Question

Four particles each of mass m are placed at the vertices of a square of side $l$. The potential energy of the system is

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

Answer 1

Answer: (B)

From figure
$AB = BC = CD = AD = l$
$\therefore AC = BD = \sqrt{l^2 + l^2} = l\sqrt{2}$
Total potential energy of the system of four particles each of mass $m$ placed at the vertices $A,B,C$ and $D$ of a square is
$U = \left(-\frac{ G\times m\times m}{AB}\right)+\left(- \frac{G\times m\times m}{AC}\right) $
$+ \left(-\frac{G\times m\times m}{AD}\right)+\left(-\frac{G\times m\times m}{BC}\right) $
$+ \left(-\frac{G\times m\times m}{BD}\right)+\left(-\frac{G\times m\times m}{CD}\right)$
$ = \left(- \frac{Gm^{2}}{l}\right)+\left(- \frac{Gm^{2}}{l\sqrt{2}}\right) + \left(- \frac{Gm^{2}}{l}\right)$
$ +\left(-\frac{Gm^{2}}{l}\right)+\left(- \frac{Gm^{2}}{l\sqrt{2}}\right) + \left(- \frac{Gm^{2}}{l}\right) $
$ = - \frac{4Gm^{2}}{l} - \frac{2Gm^{2}}{l\sqrt{2}}$
$ = -\frac{2Gm^{2}}{l}\left(2 + \frac{1}{\sqrt{2}}\right)$