Question

Four particles $\text{A, B, C}$ and $\text{D}$ each of mass $m$ are kept at the corners of a square of side $L$. Now the particle $\text{D}$ 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 $\text{D}$ during its movement is

• A. $\frac{G{m}^2}{L}$
• B. $\frac{2G{m}^2}{L}$
• 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

The correct option is D $-\frac{G{m}^2}{L}\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)$

Gravitational force is the internal conservative force for this system.

Work done by the gravitational force acting on the particle $\text{D}$ during its movement is equal to the negative change in gravitational potential energy of the system.

$W=-(U_{final}-U_{initial})...\left(1\right)$

Where, $U_{final}$ is the potential energy of the particle at infinity which is zero i.e., $U_{final}=0...\left(2\right)$

And $U_{initial}$ is the initial potential energy of the system.


$U_{initial}=U_{DA}+U_{DC}+U_{DB}$

Where, $U_{DA}$, $U_{DC}$ and $U_{DB}$ are the potential energies at point $\text{D}$ due to the masses at $\text{A, B}$ and $\text{C}$ respectively.

Substituting the values from figure,

$\Rightarrow U_{initial}=-\frac{G{m}^2}{L}-\frac{G{m}^2}{L}-\frac{G{m}^2}{\sqrt{2}L}$

$\Rightarrow U_{initial}=-\frac{G{m}^2}{L}(2+\frac{1}{\sqrt{2}})$

$\Rightarrow U_{initial}=-\frac{G{m}^2}{L}\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)...\left(3\right)$

From equation $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$, we have
$W=-\frac{G{m}^2}{L}\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)-0$

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

Hence, option (d) is the correct answer.

The work done by internal conservative force is equal to the negative of change in potential energy. $$W=-\Delta U$$ Why this question? The potential energy of the system will consist of $6$ combinations $\text{AB, BC, CD, AD, AC}$ and $\text{BD}$. In the question we have considered only $3$ combinations $\text{AD, BD}$ and $\text{CD}$. This is because only $\text{D}$ is being moved and hence the potential energy of the combinations $\text{AB, BC}$ and $\text{AC}$ doesn’t change and doesn’t contribute to the change in potential energy of the system.