Find the potential energy of a system of four particles each of mass 2m kept at the vertices of a square of side x. Also find the potential at the centr of the square.

Explanation

Distance of diagonal pair is `=sqrt(x^(2)+x^(2))=sqrt(2)x`
Now, potential energy of system = potential enegy of pairs of mass at the ends of the sides + potential energy of masses at diagonal ends
`=4[(-G(2m)^(2))/(x)]+2[(-G(2m)^(2))/(sqrt(2)x)]`
`=-16(Gm^(2))/(x)-(8Gm^(2))/(sqrt(2)x)`
`=-21.64(Gm^(2))/(x)`
Now, the gravitational potential at the centre will be V(centre) `= 4[(-G(2m))/((sqrt(2)(x)/(2)))]=8sqrt(2)(Gm)/(x)`