A pair of stars rotates about their centre of mass One of the stars has a mass `M` and the other has mass m such that `M =2m` The distance between the centres of the stars is d (d being large compared to the size of either star) .
The ratio of the angular momentum of the two stars about their common centre of mass `(L_(m)//L_(m))` is .

Options

A. 1
B. 2
C. 4
D. 9

Answer

To solve the problem, we need to find the ratio of the angular momentum of two stars rotating about their common center of mass. Let's denote the mass of the larger star as \( M \) and the smaller star as \( m \), with the relationship \( M = 2m \). The distance between the centers of the stars is \( d \).

### Step-by-Step Solution:

1. **Determine the Center of Mass**:
The position of the center of mass \( R_{cm} \) of the two stars can be calculated using the formula:
\[ R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2} \]
Here, \( m_1 = M = 2m \) and \( m_2 = m \). Let \( r_1 \) be the distance from the center of mass to the larger mass \( M \) and \( r_2 \) be the distance from the center of mass to the smaller mass \( m \).

We can express \( r_1 \) and \( r_2 \) in terms of \( d \):
\[ r_1 = \frac{m}{m + M} \cdot d = \frac{m}{m + 2m} \cdot d = \frac{1}{3}d \]
\[ r_2 = \frac{M}{m + M} \cdot d = \frac{2m}{m + 2m} \cdot d = \frac{2}{3}d \]

2. **Angular Momentum Calculation**:
The angular momentum \( L \) of a rotating body is given by:
\[ L = m v r \]
where \( v \) is the tangential velocity. The tangential velocity can be expressed in terms of angular velocity \( \omega \):
\[ v = \omega r \]

For the larger star \( M \):
\[ L_1 = M \cdot v_1 \cdot r_1 = M \cdot (\omega_1 r_1) \cdot r_1 = M \cdot \omega_1 \cdot r_1^2 \]
Substituting \( M = 2m \) and \( r_1 = \frac{1}{3}d \):
\[ L_1 = 2m \cdot \omega_1 \cdot \left(\frac{1}{3}d\right)^2 = 2m \cdot \omega_1 \cdot \frac{1}{9}d^2 = \frac{2m \omega_1 d^2}{9} \]

For the smaller star \( m \):
\[ L_2 = m \cdot v_2 \cdot r_2 = m \cdot (\omega_2 r_2) \cdot r_2 = m \cdot \omega_2 \cdot r_2^2 \]
Substituting \( r_2 = \frac{2}{3}d \):
\[ L_2 = m \cdot \omega_2 \cdot \left(\frac{2}{3}d\right)^2 = m \cdot \omega_2 \cdot \frac{4}{9}d^2 = \frac{4m \omega_2 d^2}{9} \]

3. **Finding the Angular Velocities**:
Since both stars are rotating about their common center of mass, we can use the relationship between their angular velocities derived from the gravitational force balance:
\[ \frac{M \cdot \omega_1^2 \cdot r_1}{m \cdot \omega_2^2 \cdot r_2} = \frac{G \cdot M \cdot m}{d^2} \]
This gives us the ratio of their angular velocities.

4. **Calculating the Ratio of Angular Momenta**:
Now we can find the ratio \( \frac{L_1}{L_2} \):
\[ \frac{L_1}{L_2} = \frac{\frac{2m \omega_1 d^2}{9}}{\frac{4m \omega_2 d^2}{9}} = \frac{2\omega_1}{4\omega_2} = \frac{\omega_1}{2\omega_2} \]

5. **Final Ratio**:
Since we know that \( \omega_1 \) and \( \omega_2 \) are related through their distances from the center of mass, we can conclude:
\[ \frac{L_1}{L_2} = \frac{1}{2} \]

Thus, the ratio of the angular momentum of the two stars about their common center of mass is \( \frac{L_1}{L_2} = \frac{1}{2} \).