Explanation

(a) Gravitational potential energy of mass $m$ at any point at a distance $r$ from the centre of earth is
$U = - \frac{GMm}{r}$
At the surface of earth $r = R$
$\therefore U_{s} = - \frac{GMm}{R} = - mgR\ \ \left( \because g = \frac{GM}{R^{2}} \right)$
At the height $h = nR$ from the surface of earth
$r = R + h = R + nR = R(1 + n)$
$\therefore U_{h} = - \frac{GMm}{R(1 + n)} = - \frac{mgR}{(1 + n)}$
Change in gravitational potential energy is
$\mathrm{\Delta}U = U_{h} - U_{s} = - \frac{mgR}{(1 + n)} - ( - mgR)$
$= - \frac{mgR}{1 + n} + mgR = mgR\left( 1 - \frac{1}{1 + n} \right) = mgR\left( \frac{n}{1 + n} \right)$