Chapter 13 - 13.1. Model: Model the sun (s) and the earth...

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13.1. Model: Model the sun (s) and the earth (e) as spherical masses. Due to the large difference between your size and mass and that of either the sun or the earth, a human body can be treated as a particle. Solve: sy s on you 2 se GM M F r ! " and ey e on you 2 e GM M F r " Dividing these two equations gives 2 2 30 6 s on y 4 24 11 e on y e s e 1.99 10 kg 6.37 10 m 6.00 10 5.98 10 kg 1.50 10 m F Mr F M r ! ! # $# $ # $ %% " " " % & ’& & ( )( ) ( )
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13.2. Model: Assume the two lead balls are spherical masses. Solve: (a) 11 2 2 9 12 1on 2 2 on 1 22 (6.67 10 N m /kg )(10 kg)(0.100 kg) 6.7 10 N (0.10 m) Gm m FF r ! ! "# $ $ $ $ " (b) The ratio of the above gravitational force to the gravitational force on the 100 g ball is 9 9 2 6.67 10 N 6.81 10 (0.100 kg)(9.8 m/s ) ! ! " $" Assess: The answer in part (b) shows the smallness of the gravitational force between two lead balls separated by 10 cm compared to the gravitational force on the 100 g ball.
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13.3. Model: Model the sun (s), the moon (m), and the earth (e) as spherical masses. Solve: sm s on m 2 GM M F r ! " and em e on m 2 GM M F r ! " Dividing the two equations and using the astronomical data from Table 13.2, 2 2 30 8 s on m s e m 24 11 e on m e s m 1.99 10 kg 3.84 10 m 2.18 5.98 10 kg 1.50 10 m F M r F M r ! ! # $# $ # $ %% " " " & ’& & ( )( ) ( ) Note that the sun-moon distance is not noticeably different from the tabulated sun-earth distance.
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13.4. Solve: sp sphere on particle 2 GM M F r ! " and ep earth on particle 2 e GM M F r " Dividing the two equations, 2 2 6 sphere on particle 7 se 24 earth on particle e s p 5900 kg 6.37 10 m 1.60 10 5.98 10 kg 0.50 m F Mr F M r ! ! #$ % " " " % &’ % ()
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13.5. Model: Model the woman (w) and the man (m) as spherical masses or particles. Solve: 11 2 2 7 wm w on m m on w 22 mw (6.67 10 N m /kg )(50 kg)(70 kg) 2.3 10 N (1.0 m) GM M FF r ! ! ! "# $ $ $ $ "
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13.6. Model: Model the earth (e) as a sphere. Visualize: The space shuttle or a 1.0 kg sphere (s) in the space shuttle is 66 es 6.37 10 m 0.30 10 m Rr ! " # ! # " 6 6.67 10 m # away from the center of the earth. Solve: (a) 11 2 2 24 e on s 2 6 2 (6.67 10 N m /kg )(5.98 10 kg)(1.0 kg) 9.0 N ( ) (6.67 10 m) GM M F $ # % # " " " !# (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.
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13.7. Model: Model the sun (s) as a spherical mass. Solve: (a) 11 2 2 30 2 s sun surface 2 8 2 s (6.67 10 N m /kg )(1.99 10 kg) 274 m/s (6.96 10 m) GM g R ! " # " $ $ $ " (b) 11 2 2 30 32 s sun at earth 2 11 2 se (6.67 10 5.90 10 m/s (1.50 10 m) GM g r ! ! ! " # " $ $ $ " "
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13.8. Model: Model the moon (m) and Jupiter (J) as spherical masses. Solve: (a) 11 2 2 22 2 m moon surface 2 6 2 m (6.67 10 N m /kg )(7.36 10 kg) 1.62 m/s (1.74 10 m) GM g R ! " # " $ $ $ " (b) 11 2 2 27 2 J Jupiter surface 2 7 2 J (6.67 10 N m /kg )(1.90 10 kg) 25.9 m/s (6.99 10 m) GM g R ! " # " $ $ $ "
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13.9. Model: Model the earth (e) as a spherical mass. Visualize: The acceleration due to gravity at sea level is 2 9.83 m/s (see Table 13.1) and 6 e 6.37 10 m R !" (see Table 13.2). Solve: 2 e e earth observatory 22 2 e 2 e ee (9.83 0.0075) m/s () 11 GM GM g g Rh hh R RR ! ! ! ! # $ % & % & $$ ( ( ) * ) * Here 2 earth e e g GM R ! is the acceleration due to gravity on a non-rotating earth, which is why we’ve used the value 2 9.83 m/s . Solving for h , e 9.83 1 2.43 km 9.8225 hR %& ! # !
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This note was uploaded on 02/19/2010 for the course MTHSC 104 taught by Professor Simms during the Spring '10 term at Clemson.

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Chapter 13 - 13.1. Model: Model the sun (s) and the earth...

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