Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

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1 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , GAVITATION PG 1 8. Keple s Laws Q. Wite down the statement of Keple s Laws of planetay motion? Solution : Thee ae thee laws of Keple s : (i) Law of obits o Keple s fist law : All planets move in elliptical obits with the sun situated at one of the foci (S o S ) of the ellipse as shown in figue. (ii) Laws of aeas o Keple s second law : The line that joins any planet to the sun sweeps equal aeas in equal intevals of time i.e., the aeal velocity of any planet aound the sun emains constant. (iii) Laws of peiods o Keple s thid law : The squae of the time peiod (T) of evolution of a planet is popotional to the cube of the semi-majo axis of the ellipse taced out by the planet i.e., T a, whee a is the length of semi majou axis. Q. Keple s second law is the consequence of which law? O On which law Keple s second law is based? Solution : Keple s second law is the consequence of consevation of angula momentum. Q. Deduce Keple s second law? O How Keple s second law can be undestood fom the consevation of angula momentum? Solution : The law of aeas can be undestood as a consequence of consevation of angula momentum. Let the sun be at the oigin and let the position and momentum of the planet be denoted by and p espectively. Then the aea swept out by the planet of mass m in time inteval t is as shown in figue, given by A ½( vt). A Hence A / t ½( p) / m, (since v p / m ) = L /(m ) whee v is the velocity, L is the angula momentum equal to ( p). Fo a cental foce, which is diected along, L is a constant as the planet goes aound. Hence, A / t is a constant accoding to the last equation. This is the law of aeas. Q. What is cental foce? Give one example. Solution : A cental foce is always diected towads o away fom a fixed point, i.e., along the position vecto of the point of application of the foce with espect to the fixed point. Futhe, the magnitude of a cental foce F depends on. A cental foce is such that the foce on the planet is along the vecto joining the sun and the planet. Gavitation foce is a cental foce.

2 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , Q. Which quantity is always conseved in the motion unde a cental foce? Solution : In the motion unde a cental foce the angula momentum is always conseved. Q. Wite down two impotant esults which will aise in the motion unde a cental foce? Solution : Two impotant esults which aises in the motion unde the cental foce : (i) The motion of a paticle unde the cental foce is always confined to a plane. PG (ii) The position vecto of the paticle with espect to the cente of the foce (i.e., the fixed point) has a constant aeal velocity. Q. Let the speed of the planet at the peihelion P in figue be v P and the sun-planet distance SP be P, elate { P, v P } to the coesponding quantities at the aphelion { A, v A }. Will the planet take equal times to tansvese BAC and CPB? [NCT Solved xample 8.1] Solution : The magnitude of the angula momentum at P is L p = m p p v p, since inspection tells us that p v p ae mutually pependicula. Similaly, L A = m p A v A. Fom angula momentum consevation and m p p v p = m p A v A o vp A. Since A > p, v p > v A. v A p The aea SBAC bounded by the ellipse and the adius vectos SB and SC is lage than SBPC. Fom Keple second law, equal aeas ae swept in equal times. Hence the planet will take a longe time to tansvese BAC than CPB. 8. Univesal Law of Gavitation : Q. xplain Newton s law of gavitation? O Wite down the foce between two point masses sepaated by some distance in vecto fom. Solution : The foce between two point masses sepaated by some distance is given by Newton s law of gavitation which will be stated as : the foce F on a point mass m due to anothe point mass m 1 has the magnitude F m m G 1. m1m m1m m quation can be expessed in vecto fom as F G ( ˆ) G ˆ = 1m G ˆ, whee G is the univesal gavitational constant, ˆ is the unit vecto fom m 1 to m and 1 as shown in above figue.

