The gravitational potential energy of a system of eight identical masses, each of mass m, place at the corners of a cube of side $L$ is  $-\frac{{\mathrm{Gm}}^2}{\mathrm{L}}(12+\mathrm{N}\sqrt{2}+\frac{4}{\sqrt{3}})$. Find the value of $n$.

Consider all particles are assembling on the cube take from infinity.

$\therefore {\mathrm{V}}_1={\mathrm{V}}_2={\mathrm{V}}_3=\dots \dots \dots \dots {\mathrm{V}}_8$

using $\mathrm{U}=\frac{1}{2}\sum _{ }\mathrm{mV}$

$\mathrm{U}=\left(\frac{1}{2}{\mathrm{mV}}_1\right)8\Rightarrow \mathrm{U}=4{\mathrm{mV}}_1$

${\mathrm{V}}_1=$potential at point 1,

$\therefore {\mathrm{V}}_1=\frac{-3\mathrm{Gm}}{\mathrm{L}}-\frac{3\mathrm{Gm}}{\sqrt{2}\mathrm{L}}-\frac{\mathrm{Gm}}{\sqrt{3}\mathrm{L}}$

${\mathrm{V}}_1=\frac{-\mathrm{Gm}}{\mathrm{L}}[3+\frac{3}{\sqrt{2}}+\frac{1}{\sqrt{3}}]$

$\therefore \mathrm{U}=\frac{-4{\mathrm{Gm}}^2}{\mathrm{L}}[3+\frac{3}{\sqrt{2}}+\frac{1}{\sqrt{3}}]$

$\mathrm{U}=\frac{-{\mathrm{Gm}}^2}{\mathrm{L}}[12+6\sqrt{2}+\frac{4}{\sqrt{3}}]$