arrangement of stars before the sky? 20. Let us visualize that in the northern pole of the ecliptic is opened the belonging to the solar system object, which they observed exactly during the year. In this case it was explained that, in contrast to all adjacent stars, the object does not show annual aberrational displacement. Why it is possible to assert that the period of revolution of this object around the sun is not less??? years? 21. In 2098 the astronomers Of futurlandii, using old ground-based dvadtsatimetrovym telescope, opened in the solar system object remarkable in many respects. It turned out that the visible way, passed by object before the sky in five years of observations, takes the form of the piece of the stretched chain with five by elongated components, as if assumed on top on the rope.

Version 1 (complex): To estimate the amplitude of changes in the luster of the object opened by futurlandtsami. Version 2 (simpler): To estimate angular dimension of "major axis" is sectional this chain and period of revolution around the sun of opened by our futurlandskimi associates asteroid. To estimate also its sizes, if it is known that the visible stellar magnitude of object of approximately 31m. To find the amplitude of its changes. To mark on the "chain" of the points, in which the luster of object reaches extrema. 22. Futurlandskiye astronomers conduct astrometric observations on the small expeditionary telescope, which they temporarily installed at one of the bodies of Kuiper's belt. Body has a period of revolution around the sun of approximately 730 years and it dvizhetsya along the circular orbit. To what is there equal aberration constant? 23. We will consider that the Earth dvizhetsya along the circular orbit with a velocity of 30 km/s, or , where c - speed of light. Let us assume that there is a star, which is been located at a distance of parsecs, which is visible in the direction of the spring equinox point. More precisely saying, it accomplishes small harmonic oscillations along the ecliptic relative to the mean vernal equinox, i.e., by such, whose position is corrected for the influence of precession and nutation. The oscillatory period - it is exact yr. Twice in the year star is visible accurately in the direction (of average) point of spring. It is understandable what to observe this star conveniently in autumn (why? When it culminates in autumn?). After how many days to (or after) the day of the autumnal equinox the star does penetrate (average) spring equinox point? To what is equal the amplitude of the annual fluctuations of the right ascension of this star?

26

24. The equation of synodic motion for the outer space planet takes the form where S - sidereal period, P - synodic (in the years). It is possible to rewrite it thus:

To give visual, so to say the "physical" interpretation of this expansion S in the series (type "Achilles never it will overtake tortoise"). For the inner planets analogous decomposition takes the form Try to give its visual interpretation. 25. The period of revolution of Venus around the sun is 0.61521 tropical years. It accomplishes thirteen revolutions around the sun almost accurately in 8 terrestrial years: . This generates specific feature in the apparent motion of Venus. Of what it does consist?

During November 2005 there will be the opposition of Mars. The visible way of Mars among the stars near this opposition will take the form, schematically depicted in figure. In what constellation is located the ascending node of the orbit of Mars?

26. In which of two planets - Neptune or Pluto - the fraction of year, during which they dvizhutsya backwards, is more? To estimate it for Pluto. 27.

27

The visible way of comet among the stars takes the form of the rapidly being untwisted spiral. One turn is accomplished exactly in yr, motion occurs clockwise. To find the coordinates of the center of spiral. What fate is prepared to this uncommon comet?

28. Flying through the solar system, galactic newcomer for some purpose carried off with himself Venus and Mars, and in the Earth eccentricity of its orbit increased twenty, after leaving all the remaining orbital parameters constant. How as a result did change climate on the Earth? 29. What such VLA, VLBI and VLBA? 30. The oppositions of Mars, proceeding during August or during September, are considered great, the same as they occur during February or during March, it is possible to name anti-great. If we calculate all oppositions of Mars from the times, let us say, Hipparchus, then it will be explained that, in to the disappointment of astronomers, then loving the closest approaches, them it was substantially less than anti-great. Why? To estimate the ratio of the numbers occurred in the last 3000 years of the "greatest" oppositions (occurred into one of ten nights - from 25 August through 3 September) to the number of unfavorable, that were being observed into one of ten nights half a year later. (calendar we consider ideal, so that the "slidings down of dates" in 10 000 years does not occur.) 31. Comet dvizhetsya on the ellipsis, which has eccentricity e = 0.5. By how many times its speed in the perihelion is greater than in the aphelion?

28

32. One planet dvizhetsya along the circular orbit of radius a, another - along the elliptic orbit with perigel'nym distance of a and eccentricity e. To show that in the perihelion the second planet dvizhetsya once of more rapid than the first. 33. Asteroid dvizhetsya on the ellipsis, in by right focus of which is located the sun (see figure). Times and passages of semi-ellipses BPB' and B'AB relate as by 1:3. To find eccentricity. Answer:

.

34. To what the equally time average value of distance during the motion along the Keplerian ellipsis? 35. The Earth - blue planet, and Mars red. To what are approximately equal their color indices B- b-v? After recalling, as appear Jupiter and Saturn, to give the estimation also of their color indices. 36. Where it is more than carbon dioxide - in the atmosphere of Mars, which almost wholly does consist of , or in the atmosphere of the Earth, where its content do compose about 0.03%? 37. Observer is located on the surface of Venus. If in its atmosphere not there would be clouds, then A) would be visible there at night stars (by naked eye)? b) what color there would be there sky in the daytime? c) there would be evidently the sun before the sky? 38.

Before you the usual photograph of Saturn with its rings. You will estimate the mass of rings, if it is known that they consist of small bodies with the significant dimension of R. Radius of Saturn of approximately 60 000 km, its albedos high (why?). 39. To estimate the temperature of star, which half of all photons emits in the Lyman continuum. To find also the temperature of such star, in which in the Lyman continuum is emitted half of the entire emitted by it energy. In both cases it is considered that the emission is blackbody. 40. Lyman and Balmer series do not overlap: the wavelength of the boundary of a Balmer series is more than the wavelength of line (1216 A). Do not overlap also Balmer and Paschen series. Many naively assume that does not happen to have completely hydrogen of the overlap of series. To ascertain that this not thus: already in a

29

Paschen series its first line (passage between levels n= 3 and n= 4) has a wavelength, greater than at the boundary of the following (brekketovskoy) series (passages to level n= 4). To estimate, how many lines n- oy series, n"1, are assigned on (na +y)-uyu series. [ Otvet: if 4 it did not be mistaken in the calculation, then

.]

41. After using the Saha formula, to obtain the exponential approximation of the dependence of value from temperature T in the environment . 42. The fact that they usually call energy distribution in the spectrum (let us say, the suns), in reality is a flow distribution along the wavelengths, but not on the energies. At what wavelength is located the maximum in the distribution of solar flow on the energies? To consider that the sun emits as blackbody with K. 43. To show that for the star, which is been in mechanical equilibrium, the following useful relationship occurs:

where P=P(R) - pressure, M and R - mass and a radius of star, respectively. In the course (of hydrostatic) evolution of the star of a constant mass both its radius Rand form of dependence P=P(R) change. However, to the left the integral confronting remains constant. 44. How you do think, there is much hydrogen in the center ? But in the center of Betelgeuse? 45. You, probably, repeatedly read about the fact that the neutrino is capable of flying, without being absorbed, through the inconceivable thick layers of the substance. To show that the protons, which are located in the center of the sun, far exceed the neutrino: from the generation of the sun to that moment, when proton perishes with the proton-proton reaction, pass about 1010 years. In this time it manages to fly in the substance with density on the order 150 g/sm3 - it is such density in the center of the sun - much larger way. You will estimate this way and will comprehend the obtained result. 46. In one book, dedicated to starry evolution, there is this phrase: "the sun, deprived of the energy sources, will be approximately one years compressed... to the degenerate state". To verify this assertion. To consider that the luminosity of the sun in the compression stroke does not change. 47. What it is more in the center of the sun - carbon or nitrogen? 48. If suddenly convection envelopped entire today's sun from the center to the surface, the chemical composition of its atmosphere noticeably would change: hydrogen it would become less, and helium it is more. In reality in the solar atmosphere the part by weights of hydrogen and helium are equal respectively X= 0.70, Y= 0.27. What they would become they were equal after mixing? What still perceptible changes would occur in the chemical composition of solar atmosphere? We recall that the age of the sun - about 5 billion years, of the same order and the age of life on the Earth. 49. According to calculations, in the depths of today's sun contents C, N and O change from the center to the surface, as shown in figure [ to give figure ]. How do appear the

