Four particles, each of mass $$M$$ and equidistant from each other, move along a circle of radius $$R$$ under the action of their mutual gravitational attraction. The speed of each particle is:

We begin by fixing our picture. The four identical particles, each of mass $$M$$, are placed at the four vertices of a square that is inscribed in a circle of radius $$R$$. The centre of the circle is also the centre of the square. Because the particles are equally spaced (every $$90^{\circ}$$), the arrangement is

A: $$\left(R,0\right)$$, B: $$\left(0,R\right)$$, C: $$\left(-R,0\right)$$, D: $$\left(0,-R\right)$$.

The side length of the square is obtained from the well-known relation “diagonal of a square $$= \sqrt{2}\,$$ (side)”. Here the diagonal $$AC$$ equals the diameter $$2R$$, so

$$\text{side}\;s=\frac{2R}{\sqrt{2}}=\sqrt{2}\,R.$$

Thus the distance between adjacent particles is $$s=\sqrt{2}\,R$$ and the distance between opposite particles is the diameter $$2R$$.

We now calculate, one by one, the gravitational forces on particle $$A$$ due to the other three particles.

1. Magnitude of the force exerted by each adjacent particle (for example, by $$B$$ or $$D$$): we use Newton’s law of gravitation,

$$F=\frac{G\,M\,M}{r^{2}},$$

and insert $$r=s=\sqrt{2}\,R$$, giving

$$F_{\text{adj}}=\frac{G M^{2}}{\left(\sqrt{2}\,R\right)^{2}} =\frac{G M^{2}}{2R^{2}}.$$

2. Magnitude of the force exerted by the opposite particle $$C$$ (distance $$2R$$):

$$F_{\text{opp}}=\frac{G M^{2}}{(2R)^{2}} =\frac{G M^{2}}{4R^{2}}.$$

Next we resolve these vector forces into radial (towards the centre) components. Choosing the positive $$x$$-axis to point from the centre to particle $$A$$, the radial direction for $$A$$ is along the negative $$x$$-axis.

• Force on $$A$$ due to $$B$$. The displacement vector from $$A(R,0)$$ to $$B(0,R)$$ is $$(-R,\,R)$$. Its length is $$s=\sqrt{2}\,R$$, so the unit vector is $$\left(-\tfrac{1}{\sqrt{2}},\,\tfrac{1}{\sqrt{2}}\right).$$ Multiplying by the magnitude $$F_{\text{adj}}$$ we obtain

$$\vec F_{AB}=F_{\text{adj}} \left(-\frac{1}{\sqrt{2}},\,\frac{1}{\sqrt{2}}\right).$$

Its radial (negative $$x$$) component is therefore

$$F_{AB,x}=-\frac{F_{\text{adj}}}{\sqrt{2}}.$$

• Force on $$A$$ due to $$D$$ has a displacement vector $$(-R,-R)$$ and hence

$$F_{AD,x}=-\frac{F_{\text{adj}}}{\sqrt{2}}.$$

Adding the two adjacent contributions, the transverse $$y$$-components cancel and the combined radial contribution becomes

$$F_{\text{adj,\,rad}}=F_{AB,x}+F_{AD,x} =-\,\frac{F_{\text{adj}}}{\sqrt{2}} -\,\frac{F_{\text{adj}}}{\sqrt{2}} =-\,\sqrt{2}\,F_{\text{adj}}.$$

• Force on $$A$$ due to the opposite particle $$C$$ acts exactly along the diameter toward the centre, so its entire magnitude contributes radially:

$$F_{\text{opp,\,rad}}=-\,F_{\text{opp}}.$$

Hence the total radial (inward) gravitational force on particle $$A$$ is the algebraic sum of the two inward negatives, i.e. an inward magnitude

$$F_{\text{net}} =\sqrt{2}\,F_{\text{adj}}+F_{\text{opp}} =\sqrt{2}\left(\frac{G M^{2}}{2R^{2}}\right) +\left(\frac{G M^{2}}{4R^{2}}\right).$$

Simplifying term by term, first multiply out the coefficient of the adjacent term:

$$\sqrt{2}\,F_{\text{adj}} =\sqrt{2}\left(\frac{G M^{2}}{2R^{2}}\right) =\frac{\sqrt{2}\,G M^{2}}{2R^{2}} =\frac{2\sqrt{2}\,G M^{2}}{4R^{2}}.$$

Now place both terms over the common denominator $$4R^{2}$$:

$$F_{\text{net}} =\frac{2\sqrt{2}\,G M^{2}}{4R^{2}} +\frac{G M^{2}}{4R^{2}} =\frac{G M^{2}}{4R^{2}} \left(1+2\sqrt{2}\right).$$

This net gravitational attraction towards the centre must supply the necessary centripetal force for uniform circular motion. The centripetal force law is

$$\frac{M v^{2}}{R}=F_{\text{net}}.$$

Substituting the expression for $$F_{\text{net}}$$ gives

$$\frac{M v^{2}}{R} =\frac{G M^{2}}{4R^{2}}\left(1+2\sqrt{2}\right).$$

We now solve for $$v^{2}$$ by multiplying both sides by $$R/M$$:

$$v^{2} =\frac{G M}{4R}\left(1+2\sqrt{2}\right).$$

Taking the positive square root (speed is positive) we obtain the required speed of each particle:

$$v =\frac{1}{2}\sqrt{\frac{G M}{R}\left(1+2\sqrt{2}\right)}.$$

This expression exactly matches Option D.

Hence, the correct answer is Option D.