Fisic

Física II – Sears, Zemansky, Young & Freedman.

PHYSICS ACT. http//physicsact.wordpress.com Capítulo 12

Note: to obtain the numerical results given in this chapter, the following numerical values of certain physical quantities have been used;
G = 6.673 × 10−11 N ⋅ m 2 kg 2 , g = 9.80 m s 2 and mE = 5.97 × 1024 kg. Use of other tabulated values for these quantities may result in an answer that differs in the third significant figure. 12.1: The ratio will be the product of the ratio of the mass of the sun to the mass of the earth and the square of the ratio of the earth-moon radius to the sun-moon radius. Using the earth-sun radius as an average for the sun-moon radius, the ratio of the forces is

 3.84 × 10 8 m    1.50 × 1011 m    

2

 1.99 × 10 30 kg    5.97 × 10 24 kg  = 2.18.   

12.2:
Fg = G

Use of Eq. (12.1) gives

m1 m2 (5.97 × 10 24 kg)(2150 kg) = (6.673 × 10 −11 N ⋅ m 2 kg 2 ) = 1.67 × 10 4 N. 2 5 6 2 r (7.8 × 10 m + 6.38 × 10 m) The ratio of this force to the satellite’s weight at the surface of the earth is (1.67 × 10 4 N) = 0.79 = 79%. (2150 kg)(9.80 m s 2 ) (This numerical result requires keeping one extra significant figure in the intermediate calculation.) The ratio, which is independent of the satellite mass, can be obtained directly as
GmE m r 2 GmE  RE  = 2 =  , mg r g  r 
2

yielding the same result.

12.3:

G

(nm1 )(nm2 ) mm = G 1 2 2 = F12 . 2 (nr12 ) r12

12.4: The separation of the centers of the spheres is 2R, so the magnitude of the gravitational attraction is GM 2 (2 R ) 2 = GM 2 4 R 2 .

12.5: a) Denoting the earth-sun separation as R and the distance from the earth as x, the distance for which the forces balance is obtained from GM S m GM E m = , x2 ( R − x) 2 which is solved for R x= = 2.59 × 108 m. MS 1+ ME

b) The ship could not be at equilibrium for long, in that the point where the forces balance is moving in a circle, and to move in that circle requires some force.