The change in potential energy when a body of | vedclass

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(C) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the earth is given by $U = -\frac{G M m}{r}$.At the

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$m g R \left( \frac{n}{n+1} \right)$

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(C) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the earth is given by $U = -\frac{G M m}{r}$.At the earth's surface,the distance from the center is $r_1 = R$. Therefore,the potential energy at the surface is $U_i = -\frac{G M m}{R}$.When the body is raised to a height $h = n R$ from the surface,the distance from the center of the earth becomes $r_2 = R + n R = (n+1) R$. Therefore,the potential energy at this height is $U_f = -\frac{G M m}{(n+1) R}$.The change in potential energy is $\Delta U = U_f - U_i = -\frac{G M m}{(n+1) R} - \left( -\frac{G M m}{R} \right)$.$\Delta U = \frac{G M m}{R} - \frac{G M m}{(n+1) R} = \frac{G M m}{R} \left( 1 - \frac{1}{n+1} \right)$.$\Delta U = \frac{G M m}{R} \left( \frac{n+1-1}{n+1} \right) = \frac{G M m}{R} \left( \frac{n}{n+1} \right)$.Since the acceleration due to gravity at the surface is $g = \frac{G M}{R^2}$,we can write $G M = g R^2$.Substituting this into the expression for $\Delta U$:$\Delta U = \frac{(g R^2) m}{R} \left( \frac{n}{n+1} \right) = m g R \left( \frac{n}{n+1} \right)$.

The change in potential energy when a body of mass $m$ is raised to a height $n R$ from the earth's surface is (where $R$ is the radius of the earth).

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