Two stars each of mass `M` and radius `R` are approaching each other for a head-on collision. They start approaching each other when their separation is `rgt gtR`. If their speed at this separation are negligible, the speed `v` with which they collide would be

Options

A. `v=sqrt(GM(1/R-1/r))`
B. `v=sqrt(GM(1/(2R)-1/r))`
C. `v=sqrt(GM(1/R+1/r))`
D. `v=sqrt(GM(1/(2R)+1/r))`

Explanation

To solve the problem of two stars colliding, we will use the principle of conservation of mechanical energy. Here's a step-by-step solution:

### Step 1: Understand the Initial Conditions
- We have two stars, each with mass \( M \) and radius \( R \).
- They start approaching each other from a separation \( r \) (where \( r \gg R \)).
- Their initial speeds are negligible, meaning their initial kinetic energy is zero.

**Hint:** Identify the initial conditions clearly, including the masses, radii, and initial separation.

### Step 2: Write the Initial Mechanical Energy
The initial mechanical energy \( E_i \) consists of gravitational potential energy and kinetic energy. Since the stars are at rest initially, the kinetic energy is zero.

\[ E_i = \text{Potential Energy} + \text{Kinetic Energy} = U_i + K_i = U_i + 0 \]

The gravitational potential energy \( U_i \) at separation \( r \) is given by:

\[ U_i = -\frac{G M^2}{r} \]

where \( G \) is the gravitational constant.

**Hint:** Remember that gravitational potential energy is negative and depends on the distance between the two masses.

### Step 3: Write the Final Mechanical Energy
When the stars collide, they will be at a separation of \( 2R \) (the distance between their centers). The final mechanical energy \( E_f \) consists of kinetic energy and potential energy at this point.

1. **Kinetic Energy**: If both stars have the same speed \( v \) just before collision, their total kinetic energy is:

\[ K_f = \frac{1}{2} M v^2 + \frac{1}{2} M v^2 = M v^2 \]

2. **Potential Energy**: At separation \( 2R \):

\[ U_f = -\frac{G M^2}{2R} \]

Thus, the final mechanical energy is:

\[ E_f = K_f + U_f = M v^2 - \frac{G M^2}{2R} \]

**Hint:** Ensure to account for both kinetic and potential energy in the final state.

### Step 4: Apply Conservation of Mechanical Energy
According to the conservation of mechanical energy, the initial energy equals the final energy:

\[ E_i = E_f \]

Substituting the expressions we derived:

\[ -\frac{G M^2}{r} = M v^2 - \frac{G M^2}{2R} \]

**Hint:** Set up the equation carefully, ensuring both sides represent the same physical quantity.

### Step 5: Rearrange the Equation
Rearranging the equation to solve for \( v^2 \):

\[ M v^2 = -\frac{G M^2}{r} + \frac{G M^2}{2R} \]

Factoring out \( G M^2 \):

\[ M v^2 = G M^2 \left(-\frac{1}{r} + \frac{1}{2R}\right) \]

Dividing both sides by \( M \):

\[ v^2 = G M \left(-\frac{1}{r} + \frac{1}{2R}\right) \]

**Hint:** Isolate \( v^2 \) to find the expression for velocity.

### Step 6: Solve for \( v \)
Taking the square root gives us the final speed \( v \):

\[ v = \sqrt{G M \left(\frac{1}{2R} - \frac{1}{r}\right)} \]

**Hint:** Remember to take the square root and ensure the expression is simplified correctly.

### Final Result
The speed \( v \) with which the two stars collide is:

\[ v = \sqrt{G M \left(\frac{1}{2R} - \frac{1}{r}\right)} \]