32. (a) The potential energy at the surface is (according to the graph) –5.0 ×109J, so (sinceUis 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 Umust be a factor of 4/5 what it was. Thus, at r= 1.25RsU= – 4.0 ×109J. Since mechanical energy is assumed to be conserved in this problem, we have K + U =–2.0 ×109J at this point. Since U= – 4.0
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