Two particles of same mass m go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is ,

Options

A. `V=1/(2R)sqrt(1/(Gm))`
B. `V = sqrt((GM)/(2R))`
C. `V =1/2sqrt((Gm)/R)`
D. `V = sqrt((4Gm)/R)`

Answer

To solve the problem of two particles of mass \( m \) revolving around a circle of radius \( R \) under their mutual gravitational attraction, we can follow these steps:

### Step 1: Understand the system
We have two particles, each of mass \( m \), revolving around a common center due to their mutual gravitational attraction. The distance between the two particles is \( 2R \) since they are at opposite ends of the diameter of the circle.

### Step 2: Calculate the gravitational force
The gravitational force \( F \) between the two particles can be calculated using Newton's law of gravitation:
\[ F = \frac{G m^2}{(2R)^2} = \frac{G m^2}{4R^2} \]
where \( G \) is the gravitational constant.

### Step 3: Relate gravitational force to centripetal force
For each particle to move in a circular path, the gravitational force must provide the necessary centripetal force. The centripetal force \( F_c \) required for a mass \( m \) moving with speed \( v \) in a circle of radius \( R \) is given by:
\[ F_c = \frac{m v^2}{R} \]

### Step 4: Set the forces equal
Since the gravitational force provides the centripetal force, we can set the two forces equal to each other:
\[ \frac{G m^2}{4R^2} = \frac{m v^2}{R} \]

### Step 5: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G m}{4R^2} = \frac{v^2}{R} \]
Now, multiply both sides by \( R \):
\[ \frac{G m}{4R} = v^2 \]

### Step 6: Solve for \( v \)
To find the speed \( v \), we take the square root of both sides:
\[ v = \sqrt{\frac{G m}{4R}} = \frac{1}{2} \sqrt{\frac{G m}{R}} \]

### Conclusion
The speed of each particle is:
\[ v = \frac{1}{2} \sqrt{\frac{G m}{R}} \]