1. Geometry of the configuration

The four equal masses $M$ are equally spaced on a circle of radius $R$, so they are at the vertices of a square inscribed in the circle.

• Distance between adjacent particles: $a = \sqrt{2}R$ because the side of the square inscribed in a circle of radius $R$ is $\sqrt{2}R$.

• Distance between opposite particles: $d = 2R$

Each particle moves in a circle of radius $R$, so the required centripetal force on each particle is $F_c = \frac{Mv^2}{R}.$

2. Forces on one particle

Consider one particle at a vertex of the square. It is attracted by:

• two adjacent particles,
• one opposite particle.

We resolve all gravitational forces along the radius toward the center, because tangential components cancel by symmetry.

3. Force due to each adjacent particle

Magnitude of gravitational force from one adjacent particle: $F_1 = \frac{G M^2}{(\sqrt{2}R)^2} = \frac{G M^2}{2R^2}.$

The line joining adjacent vertices makes $45^\circ$ with the radius toward the center, so radial component is $F_{1r} = F_1 \cos 45^\circ = \frac{G M^2}{2R^2}\cdot \frac{1}{\sqrt{2}} = \frac{G M^2}{2\sqrt{2}R^2}.$

Since there are two adjacent particles, total radial contribution from them is $2F_{1r} = \frac{G M^2}{\sqrt{2}R^2}.$

4. Force due to opposite particle

Magnitude of gravitational force from the opposite particle: $F_2 = \frac{G M^2}{(2R)^2} = \frac{G M^2}{4R^2}.$

This force acts exactly along the radius toward the center, so its full value is radial.

5. Net inward force

Thus total inward gravitational force on one particle is $F_{\text{net}} = \frac{G M^2}{\sqrt{2}R^2} + \frac{G M^2}{4R^2}.$

Now, $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2},$ so $F_{\text{net}} = \frac{G M^2}{R^2}\left(\frac{1}{\sqrt{2}} + \frac14\right).$

Equating this to centripetal force: $\frac{Mv^2}{R} = \frac{G M^2}{R^2}\left(\frac{1}{\sqrt{2}} + \frac14\right).$

Cancel $M$ and multiply by $R$: $v^2 = \frac{GM}{R}\left(\frac{1}{\sqrt{2}} + \frac14\right).$

Take LCM: $\frac{1}{\sqrt{2}} + \frac14 = \frac{2\sqrt{2}+1}{4}.$

Hence, $v^2 = \frac{GM}{R}\cdot \frac{2\sqrt{2}+1}{4}$ $v = \frac12\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}.$

6. Match with options

This matches Option B: $\boxed{\frac12\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}}$

7. Comparison with stored answer

Stored correct answer: B

Our derived answer is also B, so they agree.