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. 1. `v=sqrt(GM((1)/( R)-(1)/(r )))`
B. 2. `v=sqrt(GM((1)/( 2R)-(1)/(r )))`
C. 3. `v=sqrt(GM((1)/( R)+(1)/(r )))`
D. 4. `v=sqrt(GM((1)/( 2R)+(1)/(r )))`

Explanation

To solve the problem of finding the speed \( v \) with which two stars collide, we can use the principle of conservation of energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Conditions
The two stars, each of mass \( M \) and radius \( R \), start approaching each other from a separation \( r \) where \( r \gg R \). At this point, their speeds are negligible, meaning their initial kinetic energy is zero.

**Hint:** Remember that when speeds are negligible, kinetic energy is zero.

### Step 2: Calculate Initial Potential Energy
The gravitational potential energy \( U \) between two masses \( M \) separated by a distance \( r \) is given by:
\[ U_i = -\frac{G M^2}{r} \]
where \( G \) is the gravitational constant.

**Hint:** Use the formula for gravitational potential energy to find the initial energy.

### Step 3: Calculate Final Energy at Collision
When the stars collide, their centers are separated by a distance equal to \( 2R \) (since each star has a radius \( R \)). At this point, they have kinetic energy due to their motion. The final potential energy \( U_f \) when they are at the point of collision is:
\[ U_f = -\frac{G M^2}{2R} \]
The total kinetic energy \( K \) of the two stars when they collide is:
\[ K = \frac{1}{2} M v^2 + \frac{1}{2} M v^2 = M v^2 \]

**Hint:** Remember to account for both stars when calculating the total kinetic energy.

### Step 4: Apply Conservation of Energy
By the conservation of energy, the initial potential energy must equal the sum of the final potential energy and the kinetic energy:
\[ U_i + 0 = U_f + K \]
Substituting the expressions we found:
\[ -\frac{G M^2}{r} = -\frac{G M^2}{2R} + M v^2 \]

**Hint:** Set up the equation based on conservation of energy to relate initial and final energies.

### Step 5: Rearranging the Equation
Rearranging the equation gives:
\[ M v^2 = -\frac{G M^2}{r} + \frac{G M^2}{2R} \]
Factoring out \( G M^2 \):
\[ M v^2 = G M^2 \left( \frac{1}{2R} - \frac{1}{r} \right) \]
Dividing both sides by \( M \):
\[ v^2 = G M \left( \frac{1}{2R} - \frac{1}{r} \right) \]

**Hint:** Isolate \( v^2 \) to find the expression in terms of \( G \), \( M \), \( R \), and \( r \).

### Step 6: Solve for \( v \)
Taking the square root of both sides gives:
\[ v = \sqrt{G M \left( \frac{1}{2R} - \frac{1}{r} \right)} \]

**Hint:** Remember to take the square root to find the speed \( v \).

### Final Answer
Thus, the speed \( v \) with which the stars collide is:
\[ v = \sqrt{G M \left( \frac{1}{2R} - \frac{1}{r} \right)} \]