# Ch6Word - 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

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Chapter 6 Page 1 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 . 6 7 × 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 g = 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 g = 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 g = 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 r from the center of the Earth is g = 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 r from the center of the Earth is g = 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 . 6 7 × 10 –11 N · m 2 /kg 2 )(8.5 kg)(8.5 kg)/(0.70 m) 2 = 9 . 8 3 × 10 –9 N; F 2 = Gmm / r 2 2 = ( 6 . 6 7 × 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 . F 1 F 2 F 3 4 2 1 x y L 3 L

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Chapter 6 Page 2 8. For the magnitude of each attractive gravitational force, we have F V = Gm E m V / r V 2 = Gf V m E 2 / r V 2 = ( 6 . 6 7 × 10 –11 N · m 2 /kg 2 )(0.815)(5.98 × 10 24 kg) 2 /[(108 – 150) × 10 9 m] 2 = 1.10 × 10 18 N; F J = Gm E m J / r J 2 = Gf J m E 2 / r J 2 = ( 6 . 6 7 × 10 –11 N · m 2 /kg 2 )(318)(5.98 × 10 24 kg) 2 /[(778 – 150) × 10 9 m] 2 = 1.92 × 10 18 N; F Sa = Gm E m Sa / r Sa 2 = Gf Sa m E 2 / r Sa 2 = ( 6 . 6 7 × 10 –11 N · m 2 /kg 2 )(95.1)(5.98 × 10 24 kg) 2 /[(1430 – 150) × 10 9 m] 2 = 1.38 × 10 17 N. The force from Venus is toward the Sun; the forces from Jupiter and Saturn are away from the Sun. For the net force we have F net = F J + F Sa F V = 1.92 × 10 18 N + 1.38 × 10 17 N – 1.10 × 10 18 N = 9.6 × 10 17 N away from the Sun .
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## This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch6Word - 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

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