Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is :

Options

A. `(1)/(2pi) sqrt((d^3)/(3Gm))`
B. `2pi sqrt((d^3)/(3Gm))`
C. `(1)/(2pi) sqrt((3Gm)/(d^3))`
D. `2pi sqrt((3Gm)/(d^3))`

Answer

To find the period of revolution of two stars of masses \( m \) and \( 2m \) rotating about their common center of mass at a distance \( d \), we will follow these steps:

### Step 1: Determine the position of the center of mass (CM)

The center of mass \( R \) of a system of two masses can be calculated using the formula:

\[ R = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]

Here, let’s assume the star of mass \( m \) is at position \( x_1 = 0 \) and the star of mass \( 2m \) is at position \( x_2 = d \).

Substituting the values:

\[ R = \frac{m \cdot 0 + 2m \cdot d}{m + 2m} = \frac{2md}{3m} = \frac{2d}{3} \]

### Step 2: Calculate the distances of each mass from the center of mass

The distance of the mass \( m \) from the center of mass is:

\[ r_1 = R - x_1 = \frac{2d}{3} - 0 = \frac{2d}{3} \]

The distance of the mass \( 2m \) from the center of mass is:

\[ r_2 = x_2 - R = d - \frac{2d}{3} = \frac{d}{3} \]

### Step 3: Apply the formula for the period of revolution

The gravitational force provides the necessary centripetal force for circular motion. The gravitational force \( F \) between the two masses is given by:

\[ F = \frac{G \cdot m \cdot 2m}{d^2} = \frac{2Gm^2}{d^2} \]

This force acts as the centripetal force for the mass \( m \):

\[ F = m \cdot \frac{v^2}{r_1} \]

Substituting \( r_1 \):

\[ \frac{2Gm^2}{d^2} = m \cdot \frac{v^2}{\frac{2d}{3}} \]

This simplifies to:

\[ \frac{2Gm^2}{d^2} = \frac{3mv^2}{2d} \]

### Step 4: Solve for \( v^2 \)

Rearranging gives:

\[ v^2 = \frac{4Gm}{3d} \]

### Step 5: Relate linear velocity to angular velocity

The linear velocity \( v \) is related to the angular velocity \( \omega \) by:

\[ v = r_1 \cdot \omega \]

Substituting \( r_1 \):

\[ v = \frac{2d}{3} \cdot \omega \]

Squaring both sides:

\[ v^2 = \left(\frac{2d}{3}\right)^2 \cdot \omega^2 = \frac{4d^2}{9} \cdot \omega^2 \]

### Step 6: Set the two expressions for \( v^2 \) equal

Equating the two expressions for \( v^2 \):

\[ \frac{4Gm}{3d} = \frac{4d^2}{9} \cdot \omega^2 \]

### Step 7: Solve for \( \omega^2 \)

Cancelling \( 4 \) from both sides and rearranging gives:

\[ \omega^2 = \frac{3Gm}{d^3} \]

### Step 8: Find the period \( T \)

The period \( T \) is related to angular velocity by:

\[ T = \frac{2\pi}{\omega} \]

Substituting for \( \omega \):

\[ T = 2\pi \sqrt{\frac{d^3}{3Gm}} \]

### Final Result

Thus, the period of revolution \( T \) is:

\[ T = 2\pi \sqrt{\frac{d^3}{3Gm}} \]