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(A) The gravitational potential energy $U$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.At the surface of the Earth

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$\frac{1}{2}mgR$

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

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 earth to a height equal to the radius of the earth $R$,is

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