Three particles of equal mass `M` each are moving on a circular path with radius `r` under their mutual gravitational attraction. The speed of each particle is

Options

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

Answer

To find the speed of each particle moving on a circular path under mutual gravitational attraction, we can follow these steps:

### Step 1: Understand the System
We have three particles of equal mass \( M \) moving in a circular path of radius \( r \). The gravitational attraction between these particles provides the necessary centripetal force for their circular motion.

### Step 2: Calculate the Gravitational Force
The gravitational force \( F \) between any two particles can be expressed using Newton's law of gravitation:
\[ F = \frac{G M^2}{d^2} \]
where \( G \) is the gravitational constant and \( d \) is the distance between the two particles. For our configuration, the distance \( d \) between any two particles is \( \sqrt{3} r \) (since they form an equilateral triangle).

### Step 3: Substitute the Distance into the Gravitational Force Equation
Substituting \( d = \sqrt{3} r \) into the gravitational force equation gives:
\[ F = \frac{G M^2}{(\sqrt{3} r)^2} = \frac{G M^2}{3 r^2} \]

### Step 4: Determine the Resultant Force
Each particle experiences gravitational attraction from the other two particles. The angle between the forces exerted by the two other particles on one particle is \( 60^\circ \). The resultant gravitational force \( F_{\text{resultant}} \) acting towards the center can be calculated using vector addition:
\[ F_{\text{resultant}} = \sqrt{F^2 + F^2 + 2F \cdot F \cdot \cos(60^\circ)} = \sqrt{2F^2 + F^2} = \sqrt{3}F \]
Thus,
\[ F_{\text{resultant}} = \sqrt{3} \cdot \frac{G M^2}{3 r^2} = \frac{\sqrt{3} G M^2}{3 r^2} \]

### Step 5: Relate the Resultant Force to Centripetal Force
The resultant gravitational force provides the necessary centripetal force \( F_c \) for circular motion:
\[ F_c = \frac{M V^2}{r} \]
Setting the resultant gravitational force equal to the centripetal force gives:
\[ \frac{M V^2}{r} = \frac{\sqrt{3} G M^2}{3 r^2} \]

### Step 6: Solve for the Speed \( V \)
We can simplify the equation:
\[ V^2 = \frac{\sqrt{3} G M}{3 r} \implies V = \sqrt{\frac{\sqrt{3} G M}{3 r}} \]

### Final Answer
Thus, the speed of each particle is:
\[ V = \sqrt{\frac{G M}{\sqrt{3} r}} \]