homework 13 solutions - CHAPTER I 3 3 The magnitude of the...

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Unformatted text preview: CHAPTER I 3 3. The magnitude of the force of one particle on the other is given by F = Gmlmz/rz, where m and m are the masses, r is their separation, and G is the universal gravitational constant. We solve for r: r: fi—Gngmz = (6.67X10—11Nqn2/kg2)(5.2kg)(2.4kg)=19m 2.3 x10”12N 17. (a) The gravitational acceleration at the surface of the Moon is gmoon = 1.67 rn/s2 (see Appendix C). The ratio of weights (for a given mass) is the ratio of g—values, so Wmoon = (100 N)(1.67/9.8) = 17 N. (b) For the force on that object caused by Earth’s gravity to equal 17 N, then the free-fall acceleration at its location must be ag = 1.67 m/sz. Thus, GmE - GmE r (lg 21.5 x107m so the object would need to be a distance of r/RE = 2.4 “radii” from Earth’s center. 19. The acceleration due to gravity is given by ag = GM/rz, where M is the mass of Earth and r is the distance from Earth’s center. We substitute r = R + h, where R is the radius of Earth and h is the altitude, to obtain _ GM __ GM g r2 (RE + h)2 ' a We solve for h and obtain h = GM / ag —RE. From Appendix C, RE = 6.37 x 106 m and M: 5.98 x 1024 kg, so (6.67x10"“m3 /s2 -kg)(5.98x1024kg) 2 , —6.37x106m=2.6x106m. (4.9 m/s ) h: Note: We may rewrite ag as a = GM = M = __g___ g r2 (1+h/RE)2 (1+h/RE)2 where g =9.83 m/s2 is the gravitational acceleration on the surface of the Earth. The plot below depicts how ag decreases with increasing altitude. ag/g 1 0.8 0.6 0.4 0.2 "w 24. (a) What contributes to the GmIWV2 force on m is the (spherically distributed) mass M contained within r (where r is measured from the center of M). At point A we see that M1 + M2 is at a smaller radius than r = a and thus contributes to the force: IEm m|=G(Ml +M 2)m. a2 (b) In the case r— — b, only M 1s contained Within that radius, so the force on m becomes GMm/bz. (c) If the particle is at C, then no other mass is at smaller radius and the gravitational force on it is zero. 31. The density of a uniform sphere is given by p= 3M41ER3, where M is its mass and R is its radius. On the other hand, the value of gravitational acceleration ag at the surface of a planet IS given by ag= GM/Rz. For a particle of mass 'm, its escape speed 1s given by 1 mM ZGM ——mv2 =G— :> v: —. 2 R R (a) From the definition of density above, we find the ratio of the density of Mars to the density of Earth to be gamma“ M2071 3.45x102km (b) The value of gravitational acceleration for Mars is GMM _MM R_2_ GME _MM R2 a“: R2 R2 ME R2 ME R2 “2 4 2 =0.11(0.65x10 km 345 103 km) (9.8 m/s2)=3.8 m/s2. . X (c) For Mars, the escape speed is 26M. _ W 6 =5.0><103 m/s. 3.45x10 m Note: The ratio of the escape speeds on Mars and on Earth is J2 M . 3 _____M_M.G= ”3 M_.R—= (011) 65X“) km ————=o.455. vE ZGME /RE ME RM 3 45x102 km 44. Kepler’s law of periods, expressed as a ratio, is r 3 T 2 l 3 TE 2 _s = .5. :> _ = —_——_ rm Tm 2 llunar month which yields T3 = 0.35 lunar month for the period of the satellite. ...
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