Model: 13.6. Model the earth (e) as a sphere.
Visualize:
The space shuttle or a 1.0 kg sphere (s) in the space shuttle is Re ! rs " 6.37 # 106 m ! 0.30 # 106 m " 6.67 # 106 m away from the center of the earth. GM e M s (6.67 # 10$11 N % m 2 /kg 2 )(5.98 # 1024 kg)(1.0 kg) " " 9.0 N Solve: (a) Fe on s " ( Re ! rs ) 2 (6.67 # 106 m)2 (b) Because the sphere and the shuttle are in free fall with the same acceleration around the earth, there cannot be any relative motion between them. That is why the sphere floats around inside the space shuttle.
13.10. Model: Model the earth (e) as a spherical mass.
Solve: Let the acceleration due to gravity be 3gsurface when the earth is shrunk to a radius of x. Then,
gsurface !
"3
GM e Re2
and
3 gsurface !
GM e x2
GM e GM e R ! 2 " x ! e ! 0.577 Re Re2 x 3
The earths radius would need to be 0.577 times its present value.
13.14. Model: Model the earth (e) as a spherical mass. Compared to the earths size and mass, the rocket (r)
is modeled as a particle. This is an isolated system, so mechanical energy is conserved. Visualize:
Solve:
The energy conservation equation K 2 ! U 2 " K1 ! U1 is
1 GM e mr 1 GM e mr 2 mr v2 # " mr v12 # 2 r2 2 Re In the present case, r2 $ %, so
1 1 GM e mr 2GM e 2 2 mr v2 " mr v12 # & v2 " v12 # Re Re 2 2 & v2 " (1.50 ' 104 m/s) 2 # " 9.99 ' 103 m/s 2(6.67 ' 10#11 N m 2 /kg 2 )(5.98 ' 1024 kg) 6.37 ' 106 m
13.20. Model: Model the planet and satellites as spherical masses.
Visualize: Please refer to Figure EX13.20. Solve: (a) The period of a satellite in a circular orbit is T " [(4! 2 /GM ) r 3 ]1/ 2 . This is independent of the satellites mass, so we can find the ratio of the periods of two satellites a and b:
#r $ Ta " % a& Tb ' rb (
3
Satellite 2 has r2 " r1 , so T2 " T1 " 250 min. Satellite 3 has r3 " (3/2)r1 , so T3 " (3/2)3/ 2 T1 " 459 min. (b) The force on a satellite is F " GMm/r 2 . Thus the ratio of the forces on two satellites a and b is
2
Fa # rb $ # ma $ "% & % & Fb ' ra ( ' mb ( Satellite 2 has r2 " r1 and m2 " 2m1 , so F2 " (1) 2(2) F1 " 20,000 N. Similarly, satellite 3 has r3 " (3/2) r1 and m3 " m1 , so F3 " (2/3) 2 (1) F1 " 4440 N.
(c) The speed of a satellite in a circular orbit is v " (GM/r ) 2 , so its kinetic energy is K " mv 1 2 " GMm/2r. Thus 2 the ratio of the kinetic energy of two satellites a and b is
K a # rb $# ma $ " % &% & K b ' ra (' mb (
Satellite 3 has r3 " (3/2)r1 and m3 " m1 , so K1/K 3 " (3/2)(1/1) " 3/2 " 1.50.
13.24. Model: Model Mars (m) as a spherical mass and the satellite (s) as a point particle.
Visualize: The geosynchronous satellite whose mass is ms and velocity is vs orbits in a circle of radius rs around Mars. Let us denote mass of Mars by M m . Solve: The gravitational force between the satellite and Mars provides the centripetal acceleration needed for circular motion:
" GM mTs2 # GM m ms ms vs2 ms (2! rs ) 2 $ $ % rs $ & ' 2 rs2 rs rs (Ts ) 2 ( 4! ) Using G $ 6.67 + 10*11 N , m 2 /kg 2 ,
1/ 3
M m $ 6.42 + 1023 kg, and Ts $ (24.8 hrs) $ (24.8)(3600) s $ 89,280 s, we
obtain rs $ 2.05 + 107 m. Thus, altitude $ rs * Rm $ 1.72 + 107 m.
13.28. Visualize: We placed the origin of the coordinate system on the 20.0 kg mass (m1 ) so that the 5.0 kg
mass (m3 ) is on the x-axis and the 10.0 kg mass (m2 ) is on the y-axis.
Solve:
(a) The forces acting on the 20 kg mass (m1 ) are
! Gm1m2 (6.67 " 10!11 N # m 2 /kg 2 )(20.0 kg)(10.0 kg) Fm2 on m1 $ j$ j $ 3.335 " 10!7 N j 2 r12 (0.20 m) 2 ! Gm1m3 (6.67 " 10!7 N # m 2 /kg 2 )(20.0 kg)(5.0 kg) Fm3 on m1 $ i$ i $ 6.67 " 10!7 i N 2 r13 (0.10 m)2 ! j Fon m $ 6.67 " 10!7 i N % 3.335 " 10!7 N ' Fon m $ 7...