5.2 Gravitational potential energy of a body on earth's surface

Proof: Consider a body of mass $m$ on the surface of the earth.

Gravitational potential is given by, $${U_1} = - \frac{{GMm}}{R}$$

Now, the gravitational energy of a body of mass $m$ at height above the earth's surface is given by,

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

$$\Delta U = {U_2} - {U_1}$$$$\Delta U = - \frac{{GMm}}{{\left( {R + h} \right)}} - \left( { - \frac{{GMm}}{R}} \right)$$$$\Delta U = + GMm\left[ { - \frac{1}{{R + h}} + \frac{1}{R}} \right]$$$$\Delta U = GMm\left[ {\frac{{ - R + R + h}}{{R(R + h)}}} \right]$$$$\Delta U = \frac{{GMm}}{R}\left[ {\frac{h}{{R + h}}} \right]$$ or $$\Delta U = \frac{{GMm}}{{{R^2}}}\left[ {\frac{h}{{1 + \frac{h}{R}}}} \right]$$ As $\left( {h < < R} \right)$. So, $$\Delta U \approx \frac{{GMm}}{{{R^2}}}h$$ Also, $$\frac{{GMm}}{{{R^2}}} = g$$ So, $$\Delta U \approx mgh$$