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 kinetic energies of the two stars `(K_(m) //K_(M))` is .

Options

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

Answer

To find the ratio of the kinetic energies of two stars rotating about their center of mass, we can follow these steps:

### Step 1: Define the masses and distances
Let the mass of the first star be \( M = 2m \) and the mass of the second star be \( m \). The distance between the centers of the two stars is \( d \).

### Step 2: Find the center of mass
The center of mass \( R_{cm} \) of the system can be found using the formula:
\[ R_{cm} = \frac{M \cdot r_1 + m \cdot r_2}{M + m} \]
where \( r_1 \) is the distance from \( M \) to the center of mass and \( r_2 \) is the distance from \( m \) to the center of mass.

Given that \( M = 2m \), we can set up the distances:
- Let \( r_1 \) be the distance from \( M \) to the center of mass.
- Let \( r_2 \) be the distance from \( m \) to the center of mass.

From the mass ratio, we can find:
\[ \frac{r_1}{r_2} = \frac{m}{M} = \frac{m}{2m} = \frac{1}{2} \]

Since the total distance \( d = r_1 + r_2 \), we can express \( r_1 \) and \( r_2 \) in terms of \( d \):
\[ r_1 = \frac{d}{3}, \quad r_2 = \frac{2d}{3} \]

### Step 3: Relate gravitational force to centripetal force
The gravitational force acting between the two stars provides the necessary centripetal force for their circular motion. The gravitational force \( F_g \) is given by:
\[ F_g = \frac{G \cdot M \cdot m}{d^2} \]

This force acts as the centripetal force for both stars:
\[ F_c = \frac{m v_m^2}{r_2} \quad \text{for mass } m \]
\[ F_c = \frac{M v_M^2}{r_1} \quad \text{for mass } M \]

### Step 4: Set the gravitational force equal to the centripetal force
For the smaller mass \( m \):
\[ \frac{G \cdot (2m) \cdot m}{d^2} = \frac{m v_m^2}{\frac{2d}{3}} \]
This simplifies to:
\[ \frac{2Gm^2}{d^2} = \frac{3v_m^2}{2} \]
Thus, we can solve for \( v_m^2 \):
\[ v_m^2 = \frac{4G}{3d^2} \cdot m \]

For the larger mass \( M \):
\[ \frac{G \cdot (2m) \cdot m}{d^2} = \frac{(2m) v_M^2}{\frac{d}{3}} \]
This simplifies to:
\[ \frac{2Gm^2}{d^2} = 6v_M^2 \]
Thus, we can solve for \( v_M^2 \):
\[ v_M^2 = \frac{G}{3d^2} \cdot m \]

### Step 5: Calculate the kinetic energies
The kinetic energy \( K \) for each star is given by:
\[ K_m = \frac{1}{2} m v_m^2 \]
\[ K_M = \frac{1}{2} M v_M^2 \]

Substituting the values of \( v_m^2 \) and \( v_M^2 \):
\[ K_m = \frac{1}{2} m \left(\frac{4Gm}{3d^2}\right) = \frac{2Gm^2}{3d^2} \]
\[ K_M = \frac{1}{2} (2m) \left(\frac{Gm}{3d^2}\right) = \frac{Gm^2}{3d^2} \]

### Step 6: Find the ratio of kinetic energies
Now, we can find the ratio \( \frac{K_m}{K_M} \):
\[ \frac{K_m}{K_M} = \frac{\frac{2Gm^2}{3d^2}}{\frac{Gm^2}{3d^2}} = 2 \]

Thus, the ratio of the kinetic energies of the two stars is:
\[ \frac{K_m}{K_M} = 2 \]