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Unformatted text preview: 575 Chapter 13 1. The magnitude of the force of one particle on the other is given by F = Gm 1 m 2 / r 2 , where m 1 and m 2 are the masses, r is their separation, and G is the universal gravitational constant. We solve for r : ( 29 ( 29( 29 11 2 2 1 2 12 6.67 10 N m / kg 5.2kg 2.4kg 19m 2.3 10 N Gm m r F × ⋅ = = = × . 2. We use subscripts s, e, and m for the Sun, Earth and Moon, respectively. Plugging in the numerical values (say, from Appendix C) we find 2 2 2 30 8 2 24 11 / 1.99 10 kg 3.82 10 m 2.16. / 5.98 10 kg 1.50 10 m sm s m sm s em em e m em e sm F Gm m r m r F Gm m r m r × × = = = = × × 3. The gravitational force between the two parts is ( 29 ( 29 2 2 2 = = Gm M m G F mM m r r which we differentiate with respect to m and set equal to zero: ( 29 2 = 0 = 2 = 2 dF G M m M m dm r ⇒ . This leads to the result m / M = 1/2. 4. The gravitational force between you and the moon at its initial position (directly opposite of Earth from you) is 2 ( ) m ME E GM m F R R = + where m M is the mass of the moon, ME R is the distance between the moon and the Earth, and E R is the radius of the Earth. At its final position (directly above you), the gravitational force between you and the moon is 1 2 ( ) m ME E GM m F R R = . CHAPTER 13 576 (a) The ratio of the moon’s gravitational pulls at the two different positions is 2 2 2 8 6 1 2 8 6 /( ) 3.82 10 m 6.37 10 m 1.06898. /( ) 3.82 10 m 6.37 10 m m ME E ME E m ME E ME E GM m R R F R R F GM m R R R R  + × + × = = = = + × × Therefore, the increase is 0.06898, or approximately, 6.9%. (b) The change of the gravitational pull may be approximated as 1 2 2 2 2 3 4 1 2 1 2 . ( ) ( ) m m m m m E E E ME E ME E ME ME ME ME ME GM m GM m GM m GM m GM mR R R F F R R R R R R R R R  = ≈ + =  + On the other hand, your weight, as measured on a scale on Earth is 2 E g E E GM m F mg R = = . Since the moon pulls you “up,” the percentage decrease of weight is 3 3 22 6 7 5 1 24 8 7.36 10 kg 6.37 10 m 4 4 2.27 10 (2.3 10 )%. 5.98 10 kg 3.82 10 m m E g E ME F F M R F M R  × × = = = × ≈ × × × 5. We require the magnitude of force (given by Eq. 131) exerted by particle C on A be equal to that exerted by B on A . Thus, Gm A m C r 2 = Gm A m B d 2 . We substitute in m B = 3 m A and m B = 3 m A , and (after canceling “ m A ”) solve for r . We find r = 5 d . Thus, particle C is placed on the x axis, to left of particle A (so it is at a negative value of x ), at x = –5.00 d ....
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 Summer '10
 Prof.Yang
 Physics, Force, Mass

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