3 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , Q. Wite down the unit and dimensional fomula of gavitational constant G. Solution : The unit of G is Nm /kg and dimensional fomula [M 1 L T ]. Q. What is the natue of the gavitational foce? Solution : The gavitational foce is attactive cental foce. Q. Find the acceleation of the moon evolving in an obit of adius m? PG Solution : The moon evolving in an obit of adius m was subject to a centipetal acceleation due to eath s gavity of magnitude. a m V m 4 T m whee V is the speed of the moon elated to the time peiod T by the elation V = m /T. Q. Wite down the expession of total foce on m 1 due to othe point masses m, m and m 4? Solution : If we have a collection of point masses, the foce on any one of them is the vecto sum of the gavitational foces exeted by the othe point masses as shown. Gm m1 Gm m1 Gm 4m1 The total foce on m 1 is F1 ˆ 1 ˆ 1 ˆ Q. Thee equal masses of m kg each ae fixed at the vetices of an equilateal tiangle ABC. (a) What is the foce acting on a mass m placed at the centoid G of the tiangle? (b) What is the foce if the mass at the vetex A is doubled? [NCT Solved xample 8.] Take AG = BG = CG = 1 m Solution : (a) zeo (b) Gm ĵ Q. How will you detemine the foce between a hollow spheical shell of unifom density and a point mass situated outside? Solution : The foce of attaction between a hollow spheical shell of unifom density and a point mass situated outside is just as if the entie mass of the shell is concentated at the cente of the shell. Q. What is the foce on point mass which is situated inside a hollow spheical shell of unifom density? Solution : The foce of attaction due to a hollow spheical shell of unifom density, on a point mass situated inside it is zeo. 8.4 The Gavitational Constant : Q. Descibe Cavendish s expeiment which is used to detemine the value of G? Solution : The value of the gavitational constant G enteing the Univesal law of gavitation can be detemined expeimentally and this was fist done by nglish scientist Heny Cavandish. The appaatus used by him is shown in figue. 41

4 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , PG 4 The ba AB has two small lead sphees attached at its ends. The ba is suspended fom a igid suppot by a fine wie. Two lage lead sphees ae bought close to the small ones but on opposite sides as shown. The big sphees attact the neaby small ones by equal and opposite foce as shown. Thee is no net foce on the ba but only a toque which is clealy equal to F times the length of the ba, whee F is the foce of attaction between a big sphee and its neighbouing small sphees. Due to this toque, the suspended wie gets twisted till such time as the estoing toque of the wie equals the gavitational toque. If is the angle of twist of the suspended wie, the estoing toque is popotional to, equal to, whee is the estoing couple pe unit angle of twist. can be measued independently e.g., by applying a known toque and measuing the angle of twist. The gavitation foce between the spheical balls is the same as if thei masses ae concentated at thei centes. Thus if d is the sepaation between the centes of the big and its neighbouing small ball, M and m thei masses, the gavitational foce between the big sphee and its neighbouing small Mm ball is, F G. If L is the length of the ba AB, then the toque aising out of F is F multiplied by L. At d Mm equilibium, this is equal to the estoing toque and hence G L d. Obsevation of this enables one to calculate G fom this equation. Since Cavandish s expeiment, the measuement of G has been efined and the cuently accepted value is G = N m /kg 8.5 Acceleation due to Gavity of the ath : Q. A paticle mass m is situated inside the eath of mass M and adius at the distance fom the cente of eath. Find the foce on this paticle due to eath. Solution : Conside the eath to be made up of concentic shells and the point mass m situated at a distance fom the cente. Fo the shells of adius geate than, the point P lies inside, they exet no gavitational foce on mass m kept at P. The shells with adius less than o equal to make up a sphee of adius fo which the point P lies on the suface. This smalle sphee theefoe exets a foce on a mass m at P as if its mass M is Gm(M ) concentated at the cente. Thus the foce on the mass m at P has a magnitude F. We assume 4 that the entie eath is of unifom density and hence its mass is M whee M is the mass of the eath is its adius and is the density. On the othe hand the mass of the sphee M of adius is 4

