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Q.
The change in the gravitational potential energy when a body of mass $m$ is raised to a height $nR$ above the surface of the earth is (here $R$ is the radius of the earth)
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,
$\therefore \quad U_{s}=-\frac{GMm}{R}=-mgR\quad\quad\left(\because \,\,g=\frac{GM}{R^{2}}\right)$
At the height h = nR from the surface of earth
r = R + h = R + nR = R(1 + n)
$\therefore \quad U_{h}=-\frac{GMm}{R\left(1+n\right)}=-\frac{mgR}{\left(1+n\right)}$
Change in gravitational potential energy is
$\Delta U=U_{h}-U_{s}=-\frac{mgR}{\left(1+n\right)}-\left(-mgR\right)$
$=-\frac{mgR}{1+n}+mgR$
$=mgR\left(1 -\frac{1}{1+n}\right)=mgR\left(\frac{n}{1+n}\right)$