30

corresponding curves, which relate to the moment, when the sun was twice younger than now? But which there will be their form for the moment, when the sun does complete its life on the main sequence? 50. How would appear solar spectrum, if the temperature in the photosphere did diminish in depth? 51. Is it possible according to the form of the region of the spectrum of the center of solar disk to estimate the darkening of tsentr/kray in this spectral region? 52. Why in the long-wave part of solar spectrum there is not one very deep spectral line? 53. In the atmosphere of the sun the radiation absorption in the optical continuum is caused in essence by negative hydrogen ion . Corresponding coefficient of absorption (taking into account one ion ) of order cm 2. Using this, to estimate in order of magnitude pressure in the solar photosphere. The acceleration of gravity on the "surface" of the sun to you is known. 54. Which must be the tension of turbulent magnetic field in the photosphere, so that it would create a pressure, equal to gas? 55. What in order of magnitude fraction of the mass of the sun is fallen to its atmosphere? 56. They indicate that the star possesses the gray atmosphere, if the coefficient of absorption of its gases in the continuum does not depend on wavelength. If the substance of the atmosphere is located in the local thermodynamic equilibrium, then in the so-called approximation of Eddington- Barbier energy distribution in the spectrum of star is given by the formula of Planck s , where - effective temperature. The dependence of the temperature in this atmosphere on the optical depth takes the form . Let us assume that the star with the gray atmosphere has with tens of the spectral lines, scattered along entire spectrum. The absorption coefficient in the center of all these lines is one and the same; moreover it is 1000 times more than the absorption coefficient in the continuum. Then spectrum will take approximately this form, as shown in figure [ to give figure ]. Dotted line showed envelope the line of centerss of absorption, existing in the spectrum. By how many times area under this curve is less than under the curve, describing energy distribution in the continuum? Answer: 3 times. By what formula is described broken curve? 57. To estimate, what maximum depth they can have lines of absorption in the spectrum of the center of solar disk in A the sun gray and in it occurs FOR LTR.

region, assuming that the atmosphere of

58. Why in the spectrum of the sun in A

region there is not one line

with the residual flow of less than 0.3, whereas in region A of such lines ochen'-ochen' there are many? 59. At a distance of 10 PK from the sun flared up II type supernova. Prehistoric boy began to observe it. What he could describe to his grandsons about that seen by them before the sky?

31

60. What is higher - air density at the apex of everest or the average density

?

Relationship period- density for the classical cepheids takes the form

, where

P - period a day and - average density into g/sm . 61. Star emits as blackbody with temperature T. What portion of all emitted by it photons is fallen to the Lyman continuum? 62. To estimate the sizes of the specks, which the light pressure "vymetayet" from the environments of the star of the main sequence of mass M. 63. How the mass of dust on the line of sight, which does cause a certain fixed interstellar absorption, it does depend on the size of specks? To consider that is applicable geometric optics, i.e., the sizes of specks are great in comparison with the light wavelength. 64. To estimate the mass of interstellar dust on the line of sight (g/sm ), if it is known that the absorption in the visible part of the spectrum (let us say, at the wavelength ) comprises . The size of the specks of order see because of the diffraction their effective area it approximately twice exceeds geometric cross section. 65. To estimate the total mass of that interstellar dust, which causes attenuation of the 3

emission of the star of a radius on . It is considered that dust is distributed in the space between us and star evenly. 66. In one assignment book there is this interesting task. Speck with a density

of g/smof 3 and size of a= 1 m was formed at a distance

from the star, which possesses luminosity erg/s and mass To determine the speed of speck at infinity, if it moved in the vacuum.

.

Its following solution further is there given. The equation of motion of speck in the absence of resistance of medium takes the form

Integrating, we obtain

In this solution to eat an error. Where it? To give the correct solution. 67. Kh.Gyuygens in the middle THE XVII century for the first time measured the angular dimensions of planetary discs, in particular, Saturn

. Was known also the period

of its revolution around the sun of years) and, of course, then that the luster of Saturn much the same as in bright stars. Hence Newton knew how to obtain the estimation of distances to the nearest stars (in the astronomical units) correct in order of magnitude. By the way, to what is equal astronomical unit itself, at that time it was known still very bad.

32

This - historically the first method, which reliably evaluated distances to the nearest stars. Measurements of annual parallaxes in those times gave only the very rough lower estimates these distances, quantitative photometry yet not did not exist, and about which the visible stellar magnitude of the sun, not there was idea whatever. Try to reproduce Newton's reasonings even you will ascertain that they actually give a good estimation of interstellar distances. 68. After solving the previous task , glance again at the Ursa Major. Could not you now say, at what approximately distance from each other are located those five stars of ladle, which do have the general proper motion? 69. If you solved task , then easily you will answer this question: to what are approximately equal the masses of five "internal" stars of ladle? 70. What does shine more brightly in the sky of Sirius - sun or the star of the ladle of Ursa Major? 71. < to give the diagram GR of shzs 47 tuc (contemporary, with long GP). > To estimate distance to the accumulation: A) on the position of horizontal branch; b) on the position of turning point. 72. The star drifts of the open cluster Of giady are directed to one point of celestial sphere apex, and therefore angular distances between the stars of accumulation are gradually decreased. To what extent will increase the stellar magnitude of the brightest stars Of giad, when the angular dimension of accumulation before the sky does decrease 10? 73. With what astronomical object is associated in you number 273 ? Describe it on the memory in all details. But than number 273 it is noticeable in physics?

33

Solutions

1. First acquaintance 1.1 we make level the volumes of the terrestial globe and wire: where d - wire diameter, R - distance to one of the objects indicated under the condition. Hence

Substituting the numbers, we find the diameters of the "wires", lengthened to the sun and to Cen: 100 km and 200 m, respectively. But into the Andromeda nebula it would be possible to poke with the rod with a thickness of 25 cm! You will agree, you expected that the "wires" will prove to be considerably thinner. We focus attention on the fact that . Wire with thickness in several millimeters it would be possible to suffice to the most distant quasars. As you see, the universe is entirely small! But here other, in a sense the more correct estimation of that, is great in reality the thickness of our "wire", sufficed to the sun. Natural scale in the solar system is assigned by its own size ( cm), but completely not with the size of human body ( cm). The diameter of our "wire" (100 km = cm) composes in all from the size of the solar system. Therefore it, if it is convenient, is thinner than the finest cobweb. Actually, let us take for the comparison human hair. Its thickness mm = see this from the significant dimension of the human body. Therefore in the relative units of hair to four orders to the thickness of our imaginary space cord with the length of 1 AU. The intuition, which prompts to us, that the wire, sufficed to the sun, must be very and very thin, does not nevertheless us bring.

34

1.2 on the Earth lives

man. Number of stars in the galaxy

. Number of galaxies in

the observed part of the universe of the same order: (as to obtain this estimation, using the data by the "universe in the numbers"?). Therefore in our galaxy to one person a total of several ten stars is fallen, but in the entire universe - not more than on hundreds of galaxies.

1.3 to each earthling in the galaxy comes several ten stars (see the previous task ). It is not necessary to be by entomologist in order to say: gnats to each inhabitant of the Earth it is fallen much more. If you this doubt, you will take a trip by summer into the taiga. But, surely those aren't gnats! It is easy to estimate, what trees in the Siberian taiga - and is the more than stars in the galaxy. Area of Siberian taiga km2. It is obvious that less than 100 m 2 come to one tree inthe taiga, and it means, trees it is more than . By the way, this makes it possible to ascertain that gnats on the Earth much morethan stars in the galaxy - indeed under each tree in the taiga of their oy-oy-oy it is how much. In winter in Siberia frost, and there are no gnats - but they be sufficient in the moist tropical forests.

1.4 v for the triatomic molecule H2O to each atom it is necessary on the average on 6 nucleons, or at g. Therefore 1 cmof 3 waters contains atoms. Assuming that in the universe

of the stars (see task

the same number of atoms occurs equal

), then the diameter of the drop of water with to mm.

1.5 Newtonian force is nevertheless more than coulomb, but in all several times. You will be convinced of this, after considering that the mass of cosmonauts - at 100 kg (they indeed in the diving-dresses!), and the distance between them of 1 m. the calculation of no labor comprises, but result impresses. You, of course, knew that gravitational interaction - weakest of existing in nature four fundamental interactions. But did visualize you that it is so weaker than the electromagnetic?

1.6 we live in several inserted in each other as matryoshkas potential pits. Any person forms with the Earth the gravitational- connected system with the negative energy, taking into account the unit of the mass of equal erg/g. Here

. The depth of this potential pit is and

velocities.

35

- respectively orbital and planet escape

Bearing in mind that in space gravitational energy frequently (for example, with the accretion) passes into the ionization energy and atomic excitation, it is instructive to estimate the depth of terrestrial gravitational potential pit in other units - into eV/nuklon: (prover'te!).

eV/nuklon

But here is view on the same things from entirely other side. In order to send man to the Moon, it is necessary to overcome gravitational energy of his binding with the Earth. With the mass of body in 70 kg (as in "average" cosmonaut) this energy is erg, or 4 GJ. This about 1000 kilowatt-hours. The cost of kilowatt-hours rub (price of the end of 1996 for the organizations - hardly you you will fly to the Moon as a private individual). Therefore it occurs that energy, which must be spent in order to derive man on the spaces of the solar system, cost in all rubles, or about 35 dollars. Conclusions: 1) "overhead expenses" in cosmonautics are colossal; 2) the economic pit, in which we now proved to be, is much deeper than gravitational. Further, the Earth with all its inhabitants forms the gravitational- connected system with the sun, and the sun - with the galaxy. Find the depths of the corresponding potential pits independently. Here are the answers: "solar" pit into 14, while "galactic" - are 1000 times deeper than the "terrestrial". During the estimation of energy, which must be reported to particle for the output from these potential pits, do not forget to consider that the Earth dvizhetsya around the sun, but the solar system - on its close one to the circular orbit in the galaxy. Because of this the required energies half the depths of the corresponding potential pits (prove!). Thus, output from the earth's orbit to the galactic spaces requires approximately 7 times of larger energy than it is necessary for the flight to the Moon. For the departure into the metagalaxy from our galactic orbit it will be necessary additionally to spend still "lunar" energy units.

