Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is :

Options

A. `sqrt((G)/(2R^(3)))`
B. `1/2sqrt((G)/(R^(3)))`
C. `1/(2R)sqrt(1/G)`
D. `sqrt((2G)/(R^(3)))`

Answer

To solve the problem of two identical particles of mass 1 kg each rotating around a circle of radius R under their mutual gravitational attraction, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two identical particles, each with a mass \( m = 1 \, \text{kg} \).
- They are rotating around a common center with a distance \( r \) between them. Since they are identical and rotating around each other, the distance between them is \( 2R \) (where \( R \) is the radius of the circular path of each particle).

2. **Gravitational Force Calculation**:
- The gravitational force \( F \) between the two particles is given by Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{d^2} \]
- Here, \( d = 2R \) (the distance between the two particles), and \( m_1 = m_2 = 1 \, \text{kg} \).
- Thus, the gravitational force becomes:
\[ F = \frac{G \cdot 1 \cdot 1}{(2R)^2} = \frac{G}{4R^2} \]

3. **Centripetal Force Requirement**:
- The gravitational force acts as the centripetal force required for circular motion. The centripetal force \( F_c \) for one particle is given by:
\[ F_c = \frac{m v^2}{R} \]
- Since \( m = 1 \, \text{kg} \), we have:
\[ F_c = \frac{v^2}{R} \]

4. **Setting Forces Equal**:
- Setting the gravitational force equal to the centripetal force:
\[ \frac{G}{4R^2} = \frac{v^2}{R} \]

5. **Solving for Velocity \( v \)**:
- Rearranging the equation gives:
\[ v^2 = \frac{G}{4R} \]
- Taking the square root:
\[ v = \sqrt{\frac{G}{4R}} = \frac{1}{2} \sqrt{\frac{G}{R}} \]

6. **Finding Angular Speed \( \omega \)**:
- The relationship between tangential speed \( v \) and angular speed \( \omega \) is given by:
\[ v = R \omega \]
- Therefore, we can express \( \omega \) as:
\[ \omega = \frac{v}{R} \]
- Substituting the expression for \( v \):
\[ \omega = \frac{1}{2} \sqrt{\frac{G}{R}} \cdot \frac{1}{R} = \frac{1}{2} \sqrt{\frac{G}{R^3}} \]

7. **Final Result**:
- Thus, the angular speed \( \omega \) of each particle is:
\[ \omega = \frac{1}{2} \sqrt{\frac{G}{R^3}} \]