The potential energy of an object at the surface of the earth

$U_1=-\frac{GMm}{R} \dots .\left(\mathrm{i}\right)$

The potential energy of the subject at a height $h=R$ from the surface of the earth

$U_2=-\frac{GMm}{R+h}=-\frac{GMm}{R+R} \dots \left(\mathrm{i}\mathrm{i}\right)$

Hence, the gain in potential energy of the object

$∆U=U_2-U_1$

$∆U=-\frac{GMm}{R+R}+\frac{GMm}{R}$

$∆U=-\frac{GMm}{2R}+\frac{GMm}{R}$

$∆U=\frac{1}{2}\frac{GMm}{R}$

But we know that $GM={gR}^2$

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e},\mathrm{ }\mathrm{ }∆U=\frac{1}{2}\frac{{gR}^{2}m}{R}$

$\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }∆U=\frac{1}{2}mgR$