1.7 po.opredelenih, distance modulus m - M to the object, which is been located at a distance of R (in the parsecs), is called the value Taking into account the fact that cm, and distance from Saint Petersburg to Moscow cm that in the parsecs it comprises , we find that the distance modulus corresponding to it is equal

.

Prover'te independently, that the distance modulus in 1 cm is equal

.

The main thing, that it is necessary to clarify: distance modulus, although it is expressed in the stellar magnitude, no to emission refers. This is simple one of the possible, and besides convenient for astronomy extrasystemic units of the measurement of distances. (cf. with the light year, where the distance is measured in the units of time.) Convenience in the use of the logarithmic scale of stellar magnitude for measuring the distances in the universe is explained by the colossal spread of distances to different astronomical objects.

36

1.8

The brightest stars of pleiads are hot - indeed the star of the ladle of pleiads clearly white, even bluish. Since pleiads - this scattered star cluster, these hot stars lie on the main sequence. The stars of the late sub-classes B have the absolute stellar magnitude, equal to zero, and they "white". Blue stars have even larger luminosity - and therefore the negative absolute stellar magnitude, which, thus, is lower than the solar at least per 5 units. But the stars of pleiads with the observation from the environments of the sun have the 4th visible stellar magnitude. It means, the sun with the observation from the pleiads will have, as a minimum, a 9th it will evidently not be its naked eye. Although the answer is obtained, it is worthwhile to make one additional step and to refine the obtained estimation. According to Allen [ 1 ], the distance to the pleiads is 127 PK, and according to newest data ( HIPPARCOS) it is equal to 117 PK. Therefore the visible stellar magnitude of the sun with its observation will from there be m = M - 5 + 5 lg R = 4.8 - 5 + 5 lg 117 10.1m, which confirms the conclusion made earlier. Sun not to see from the pleiads not only with naked eye, but also into the binoculars.

1.9 first answer, which occurs, "certainly, the angular dimension of the disk of the Moon is greater". It is clear that it must be incorrect, otherwise task would not be included in collector. Thus, let us estimate the angular dimension of the Andromeda nebula. Distance to this galaxy 700 KPK. As the value of its diameter let us take the diameter of our galaxy, 30 KPK. Then the unknown value is equal

radian, or

.

1.10 if stars filled infinite Euclidean space on the average evenly, then entire sky would have the same brightness as in solar disk. Actually, the solid angle, at which are visible the disks of stars, which are been located between R and R+dr, it is equal to the volume of ball layer , multiplied by the number of stars per unit of volume and by the solid angle, at which is visible each of the stars (we we taciturn assume that they all are identical; refusal of this assumption does not change final conclusion). The latter is equal, obviously,

where

- a radius of star. Thus, stars from the ball layer

37

occupy before the sky solid

angle . Infinite solid angle is obtained with the integration for R, so that the disks of stars must overlap, completely covering with themselves sky. If peace was then it was arranged, then day would not be differed from night, but solar disk to examine before the sparkling sky would be impossible. This is the famous photometric paradox of Olbers (in actuality it it was for the first time formulated in the century to Olbers by the contemporary of Newton by Edmund Halley). Thus, from the fact that at night is dark, is possible to draw the very important conclusion about the structure of the universe, precisely, that the assumptions made above are not carried out. The permission of the photometric paradox of Halley - Olbers in the fact that the universe is enlarged, and therefore it is necessary to consider red shift. To in greater detail explain here not place.

2. Kinematics of the sky 2.1 it is sufficient, lighting up, to trace the direction of the daily motion of the sun. If in the morning (you recently they awoke!) it is moved, as usual, upward and to the right, you in the northern hemisphere. But if upward and to the left - that in south. But if directly upward - you somewhere in equator itself. Why under the condition of task it is mentioned that the matter was 21 March?

2.2 if in the northern hemisphere winter, then the declination of the sun is negative, and with the observation from the equator its diurnal motion occurs to customary to us by means - from left to right. However, for the Europeans the season, and the daily displacement of the sun over the sky with the observation from the equator occurs from right to left in summer. Therefore to establish that in Petersburg - winter or summer - special labor it will not comprise.

2.3 you will be convinced of the fact that primitive people are not very aggressive, is understood your speech and they are ready you to attentively listen to. Then you will recall simple evidence of the sphericity of the Earth, known already to Aristotle. For example, the hiding themselves under the horizon silhouettes of ships. Or wait for the lunar eclipse, you will explain, why it proi.chkhodit and show that the earth's shadow is round. With the axial rotation of the Earth the matter is more complexly. The change of day and night, the daily rotation of celestial sphere and other similar phenomena will not convince collocutors 38

that revolves precisely the Earth. Mechanical experiments can serve as proofs and the observations, which indicate the presence of the inertial forces, which act in the noninertial frame of references. Without pronouncing such odd terms, climb up to the tree and try, after tying stone to the rope, to make the Foucault pendulum. If the matter occurs not very closely to the equator, experiment, possibly, it will succeed.

2.4 you will recall that the precession is otherwise called precession of the equinoxes (Latin word "praecessio" it indicates "forewarning"). The following equinox begins earlier than previous. It means, the point of spring dvizhetsya towards the sun. But the sun in its annual motion is moved from the West to the east. Consequently, the point of spring dvizhetsya from the east to the West, i.e., from left to right in the northern hemisphere of the Earth and from right to left - in south.

2.5 v of Russia the sun always culminates to the south from the zenith. Therefore it is possible to use the following formula for the celestial altitude in the upper culmination: where - the latitude of place, star declination. Let us consider also that the declination of the sun during the year changes in the limits where

- obliquity to the equator. (we focus attention, that in reality , but completely not ). We have therefore

For example, in Petersburg (

)

During what days are reached these the limits? But which will be, if fate did throw us from Russia, let us say, to the south of India?

2.6 let us estimate the time of approach with an accuracy to half-hour. The hour angle of the sun with the approach is determined from the formula The latitude of place must be known. The declination of the sun

either you will estimate

themselves on the date or take from the annual. For example, for Petersburg say, on 5 November , so that , and visit 4 hours after apparent noon. True noon will begin near Consequently, the sun will visit near .

39

and, let us

. It means, the sun will Moscow time (see task ).

2.7 if you man dodgy and a little know geography, then almost nothing being able to reason in spherical astronomy, you will immediately say that the matter was in Armavire. Actually, it is clear that to us it is proposed to derive the certain formula, into which will enter the latitude of observation point . It is understandable that the latitude will be argument in some of trigonometric function - sine, cosine or tangent. It is clear from the style of assignment book that the value must be such so that it would be possible to make all calculations without the calculator. Mean, either 45oor 60o. But 60o do not approach - this is the latitude of Petersburg, and not Arzamasa (and all the more not Armavira). It means , but this is clear not Arzamas - it is noticeably northern. Thus, answer must be - Armavir. This is correct answer. All this, of course, it is unimportant, although sometimes in the life a similar mental acuity strongly helps. However, almost no one is solved to such impudence, when the discussion deals with the formulas, and in vain, as we recently were convinced. Here is the present solution. We have regarding the spring equinox point, on 21 March . We will use formula from the solution of the previous problem Let us find how it changes

with the small change

But during the day of the vernal equinox that

, and the sun goes accurately in the West, so

. Therefore about 21 March

are connected as follows:

In one day the sun is passed on the ecliptic have about 21 March

(in reality only less

). Therefore we

for the change in 1 day

But according to the condition of task

so that

. Differentiating, we obtain

. Therefore

. This is - Armavir, it is located practically accurately on

. However,

Arzamas lies to 10 degrees north, it is more precise, on , so that for it . The sun in Arzamase on 22 March goes to later than the day before - very noticeable difference in comparison with Armavirom.

2.8 nonsetting heavenly bodies - those, which have the height of the lower culmination i.e. star with the declinations

Figure shows the projection of celestial sphere on the plane of the celestial meridian.

40

The nonsetting stars fill segment

. The area of this segment exists

, while the area of entire celestial sphere comprises the nonsetting stars is equal to the ratio of these areas, i.e.

. The share of

.

The special cases: 

half of stars never



all stars raise



in Petersburg (

goes at the pole (); and go at the equator (); ) the share of the nonsetting stars is equal to 1/4.

On 2.9 22 June at the north pole the sun around the clock is located at height ~231/2o above the horizon. Since into the plenilune the Moon is located almost at the point of celestial sphere opposite to the sun, it there will be always approximately at the same height under the horizon and, therefore, not it will be visible. In order to base answer more accurately, let us recall that the orbit inclination of the Moon to the plane of ecliptic is approximately . Therefore the Moon on 22 June will be as the minimum into 231/2o - 5o = 181/2o under the horizon. If plenilune it is necessary even on 1 June, the Moon is nevertheless visible not will be visible (why?).

