Assuming that the gravitational potential energy of an object at...

The gravitational potential energy of an object placed at earth's surface is $$\mathrm{U}_1=-\frac{\mathrm{GMm}}{\mathrm{R}}$$ where, $G=$ gravitational constant, $M=$ Mass of earth, $m=$ mass of object and $\mathrm{R}=$ radius of the earth The negative sign in the above relation indicates that it is the work done in bringing the object from infinity to a distance $R$. The gravitational potential energy of object at a height habove the surface of earth is $$\mathrm{U}_2=-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$$ So, the change in potential energy is \begin{aligned} \Delta U & =\mathrm{U}_2-\mathrm{U}_1=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right) \\ & =-\mathrm{GMm}\left[\frac{1}{(\mathrm{R}+\mathrm{h})}-\frac{1}{\mathrm{R}}\right]=-\mathrm{GMm}\left(-\frac{\mathrm{h}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right) \\ & =\frac{\mathrm{GMmh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})} \end{aligned}