Question
A body of mass $$m$$ is placed on the earth’s surface. It is then taken from the earth's surface to a height $$h = 3\,R,$$ then the change in gravitational potential energy is
A.
$$\frac{{mgh}}{R}$$
B.
$$\frac{2}{3}mgR$$
C.
$$\frac{3}{4}mgR$$
D.
$$\frac{{mgR}}{2}$$
Answer :
$$\frac{3}{4}mgR$$
Solution :
Potential energy, $$U = - \frac{{GMm}}{r}$$
At the earth’s surface, $$r = R$$
$$\therefore {U_e} = - \frac{{GMm}}{R}$$
Now, if a body is taken to height $$h = 3R,$$ then the potential energy is given by
$$\eqalign{
& {U_h} = - \frac{{GMm}}{{R + h}}\,\,\left( {\because r = h + R} \right) \cr
& = - \frac{{GMm}}{{4R}} \cr} $$
Thus, change in gravitational potential energy,
$$\eqalign{
& \Delta U = {U_h} - {U_e} \cr
& = - \frac{{GMm}}{{4R}} - \left( { - \frac{{GMm}}{R}} \right) \cr
& = - \frac{{GMm}}{{4R}} + \frac{{GMm}}{R} = \frac{3}{4}\frac{{GMm}}{R} \cr
& \therefore \Delta U = \frac{3}{4}\frac{{g{R^2}m}}{R}\,\,\left( {\because GM = g{R^2}} \right) \cr
& = \frac{3}{4}mgR \cr} $$