91Ó°ÊÓ

An object of mass m is placed initially at position r₁ = 3 R_E and then moved to other position r₂ = 4R_E along the radial line.

We can use the definition of work done as integral force times displacement. Then we can find work done by the applied force during the move by integrating the gravitational force along the given radial distance.

Formulae:

$\begin{array}{c}W=\stackrel{→}{F}\stackrel{→}{S}\\ \underset{1}{\overset{2}{=∫}}\stackrel{→}{dW}\\ =\underset{1}{\overset{2}{}}\stackrel{→}{F}\stackrel{→}{dS}\end{array}$

Work done by an applied force during the displacement S is given by$W=\stackrel{→}{F}\stackrel{→}{S}$

In a given situation, the force and displacement both are along the radial direction of the earth. Hence, the angle between$\stackrel{→}{F}$and$\stackrel{→}{S}$is zero. Therefore, the work done is W = FS.

When the displacement is considered through a small magnitude dS, the differential work done is dW = F ds and the total work done is

$\begin{array}{c}W=\stackrel{→}{F}\stackrel{→}{S}\\ \underset{1}{\overset{2}{=∫}}\stackrel{→}{dW}\\ =\underset{1}{\overset{2}{}}\stackrel{→}{F}\stackrel{→}{dS}\end{array}$

In the given problem, F =$\frac{−G{M}_{E}m}{r^2}$and the displacement is dr = distance between position r₁ = 3 R_E and r₂ = 4R_E.

Hence

$\begin{array}{l}W=\underset{1}{\overset{2}{}}\stackrel{→}{dW}\\ =\underset{1}{\overset{2}{}}\stackrel{→}{F}\stackrel{→}{dS}\\ =\underset{3R_E}{\overset{4R_E}{}}\frac{−G{M}_{E}m}{r^2}dr\\ =−G{M}_{E}m(\frac{1}{4R_E}−\frac{1}{3R_E})\\ =\frac{G{M}_{E}m}{12R_E}\end{array}$

The work done by the applied force is$\frac{G{M}_{E}m}{12R_E}.$