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

Options

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

Explanation

To solve the problem of two particles of equal mass moving in a circle under their mutual gravitational attraction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**:
We have two particles, both of mass \( m \), moving in a circular path of radius \( R \). The distance between the two particles is \( 2R \) since they are on opposite sides of the circle.

2. **Gravitational Force Calculation**:
The gravitational force \( F \) between the two particles can be calculated using Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{d^2} \]
where \( G \) is the gravitational constant, \( m_1 = m \), \( m_2 = m \), and \( d = 2R \) (the distance between the two particles). Thus, we have:
\[ F = \frac{G m \cdot m}{(2R)^2} = \frac{G m^2}{4R^2} \]

3. **Centripetal Force Requirement**:
For each particle to move in a circle, 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:
\[ F_c = \frac{m v^2}{R} \]

4. **Equating Forces**:
Set the gravitational force equal to the centripetal force:
\[ \frac{G m^2}{4R^2} = \frac{m v^2}{R} \]

5. **Simplifying the Equation**:
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G m}{4R^2} = \frac{v^2}{R} \]
Multiply both sides by \( R \):
\[ \frac{G m}{4R} = v^2 \]

6. **Solving for Speed \( v \)**:
Taking the square root of both sides gives us the speed \( v \):
\[ v = \sqrt{\frac{G m}{4R}} = \frac{1}{2} \sqrt{\frac{G m}{R}} \]

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