Two stars of masses `m_(1)` and `m_(2)` distance r apart, revolve about their centre of mass. The period of revolution is :

Options

A. `2pi sqrt((r^(3))/(2G (m_(1)+m_(2)))`
B. `2pi sqrt((r^(3)(m_(1)+m_(2)))/(2G m_(1)m_(2)))`
C. `2pi sqrt((2r^(3))/(G (m_(1)+m_(2)))`
D. `2pi sqrt((r^(3))/(G (m_(1)+m_(2)))`

Answer

To find the period of revolution of two stars of masses \( m_1 \) and \( m_2 \) that are \( r \) distance apart and revolve around their center of mass, we can follow these steps:

### Step 1: Determine the position of the center of mass
The center of mass \( r' \) of the two stars can be calculated using the formula:
\[ r' = \frac{m_2 r}{m_1 + m_2} \]
where \( r \) is the distance between the two stars.

### Step 2: Write the expression for gravitational force
The gravitational force \( F_g \) between the two stars is given by Newton's law of gravitation:
\[ F_g = \frac{G m_1 m_2}{r^2} \]
where \( G \) is the gravitational constant.

### Step 3: Write the expression for centripetal force
The centripetal force \( F_c \) required to keep the star of mass \( m_1 \) in circular motion around the center of mass is given by:
\[ F_c = m_1 \omega^2 r' \]
where \( \omega \) is the angular velocity.

### Step 4: Set the gravitational force equal to the centripetal force
Since the gravitational force provides the necessary centripetal force, we can set these two forces equal:
\[ \frac{G m_1 m_2}{r^2} = m_1 \omega^2 r' \]

### Step 5: Substitute \( r' \) into the equation
Substituting \( r' \) from Step 1 into the equation gives:
\[ \frac{G m_1 m_2}{r^2} = m_1 \omega^2 \left(\frac{m_2 r}{m_1 + m_2}\right) \]

### Step 6: Simplify the equation
Cancelling \( m_1 \) from both sides (assuming \( m_1 \neq 0 \)):
\[ \frac{G m_2}{r^2} = \omega^2 \left(\frac{m_2 r}{m_1 + m_2}\right) \]
Now, we can cancel \( m_2 \) (assuming \( m_2 \neq 0 \)):
\[ \frac{G}{r^2} = \omega^2 \left(\frac{r}{m_1 + m_2}\right) \]

### Step 7: Solve for \( \omega^2 \)
Rearranging gives:
\[ \omega^2 = \frac{G}{r^2} \cdot \frac{m_1 + m_2}{r} \]
\[ \omega^2 = \frac{G (m_1 + m_2)}{r^3} \]

### Step 8: Relate angular velocity to the period
The angular velocity \( \omega \) is related to the period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the equation gives:
\[ \left(\frac{2\pi}{T}\right)^2 = \frac{G (m_1 + m_2)}{r^3} \]

### Step 9: Solve for the period \( T \)
Rearranging to find \( T \):
\[ T^2 = \frac{4\pi^2 r^3}{G (m_1 + m_2)} \]
Taking the square root gives:
\[ T = 2\pi \sqrt{\frac{r^3}{G (m_1 + m_2)}} \]

### Final Answer
Thus, the period of revolution \( T \) of the two stars is:
\[ T = 2\pi \sqrt{\frac{r^3}{G (m_1 + m_2)}} \]