Question

Four particles each of mass $1kg$ are at the four corners of a square of side $1m$. The work done to move one of the particles to infinity.

Answer 1

Given,

$m=1kg$ mass of each particle

$a=1m$ side of square.

$d=\sqrt{2}a=\sqrt{2}m$ diagonal of square.

From the figure,

The initial potential energy of the system,

$U_i=\left(\frac{-4G{m}^2}{a}\right)+\left(\frac{-2G{m}^2}{\sqrt{2}a}\right)$

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

When one mass is removed,

The final potential energy of the system,

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

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

The work done, $W=U_f-U_i$

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

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

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

$W=\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)G$