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

Options

A. `(a)sqrt((GM)/(R )sqrt((2)/(3)))`
B. `(b)sqrt((GM)/(sqrt(3)R ))`
C. `(c)sqrt((GM)/(3R))`
D. `(d)sqrt((2)/(3) (GM)/(R))`

Explanation

To solve the problem of three identical particles each of mass \( M \) moving along a common circular path of radius \( R \) under mutual interaction, we can follow these steps:

### Step 1: Understand the Configuration
The three particles are positioned at the vertices of an equilateral triangle inscribed in a circle of radius \( R \). The center of the triangle is the center of the circular path.

### Step 2: Determine the Distance Between Particles
The distance between any two particles can be calculated using the properties of an equilateral triangle. The length of each side \( a \) of the triangle can be expressed in terms of the radius \( R \):
\[ a = R \sqrt{3} \]
This is derived from the relationship between the radius of the circumcircle and the side length of the equilateral triangle.

### Step 3: Calculate the Gravitational Force Between Particles
The gravitational force \( F \) between any two particles (say particles B and C) is given by Newton's law of gravitation:
\[ F = \frac{G M^2}{a^2} = \frac{G M^2}{(R \sqrt{3})^2} = \frac{G M^2}{3R^2} \]

### Step 4: Determine the Net Force Acting on Each Particle
Each particle experiences gravitational attraction from the other two particles. The resultant force \( F_{\text{net}} \) acting on any particle can be calculated by considering the vector sum of the forces from the other two particles. Since these forces are equal and form a \( 60^\circ \) angle with each other, we can use the formula for the resultant of two forces:
\[ F_{\text{net}} = 2F \cos(30^\circ) = 2 \left(\frac{G M^2}{3R^2}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{G M^2 \sqrt{3}}{3R^2} \]

### Step 5: Relate the Net Force to Centripetal Force
For circular motion, the centripetal force required to keep each particle in circular motion is given by:
\[ F_{\text{centripetal}} = \frac{M v^2}{R} \]
Setting the net gravitational force equal to the centripetal force gives:
\[ \frac{M v^2}{R} = \frac{G M^2 \sqrt{3}}{3R^2} \]

### Step 6: Solve for Velocity \( v \)
Now, we can solve for the velocity \( v \):
\[ M v^2 = \frac{G M^2 \sqrt{3}}{3R} \]
Dividing both sides by \( M \):
\[ v^2 = \frac{G M \sqrt{3}}{3R} \]
Taking the square root:
\[ v = \sqrt{\frac{G M \sqrt{3}}{3R}} \]

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