Explanation

(c) The potential energy of an object at the surface of the earth
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U_{1} = - \frac{GMm}{R}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
The potential energy of the subject at a height $h = R$ from the surface of the earth
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U_{2} = - \frac{GMm}{R + h} = - \frac{GMm}{R + R}\ \ \ \ldots(ii)$
Hence, the gain in potential energy of the object
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}U = U_{2} - U_{1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}U = - \frac{GMm}{R + R} + \frac{GMm}{R}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}U = - \frac{GMm}{2R} + \frac{GMm}{R}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}U = \frac{1}{2}\frac{GMm}{R}$
But we know that $GM = {gR}^{2}$
$Hence,\ \ \mathrm{\Delta}U = \frac{1}{2}\frac{{gR}^{2}m}{R}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}U = \frac{1}{2}mgR$