5 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , PG 5 4 and hence F Gm M Gm GmM If the mass m is situated on the suface of eath, then = and the gavitational foce on it is given by M m F G. Q. Deive the expession of acceleation due to gavity due to eath on its suface? Solution : The acceleation expeienced by the mass m, which is usually denoted by the symbol g is elated to F by Newton s nd law by elation F = mg. Thus g F m GM 8.6 Acceleation Due to Gavity below and above the Suface of ath : Q. Discuss the vaiation of g with height and depth? Solution : Vaiation of g with height above the suface of eath : Conside a point mass m at a height h above the suface of the eath as shown in figue. The adius of the eath is denoted by. Since this point is outside the eath, its distance fom the cente of the eath is GM m ( + h). If F(h) denoted the magnitude of the foce on the point mass m, we get F(h). ( h) The acceleation expeienced by the point mass is F(h)/m g(h) and we get F(h) GM g(h) = m ( h) GM (1 h / ) g(1 h / ) GM, whee g. h h Fo 1 (nea the suface of eath), using binomial expession, g (h) g 1. Vaiation of g with depth : Now, conside a point mass m at a depth d below the suface of the eath, so that its distance fom the cente of the eath is ( d) as shown in figue. The eath can be thought of as being composed of a smalle sphee of adius ( d) and a spheical shell of thickness d. The foce on m due to the oute shell of thickness d is zeo. As fa as the smalle sphee of adius ( d) is concened, the point mass is outside it and hence the foce due to this smalle sphee is F(d) = G M S m/( d), hee M s is the mass of the smalle sphee.

6 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , PG 6 The value of M s is given by, M s 4 d = M 4 4 ( d) Subtituting fo M s in F(d) = G M S m/( d), we get F(d) = G M m( d)/. Hence the acceleation due to gavity at a depth d is given by M ( = F(d) GM g(d) d ( d) = g g(1 d / ) GM m g Hence acceleation due to eath s gavity is maximum on its suface and deceases whethe you go up o down fom the suface. 8.7 Gavitational Potential negy : Q. What is gavitational potential enegy? Deive the expession of gavitational potential enegy fo two point masses m 1 and m sepaated by distance? Solution : The foce of gavity is a consevative foce and we can calculate the potential enegy of a body aising out of this foce, called the gavitational potential enegy. The gavitational potential enegy at a point is the amount of wok done in displacing the paticle fom infinity to that point. We know that a point outside the eath, the foce of gavitation on a paticle diected towads the cente of GM m the eath is F, whee M = mass of eath, m = mass of the paticle and its distance fom the cente of the eath. Now we calculate the wok done in lifting a paticle fom to, we get W = V = Fd GMm d = GM 1 m 1 1 GM m GM m. Hence the gavitational potential enegy associated with two paticles of masses m 1 and m sepaated by Gm m distance by a distance is given by V = 1 (if we choose V = 0 as ). Q. Find the potential enegy of a system of fou paticles placed at the vetices of a squae of side l. Also obtain the potential at the cente of the squae. [NCT Solved xample 8.] Solution : Conside fou masses each of mass m at the cones of a squae of side l. We have fou mass pais at distance l and two diagonal pais at distance l. d). Hence, Gm Gm W() 4 l l Gm l 1 Gm 5.41 l The gavitational potential at the cente of the squae ( = l/) is 8.8 scape Speed : U() Gm 4. l Q. Define escape speed? Deive the expession of escape speed fom the suface of eath? Solution : scape speed on eath (o any othe planet) is defined as the minimum speed with which a body has to be pojected fom the suface of eath (o any othe planet) so that it just cosses the gavitational field of eath (o of that planet) and neve etuns. Deivation of escape speed : Let V e is the escape speed of an object of mass m fom the suface of spheical astonomical body of mass M and adius then fom consevation of mechanical enegy pinciple,

