A binary star system is revolving in a circular path with angular speed `'omega'` and mass of the stars are 'm' and '4m' respectively. Both stars stop suddenly, then the speed of havirer star when the separation between the stars when the seperation between the stars becomes half of initial value is :

Options

A. a.`sqrt(2[((Gmomega)^(2))/(2)]^(1//3))`
B. b. `sqrt(2/(5)[((Gmomega)^(2))/(5)]^(1//3))`
C. c.`sqrt([((Gmomega)^(2))/(5)]^(1//3))`
D. d.`sqrt(2[((Gmomega)^(2))/(5)]^(1//3))`

Answer

To solve the problem, we need to analyze the binary star system and apply the principles of physics, particularly the concepts of centripetal force and conservation of energy. Here’s a step-by-step solution:

### Step 1: Identify the System
We have a binary star system with two stars:
- Star 1 (mass = m)
- Star 2 (mass = 4m)

The stars are revolving around their common center of mass with an angular speed of \( \omega \).

### Step 2: Determine the Center of Mass
The center of mass (CM) of the system can be determined using the formula:

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

Let the distance from Star 1 to the center of mass be \( r_1 \) and from Star 2 be \( r_2 \). Since \( m_2 = 4m \), we can express the distances in terms of the total distance \( R \) between the stars:

\[ r_1 = \frac{4m \cdot R}{m + 4m} = \frac{4R}{5} \]
\[ r_2 = \frac{m \cdot R}{m + 4m} = \frac{R}{5} \]

### Step 3: Calculate the Centripetal Force
The centripetal force acting on each star can be expressed as:

For Star 1:
\[ F_{c1} = m \cdot \frac{R}{5} \cdot \omega^2 \]

For Star 2:
\[ F_{c2} = 4m \cdot \frac{4R}{5} \cdot \omega^2 \]

### Step 4: Apply Conservation of Energy
When the stars suddenly stop, the gravitational potential energy changes as the separation between the stars becomes half of the initial value. The initial gravitational potential energy \( U_i \) is given by:

\[ U_i = -\frac{G \cdot m \cdot 4m}{R} \]

When the separation becomes \( \frac{R}{2} \), the potential energy \( U_f \) is:

\[ U_f = -\frac{G \cdot m \cdot 4m}{\frac{R}{2}} = -\frac{8Gm^2}{R} \]

### Step 5: Set Up the Energy Conservation Equation
The change in potential energy will equal the kinetic energy gained by the stars:

\[ U_f - U_i = K.E. \]

The kinetic energy of both stars when the separation is \( \frac{R}{2} \) is:

\[ K.E. = \frac{1}{2} m v_1^2 + \frac{1}{2} (4m) v_2^2 \]

### Step 6: Solve for Velocities
From the conservation of momentum, since the system is isolated, we have:

\[ m v_1 = 4m v_2 \implies v_1 = 4 v_2 \]

Substituting \( v_1 \) in the kinetic energy equation gives:

\[ -\frac{8Gm^2}{R} + \frac{4Gm^2}{R} = \frac{1}{2} m (4v_2)^2 + \frac{1}{2} (4m) v_2^2 \]

### Step 7: Simplify and Solve for \( v_2 \)
After substituting and simplifying, we can solve for \( v_2 \):

\[ -\frac{4Gm^2}{R} = \frac{8mv_2^2}{2} \]

This leads to:

\[ v_2 = \sqrt{\frac{4G}{R}} = \frac{2}{5} \sqrt{g m \omega^2} \]

### Final Result
Thus, the speed of the heavier star when the separation becomes half of the initial value is:

\[ v_2 = \frac{2}{5} \sqrt{g m \omega^2} \]