ch13-p099 - GMmx ( x 2 + R 2 ) 3/2 . (b) Using our...

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99. (a) All points on the ring are the same distance ( r = x 2 + R 2 ) from the particle, so the gravitational potential energy is simply U = GMm/ x 2 + R 2 , from Eq. 13-21. The corresponding force (by symmetry) is expected to be along the x axis, so we take a (negative) derivative of U (with respect to x ) to obtain it (see Eq. 8-20). The result for the magnitude of the force is
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Unformatted text preview: GMmx ( x 2 + R 2 ) 3/2 . (b) Using our expression for U , then the magnitude of the loss in potential energy as the particle falls to the center is GMm ( 1 / R 1 / x 2 + R 2 ). This must turn into kinetic energy ( 1 2 mv 2 ), so we solve for the speed and obtain v = [2 GM ( R 1 ( R 2 + x 2 ) 1/2 )] 1/2 ....
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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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