2.10 seen color and the brightness of the Moon - this, it goes without saying, the effect of the passage of rays from the Moon through the earth's atmosphere: on the horizon atmospheric weakening is great (especially for the dark-blue rays). The Moon is visible with dim and yellowish. So that is sufficient to explain, why in winter into the plenilune the Moon culminates highly, and in summer - low- above the horizon. In the plenilune the Moon - in the direction, opposite to direction in the sun (if we disregard small , by the inclination of its orbit to the plane of ecliptic). Means, full moon in midnight in winter approximately, where the sun - in summer at noon. Everyone knows that in summer the sun at noon stands highly, and in winter low-. But the Moon, it is understandable, vice versa.

41

2.11 since the orbit inclination of the Moon to the plane of ecliptic , the declination of the Moon during the period of the precession of nodal line (18.6 years) it varies within the limits

where = 231/2o - obliquity to the equator, so that . Further the solution is analogous with solution of problem . For example, in Petersburg ( = 60o) for the height of the Moon in the upper culmination we have inequalities There are, therefore, the periods of the time, when the Moon in Petersburg does not rise above one-and-a-half degrees above the horizon. When this is - in summer or in winter? But the "lunar Arctic Circle" pass on the latitude of approximately 611/2o. The zone, where at least once every 18.6 years the Moon into the plenilune does not raise completely, north lies.

2.12 v zenith culminate heavenly bodies s . Since the declination of the Moon is always concluded within the limits (see solution of problem )

and the very southern point of Russia has a latitude near observed in the zenith.

, the Moon in Russia not can to be

2.13 when the Moon in last fourth, it is visible before the sky near the apex of the orbital motion of the Earth (you will explain this by drawing). Consequently, stars, which are located on the celestial sphere not far from the Moon, on the average (only on the average!) they approach us with a speed of the orbital motion of the Earth, 30 km/s. However, we were a little inaccurate. By term "radial velocity of star" usually is understood its radial velocity relative to the barycenter of the solar system. Therefore in the formulation of task it was to be indicated that the discussion deals with the directly measured radial velocities, not corrected for the motion of the Earth.

2.14 date on 7 February "are exist equidistantlyed" from the days of the winter solstice and the vernal equinox, so that the right ascension of the sun is equal . Since the Moon in last fourth, its right ascension to less and, which means, equally .

42

2.15 let us explain the visibility conditions of constellations in Petersburg at present. In Orion , which indicates upper midnight culmination in winter. Height in the upper culmination is the same as in the sun at the end of March, i.e. . In southern crown , which indicates upper midnight culmination in summer, and height in the upper culmination will comprise approximately . Thus, the constellation of Orion one can see well in winter, but southern crown is not visible completely. 13 000 years - half of the period of precession. In this time the equatorial plane, which constitutes angle with the plane of ecliptic, will turn itself on the half turn around the axis of ecliptic. (make a drawing of celestial sphere and apply to it ecliptic, positions of equator now, also, in 13 000 years; note also position of both constellations.) As a result of the precession the equatorial coordinates of Orion in 13 000 years will comprise and the coordinates of southern crown will become (you will be convinced of the correctness of these formulas according to drawing.) Therefore in 13 000 years the constellation of Orion will be in the upper midnight culmination in summer, and its height in the culmination will be -17o, so that the famous belt of Orion and other beauties of this constellation will become inaccessible to peterburzhtsam. Then the southern crown, invisible now, it will be well visible in Petersburg in winter, culminating at height 37o.

2.16 began, if the matter occurs in the northern hemisphere on the latitude

, and

end otherwise. Actually, on 22 June in the northern hemisphere on the latitudes and the sun, and the Moon they are moved on the ecliptic from right to left. With this sun it goes around entire ecliptic in yr, and the Moon - in the month. Therefore the Moon overtakes the sun, and in the beginning of eclipse its disk creeps in the sun to the right. By the way, a similar naive "geocentric" description of the solar eclipse makes it possible to easily estimate its greatest possible duration (from the first to the fourth contact): if in the month, it is more precise, in the synodic month, i.e., after , the Moon displaces on the ecliptic relative to the sun on , then - the sum of the angular diameters of the solar disks and Moon - it will pass after

. With such crude

estimate the inclination of lunar orbit to the plane of ecliptic completely can be disregarded. One essential inaccuracy in the recently made estimation, however, nevertheless exists: we did not consider the rotation of the Earth. "touch up" our solution independently.

43

2.17 for the offensive at least of particular solar eclipse it is necessary that the angular distance of the center of the disk of the Moon from the center of solar disk would not exceed the sum of angular radii of the Moon and sun, i.e. . Since the center of solar disk dvizhetsya on the ecliptic, this condition can be reformulated thus: topcentrical ecliptic latitude of the Moon must be not more . Since horizontal daily lunar parallax comprises , and solar parallax can be disregarded, the geo-centrical ecliptic latitude of the Moon must be not more

.

Now let us examine the right spherical triangle, as apexes of which serve the center of the disk of Moon M, the center of solar disk S and node of moon's orbit (see Fig.). The arcs of ecliptic, orbit of the Moon and the great circle, passing through the centers of the disks of the Moon and sun and pole of the ecliptics, are the sides of triangle. Angle with the center of the sun of straight line.

Sharp angle with the node of moon's orbit is an orbit inclination to the plane of ecliptic . The catheti of triangle are equal to the distance of the sun from the knot and to the geocentric ecliptic latitude of the Moon . Triangle s grow prettier by accuracy it is possible to consider it flat and narrow, so that . But rad. Therefore

.

2.18 task are interesting in the "everyday" sense: it is interesting, on how much is differed from 12 it is hour the moment of the offensive of the "present" of noon - the moment, when the sun above all, shadow of the objects are oriented strictly to the north or to the south (where precisely - to the north or to the south - it depends on the latitude of places and time years; you will be examined in this independently) and t. d. Let us assume that for the certainty today on 10 November, and you be situated in Petersburg.

44

True noon - moment, when apparent solar time

. Mean time will comprise

, where - equation of time. After using the graph, which gives the equation of time to the different of date (for example, see [ 2 ], fig. 14), let us find that on 10 November Universal time that

. Consequently

.

, and the length of the center of Petersburg

, so

.

Following step - calculation of standard time. Petersburg is located in the second time zone; therefore

.

Finally, let us recall that in Russia daylight saving time, which acts for one hour in front of the waist. The summer time, which adds another hour, 1st November does not act; therefore Moscow time coincides with the decree. Finally we obtain

2.19 since solar time is counted off from midnight, then when in Greenwich 0h, there midnight, i.e. Sun - in the lower culmination. However, in the upper culmination it will be at this moment on the opposite side of the terrestial globe, so that the length of observation point is equal to 12h.

2.20 a) the period of swing of the pendulums with a length of l is given by Huygens's formula

where - the acceleration due to gravity, G - gravitational constant, M - mass of the Earth, R - distance from the center of the earth at the point, where is located pendulum. Thus,

so that

from where

45

From the equator to the pole

, and it means

twenty-four hours the hours will leave on

. In the .

b) the change in the period, caused by change l, is approximately equal Therefore required for compensating the change in the clock rate a change in the length of the pendulum With the length of pendulum 1.5 m at its pole one should lengthen on 1 cm.

2.21 answer is striking: hours will leave for the entire hour! Actually, the Earth revolves ravnouskorenno (more precise, ravnozamedlenno). Let - its initial angular velocity, angular acceleration, the angular velocity at moment t. Then But where

- the angle of rotation of the Earth. Integrating, we obtain:

If the Earth revolved without the acceleration, then we would have, obviously, so that complementary angle exists It remained to calculate the angular acceleration of the Earth. In 24 hrs the duration day grows by Therefore the angular velocity, initially equal it will become in twenty-four hours Increase in the angular velocity in 24 hrs exists and the angular acceleration Hence

46

In this task not only the answer is striking. Is even more amazing that so small an angular acceleration ( ) nevertheless was possible to reveal and to measure. This was done via the analysis of the ancient chronicles (!), which contain the descriptions of the solar eclipses (understand, in what here the matter, independently).

3. Tools 3.1 during the detection by the camera of obstacle on the way of Mars rover transmitter will report this to the Earth, and control center as the answer will send signal to the engine of apparatus. Signal must arrive before Mars rover will reach obstacle. Since 1 AU. - this 500 light seconds, to overcoming of distance in 3 AU. (doubled average distance from the Earth to Mars) departs 1500 s. Therefore the safe speed of the motion of Mars rover not more than 10/1500 m/s = of 7 mm/s, or about 40 cm/min. Truly tortoise speed! You will recall now, in what limits the geocentric distance of Mars changes and you will refine the obtained estimation. Most rapidly Mars rover it could move with the closest approaches of Mars, when R decrease to 55 million km they they are always at the same time of year (during August - September). How you do think, why?

