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) $
$=-\frac{G m^{2}}{L}\left(\frac{2 \sqrt{2}+1}{\sqrt{2}}\right)$