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(A) The distance from each vertex of the square to its centre is $r = \frac{L}{\sqrt{2}}$.Gravitational potential $V$ due to a single mass $M$ at a di

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$ - \sqrt{32} \frac{GM}{L} $

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(A) The distance from each vertex of the square to its centre is $r = \frac{L}{\sqrt{2}}$.Gravitational potential $V$ due to a single mass $M$ at a distance $r$ is given by $V = -\frac{GM}{r}$.Substituting $r = \frac{L}{\sqrt{2}}$,the potential due to one particle is $V_1 = -\frac{GM}{L/\sqrt{2}} = -\frac{\sqrt{2}GM}{L}$.Since there are four such particles,the total gravitational potential at the centre is $V_{total} = 4 	imes V_1$.$V_{total} = 4 	imes \left( -\frac{\sqrt{2}GM}{L} ight) = -\frac{4\sqrt{2}GM}{L}$.Since $4\sqrt{2} = \sqrt{16 	imes 2} = \sqrt{32}$,the total potential is $V_{total} = -\sqrt{32} \frac{GM}{L}$.

Four particles,each of mass $M$,are located at the vertices of a square with side $L$. The gravitational potential due to this system at the centre of the square is:

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