Ch 6Word - Chapter 6 CHAPTER 6 Gravitation and Newton's...

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Chapter  6 CHAPTER 6 - Gravitation and Newton’s Synthesis 1. Because the spacecraft is 2 Earth radii above the surface, it is 3 Earth radii from the center.  The  gravitational force on the spacecraft is F   GM E M / r 2 = (6.67 × 10 –11  N     m 2 /kg 2 )(5.98 × 10 24  kg)(1400 kg)/[3(6.38 × 10 6  m)] 2  =       1.52 × 10 3  N . 2. The acceleration due to gravity on the surface of a planet is F / M  =  GM planet / r 2 . For the Moon we have g Moon  = (6.67 × 10 –11  N     m 2 /kg 2 )(7.35 × 10 22  kg)/(1.74 × 10 6  m) 2  =       1.62 m/s 2 . 3. The acceleration due to gravity on the surface of a planet is F / M  =  GM planet / r 2 . If we form the ratio of the two accelerations, we have g planet / g Earth  = ( M planet / M Earth )/( r planet / r Earth ) 2 ,   or g planet  =  g Earth ( M planet / M Earth )/( r planet / r Earth ) 2  = (9.80 m/s 2 )(1)/(2.5) 2  =        1.6 m/s 2 . 4. The acceleration due to gravity on the surface of a planet is F / M  =  GM planet / r 2 . If we form the ratio of the two accelerations, we have g planet / g Earth  = ( M planet / M Earth )/( r planet / r Earth ) 2 ,   or g planet  =  g Earth ( M planet / M Earth )/( r planet / r Earth ) 2  = (9.80 m/s 2 )(3.0)/(1) 2  =        29 m/s 2 . 5. The acceleration due to gravity at a distance  from the center of the Earth is F / M  =  Gm Earth / r 2 . If we form the ratio of the two accelerations for the different distances, we have g h / g surface  = [( r Earth )/( r Earth  +  h )] 2  = [(6400 km)/(6400 km + 300 km)] 2   which gives        g h  = 0.91 g surface . 6. The acceleration due to gravity at a distance  from the center of the Earth is F / M  =  Gm Earth / r 2 . If we form the ratio of the two accelerations for the different distances, we have g / g surface  = [( r Earth )/( r Earth  +  h )] 2  ; ( a ) g  = (9.80 m/s 2 )[(6400 km)/(6400 km + 3.20 km)] 2  =         9.80 m/s 2 . ( b ) g  = (9.80 m/s 2 )[(6400 km)/(6400 km + 3200 km)] 2  =         4.36 m/s 2 . 7. We choose the coordinate system  shown  in the figure and   find the force on the mass in the lower left corner.   Because the masses are equal, for the magnitudes  of the  forces from the other corners we have F 1  =  F 3 Gmm / r 1 2 = (6.67 × 10 –11  N     m 2 /kg 2 )(8.5 kg)(8.5 kg)/(0.70 m) 2   = 9.83 × 10 –9  N; F 2   =  Gmm / r 2 2 = (6.67 × 10 –11  N     m 2 /kg 2 )(8.5 kg)(8.5 kg)/[(0.70 m)/cos  45 ° ] 2   = 4.92 × 10 –9  N. From the symmetry  of the forces we see that the resultant  will be  along the diagonal.  The resultant  force is F   = 2 F 1  cos 45 °  +  F 2 = 2(9.83 × 10 –9  N) cos 45 °  + 4.92 × 10 –9  N =         1.9 × 10 –8  N toward  center of the square .
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