Two particles of masses m and 3m are moving under their mutual gravitational force, around their centre of mass, in circular orbits. The separation between the masses is r. The gravitational attraction the two provides nessary centripetal force for circular motion
The ratio of centripetal forces acting on the two masses will be

Options

A. A. 1
B. B. 2
C. C. 3
D. D. None of these

Answer

To solve the problem, we need to find the ratio of the centripetal forces acting on two masses \( m \) and \( 3m \) that are moving under their mutual gravitational force. Let's break down the solution step by step.

### Step-by-Step Solution

1. **Identify the Masses and Separation**:
- Let the two masses be \( m_1 = m \) and \( m_2 = 3m \).
- The separation between the two masses is given as \( r \).

2. **Determine the Position of the Center of Mass**:
- The position of the center of mass \( x_{cm} \) can be calculated using the formula:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
- Assuming \( m_1 \) is at the origin \( (0) \) and \( m_2 \) is at \( (r) \):
\[ x_{cm} = \frac{m \cdot 0 + 3m \cdot r}{m + 3m} = \frac{3mr}{4m} = \frac{3r}{4} \]

3. **Calculate the Distances from the Center of Mass**:
- The distance of mass \( m \) from the center of mass:
\[ d_1 = x_{cm} - 0 = \frac{3r}{4} \]
- The distance of mass \( 3m \) from the center of mass:
\[ d_2 = r - x_{cm} = r - \frac{3r}{4} = \frac{r}{4} \]

4. **Determine the Gravitational Force**:
- The gravitational force \( F \) between the two masses is given by:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} = \frac{G \cdot m \cdot 3m}{r^2} = \frac{3Gm^2}{r^2} \]

5. **Calculate the Centripetal Forces**:
- The centripetal force \( F_1 \) acting on mass \( m \):
\[ F_1 = \frac{m v_1^2}{d_1} \]
- The centripetal force \( F_2 \) acting on mass \( 3m \):
\[ F_2 = \frac{3m v_2^2}{d_2} \]

6. **Relate the Velocities**:
- Since both masses are in circular motion around the center of mass, the velocities are related by:
\[ v_1 \cdot d_1 = v_2 \cdot d_2 \]
- This implies:
\[ v_2 = \frac{d_1}{d_2} v_1 = \frac{\frac{3r}{4}}{\frac{r}{4}} v_1 = 3v_1 \]

7. **Substitute Velocities into Centripetal Forces**:
- Substitute \( v_2 \) into \( F_2 \):
\[ F_2 = \frac{3m (3v_1)^2}{\frac{r}{4}} = \frac{3m \cdot 9v_1^2}{\frac{r}{4}} = \frac{27mv_1^2}{\frac{r}{4}} = \frac{108mv_1^2}{r} \]

8. **Ratio of Centripetal Forces**:
- Now, we can find the ratio \( \frac{F_1}{F_2} \):
\[ \frac{F_1}{F_2} = \frac{F_1}{\frac{108mv_1^2}{r}} = \frac{\frac{mv_1^2}{\frac{3r}{4}}}{\frac{108mv_1^2}{r}} = \frac{4}{108} = \frac{1}{27} \]

### Final Answer
The ratio of centripetal forces acting on the two masses is:
\[ \frac{F_1}{F_2} = \frac{1}{27} \]