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GRAVITATION 137 2 G M 4 1 traverses a distance 2 (R + h) with speed V. Its v     E R 5 2  time period T therefore is 2 (R  h) 2 (R h)3/2 3G M /2 T  E  E (7.37) v   ⊳ V G M E  5 R  on substitution of value of V from Eq. (7.35). Apointtonoteisthatthespeedoftheprojectile Squaring both sides of Eq. (7.37), we get is zero at N, but is nonzero when it strikes the T = k ( R + h) (where k = 4 2/ GM ) (7.38) heavier sphere 4 M. The calculation of this speed E E is left as an exercise to the students. which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a 7.9 EARTH SATELLITES satellite very close to the surface of earth h can be neglected in comparison to R in Eq. (7.38). Earth satellites are objects which revolve around E the earth. Their motion is very similar to the Hence, for such satellites, T is o , where motion of planets around the Sun and hence T 0 2  R Eg (7.39) Kepler’s laws of planetary motion are equally If we substitute the numerical values applicable to them. In particular, their orbits -2 g ≃ 9.8 m s and R = E400 km., we get around the earth are circular or elliptic. Moon is 6 the only natural satellite of the earth with a near 6.4 10 circular orbit with a time period of approximately T 0 2 9.8 s 27.3 days which is also roughly equal to the Which is approximately 85 minutes. rotational period of the moon about its own axis. Since,1957,advancesintechnologyhaveenabled Example 7.5 The planet Mars has two moons, phobos and delmos. (i) phobos has manycountriesincludingIndiatolaunchartificial earth satellites for practical use in fields like a period 7 hours, 39 minutes and an orbital telecommunication, geophysics and meteorology. radius of 9.4 10 km. Calculate the mass of mars. (ii) Assume that earth and mars We will consider a satellite in a circular orbit of a distance (E + h) from the centre of the earth, move in circular orbits around the sun, where R = radius of the earth. If m is the mass with the martian orbit being 1.52 times E the orbital radius of the earth. What is of the satellite and V its speed, the centripetal force required for this orbit is the length of the martian year in days ? mV 2 Answer (i) We employ Eq. (7.38) with the Sun’s F(centripetal) = (7.33) (R Eh) mass replaced by the martian mass M m directedtowardsthecentre. Thiscentripetalforce 2 T 2  4 R3 is provided by the gravitational force, which is GM 2 3 Gm M E 4 R F(gravitation) = (R  h) 2 (7.34) M m  2 E G T where M iE the mass of the earth. 18 Equating R.H.S of Eqs. (7.33) and (7.34) and 4  3.14 2 9.43 10  -11 2 cancelling out m, we get 6.67  10  459  60  G M V  E 4 3.142 9.4  1018 (REh) (7.35) M  m 6.67 4.59 6 210 -5 Thus V decreases as h increases. From = 6.48 10 kg. equation (7.35),the speed V for h = 0 is 2 (ii) Once again Kepler’s third law comes to our V (h  0)  GM R/ ER E (7.36) aid, 2 3 where we have used the relation TM R MS 2 2  3 g = GM /R E . In every orbit, the satellite TE R ES Reprint 2026-27