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

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

$U=\left(-\frac{G\times m\times m}{AB}\right)+\left(-\frac{G\times m\times m}{AC}\right)$

$+\left(-\frac{G\times m\times m}{AD}\right)+\left(-\frac{G\times m\times m}{BC}\right)$

$+\left(-\frac{G\times m\times m}{BD}\right)+\left(-\frac{G\times m\times m}{CD}\right)$

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

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

$=-\frac{4G{m}^2}{l}-\frac{2G{m}^2}{l\sqrt{2}}$

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