Question

Four particles $A,B,C$ and $D$ each of mass $m$ are kept at the corners of a square of side $L$. Now the particle $D$ is taken to infinity by an external agent keeping the other particles fixed at their respective positions. The work done by the gravitational force acting on the particle $D$ during its movement is

• A. $2\frac{G{m}^2}{L}$
• B. $-2\frac{G{m}^2}{L}$
• C. $\frac{G{m}^2}{L}\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)$
• D. $-\frac{G{m}^2}{L}\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)$

Answer 1

Answer: (D)

Work done by the gravitational force acting on the particle $D$ during its movement
$=-\Delta U$
$=-\left(U_{\text{final }}-U_{\text{inilial }}\right)$
$=U_{\text{intial }}-U_{\text{final }}$
Now, when the particle is at infinity, $U=0$
$\Rightarrow U_{\text{final }}=0$
$\Rightarrow$ Work done $=U_{\text{inital }}$
$U_{\text{inital }}=-\frac{G{m}^2}{L}-\frac{G{m}^2}{L}-\frac{G{m}^2}{\sqrt{2}L}$
$=-\frac{G{m}^2}{L}\left(2+\frac{1}{\sqrt{2}}\right)$
$=-\frac{G{m}^2}{L}\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)$