After the manuscript with the text of the solution given recently was returned into the publishing house, occurred the remarkable events, which directly relate to the theme of this task. On 4 July, 1997, automatic spacecraft "Pathfinder" ("pathfinder") completed landing on the surface of Mars. The soon small controlled from the Earth six-wheel Mars rover with size large children's toy (its length - 65 cm) it crawled along Mars with a speed of cm/s in the complete agreement with the estimation obtained recently.

The meteorological station, established on the "pathfinder", conducts the measurements of the temperature of Martian "air" in the landing place. In the daytime it rises to to same (

C, at night it falls

C. It is interesting that the lowest temperature of air, fixed on the Earth, practically the C, station "Vostok", Antarctica, on 21 July, 1983 ).

47

In order to ensure with all rapid and reliable access to information about Pathfinder'e, that flows and archive, NASA - NATIONAL AERONAUTICS AND SPACE ADMINISTRATION places it in the Internet simultaneously on many servers ("mirrors"), scattered on entire terrestial globe. If to you it proves to be difficult "to be opened" directly in NASA - NATIONAL AERONAUTICS AND SPACE ADMINISTRATION for one of the following addresses: http://mpfwww.jpl.nasa.gov or http://mpfwww.arc.nasa.gov can try to use by the server of Institute of Space Research (IKI) in Moscow: http:/www.iki.rssi.ru/jplmirror/mars or, let us say, to following Danish: http://sunsite.auc.dk/mars. During the first days after the embarkation of "pathfinder" on Mars after the information about it on the Internet daily it acted to 100 million rotation! The system of "mirrors" even during the hottest first days and in the hours made it possible to obtain the current information about the events on Mars without any serious delays. "pathfinder" - first of the series of the sitting down themselves on Mars research apparatuses, whose starting will be carry ouied BY NASA - NATIONAL AERONAUTICS AND SPACE ADMINISTRATION within the next few years. It is assumed that in 2005 the models of Martian soil will be delivered to the earth.

3.2 any equatorial installation, including German, will be simultaneous and azimuthal, if telescope is established at the pole - it is unimportant, southern or northern.

3.3 in all by the eye evidently of stars, their luster is concluded between (Sirius) and . Eye is capable in principle of noting difference in , so that in all is located the gradations of brightness. This gives 7 bits of information to the star. Furthermore, stars are distinguished by the color - let us say, different it is color (although hardly and this of eyes it is capable to distinguish in the weak stars). We obtain the three additional bits of information. The position of star before the sky in the doteleskopicheskuyu epoch could be determined with the accuracy at best in . In the radian it is contained the ang. of minutes. Coordinates two, but the second (let us say ) changes in all within the limits , and therefore its task with an accuracy to allows (but not ) possibilities. In all - 27 bits for the position of star. Thus, about each star it was possible to obtain the following quantity of information: 7 bits luster, 3 bits - color, 27 bits - condition; altogether bits to the star. The total quantity of information about the stars, accessible to people in doteleskopicheskuyu (but more right it will, perhaps, even say - and into entire dospektroskopicheskuyu) epoch of development astronomy, literally strikes with its scantiness: the bit of kilos-byte.

48

Certainly, any, even simplest disketki to 360 kilos-byte, about which all had time already long ago to forget, it would be sufficient for the eyes for the recording of all these data. Actually up to the moment of the invention of telescope were assembled less than 10% this information - the catalogs of weak on the measures of that time stars were evaluated roughly.

there did not exist, luster and color

But as does be matter today? Here are the data about the quantity of information, assembled by the astrometric space observatory HIPPARCOS. It worked from the end of 1989 through June 1993 and transmitted in this time to the earth of the gigas-byte of the bytes of data. Result of processing these data - catalog HIPPARccOS'a (June 1997 ). For the stars their positions before the sky are given in it with the millisecond accuracy, even for the stars - with the accuracy of milliseconds. HIPPARCOS were obtained also the most valuable data about the proper motions and about the parallaxes of stars. In particular, they for the first time reliably measured the trigonometric parallaxes of cepheids, and this - basis of the construction of the scale of distances in the universe.

3.4 distance from the Earth to the Moon is equal approximately to 400000 km, and the permission of the eye

of radian. Therefore the smallest linear dimension of

the formations on the Moon, distinguished by the naked eye, is the greatest craters they have only larger size and they are close to the resolving limit. The craters are already well visible in the binoculars with a six-fold increase.

km

3.5 size of the smallest details on Earth l, which can be photographed from space, is determined by the angular dimension of the disk of the vibration of star. Let us accept its value equal to one second of arc. Then with the distance in 200 km from the earth's surface we have m. using contrivances, they attain larger permission, but as - this already, is probable, the military secret of Punchinello. Not we will reveal it. But how is determined the minimum size of the details on the Moon, which can be photographed from space?

3.6 let us estimate radius R of body, which at the heliocentric distance of R = of 40 AU. (middle of Kuiper's belt) it has the stellar magnitude (penetrating force of the ground-based 10meter telescopes of cake and space telescope of Hubble). Illumination from the body on the Earth (or to the sun, up to the distance from the Earth to the sun much less than the distance to the body) is equal

49

where of body,

- the "luminosity" of the inverted to us hemisphere of body, A - the albedo illumination from the sun on the body surface. (as you you think, why in this case instead of it would seem obvious precise equality

from the sun on the Earth

and to the tele- belts of Kuiper

?) Illumination

relate as

Therefore

You will focus attention, that the illumination diminishes here as the fourth degree, but not as the square of distance! From other side,

where - the visible stellar magnitude of the sun ( ), m - the visible stellar magnitude of the body ( - the limit, accessible to Hubble telescope). We obtain from two last expressions

Assuming albedo equal to 0.2 and substituting R = of 40 AU., we find km you will focus attention that , so that on the neighbor and on the distant the edges of Kuiper's belt values R are distinguished in

time.

At the beginning of 1997 in Kuiper's belt were discovered 46 objects, in eight months of 1997 it is opened 9 more. It is assumed that there there are tens of thousands of bodies of larger than 100 km. The extensive program of the search for trans-Neptune objects (TNO) is carried out on the twometer telescope of the Hawaiian university, on which is opened the lion's share of these objects. The stellar magnitude detected on it moving relative to surrounding stars TNO (on what and is based the method of their search) - near . You will estimate the independently penetrating force of this telescope; the diameter of mirror 2.2 m. the size of accessible to it tno it is easy to estimate according to the brought-out recently formula (make this!). The electronic catalog of the bodies of Kuiper's belt is accessible in the Internet with the addresses http://www.ifa.hawaii.edu/faculty/jewitt/kb.html and http://cfa.www.harvard.edu/cfa/ps/lists/TNOs.html Of all TNO, about detection of which were communications toward the end of August of 1997 , greatest semiaxis a = of 84 AU. has the body, which moves along the strongly elongated orbit (e

50

= 0.58). In the aphelion it is moved away from the sun on 133 AU. It is amazing that in approximately 40% of known at the present time TNO the semimajor axis is the same as in Pluto (although the three-dimensional arrangement of orbits another). These objects was called name plutino (plutino), or in Russian - plutonchiki. The revolution period plutino as Pluto itself, find in resonance 2:3 with Neptune. The arrangement of Pluto and plutino in orbits is such, that the close encounters with Neptune it does not occur.

3.7 cen - star . It means, photon flow from this star composes approximately fotonov/(sm s). Photon flow from the sun in once more composes approximately fotonov/(sm s). It is analogous, photon flow from the star once less is equal fotonov/(sm s) or 1 foton/(m s).

3.8 first - several words about the telescopes of cake. U. Cake (W.M. Keck) - American rich person, who endowed 130 million dollars to the development and buildings of these telescopes. They are established at the peak of the mountain of Mauna Kea (are more right - Valerian- Ci, but accepted to tell in Mauna Kea) on the main island of Hawaiian archipelago at the height of 4200 m above sea level. The first telescope entered into the system in 1993 , the first trial photograph on the second telescope was obtained on 27 April, 1996 , and into the regular operation it was transmitted on 1 October, 1996, it is assumed that in the near future these two telescopes will begin to work in the regime of optical interferometer with the base m, which will have to make it possible to obtain permission in (at the wavelength of 2 microns). But now - directly to the task. According to task , photon flow from the star is equal fotonov/(sm s). Area of the mirror of the telescope of cake cm2. Therefore from Vegi each second on this mirror it falls photons. From the star the telescope of cake obtains once of less photons than from Vegi, or approximately 1 photon per second.

4. Kinematics of the solar system

4.1 attempting to define the distances of planets from the sun and their revolution periods from the observations, you actually occur in the position of Johann Kepler, at disposal of whom exactly there were only "damp" data about the position of planets on the celestial sphere, and

51

which determined according to these data of distance and periods in order to establish the laws of the motion of planets. Thus, let us examine first lower planet - Venus. Should be wait for the elongation of Venus and measured the greatest angle, to which the planet is moved away from the sun. You will obtain . Draw the simple figure, which depicts the circular orbits of the Earth and Venus, arbitrary position of the Earth and Venus in the elongation. The straight Earth - Venus in this case is tangent to the orbit of Venus. It is obvious from the figure that the sine of the angle of elongation, i.e. , it is equal to the desired radius of the orbit of Venus in the astronomical units. Distance is found, let us determine now from the observations the period of revolution ("after forgetting" about the third law of Kepler). It follows to wait for repetition of one of the configurations of Venus - for example, eastern elongation. This will give the synodic period of revolution of Venus, 590 days. Using the equation of synodic motion, let us find the desired sidereal period P: from where P= of 225 days.

