Animated Solution for Physics - Gravitation: Four particles each of mass $M$, move along a circle of radius $R$ under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is

## The Cosmic Dance of Four Masses

Imagine four identical particles, each of mass $M$, placed perfectly at the corners of a square inscribed in a circle of radius $R$. They are moving along this circular path in a delicate cosmic dance. Why? Because their mutual gravitational attraction is pulling them together, and this inward pull is exactly what keeps them moving in a circle.

To understand the forces at play, let's focus on just one particle. Because the system is perfectly symmetric, the physics will be identical for all four.

Our chosen particle is being pulled by its three companions. Two of them are its immediate neighbors, forming the sides of the square. The distance between adjacent particles in a square inscribed in a circle of radius $R$ is $\sqrt{2}R$. The third particle is diagonally opposite, sitting straight across the diameter of the circle, at a distance of $2R$.

For any object to move in a circle, it requires a centripetal force directed towards the center. In this scenario, the net gravitational force acting on the particle provides this necessary centripetal force.

Therefore, our governing equation is:

Let's calculate the individual gravitational pulls using Newton's Law of Gravitation. The force from each of the two adjacent neighbors, let's call it $F$, is:

The force from the diagonally opposite particle, let's call it $F_1$, is:

Now, we need the net force directed towards the center. The two adjacent forces $F$ are at a $90^{\circ }$ angle to each other. Their resultant points exactly towards the center and has a magnitude of $\sqrt{F^2+F^2}=\sqrt{2}F$.

The diagonal force $F_1$ is already pointing directly towards the center. So, the total net force is simply the algebraic sum of these two central components:

Substituting the values we found:

Factoring out the common terms, we get our simplified net gravitational force:

We bring back our master equation and equate the net gravitational force to the required centripetal force:

Notice how one mass $M$ and one radius $R$ elegantly cancel out from both sides. Isolating $v^2$ and taking a common denominator of $4$ inside the bracket, we get:

Taking the square root yields our final orbital speed:

This perfectly matches option (b). The beauty of this problem lies in how perfectly symmetric geometry translates into elegant vector algebra!