Geometry of motion

Four identical masses are at the corners of a square of side $a$. If they revolve maintaining the square configuration, each particle moves in a circle about the center of the square.

The radius of this circular path is the distance from the center of the square to a corner: $R=\frac{a}{\sqrt{2}}.$

Forces on one particle

Consider one corner mass. The other three masses exert gravitational forces on it.

We resolve the net force along the diagonal toward the center, since by symmetry the resultant must point toward the center.

(i) Force due to each adjacent corner mass

Distance to each adjacent mass $=a$.

So magnitude of each force is $F_1=\frac{GM^2}{a^2}.$

There are two such forces, perpendicular to each other. Their resultant is along the diagonal toward the center: $F_{12}=\sqrt{F_1^2+F_1^2}=\sqrt{2}F_1=\sqrt{2}\frac{GM^2}{a^2}.$

(ii) Force due to the opposite corner mass

Distance to opposite corner $=a\sqrt{2}$.

Hence, $F_2=\frac{GM^2}{(a\sqrt{2})^2}=\frac{GM^2}{2a^2}.$

This force is also along the same diagonal toward the center.

Therefore total inward force on one particle is

So, $F=\frac{GM^2}{a^2}\left(\sqrt{2}+\frac12\right).$

Use centripetal force condition

Since each mass moves in a circle of radius $R=a/\sqrt{2}$, $\frac{Mv^2}{R}=F.$

Substituting $R=a/\sqrt{2}$: $\frac{Mv^2}{a/\sqrt{2}}=\frac{GM^2}{a^2}\left(\sqrt{2}+\frac12\right).$

Cancel $M$: $v^2=\frac{a}{\sqrt{2}}\cdot \frac{GM}{a^2}\left(\sqrt{2}+\frac12\right).$

$v^2=\frac{GM}{a}\cdot \frac{1}{\sqrt{2}}\left(\sqrt{2}+\frac12\right).$

Simplify: $v^2=\frac{GM}{a}\left(1+\frac{1}{2\sqrt{2}}\right).$

Now, $\frac{1}{2\sqrt{2}}\approx 0.3536$ so $v^2\approx 1.3536\frac{GM}{a}.$

Therefore, $v\approx \sqrt{1.3536}\sqrt{\frac{GM}{a}}\approx 1.16\sqrt{\frac{GM}{a}}.$

Match with options

This corresponds to: $\boxed{1.16\sqrt{\frac{GM}{a}}}$

Hence the correct option is B.