Ch 6Word - 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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/01/2008 for the course PHYSICS 7A taught by Professor Lanzara during the Spring '08 term at University of California, Berkeley.

Page1 / 14

Ch 6Word - 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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online