problem12_56

University Physics with Modern Physics with Mastering Physics (11th Edition)

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12.56: a) Following the hint, use as the escape velocity , 2 gh v = where h is the height one can jump from the surface of the earth. Equating this to the expression for the escape speed found in Problem 12.55, , 4 3 or , 3 8 2 2 2 G gh R GR π gh ρ π = = where 2 s m 80 . 9 = g is for the surface of the earth, not the asteroid. Using m 1 = h (variable for different people, of course), km. 7 . 3 = R As an alternative, if one’s
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Unformatted text preview: jump speed is known, the analysis of Problem 12.55 shows that for the same density, the escape speed is proportional to the radius, and one’s jump speed as a fraction of s m 60 gives the largest radius as a fraction of km. 50 b) With . m kg 10 03 . 3 , 3 3 4 3 2 × = = = GR a R v a...
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This document was uploaded on 02/05/2008.

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