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 in a circular path under mutual gravitational attraction, we can follow these steps:

### Step 1: Understand the Configuration
We have three particles, each of mass \( M \), positioned symmetrically on a circular path of radius \( r \). The particles are located at equal angles of \( 120^\circ \) from each other.

### Step 2: Determine the Forces Acting on One Particle
Consider one particle (let's call it Particle A). It experiences gravitational attraction from the other two particles (Particles B and C). The gravitational force between any two particles is given by:

\[ F = \frac{G M^2}{d^2} \]

where \( G \) is the gravitational constant and \( d \) is the distance between the two particles.

### Step 3: Calculate the Distance Between Particles
The distance \( d \) between any two particles (for example, A and B) can be found using the law of cosines in the triangle formed by the three particles. The angle between the lines connecting the center of the circle to the particles is \( 120^\circ \). Thus, we can use:

\[ d = \sqrt{r^2 + r^2 - 2r^2 \cos(120^\circ)} = \sqrt{2r^2(1 + \frac{1}{2})} = \sqrt{3r^2} = r\sqrt{3} \]

### Step 4: Calculate the Gravitational Force
Now we can calculate the gravitational force \( F \) between Particle A and Particle B:

\[ F = \frac{G M^2}{(r\sqrt{3})^2} = \frac{G M^2}{3r^2} \]

### Step 5: Find the Net Force Acting on Particle A
Since Particle A is attracted by both Particle B and Particle C, the net force \( F_{\text{net}} \) acting on Particle A towards the center of the circle can be calculated as follows:

The components of the forces due to B and C that act towards the center can be resolved. The vertical components of the forces will cancel out, and the horizontal components will add up. Since the angle between the forces is \( 120^\circ \), the net force is:

\[ F_{\text{net}} = 2F \sin(30^\circ) = 2 \left(\frac{G M^2}{3r^2}\right) \cdot \frac{1}{2} = \frac{G M^2}{3r^2} \]

### Step 6: Relate the Net Force to Centripetal Force
The net force acting on Particle A must provide the necessary centripetal force for it to move in a circle:

\[ F_{\text{net}} = \frac{M v^2}{r} \]

Setting the two expressions for the net force equal gives:

\[ \frac{G M^2}{3r^2} = \frac{M v^2}{r} \]

### Step 7: Solve for the Speed \( v \)
Rearranging the equation to solve for \( v^2 \):

\[ v^2 = \frac{G M}{3r} \]

Taking the square root gives:

\[ v = \sqrt{\frac{G M}{3r}} \]

### Final Answer
The speed of each particle is:

\[ v = \sqrt{\frac{G M}{3r}} \]

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