Find the potential energy of a system of four particles,each of mass $m$,placed at the vertices of a square of side $l$. Also,obtain the gravitational potential at the centre of the square.

(N/A) Consider four masses,each of mass $m$,placed at the corners of a square of side $l$. The total gravitational potential energy $U$ is the sum of the potential energies of all distinct pairs of masses.
There are $4$ pairs of masses separated by distance $l$ (the sides of the square) and $2$ pairs of masses separated by distance $\sqrt{2}l$ (the diagonals of the square).
The potential energy of a pair of masses $m_1$ and $m_2$ at distance $r$ is $U = -\frac{G m_1 m_2}{r}$.
Thus,the total potential energy is:
$U = -4 \left( \frac{G m^2}{l} \right) - 2 \left( \frac{G m^2}{\sqrt{2} l} \right)$
$U = -\frac{G m^2}{l} \left( 4 + \frac{2}{\sqrt{2}} \right) = -\frac{G m^2}{l} (4 + \sqrt{2}) \approx -5.414 \frac{G m^2}{l}$.
The gravitational potential $V$ at the centre of the square is the sum of the potentials due to each of the four masses. The distance of each mass from the centre is $r = \frac{\sqrt{2}l}{2} = \frac{l}{\sqrt{2}}$.
$V = 4 \times \left( -\frac{G m}{r} \right) = 4 \times \left( -\frac{G m}{l/\sqrt{2}} \right) = -4\sqrt{2} \frac{G m}{l}$.