Question

If the acceleration due to gravity at the surface of the earth is g, the work done in slowly lifting a body of mass m from the earth's surface to a height R equal to the radius of the earth is
(a) $\frac{1}{2}mgR$

(b) $2mgR$

(c) $mgR$

(d) $\frac{1}{4}mgR$.

Answer 1

(a) $\frac{1}{2}mgR$

Work done = $-$(final potential energy $-$ initial potential energy)
$$\begin{gathered} \Rightarrow W=-\left(\frac{GMm}{2R}-\frac{GMm}{R}\right) \\ \Rightarrow W=\frac{1}{2}\frac{GMm}{R}=\frac{1}{2}mR\times \left(\frac{GM}{R^2}\right) \\ \Rightarrow W=\frac{1}{2}mRg\left[\because g=\frac{GM}{R^2}\right] \end{gathered}$$