Question

If $g$ is the acceleration due to gravity on the Earth' surface, find the gain in potential energy of a body of mass $m$ when taken from the surface of Earth at a height equal to the radius $R$ of the Earth.

Answer 1

Let

$R\to$ Radius of the earth

$M\to$ Mass of the earth

$m\to$ Mass of the body

$g\to$ Acceleration due to gravity on earth
when the body is on earth surface, considering the force of attraction of earth on it we can write

$mg=\frac{GMm}{R^2}$ Where, G= is gravitational constant.

$E_s=-\frac{GmM}{R}\dots \left(2\right)$

Again the PE of the system when the body is at height h from the the earth surface is given by

$E_h=-\frac{GmM}{R+h}$

If $h=R$ then

$E_R=-\frac{GmM}{R+R}=-\frac{GmM}{2R}\dots \left(3\right)$

So increase in PE due to shift of the body of mass m from surface to the height equal to the radius (R) of the earth is given by-

$\Delta E_p=E_R-E_s=-\frac{GmM}{2R}+\frac{GmM}{R}=\frac{GmM}{2R}$$=\frac{mg{R}^2}{2R}=\frac{1}{2}mgR$

Hence, increase in potential energy is $\frac{1}{2}mgR$