Let us switch over to outer space planet - Jupiter. Observations show that after opposition S-T-J (see fig.) Jupiter dvizhetsya 2 months by antecedence. Then direct motion occurs in the course of 9 months. The antecedence again begins after this, and the following opposition begins in 2 months. Thus, the synodic period of revolution of planet, i.e., time interval from one opposition to another, is equal TO T = 2+9+2 = 13 to months. Let us find the desired sidereal period P from the equation of synodic motion for the outer space planet: where the time is measured in the years, from where

(more accurate observations they will give more precise value, 12 years.)

52

Having again choked in itself temptation to use the third law of Kepler, let us determine now distance from Jupiter to the sun from the observations. To make this is somewhat more difficult than in the case of Venus. Let us examine again the moment of opposition, S-T-J. 2 months after this (more precise, in 59 days) will begin the standing of Jupiter occupy position

. The angle

can be measured:

calculated: in 59 days the Earth passes angle where

in

. Now we calculate the angle

; The Earth in this case will

. However, angle

can be

, and Jupiter - angle

, equal

:

according to the theorem of sines of Jupiter is found: 5.1 AU (in reality - 5.203 AU).

4.2 perihelion distance

for Pluto composes

, from

. We have . A radius of the orbit

AU. The more

precise value: AU, so that in the perihelion Pluto only is nearer to the sun than Neptune, whose almost accurately circular (e = 0.0086) orbit has a = 30.1. The close encounters of Neptune and Pluto never it occurs. The periods of their revolution find in resonance 3:2 (with what accuracy?). In the beginning THE XXII century Pluto will prove to be near the aphelion, and its distance from the sun will be close to AU. Therefore, if we consider that the instantaneous size of the solar system is determined by distance from the sun to the planet outermost from it at the given moment, then it is possible to say that it periodically varies from 30 to 50 AU. See, however task . Period of revolution of Pluto around the sun of 250 years. It was open By klaydom Tombo in 1930 , i.e. 67 years ago. In this time it was displaced along the orbit to the angle . In reality displacement somewhat more (why?).

4.3 according to the third law of Kepler the semimajor axis of the orbit of Neptune is equal to AU, i.e. Neptune is located 30 times further from the sun, than the Earth. The angular diameter of the sun, seen from the Earth, is equal approximately . Consequently, with the observation from Neptune solar disk will be visible at angle , i.e., on the resolving limit of eye. Cannot be actually seen disk will be - the sun "will blind eyes", and maximum permission not will be be reached.

4.4 here is the corresponding figure:

53

4.5 since the semimajor axis of the orbit of Jupiter is equal to 5 AU, the question, set in the task, can be reformulated thus: at what angle is visible 1 AU, located perpendicular to line of sight, from the distance in 5 AU? Answer is obvious: this angle is equal to approximately 1/5 radians, i.e., near .

4.6 distance to Cen is equal approximately to 1.3 PK. Regarding the parsec, this means that the semimajor axis of the orbit of the Earth, i.e. 1 AU, located perpendicular to line of sight, is visible with Cen at angle the ang. of s. Since the semimajor axis of the orbit of Jupiter is equal to 5 AU, and its orbit itself is close to the circular, the greatest angular distance from the sun, at which Jupiter occurs visible with Cen, it composes arcs.

second of

4.7 synodic period of the rotation of the sun for the observer in mercury is computed according to the formula of the synodic motion: the twenty-four hours (merkurianskiy yr it is equal ). However, Pluto dvizhetsya extremely slowly, so that the synodic period of the rotation of the sun practically coincides with the sidereal, 25 days. You will calculate synodic period with the observation from the Earth independently.

4.8 the angular diameter of solar disk comprises . Distance from the sun to Venus 0.7 AU, distance from the Earth to Venus in the lower connection 0.3 AU. Therefore, intersecting along diameter solar disk, Venus passes to its synodic motion the arc Fig.). For this it is required its synodic period. The latter is equal task ). Hence we find the desired time: are hour about 8.

54

(see (see

In contrast to the task about the solar eclipse, for the answer to a question about the direction of the displacement of Venus over the solar disk we will be geliotsentristami. If we look at the solar system from the side of northern terrestrial pole, then Venus, and Earth dvizhutsya around the sun counterclockwise; moreover Venus is more rapid than the Earth. Therefore near the lower connection Venus is moved on the sky from left to right. The same will be its motion along the solar disk.

4.9 covered star is located to many orders further from the Earth, than Pluto. Therefore the cone of the shadow, rejected by Pluto to the earth with the coating, can be considered the cylinder, the diameter of section of which is equal to the diameter of Pluto, 2300 km this and there is an estimation of zone width on the earth's surface, in limits of which it is possible to observe coating. [ in reality one must take into account that the Earth is not flat, but is ball-shaped. Because of this zone width can reach 5600 km; show this independently. ] The duration of coating is determined by the diameter of shadow and by the speed of its motion by the earth's surface. The orbital speed of the Earth is equal to 30 km/s, Pluto once it is less, since speed is inversely proportional to root from a radius of orbit. [ evaluating the speed of Pluto, we disregarded the ellipticity of its orbit. It is not difficult to consider it and to find that the speed of Pluto in the perihelion of elliptic orbit with a = of 40 AU and e = 0.25 approximately once higher than speed of motion along the circular orbit of radius 30 AU. ] If during the coating the velocity vector of the Earth is perpendicular to the axis of the cylinder of shadow, then shadow dvizhetsya on the earth's surface with the speed of the Earth relative to Pluto, km/s; if it is parallel, then with a speed of Pluto, km/s. Hence - the estimation of the duration of coating in that place, where the observer intersects shadow along the diameter: c min in the first case and other places the duration of coating will be less.

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min in the second. In

The duration of coating of 1988 , which was observed by eight expeditions in Australia and New Zealand and in course of which in Pluto was opened the atmosphere, it comprised on the average approximately one minute.

4.10 power of the signal, which is necessary at the sounded body, is proportional . The power of the signal, which is necessary from the body to the earth, is also proportional . Therefore the power of echo signal is proportional . Here, as in the task , the measured value diminishes as the fourth degree of distance, which in the astronomical tasks is encountered rarely. Distance from the Earth to the asteroid in the connection

AU, in the

opposition AU; ratio of the distance . It means, with the location of asteroid near the connection should be sent the signal, once more powerful than in the opposition. Not expected, you will agree, result. However, illumination from the asteroid in the opposition only once is more than in the connection. The corresponding difference in the stellar magnitude is close k .

5. The universal gravitation

5.1 the solar system will be destroyed. Planets will fly away from the sun along the parabolas, since the speed of their motion along the initial (circular) orbits exactly is equal to parabolic velocity with the mass of central body reduced doubly. Possibly, the sun will preserve mercury, Mars and Pluto. However, if this catastrophe happened of the sun for several next years (what definitely it will not occur), then Pluto would be also for sure lost - it now is located near the perihelion of its noticeably noncircular orbit. But to previously say something is difficult about Mars and about mercury. Everything will depend on their position in orbits at that moment, when the sun "loses weight". If they prove to be near the aphelia, then they will be preserved near the sun, but if they are near the perihelions, then they will fly away from it forever.

5.2 v the moment of a sudden increase in the mass of the sun the Earth begins to experience doubly greater than before, attracting force from the side of the sun. Consequently, it will pass from the circular orbit to the elliptical, that wholly lying inside the previous orbit (see Fig. on the following page). Thus, at the moment of gathering from the circular orbit the Earth will be located in the aphelion of its new elliptic orbit.

56

The integrals of energy, which describe the motion of the Earth in the field of central body with the masses, equal TO M and 2M, take the respectively following form:

where a - initial and - the new value of the semimajor axis of the orbit of the Earth (after the mass of the sun was suddenly doubled). Comparing these two expressions between themselves, we find that

.

Let us find the period of revolution of the earth from the new orbit. According to the third law of Kepler we have

from where

Eccentricity of new orbit let us find from relationship a = a' (1+e'), from where e' = 0.5.

5.3 of the equating of centrifugal force

to the gravity force

follow that

(constant value calculate themselves). However, the average densities of all bodies of the solar system differ less than by an order. They are concluded between 0.7 g/sm3 (Saturn; the corresponding time of circling - 4.2 hours) and 5.5 g/sm3 (Earth; after flying around the Earth in one-and-a-half hours, Gagarin established thus the first in the history and to this day only interplanetary record). The time of the circling of the sun, and therefore simultaneously and the upper estimation of the possible minimum period of the axial rotation of stars of the type of the sun, in the time of more than the minimum time of the circling of the Earth composes, thus, a total of about 3 are hour! Not truth whether, it is amazing? Way of

57

millions of km is passed in

this time, flight speed is close to 400 km/s - in

the time of less than the speed of running away from the "surface" of the sun. However, the typical white dwarf has the average densityof g/sm3, and therefore the time of the circling of the order of 10 seconds, the speed of motion with this entire km/s ("in all" - this means that although on the daily measures it is great, nevertheless

.

