ch13-p064

# ch13-p064 - 6.33 10 J. 2 2(7.87 10 m) E GM m E r = = = (c)...

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11 3 2 24 6 9 (6.67 10 m /kg s )(5.98 10 kg)(125 kg) 7.87 10 m 6.33 10 J. E AB GM m EE E r ×⋅ × = + =− =− × =− × (b) We note that the speed of the wreckage will be zero (immediately after the collision), so it has no kinetic energy at that moment. Replacing m with 2 m in the potential energy expression, we therefore find the total energy of the wreckage at that instant is 11 3 2 24 9 6 (2 ) (6.67 10 m /kg s )(5.98 10 kg)2(125 kg) 6.33 10 J. 22 ( 7
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Unformatted text preview: 6.33 10 J. 2 2(7.87 10 m) E GM m E r = = = (c) An object with zero speed at that distance from Earth will simply fall towards the Earth, its trajectory being toward the center of the planet. 64. (a) From Eq. 13-40, we see that the energy of each satellite is GM E m /2 r . The total energy of the two satellites is twice that result:...
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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