The potential energy of a satellite having mass m and rotating at a height of $6\cdot 4\times {10}^6$from the earth surface is (given R =6.4 x 10⁶ m)

The potential energy of a satellite is given by
$\mathrm{P}.\mathrm{E}.=-\left({\mathrm{GM}}_{\mathrm{e}}\mathrm{m}/\mathrm{r}\right)$
r = distance from earth's surface
Given $\mathrm{r}=\mathrm{R}+\left(6\cdot 4\times {10}^6\mathrm{m}\right)=\mathrm{R}+\mathrm{R}=2\mathrm{R}$
Further ${\mathrm{GM}}_{\mathrm{e}}={\mathrm{gR}}^2$
$\therefore \text{ P.E. }=-\frac{{\mathrm{gR}}^2\times \mathrm{m}}{2\mathrm{R}}=-\mathrm{mgR}/2$