Two uniform spherical starts made of same material have radii R and 2R. Mass of the smaller planet is m. They start mmoving from rest towards each other from a large distance under mutual force of gravity. The collision between the stars is inelastic with coefficient of restitution 1/2.
The maximum separation between their centres after their first collision

Options

A. 4R
B. 6R
C. 8R
D. 12R

Explanation

To solve the problem step-by-step, we will analyze the gravitational interaction between the two spherical stars, their collision, and the subsequent motion after the collision.

### Step 1: Determine the Masses of the Stars
Given:
- Radius of the smaller star, \( R \)
- Radius of the larger star, \( 2R \)
- Mass of the smaller star, \( m \)

Since both stars are made of the same material, the mass is proportional to the volume. The volume \( V \) of a sphere is given by:

\[ V = \frac{4}{3} \pi r^3 \]

Thus, the mass of the larger star is:

\[ M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi (2R)^3 = \rho \cdot \frac{4}{3} \pi (8R^3) = 8m \]

So, the mass of the larger star is \( 8m \).

### Step 2: Calculate the Initial Kinetic Energy
Initially, both stars are at rest and at a large distance from each other, so the initial kinetic energy \( K_0 \) is:

\[ K_0 = 0 \]

### Step 3: Calculate the Potential Energy at Collision
As they move towards each other under mutual gravitational attraction, the potential energy \( U \) when they are at a distance \( d \) apart (where \( d = R + 2R = 3R \) at the moment of collision) is given by:

\[ U = -\frac{G m M}{d} = -\frac{G m (8m)}{3R} = -\frac{8G m^2}{3R} \]

### Step 4: Apply Conservation of Energy
At the point of collision, all potential energy is converted into kinetic energy. Thus, we have:

\[ K_0 + U = K_f \]

Since \( K_0 = 0 \):

\[ -\frac{8G m^2}{3R} = K_f \]

### Step 5: Calculate the Kinetic Energy After Collision
The coefficient of restitution \( e \) is given as \( \frac{1}{2} \). The kinetic energy after the collision \( K_f' \) is:

\[ K_f' = e^2 K_f = \left(\frac{1}{2}\right)^2 K_f = \frac{1}{4} K_f \]

Substituting \( K_f \):

\[ K_f' = \frac{1}{4} \left(-\frac{8G m^2}{3R}\right) = -\frac{2G m^2}{3R} \]

### Step 6: Calculate Maximum Separation After Collision
After the collision, the stars will move apart until their kinetic energy is converted back into potential energy. At the maximum separation \( d' \), we have:

\[ K_f' = -\frac{G m (8m)}{d'} \]

Setting the kinetic energy equal to the potential energy:

\[ -\frac{2G m^2}{3R} = -\frac{8G m^2}{d'} \]

Cancelling \( -G m^2 \) from both sides gives:

\[ \frac{2}{3R} = \frac{8}{d'} \]

Cross-multiplying yields:

\[ 2d' = 24R \implies d' = 12R \]

### Step 7: Calculate the Maximum Separation Between Their Centers
The maximum separation between the centers of the stars after the collision is:

\[ d' = 12R \]

### Final Answer
The maximum separation between their centers after the first collision is \( 12R \).

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