7 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , PG 7 1 e mv GMm 0 Using the elation g = GM /, we get e V e V g. Q. xpess the escape speed in tems of the density of planet. GM Solution : Let the astonomical body is of unifom density then M = Q. Does the escape speed depend on the mass of pojected paticle? 4 8 and hence V e G. Solution : The escape speed of an object at a given point in the field is independent of its mass. Q. Does the escape speed depend on angle of pojection? Solution : The escape speed does not depend on angle of pojection. Q. What is value of escape speed fom the suface of eath? Solution : The value of escape speed is 11. km/s fom the suface of eath. Q. Moon has no atmosphee. Give eason? Solution : The escape speed fo the moon tuns out to be. km/s. This is the eason that moon has no atmosphee. Gas molecules if fomed on the suface of the moon has no atmosphee. Gas molecules if fomed on the suface of the moon having velocities lage than this will escape the gavitational pull of the moon. Q. Two unifom solid sphees of equal adii, but mass M and 4 M have a cente to cente sepaation 6, as shown in figue. The two sphees ae held fixed. A pojectile of mass m is pojected fom the suface of the sphee of mass M diectly towads the cente of the second sphee. Obtain an expession fo the minimum speed v of the pojectile so that it eaches the suface of the second sphee. [NCT Solved xample 8.4] Solution : v 8.9 ath Satellites : GM 5 1/ Q. What is eath satellite? Solution : ath satellites ae objects which evolve aound the eath. In paticula, thei obits aound the eath ae cicula o elliptic. Q. What is the use of atificial eath satellites? Solution : Atificial eath satellites fo pactical use in fields like telecommunication, geophysics and meteology. Q. Deive the expession of time peiod of a eath satellite? O Deive Keple s thid law fo cicula obit? Solution : We will conside a satellite in a cicula obit of a distance ( + h) fom the cente of the eath, whee = adius of the eath. If m is the mass of the satellite and V its speed, the centipetal foce equied fo this obit is mv F(centipetal) diected towads the cente. The centipetal foce is povided ( h) by the gavitational foce, which is GmM F(gavitation) whee M is the mass of the eath. ( h)

8 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , PG 8 quating.h.s. of both equation, we get V / GM. Its time peiod T theefoe is ( h) ( h) ( h) T. Squaing both sides of this equation, we get T = k( + h) V GM (whee k = 4 /GM ) which is Keple s law of peiods, as applied to motion of satellites aound the eath. Q. Find the time peiod of a satellite which is vey nea to the suface of eath? Solution : Fo a satellite vey close to the suface of eath h can be neglected in compaison to in equation. Hence, fo such satellites, T is T 0, whee T0 g = 9.8 m s and = 6400 km., we get T 0 6 / g. If we substitute the numeical values s which is appoximately 85 minutes. 9.8 Q. The planet Mas has two moons, phobos and delmos. (i) phobos has a peiod 7 hous, 9 minutes and an obital adius of km. Calculate the mass of mas. (ii) Assume that eath and mas move in cicula obits aound the sun, with the matian obit being 1.5 times the obital adius of the eath. What is the length of the matian yea in days? [NCT Solved xample 8.5] Solution : (i) kg (ii) Q. Weighing the ath : You ae given the following data : g = 9.81 ms, = m, the distance to the moon = m and the time peiod of the moon s evolution is 7. days. Obtain the mass of the ath M in two diffeent ways. [NCT Solved xample 8.6] Solution : kg Q. xpess the constant k (k = 4 /GM ) in days and kilometes. Give k = 10 1 s m. The moon is at a distance of km fom the eath. Obtain its time peiod of evolution in days. [NCT Solved xample 8.7] Solution : K = d km, T = 7. d 8.10 negy of an Obiting Satellite : Q. Deive the expession of kinetic enegy, potential enegy and total enegy of an obiting satellite? 1 GmM Solution : The kinetic enegy of the satellite in a cicula obit with speed v is K mv, ( h) the potential enegy at distance ( + h) fom the cente of the eath is GmM P.. The total is ( h) GmM K. P.. ( h) Q. If the total enegy is, potential enegy is U and kinetic enegy is K, then find the atio : U : K? GmM GmM GmM Solution : As K =, U =, =, hence : U : K = 1 : : 1. ( h) ( h) ( h) Q. Does the kinetic and potential enegy of a satellite in elliptic obit change fom point to point in the obit? What will happen to total enegy of satellite in this obit? Solution : When the obit of a satellite becomes elliptic, both the K.. and P.. vay fom point to point. The total enegy which emains constant is negative

