A double star is a system of two stars of masses `m` and `2m`, rotating about their centre of mass only under their mutual gravitational attraction. If `r` is the separation between these two stars then their time period of rotation about their centre of mass will be proportional to

Options

A. Heavier star revolves in orbit of radius `2r//3`.
B. Both the stars revolve with the same speed, period of which is equal to `(2pi//r^(3))(2GM^(2)//3)`
C. Kinetic energy of the heavier star is twice that of the other star.
D. None of the above

Explanation

To solve the problem of finding the time period of rotation of a double star system consisting of two stars with masses `m` and `2m`, we will follow these steps:

### Step 1: Determine the position of the center of mass
The center of mass (CM) of the system can be calculated using the formula:
\[ x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
Here, \(m_1 = m\) (mass of the first star) and \(m_2 = 2m\) (mass of the second star). Let’s assume the distance between the two stars is \(r\). The distance of the first star from the center of mass is \(r_1\) and the distance of the second star from the center of mass is \(r_2\).

Since the center of mass is closer to the heavier star:
\[ r_1 = \frac{2m}{m + 2m} \cdot r = \frac{2}{3}r \]
\[ r_2 = \frac{m}{m + 2m} \cdot r = \frac{1}{3}r \]

### Step 2: Apply Newton's law of gravitation
The gravitational force \(F_g\) between the two stars can be expressed as:
\[ F_g = \frac{G \cdot m \cdot 2m}{r^2} = \frac{2Gm^2}{r^2} \]

### Step 3: Set up the centripetal force equations
The centripetal force required for each star to maintain circular motion about the center of mass can be expressed as:
- For the star of mass \(m\):
\[ F_c = m \omega^2 r_1 = m \omega^2 \left(\frac{2}{3}r\right) \]
- For the star of mass \(2m\):
\[ F_c = 2m \omega^2 r_2 = 2m \omega^2 \left(\frac{1}{3}r\right) \]

### Step 4: Equate gravitational force to centripetal force
For the star of mass \(m\):
\[ \frac{2Gm^2}{r^2} = m \omega^2 \left(\frac{2}{3}r\right) \]
This simplifies to:
\[ \frac{2Gm}{r^2} = \omega^2 \left(\frac{2}{3}r\right) \]
\[ \omega^2 = \frac{2Gm}{\frac{2}{3}r^3} = \frac{3Gm}{r^3} \]

### Step 5: Relate angular velocity to time period
The time period \(T\) is related to angular velocity \(\omega\) by:
\[ T = \frac{2\pi}{\omega} \]
Substituting \(\omega\) from our previous result:
\[ T = 2\pi \sqrt{\frac{r^3}{3Gm}} \]

### Step 6: Determine proportionality
From the expression for \(T\), we can see that:
\[ T \propto r^{3/2} \quad \text{and} \quad T \propto \frac{1}{\sqrt{m}} \]

### Final Answer
Thus, the time period of rotation about their center of mass is proportional to:
\[ T \propto r^{3/2} \quad \text{and inversely proportional to } m^{1/2} \]