problem18_74

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

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18.74: (See Example 12.5 for calculation of the escape speeds) a) Jupiter: . (0.0221) s m 10 31 . 1 mol) kg 10 02 . 2 ( K) K)(140 mol J 3(8.3145 s e 3 3 v v = × = × = - Earth: . (0.146) s m 10 65 . 1 mol) kg 10 02 . 2 (( K) K)(220 mol J 3(8.3145 s e 3 3 v v = × = × = - b) Escape from Jupiter is not likely for any molecule, while escape from earth is possible for some and hence possible for all.
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Unformatted text preview: c) s. m 395 mol) kg 10 . 32 ( K) K)(200 mol J 3(8.3145 s 3 = × ⋅ =-v The radius of the asteroid is m, 10 68 . 4 ) 4 3 ( 5 3 1 × = = πρ M R and the escape speed is s, m 542 2 = R GM so there can be no such atmosphere....
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