9 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , Geostationay and Pola Satellites : PG 9 Q. What is Geostationay satellite? Find the height above the suface of eath of such satellites? Solution : Geostationay satellite ae those satellite fo which the cicula obit is in the equatoial plane of the eath, having the same peiod as the peiod of otation of the eath about its own axis and hence it would appea stationay viewed fom a point on eath. Height of geostationay satellite above the suface of eath : Fom ( h) ( h) T V GM /, we get 1/ T GM h = T g 4 4 1/ GM g Put T = 4 hous, = 6400 km then we get h = 5800 km. Q. Give one example of Geostationay satellite and wite down one use of this? Solution : The INSAT goup of satellites sent up by India ae one such goup of Geostationay satellites which ae widely used fo telecommunications in India. Q. What is pola satellites? What ae the uses of such satellite? Solution : Pola satellites ae low altitude ( h 500 to 800 km) satellites, they go aound the poles of the eath in a noth-south diection. Use of pola satellites : in emote sensing, meteology and as well as fo envionmental studies of the eath. 8.1 Weightlessness : Q. What is weightlessness? Solution : When an object is in fee fall, it is weightless and this phenomenon is usually called the phenomenon of weightlessness. Q. xplain the phenomenon of weightlessness in satellites aound the eath? Solution : In a satellite aound the eath, evey pat and pacel of the satellite has an acceleation towads the cente of the eath which is exactly the value of eath s acceleation due to gavity at that position. Thus in the satellite eveything inside it is in a state of fee fall. This is just as if we wee falling towads the eath fom a height. Thus, in a manned satellite, people inside expeience no gavity.

10 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , Answe the following : (a) (b) (c) NCT XCIS PG 10 You can shield a chage fom electical foces by putting it inside a hollow conducto. Can you shield a body fom the gavitational influence of neaby matte by putting it inside a hollow sphee o by some othe means? As astonaut inside a small space ship obiting aound the eath cannot detect gavity. If the space station obiting aound the eath has a lage size, can he hope to detect gavity? If you compae the gavitational foce on the eath due to the sun to that due to the moon, you would find that the Sun s pull is geate than the moon s pull. (you can check this youself using the data avail able in the succeeding execise). Howeve, the tidal effect of the moon s pull is geate than the tidal effect of sun. Why? 8. Choose the coect altenative : (a) (b) (c) (d) Acceleation due to gavity inceases/deceases with inceasing altitude. Acceleation due to gavity inceases/deceases with inceasing depth (assume the eath to be a sphee of unifom density). Acceleation due to gavity is independent of mass of the eath/mass of the body. The fomula G Mm(1/ 1/ 1 ) is moe/less accuate than the fomula mg( 1 ) fo the diffeence of potential enegy between two points and 1 distance away fom the cente of the eath. 8. Suppose thee existed a planet that went aound the sun twice as fast as the eath. What would be its obital size as compaed to that of the eath? 8.4 Io, one of the satellites of Jupite, has an obital peiod of days and the adius of the obit is m. Show that the mass of Jupite is about one-thousandth that of the sun. 8.5 Let us assume that ou galaxy consists of stas each of one sola mass. How long will a sta at a distance of 50,000 ly fom the galactic cente take to complete one evolution? Take the diamete of the Milky Way to be 10 5 ly. 8.6 Choose the coect altenative : (a) (b) If the zeo of potential enegy is at infinity, the total enegy of an obiting satellite is negative of its kinetic/potential enegy. The enegy equied to launch an obiting satellite out of eath s gavitational influence is moe/less than the enegy equied to poject a stationay object at the same height (as the satellite) out of eath s influence. 8.7 Does the escape speed of a body fom the eath depend on (a) the mass of the body, (b) the location fom whee it is pojected, (c) the diection of pojection, (d) the height of the location fom whee the body is launched? 8.8 A comet obits the sun in a highly elliptical obit. Does the comet have a constant (a) linea speed, (b) angula speed, (c) angula momentum, (d) kinetic enegy, (e) potential enegy, (f) total enegy thoughout its obit? Neglect any mass loss of the comet when it comes vey close to the Sun. 8.9 Which of the following symptoms is likely to affilict an astonaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) oientational poblem In the following two execises, choose the coect answe fom among the given ones : The gavitational intensity at the cente of a hemispheical shell of unifom mass density has the diection indicated by the aow (see in the figue) (i) a, (ii) b, (iii) c, (iv) 0.

