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(C) The gravitational potential energy of an object of mass $m$ at the surface of the earth is given by $U_{1} = -\frac{GMm}{R}$.At a height $h = R$ f

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$\frac{1}{2} mgR$

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(C) The gravitational potential energy of an object of mass $m$ at the surface of the earth is given by $U_{1} = -\frac{GMm}{R}$.At a height $h = R$ from the surface,the distance from the center of the earth is $r = R + h = R + R = 2R$.The potential energy at this height is $U_{2} = -\frac{GMm}{2R}$.The gain in potential energy is $\Delta U = U_{2} - U_{1} = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^{2}}$,we have $GM = gR^{2}$.Substituting this into the expression for $\Delta U$:$\Delta U = \frac{1}{2} \frac{(gR^{2})m}{R} = \frac{1}{2} mgR$.

If $g$ is the acceleration due to gravity on the earth's surface,the gain in the potential energy of an object of mass $m$ raised from the surface of the earth to a height equal to the radius $R$ of the earth is

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