A pair of stars rotates about a common centre of mass. One of the stars has a mass M and the other has mass m such that =2m. The distance between the centres of the stars is d ( d being large compare to the size of eithe star).
The period of rotation of the stars about their common centre of mass ( in terms of d,m,G) is

Options

A. `sqrt((4 pi^(2))/(Gm)d^(3))`
B. `sqrt((8pi^(2))/(Gm)d^(3))`
C. `sqrt((2pi^(2))/(3Gm)d^(3))`
D. `sqrt((4 pi^(2))/(3Gm)d^(3))`

Answer

To solve the problem of finding the period of rotation of a pair of stars about their common center of mass, we can follow these steps:

### Step 1: Understand the System
We have two stars: one with mass \( M = 2m \) and the other with mass \( m \). The distance between their centers is \( d \). We need to find the period of rotation \( T \) of these stars about their common center of mass.

### Step 2: Find the Center of Mass
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, let \( d_1 \) be the distance from the mass \( m \) to the center of mass, and \( d_2 \) be the distance from the mass \( 2m \) to the center of mass. Since the total distance \( d = d_1 + d_2 \), we can express this as:
\[ d = d_1 + d_2 \]

### Step 3: Relate Distances Using Masses
Using the center of mass equation:
\[ m \cdot d_2 = 2m \cdot d_1 \]
This simplifies to:
\[ d_2 = 2d_1 \]
Substituting \( d_2 \) into the total distance equation:
\[ d = d_1 + 2d_1 = 3d_1 \]
Thus, we find:
\[ d_1 = \frac{d}{3}, \quad d_2 = \frac{2d}{3} \]

### Step 4: Apply Newton's Law of Gravitation and Centripetal Force
For circular motion, the gravitational force provides the necessary centripetal force. The gravitational force \( F_g \) between the two stars is given by:
\[ F_g = \frac{G \cdot (2m) \cdot m}{d^2} \]
The centripetal force \( F_c \) acting on the mass \( m \) is:
\[ F_c = m \cdot \frac{v^2}{d_1} \]
Where \( v \) is the tangential velocity.

### Step 5: Relate Velocity to Angular Velocity
The tangential velocity \( v \) can be expressed in terms of angular velocity \( \omega \):
\[ v = d_1 \omega \]
Substituting this into the centripetal force equation:
\[ F_c = m \cdot \frac{(d_1 \omega)^2}{d_1} = m d_1 \omega^2 \]

### Step 6: Set Gravitational Force Equal to Centripetal Force
Setting the gravitational force equal to the centripetal force:
\[ \frac{G \cdot (2m) \cdot m}{d^2} = m \cdot \frac{d}{3} \cdot \omega^2 \]
Cancelling \( m \) from both sides:
\[ \frac{2Gm}{d^2} = \frac{d}{3} \omega^2 \]

### Step 7: Solve for Angular Velocity
Rearranging gives:
\[ \omega^2 = \frac{6Gm}{d^3} \]

### Step 8: Relate Angular Velocity to Period
The angular velocity \( \omega \) is related to the period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the equation gives:
\[ \left(\frac{2\pi}{T}\right)^2 = \frac{6Gm}{d^3} \]
This simplifies to:
\[ \frac{4\pi^2}{T^2} = \frac{6Gm}{d^3} \]

### Step 9: Solve for Period \( T \)
Rearranging to solve for \( T^2 \):
\[ T^2 = \frac{4\pi^2 d^3}{6Gm} \]
Thus, the period \( T \) is:
\[ T = \sqrt{\frac{2\pi^2 d^3}{3Gm}} \]

### Final Answer
The period of rotation of the stars about their common center of mass is:
\[ T = \sqrt{\frac{4\pi^2 d^3}{3Gm}} \]