The circling of the neutron star of g/sm3) would engage a total of several milliseconds and (with a radius of star km) it would occur with the speed into many tens of thousands of km/s. It is clear that, studying neutron stars, we are found at the very limit of the applicability of classical mechanics. Relativistic corrections for the neutron stars must be already very noticeable. Note (for the "scholars"). That which

, follows also from the generalized third law

of Kepler . However, not- scholars know only whereas that, to remember expression for the constant - this is "erudition".

5.4 of the integral of the energy

and expressions for the perihelion distances and in the aphelion it follows that the ratio of the corresponding velocities exists If it is equal to 3, then e= 0.5.

5.5 let P - revolution period in years, R - a radius of orbit in AU and v - the speed of orbital movement in km/s. Since the orbital speed of the Earth is equal to 30 km/s, we have, obviously, From other side, according to the third law of Kepler

, and therefore

so that, for example, Jupiter (R= 5) dvizhetsya along the orbit with a speed

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of km/s.

5.6 let us write down the integral of energy for the comet, which is located on the heliocentric distance of the Earth:

where - the semimajor axis of the orbit of the Earth, 1 AU. Circular rotation in orbit of the Earth exists

Since on the condition

, semimajor axis proves to be equal

The revolution period is located hence through the third law of Kepler:

It the same as with the drop in the attracting center on the straight line, see task . It follows from the condition of task that the comet at a distance of 1 AU is located in the aphelion of its orbit, so that aphelion distance is

AU. But

, hence eccentricity

Consequently, perihelion distance will be extremely small: This, by the way, about 750000 km, and it means, comet in the perihelion will almost "hook" the sun. Such comets, "which scratch the sun", repeatedly were observed.

5.7 of perigee distance and in the apogee they give the semimajor axis of the orbit of the satellite and its eccentricity The revolution period according to the third law of Kepler is equal

In this formula it is possible to substitute the numbers, but calculations can be substantially reduced as follows. We know that the low-flying satellite accomplishes orbit around the earth in 1.5 hours ("Gagarin time"). This means that with the semiaxis

period

. After writing down the third law of Kepler in the relative form

we obtain

59

Actually, "Molniya" - semidiurnal satellite.

5.8 semimajor axis of the orbit 1 AU 1 Earth, Mars - 1.5 AU. The period of revolution of the earth is equal to 1 year. The semimajor axis of Hohmann ellipsis is equal, obviously, to the halfsum of radii of the orbits of the Earth and Mars: a= of 1.25 AU. According to the third law of Kepler the revolution period for the Hohmann orbit in the years is equal The desired time of overflight composes half of the revolution period, i.e., about 8 months.

5.9 sidereal period of the rotation of the sun in the equator , is the same period of revolution of automatic spacecraft in geliostatsionarnoy orbit. For the Earth the period of revolution of yr, the semimajor axis of orbit AU. According to the third law of Kepler, expressing P and a in the years and in AU, correspondingly, we have , from where we find a radius of the geliostatsionarnoy orbit: Without any calculations it was possible to immediately assert that the geliostatsionarnaya orbit lies inside the orbit of mercury, the period of revolution of which around the sun is equal to 88 days, which is substantially more than the period of the axial rotation of the sun.

5.10 as follow from the law of conservation of energy, such as speed not had body on the boundary of the sphere of influence of the Moon, with the contact of lunar surface it cannot be less than the speed of running away from the surface of the Moon, 2.4 km/s.

we will note 5.11 values, which relate to Jupiter, by index J. Then the relation of luminosities is equal to the portion of the surface of the sphere of radius 5 AU, which the disk of Jupiter occupies:

where obtain

- a radius of Jupiter in the km taking into account that a radius of Jupiter

The desired rate of accretion is estimated according to the obvious formula (cf. solution of

60

, we

problem

where

)

km/s - planet escape velocity for Jupiter.

Thus, if this accretion occurred, in the time of life of the solar system of years) the mass of Jupiter noticeably would not change here appropriate to recall that the real luminosity of Jupiter approximately doubly higher than that, which is ensured by the emission incoming from the sun. However, the source of this energy should be searched for in Jupiter itself, but not in the accretion.

5.12 energy, necessary for the delivery of vacuum cleaner to the Moon, is approximately equal , where and - orbital and planet escape velocities, m - mass of vacuum cleaner. The energy, which separates with the work of vacuum cleaner, is equal TO PT, where P - power of its motor, T - operating time. If we accept P = of 500 W erg/s, of m = of 5 kg, then we will obtain: from the twenty-four hours. In all are necessary, thus, about 80 kW hour, and this it stands (in the prices of end 1996 g.) in all some rub. See also task .

5.13 answer is such: maximum radius is about km, if we jump upward, without scattering, and somewhat more, if we first scatter ourselves. Here is the corresponding calculation. It is clear that at the moment of separation from the surface of asteroid the jumper must develop the planet escape velocity

The second expression for vundoubtedly more greatly is suitable for our purposes, since the average density of asteroid to estimate does not compose the labor: are concluded between 1 g/sm3 (ice) and 8 g/sm3 (iron). We subsequently will take

g/sm3. Thus,

The vertical component of the speed of jumper with the leap on the Earth can be estimated according to the formula where g - terrestrial acceleration of gravity and h - height, to which the center of gravity rises in the leap. As the reasonable estimation let us take h= of 1 m (then jumper it will overcome lath at the height cm - cosmonaut, who proved to be himself on the asteroid, necessary to think, is well trained). 61

As a result a radius of the asteroid, from which it is possible, after running up upward, to fly away into open space, occurs equal

If, however, before the leap cosmonaut scatters, then he will know how to jump also from the body of larger size. On the asteroid the takeoff gives the unexpected effect, with which terrestrial athletes are not familiar. On the asteroid with size in several kilometers, having good shipovki, are easy to scatter to the orbital velocity (prover'te!). But then due to the leap upward will have to overcome smaller potential barrier.

5.14 in the satellite, which moves along the circular orbit, centrifugal force balances attracting force, which gives

Let us designate through and , correspondingly, the kinetic and potential energy taking into account the unit of the mass of satellite. Then last equality can be written down also thus: Let, further, E - total energy of the satellite: These relationships give

from where . This equality means that really the rate of energy loss to the friction against air (negative value - energy it is expended) it is equal to the rate of increase in the kinetic energy of satellite

(positive!). However, from where this energy is drawn? It is

obvious that from the potential energy - there is no other source. Actually, since

(see

above), then , so that satellite obtains only half of the separating gravitational energy, whereas second half passes in the heat-. Thus, in the Newtonian gravitational field acts the unique "morals", close to the Christian: returning energy into the environment, in other words, making with it "good", the moving body auto from this becomes "better", i.e. it acquires kinetic energy. This entire energy, both returned and as a result of this acquired, is drawn from the potential energy, which comes out, if it is convenient, in the role the "faith," which gives birth to "good". Although the proof given above quite right, it nevertheless can leave in the reader some feeling of dissatisfaction. Let us try to explain the surprising result, which we discuss, him they sometimes call virial paradox - entirely "on the fingers". The Moon dvizhetsya along its orbit with a speed of km/s. If it moved in the resisting medium, then it would begin slowly "to fall downward" - this seems obvious. In the course of time it would become the low-flying satellite,

62

and the speed of its motion, as they know everything, it is close to 8 km/s. Thus, kinetic energy repeatedly would grow - and this for some reason no one astonishes! However, in the essence of the matter this exactly the same, that we discussed above.

5.15 since under the condition of task is used subjunctive inclination, this means that the way of the Moon relative to the sun, i.e., its orbit in the solar system, points of inflection does not have in reality and it is everywhere inverted by convexity from the sun. This fact almost no one knows, and he seems unexpected. It is understandable that the path curvature of the Moon in the solar system changes with the synodic period, being greatest in the plenilune and smallest in the new moon. So that the convexity even in the new moon would be inverted from the sun, it is necessary that resultant force of the attraction of the Moon to the sun and to the Earth would be directed toward the sun. In other words, the lunar gravity to the sun Earth

must be more than the force of its attraction to the

. We have:

Hence

so that the Moon is attracted to the sun approximately doubly stronger than to the Earth. Not truth whether, curious fact? So that in lunar orbit in the solar system there would be points of inflection, in the new moon must be (coefficient

, so that distance to the Moon must be

thousand. km

here not precise, it it is undertaken of the obtained above estimation of value

in the "real" solar system). See also task

.

5.16 v the course of general astronomy discuss ocean tides, caused by the attraction of the Moon (and of the suns). However, if gravitational force substantially changes at the distances m, the completely perceptible flows will appear also into the tele- person. Actually, tidal acceleration is equal where M - mass of star, l - the significant dimension of the body of cosmonaut, R - distance from the automatic spacecraft to the center of star. If you do not remember this expression, you will obtain it independently, after writing down the accelerations, reported by the star of that of most

63

and least removed from it to the points of body and after calculating a difference in these accelerations disregarding by low values, beginning from square l/R. We will consider overload a = 2 g maximum, where g - surface gravity of the Earth. Then from where

Typical mass of neutron star human l = 100;

; the significant dimension of the body of

(cgs system). Hence

cm

km.

