Gravitation Solved Example

GRAVITATION Page # 8 SOLVED EXAMPLE Ex.1 Let the speed of the planet at the perihelion P be vP and the Sun-planet distance SP be rP. Relate {rP, vP} to the corresponding quantities at the aphelion {rA, vA}. Will the planet take equal times to traverse BAC and CPB ? Ans. The magnitude of the angular momentum at P is Lp = mp rp p, since inspection tells us that rp and vp are mutually perpendicular. Similarly, LA = mp rA A. From angular momentum conservation mp rp p = mp P r  A rA A or A rP Since rA > rp, p > A . The area SBAC bounded by the ellipse and the radius vectors SB and SC is larger than SBPC in. From Kepler’s second law, equal areas are swept in equal times. Hence the planet will take a longer time to traverse BAC than CPB. Alternatively, one expects on the basis of symmetry that the resultant force ought to be zero. (b) By symmetry the x-component of the force cancels out. The y-component survives ^ ^ ^ FR  4Gm2 j  2Gm2 j  2Gm2 j Ex.3 Find the potential energy of a system of four particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square. Ans. Consider four masses each of mass m at the corners of a square of side l ; See Fig. We have four mass pairs at distance l and two diagonal pairs at distance 2  Hence W(r)  4 Ex.2 Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. (a) What is the force acting on a mass 2m placed at the centroid G of the triangle? (b) What is the force if the mass at the vertex A is doubled ? Take AG = BG = CG = 1m  2Gm2  Gm2 Gm2 2  2  1  Gm2  2    5.41  2  The gravitational potential at the centre of the square Gm  Ex.4 Two uniform solid spheres of equal radii R, but mass M and 4 M have a center to centre separation 6 R, as shown in Fig. 8.10. The two spheres are held fixed. A projectile of mass m is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere. r  Ans. (a) The angle between GC and the positive xaxis is 30° and so is the angle between GB and the negative x-axis. The individual forces in vector notation are FGA  Gm(2m) ^ j 1 FGB  ^  Gm(2m)  ^   i cos 30º j sin 30º  1   FGC  ^  Gm(2m)  ^   i cos 30º j sin 30º  1    2  / 2 is U(r)  4 2 From the principle of superposition and the law of vector addition, the resultant gravitational force FR on (2m) is FR = FGA + FGB + FGC ^ ^  ^  FR  2Gm2 j  2Gm2   i cos 30º j sin 30º    ^ ^   2Gm2  i cos 30º j sin 30º   0   Ans. The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point N (see Fig. 8.10) is defined as the position where the two forces cancel each other exactly. If GRAVITATION Page # 9 GMm ON = r, we have 2  (ii)Once again Kepler’s third law comes to our aid, 4GMm 2 r (6R  r) (6R – r)2 = 4r2 6R – r = ± 2r r = 2R or –6R The neutral point r = – 6R does not concern us in this example. Thus ON = r = 2R. It is sufficient to project the particle with a speed which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface of M is Et  1 GMm 4GMm m2   2 R 5 /R At the neutral point N, the speed approaches zero. The mechanical energy at N is purely potential. EN   GMm 4GMm  2R 4R From the principle of conservation of mechanical energy 1 2 GM 4GM GM GM      2 R 5R 2R R or 2  1/2 42 3 R GMm 3 R ES Ex.6 Weighing the Earth : You are given the following data: g = 9.81 ms–2, RE = 6.37 ×106 m, the distance to the moon R = 3.84 ×108 m and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth ME in two different ways. Ans. we have ME  gRE2 G  9.81  6.37  106 6.67  10 11  2 = 5.97 × 1023 kg ME  = 42R 3 G ME 42R 3 G T2 4  3.14  3.14  (3.84)3  1024 6.67  1011  (27.3  24  60  60)2 = 6.02 × 1024 kg Ex.7 Express the constant k of Eq. (8.38) in days and kilometres. Given k = 10–13 s2 m–3. The moon is at a distance of 3.84 ×105 km from the earth. Obtain its time-period of revolution in days. Ans. Given k = 10–13 s2 m–3    1 1 =10 13  d2   2 3 3  (24  60  60)   (1 / 1000) km  42 R 3 G T2 = 1.33 ×10–14 d2 km–3 2 Mm  3 R MS where RMS is the mars -sun distance and RES is the earth-sun distance.  TM = (1.52)3/2 × 365 = 684 days We note that the orbits of all planets except Mercury, Mars and Pluto are very close to being circular. T2  Ex.5 The planet Mars has two moons, phobos and delmos. (i) phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 103 km. Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth. What is the length of the martian year in days ? Ans. (i) The sun’s mass replaced by the martian mass =  The moon is a satellite of the Earth. From the derivation of Kepler’s third law  3GM     5R  Mm  TE2  2GM  4 1     R 5 2 2 Mm T  TM2 3 18 4  (3.14)  (9.4)  10 6.67  1011  (459  60)2 4  (3.14)2  (9.4)3  1018 6.67  (459  6)2  105 = 6.48 × 10 K= 23 kg 42 , T2 GMg = K [RE = h]3 and the given value of k, the time period of the moon is T2 = (1.33 ×10-14)(3.84 ×105)3 T = 27.3 d