Three identical particles each of mass `M` move along a common circlar path of radius `R` under the mutual interaction of each other. The velocity of each particle is

Options

A. `sqrt((GM)/(R )sqrt((2)/(3)))`
B. `sqrt((GM)/(sqrt(3R )))`
C. `sqrt((GM)/(3R))`
D. `sqrt((2)/(3) (GM)/(R))`

Explanation

To find the velocity of each of the three identical particles of mass \( M \) moving in a circular path of radius \( R \) under mutual gravitational interaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
We have three identical particles, each of mass \( M \), moving in a circular path of radius \( R \). The particles exert gravitational forces on each other, and we need to find the velocity of each particle.

2. **Configuration of the Particles**:
The three particles will form the vertices of an equilateral triangle, with the center of the circle coinciding with the centroid of the triangle. The distance from the centroid to each vertex (the radius \( R \)) can be related to the side length \( A \) of the triangle.

3. **Finding the Side Length of the Triangle**:
The relationship between the radius \( R \) of the circumcircle and the side length \( A \) of the equilateral triangle is given by:
\[ R = \frac{A}{\sqrt{3}} \]
Thus, we can express \( A \) in terms of \( R \):
\[ A = R \sqrt{3} \]

4. **Calculating the Gravitational Force**:
The gravitational force between any two particles (say \( A \) and \( B \)) is given by:
\[ F = \frac{G M^2}{A^2} \]
Substituting \( A = R \sqrt{3} \):
\[ F = \frac{G M^2}{3R^2} \]

5. **Net Force Acting on a Particle**:
Each particle experiences forces due to the other two particles. The net force acting towards the center of the circle can be calculated using vector addition. The angle between the forces due to two particles is \( 60^\circ \). The net force \( F_{\text{net}} \) can be calculated as:
\[ F_{\text{net}} = \sqrt{F^2 + F^2 + 2F \cdot F \cdot \cos(60^\circ)} = \sqrt{F^2 + F^2 + F^2} = \sqrt{3} F \]

6. **Centripetal Force Requirement**:
For circular motion, the net force must equal the centripetal force required to keep the particle in circular motion:
\[ \frac{M v^2}{R} = \sqrt{3} F \]
Substituting \( F \):
\[ \frac{M v^2}{R} = \sqrt{3} \cdot \frac{G M^2}{3R^2} \]

7. **Solving for Velocity \( v \)**:
Rearranging the equation gives:
\[ v^2 = \frac{\sqrt{3} G M}{3R} \cdot R = \frac{G M}{\sqrt{3} R} \]
Taking the square root:
\[ v = \sqrt{\frac{G M}{\sqrt{3} R}} \]

### Final Answer:
The velocity of each particle is:
\[ v = \sqrt{\frac{G M}{\sqrt{3} R}} \]