11 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , PG The diection of the gavitational intensity at an abitay point P is indicated by the a o w (i) d, (ii) e, (iii) f, (iv) g. 8.1 A ocket is fied fom the eath towads the sun. At what distance fom the eath s cente is the gavitational foce on the ocket zeo? Mass of the sun = 10 0 kg, mass of the eath = kg. Neglect the effect of othe planets etc. (obital adius = m). 8.1 How will you weigh the sun, that is estimate its mass? The mean obital adius of the eath aound the sun is km A satun yea is 6.5 times the eath yea. How fa is the satun fom the sun if the eath is km away fom the sun? 8.15 A body weighs 6 N on the suface of the eath. What is the gavitational foce on it due to the eath at a height equal to half the adius of the eath? 8.16 Assuming the eath to be a sphee of unifom mass density, how much would a body weigh half way down to the cente of the eath if it weighed 50 N on the suface? 8.17 A ocket is fied vetically with a speed of 5 km s 1 fom the eath s suface. How fa fom the eath does the ocket go befoe etuning to the eath? Mass of the eath = kg; mean adius of the eath = m; G = N m kg The escape speed of a pojectile on the eath s suface is 11. km s 1. A body is pojected out with thice this speed. What is the speed of the body fa away fom the eath? Ignoe the pesence of the sun and othe planets A satellite obits the eath at the height of 400 km above the suface. How much enegy must be expended to ocket the satellite out of the eath s gavitational influence? Mass of the satellite = 00 kg; mass of the eath = kg; adius of the eath = m; G = N m kg. 8.0 Two stas each of one sola mass (= 10 0 kg) ae appoaching each othe fo a head on collision. When they ae a distance 10 9 km, thei speeds ae negligible. What is the speed with which they collide? The adius of each sta is 10 4 km. Assume the stas to emain undistoted until they collide. (Use the known value of G). 8.1 Two heavy sphees each of mass 100 kg and adius 0.10 m ae placed 1.0 m apat on a hoizontal table. What is the gavitational foce and potential at the mid point of the line joining the centes of the sphees? Is an object placed at that point in equilibium? If so, is the equilibium stable o unstable? 8. As you have leant in the text, a geostationay satellite obits the eath at a height of nealy 6,000 km fom the suface of the eath. What is the potential due to eath s gavity at the site of this satellite? (Take the potential enegy at infinity to be zeo). Mass of the eath = kg, adius = 6400 km. 8. A sta.5 times the mass of the sun and collapsed to a size of 1 km otates with a speed of 1. ev. pe second. (xtemely compact stas of this kind ae known as neuton stas. Cetain stella objects called pulsas belong to this categoy). Will an object placed on its equato emain stuck to this suface due to gavity? (mass of the sun = 10 0 kg). 8.4 A spaceship is stationed on Mas. How much enegy must be expended on the spaceship to launched it out of the sola system? Mass of the space ship = 1000 kg; mass of the sun = 10 0 kg; mass of mas = kg; adius of mas = 95 km; adius of the obit of mas = km; G = N m kg. 8.5 A ocket is fied vetically with a speed of 5 km s 1 fom the eath s suface. How fa fom the eath does the ocket go befoe etuning to the eath? Mass of the eath = kg; mean adius of the eath = m; G = N m kg.

12 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , (a) No. (b) Yes, if the size of the sphee ship is lage enough fo him to detect the vaiation in g. PG 1 (c) Tidal effect depends invesely on the cube of the distance unlike foce, which depends invesely on the squae of the distance. 8. (a) deceases; (b) deceases; (c) mass of the body; (d) moe 8. Smalle by a facto of yeas 8.6 (a) Kinetic enegy, (b) less 8.7 (a) No, (b) No, (c) No, (d) Yes 8.8 All quantities vay ove an obit except angula momentum and total enegy. 8.9 (b), (c) and (d) and 8.11 fo these two poblems, complete the hemisphee to sphee. At both P, and C potential is constant and hence intensity = 0. Theefoe, fo the hemisphee, (c) and (e) ae coect m kg m N N m fom the eath s cente km/s J m/s 8.1 0, J/kg; an object placed at the mid point is in an unstable equilibium J/kg 8. GM/ = ms, = ms ; hee is the angula speed of otation. Thus in the otating fame of the sta, the inwad foce is much geate than the outwad centifugal foce as its equato. The object will emain stuck (and not fly off due to centifugal foce). Note, if angula speed of otation inceases say by a facto of 000, the object will fly off J m fom the eath s cente ANSWS

