Two particles of mass `m` and M are initialljy at rest at infinite distance. Find their relative velocity of approach due to gravitational attraction when `d` is their separation at any instant

Options

A. `sqrt((2G(M+m))/(d))`
B. `sqrt((G(M+m))/(d))`
C. `sqrt((G(M+m))/(2d))`
D. `sqrt((G(M+m))/(4d))`

Explanation

To find the relative velocity of approach of two particles of mass `m` and `M` due to gravitational attraction when their separation is `d`, we can follow these steps:

### Step 1: Understand the Initial Conditions
The two particles are initially at rest at an infinite distance. As they move towards each other due to gravitational attraction, we need to analyze their velocities when they are at a distance `d`.

### Step 2: Apply Conservation of Energy
The total mechanical energy of the system is conserved. Initially, both particles are at rest, so the initial kinetic energy (KE) is 0, and the potential energy (PE) at infinite distance is also 0. When they are at a distance `d`, the potential energy is given by:
\[ PE = -\frac{G m M}{d} \]
The kinetic energy when they are at distance `d` is:
\[ KE = \frac{1}{2} m v_1^2 + \frac{1}{2} M v_2^2 \]
By conservation of energy, we have:
\[ 0 + 0 = \frac{1}{2} m v_1^2 + \frac{1}{2} M v_2^2 - \frac{G m M}{d} \]
Thus, we can write:
\[ \frac{1}{2} m v_1^2 + \frac{1}{2} M v_2^2 = \frac{G m M}{d} \]

### Step 3: Relate the Velocities
Using the fact that the particles are attracted towards each other, we can relate their velocities using the concept of the center of mass. The velocities are related by:
\[ \frac{v_1}{v_2} = \frac{M}{m} \]
This implies:
\[ v_2 = \frac{m}{M} v_1 \]

### Step 4: Substitute for One Velocity
Substituting \( v_2 \) in terms of \( v_1 \) into the energy equation:
\[ \frac{1}{2} m v_1^2 + \frac{1}{2} M \left(\frac{m}{M} v_1\right)^2 = \frac{G m M}{d} \]
This simplifies to:
\[ \frac{1}{2} m v_1^2 + \frac{1}{2} \frac{m^2}{M} v_1^2 = \frac{G m M}{d} \]
Factoring out \( v_1^2 \):
\[ \frac{1}{2} v_1^2 \left(m + \frac{m^2}{M}\right) = \frac{G m M}{d} \]

### Step 5: Solve for \( v_1^2 \)
Rearranging gives:
\[ v_1^2 = \frac{2 G M}{d} \cdot \frac{1}{\left(1 + \frac{m}{M}\right)} \]
This can be rewritten as:
\[ v_1^2 = \frac{2 G m}{d} \cdot \frac{M}{m + M} \]

### Step 6: Find \( v_2 \)
Using \( v_2 = \frac{m}{M} v_1 \):
\[ v_2^2 = \left(\frac{m}{M}\right)^2 v_1^2 = \frac{m^2}{M^2} \cdot \frac{2 G m}{d} \cdot \frac{M}{m + M} \]

### Step 7: Calculate Relative Velocity of Approach
The relative velocity of approach \( v_{rel} \) is given by:
\[ v_{rel} = v_1 + v_2 \]
Substituting the expressions for \( v_1 \) and \( v_2 \) gives:
\[ v_{rel} = v_1 + \frac{m}{M} v_1 = v_1 \left(1 + \frac{m}{M}\right) \]
Thus:
\[ v_{rel} = v_1 \cdot \frac{m + M}{M} \]

### Final Result
Substituting \( v_1 \) back into this equation gives us the final expression for the relative velocity of approach:
\[ v_{rel} = \sqrt{\frac{2G(m + M)}{d}} \]