This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full
Document
This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
 Summer '10
 Prof.Yang
 Physics, Force, Mass

Click to edit the document details