09-203-hwk-Ch12

# 09-203-hwk-Ch12 - Chapter 12 Gravity Solutions to Problems...

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Chapter 12: Gravity Solutions to Problems 4. Picture the Problem : You and the asteroid attract each other gravitationally. Strategy: Estimate that your mass 70 kg. Apply equation 12-1 to find the gravitational force between you and Ceres. Solution: Apply equation 12-1: ( ) ( ) ( ) ( ) 20 11 2 2 1 2 2 2 6 8.7 10 kg 70 kg 6.67 10 N m /kg 0.028 N 12 10 m m m F G r - × = = × = × Insight: If you stood on the surface of Ceres (radius 500 km) the force would be 16 N (3.6 lb). 14. Picture the Problem : The acceleration of gravity at an altitude h above the Earth’s surface is reduced due to the increased distance from the center of the Earth. Strategy: Set the acceleration of gravity at an altitude h equal to one-half the acceleration at h = 0 using equation 12-4, and solve for h . Solution: Set 0 h g g = and solve for h : ( ) ( ) ( ) ( ) ( ) E E 2 2 E E 2 2 E E 6 6 E 1 2 2 2 1 2 1 6.37 10 m 2.64 10 m GM GM R R h R h R h R = + + = = - = - × = × Insight: This altitude is over ten times higher than the 200-km altitude of the International Space Station. 19. Picture the Problem : Both the Earth and the Moon exert a force on other masses in their vicinity according to Newton’s Universal Law of Gravitation. Strategy: Use a ratio of gravitational forces to find the mass of the Moon in terms of the mass of the Earth. Note that if you solve Newton’s Universal Law of Gravitation (equation 12-1) for 2 m you find 2 2 1 . m F r Gm = Solution: Determine the ratio M E m m by solving equation 12-1 for 2 m : ( ) ( ) 2 2 2 1 1 E E 6 4 M M M 1 M M M E 2 2 2 E E E 1

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## This homework help was uploaded on 04/14/2008 for the course PHYSICS 203 taught by Professor Croft during the Spring '08 term at Rutgers.

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09-203-hwk-Ch12 - Chapter 12 Gravity Solutions to Problems...

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