13 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , ADDITIONAL QUSTIONS AND POBLMS PG 1 Q. Daw gaphs showing the vaiation of acceleation due to gavity with (a) height above the ath s suface, (b) depth below the ath s suface. A. Q. Why ae space ockets usually launched fom west to east in the equatoial line? A. We know that eath evolves fom west to east about its pola axis. Theefoe, all the paticles on the eath have velocity fom the west to east. This velocity is maximum in the equatoial line, as v =, whee is the adius of eath and is the angula velocity of evolution of eath about its pola axis. When a ocket is launched fom west to east in equatoinal plane, the maximum linea velocity is added to the launching velocity of the ocket, due to it, launching becomes easie. Q. The escape speed of the pojectile on the eath s suface is v e. A body is pojected out with thice of this speed. What is the speed of the body afte cossing the gavitational field of the eath? Ignoe the pesence of Sun and the anothe planets. A. v e Q. What is the potential enegy of a body of mass m elative to the suface of ath at a (a) height h = above its suface (b) depth d = below its suface? Q. The mass and the diamete of a planet ae twice those of ath. What will be the time peiod of that pendulum on this planet, which is a second pendulum on the ath. Q. Although gavitational pull of sun on eath is moe than that of moon, yet moon s contibution towads tide fomation on eath is geate than that due to sun. Why? A. The distance between moon and eath is vey small as compaed to the distance between eath and sun. Since, the tidal effect on oscen wate of eath is invesely popotional to the cube of the distance, theefoe, tidal effect on oscen wate due to moon is lage than that due to sun. Q. Why does a body lose weight at the cente of the eath? Q. xplain, why a tennis ball bounces highe on hills than in plains. Q. Does the foce of attaction between two bodies depend upon the pesence of othe bodies and popeties of intevening medium? A. The gavitational foce of attaction between the two bodies is independent of the pesence of othe bodies and popeties of intevening medium. Q. Why is Newton s law of gavitation called a univesal law? A. Newton s law of gavitation holds good iespective of the natue of two bodies, i.e., big o small, at all times, at all locations and fo all distances in the univese. Q. On eath value of G = Nm kg. What is its value on moon, whee g is nealy one-sixth than that of eath? A. The value of G is same on moon as on the suface of eath because G is a univesal gavitational constant. Q. The value of g on the moon is 1/6th of that of eath. If a body is taken fom the eath to the moon, then what will be the change in its (i) weight, (ii) inetial mass and (iii) gavitational mass? A. Weight of body on moon will become 1/6th of that on the eath, but thee will be no change in inetial and gavitational masses. Q. Moon tavelles tie heavy weight at thei back befoe landing on the moon. Why? A. The value of g on moon is small, theefoe, the weight of moon tavelles will also be small.

14 instein Classes, Unit No. 10, 10, Vadhman ing oad Plaza, Vikas Pui xtn., Oute ing oad New Delhi , Ph. : , PG 14 Q. Is it possible to put an atificial satellite on an obit in such a way that it always emains visible diectly ove chandigah? A. No, because to put an atificial satellite in an obit such that it always emains diectly ove a paticula place, its time peiod should be the same as that of the eath in the equatoial plane. As Chandigah does not lie on the equatoial plane, a geostationay satellite cannot be seen ove Chandigah. Q. An astonaut, while evolving in a cicula obit happens to thow a ball outside. Will the ball each the suface of eath? A. The ball will neve each the suface of eath as it will continue to move in the same cicula obit and will chase the astonaut. Q. Distinguish between g and G. A. The value of g changes with height depth and otation of the eath and is zeo at the cente of the eath. The value of G is constant at all locations. The value of G is neve zeo any whee. Q. Geneally the path of a pojectile fom the suface of eath is paabolic but it is elliptical fo pojectiles going to a vey geat height. Why? A. Upto odinay heights, the change in the distance of a pojectile fom the cente of the eath is negligible small, as compaed to the adius of the eath. Due to which the pojectile moves unde a unifom gavitational foce and its path is paabolic. But if pojectile is going to a vey geat height, the gavitational foce deceases, which is invesely popotional to the squae of the distance of the pojectile fom the cente of the eath. Unde such a vaiable foce, the path of pojectile is elliptical.