Since a radius of the sun two orders of more than this value, it is clear that with the approach to the sun the in no way tidal forces would threaten cosmonaut. Dangerous factors will be high temperature, hard radiation , etc.

5.17 it is at first glance, must be carried out the following condition: the attracting force of satellite to the asteroid must exceed attracting force its to the sun. The condition of the equality of two forces is written in the form

where M - mass of asteroid, R - the heliocentric distance of asteroid, d - the desired distance between the asteroid and its satellite. The mass of 100-kilometer asteroid (with density 2 g/sm3) is find

g. Therefore

. Assuming R = of 3 AU.

km, we

km.

However, if the same reasonings are are applied not to the satellite of asteroid, but to the Earth satellite, maximum distance will prove to be equal to 260000 km (see task ). The Moon is located at considerably great distance! Paradox easily is permitted: in reality it is necessary to examine not the acceleration, reported to satellite by the sun, but a difference in the accelerations, reported to satellite and to the body, around which it dvizhetsya. This difference, as it is easy to show, does not exceed the value (cf. solution of the previous problem), and therefore equation for determining of d it takes the form from where

With the same numerical values we obtain for our dual asteroid

km.

To you perhaps it will be interesting to know, which in reality minimum radius of the circular orbit of the satellite, with which it can leave asteroid and begin to move along the heliocentric 64

orbit. Its determination - this is complex task even for the professionals - celestial mechanics. Corresponding radius

, where

it is called Hill's radius. As is evident, our estimation entirely of neplokha. But now - from the dry theory to living today's astronomy. Space probe "Galileo" on its way to Jupiter tested rapprochement with the asteroid IDA and transmitted his image. It was unexpectedly revealed that in the ides there is a miniature satellite. Image the ides with its satellite see also in the Internet with address http://galileo.ivv.nasa.gov/idamoon.html

5.18 let us designate through the masses of the Earth and sun, through a - distance between them. Let us introduce the coordinate system, as shown in figure below. It is clear that the unknown surface possesses axial symmetry relative to X-axis. Therefore is sufficient to find the section of surface by plane XY, i.e., the equation of the plane curve of form f(x,yof)=0.

Writing the force equation of attraction to the sun and to the Earth and taking into account that after simple conversions we obtain the equation of the sphere of the gravity:

Thus, the sphere of gravity - this is actual sphere. Its radius is equal

and center is displaced along x axis from the center of the earth in the antisolar direction to the

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distance

It is numerical, km the orbit of the Moon lies on the substantially great distance, so that the sun attracts the Moon more strongly than the Earth - known paradox, see tasks

and

. Further,

km, so that the center of sphere lies inside the Earth.

By plane (passing accurately in the middle between the Earth and the sun and the perpendicular to the line the Earth - the sun) the sphere of gravity would be, if the mass of the Earth was equal to solar.

5.19 v to task are two obvious dimensional values, R and M. The third, that figures in the task implicitly, by the dimensional value, which also must enter into the solution, is the constant of gravity G - indeed motion it occurs in the Newtonian gravitational field. Of these three values "to design" value with the dimensionality of the time most simply thus. Let us agree through [Q] to designate the dimensionality of value Q. Let us find the dimensionality of the constant of gravity G, after using the law of universal gravitation , from where [ sila]=[G]gof 2/sm2, or, since [ sila]=[massa ] [ acceleration ], then

It is now clear that . Therefore, after designating after the desired time of free fall to the body surface (index s - from surface), we will find that the value

it is dimensionless. Nature is arranged so that the dimensionless combinations of the determining parameters are usually the numbers of order of one. The indeterminate word "usually" indicates here that we are not found by a "number" with the singularity of one or other kind or another. Probably, these words will seem the reader by not completely intelligible - but it is useful to memorize them. You will learn to feel in the course of time, that they mean. Thus, the first (lighter) aspect of the task is solved. We pass to the second part - obtaining estimation . According to the law of conservation of energy, in the falling material point of single mass ("stone") the sum of its kinetic and (negative) potential energy -GM/R must remain constant. Let us find the value of this constant, after noting that the stone rests at the initial moment, and therefore its kinetic energy is equal to zero, whereas potential is equal GM/(2R). Therefore

At the moment of drop on the surface, i.e., with R=R, speed of stone it proves to be equal

66

The satellite, which flies along the circular orbit directly above the surface of planet has the same speed (first space, or circular rotation). In order to estimate the time of free fall, let us enter as follows. Let us visualize that our "stone" falls from distance of 2R not on the planet of mass M and radius R, but in the attracting center, in which is concentrated mass point M. Then a drop in the stone in this center can be considered as the degenerate case of motion along the ellipsis (with eccentricity e= 1 and semiaxis a=R). The doubled time of drop is a period of the complete revolution along this rectilinear orbit. According to the third law of Kepler, it is equal to the period of the revolution of satellite, which moves along the circular orbit at a distance of R=R from the attracting center - semiaxis in two orbits they are identical. Hence we find that the time of drop in the center (index c - from center) from distance of 2R is equal

It is clear that it is more than the desired time of drop to the surface of planet, but temporarily, since the second-half of way stone flies, assuming "intermediate start" with R=R with the high initial velocity , but not with the zero as in the beginning of motion with R= 2R. Therefore, evaluating the transit time of the second-half of way, we search for the small correction, which must be deducted from for obtaining . Even having found this correction it is not very accurate, we will obtain not bad estimate of the magnitude . Let us allow to the period that from distance of R the drop in the attracting center occurs with the uniform acceleration and by the initial velocity . In reality with the drop in the attracting acceleration center grows in the course of time, and therefore true speed will be higher than during the uniformly accelerated motion, and it means, the time of drop is less. Hence will follow estimation . Way R in the uniformly accelerated motion with the acceleration and by the initial velocity is passed in time such, that

We can find from this quadratic equation . We will act not into the forehead, but let us try based on our very simple example to show that it means to conduct calculations competently. Let us recall those considerations about the dimensionality, from which we began solution of problem. They prompt the expediency of introduction instead of the dimensional time t dimensionless variable such, that

To the upper limit of change , i.e. nondimensional distance

, it corresponds

67

. Let us introduce also the

In these natural for the task of dimensionless variables in question quadratic equation for the presence takes the form

where

- the dimensionless time , which corresponds

. We find from this equation that

The second root does not approach - it is negative, but time is counted off from start and therefore negative be it cannot. Let us designate the true dimensionless transit time of the second-half of way by the stone, which falls with R= 2R, through (index t - from true). It is clear that , and therefore for the dimensionless time of a drop in the stone from R= 2R to R=R, which under the condition of task was marked through , we find This is even somewhat more strongly estimation how to us it was necessary to obtain . Precise value exists See about this the following task.

5.20 answer: It is obtained in the ballistics

.

Solution: Let us find the first greatest height, to which will rise the rocket. The initial velocity, on the condition, is equal to the first space:

where M and R - mass and radius of the earth. The distance of upper point in the trajectory from the center of the earth can be found from that condition that kinetic energy of rocket at the start occurs up to the moment of its stoppage of the potential energy spent to an increase:

It follows of two written formulas that , so that the rocket, vertically neglected with the orbital velocity, rises above the surface to the height, equal to a radius of the body, from which it started, in our "battle" history - the Earth. If entire mass of the Earth was concentrated in its center, then the falling rocket would reach this attracting center, flew around it and instantly it would fly back. It would begin to accomplish periodic motion through the section of length 2R, which can be considered as the degenerate ellipsis with eccentricity e= 1. The upper point, by which reaches the rocket, this aphelion, or is more precise the apogee of this orbit. Stoppage here occurs, after which the drop in the attracting center begins. Speed monotonically grows from zero with R= 2R to

68

km/s with R=R

and, continuing grow, it becomes (formally) infinite, when the attracting center is reached. It is clear that for the solution of problem it is necessary to find the time of free fall from the height to R=R. After doubling it, we will obtain the total flying time of rocket from its starting to destruction of target. If you a little know celestial mechanics, the presence of this time labor will not compose. It suffices to recall the geometric sense of eccentric anomaly E and to consider that at the moment, when rocket strikes target mean anomaly:

. After using Kepler's equation with e= 1, let us find the

Therefore flight time from the apogee to the earth's surface is equal

and flight time from the starting to the purpose will comprise

Let us give now another solution, which does not require the knowledge of the equation of Kepler and concepts "eccentric" and "mean anomalies". However, actually they will appear also in this solution. First of all let us note that complete cycle of motion through the rectilinear section of length 2R (considered as ellipsis with semiaxis a=R and eccentricity e= 1) it is equal to the time of the circling of the Earth on the low-flying satellite (the same semiaxis, but e= 0), i.e., it composes the Gagarin of one-and-a-half hours. More precise value is such:

It is clear that, falling in the center and being slowly accelerated, body will be located at a distance of R>R the large part of this time. The second-half of the way (R