ch13-p032 - 32. (a) The potential energy at the surface is...

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32. (a) The potential energy at the surface is (according to the graph) –5.0 × 10 9 J, so (since U is inversely proportional to r – see Eq. 13-21) at an r -value a factor of 5/4 times what it was at the surface then U must be a factor of 4/5 what it was. Thus, at r = 1.25 R s U = – 4.0 × 10 9 J. Since mechanical energy is assumed to be conserved in this problem, we have K + U = –2.0 × 10 9 J at this point. Since U = – 4.0
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