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 Concept of Reduced Mass
The concept of reduced mass (\( \mu \)) is used to simplify the two-body problem into an equivalent one-body problem. The reduced mass is defined as:
\[ \mu = \frac{mM}{m + M} \]
where \( m \) is the mass of the smaller particle and \( M \) is the mass of the larger particle.

### Step 2: Gravitational Force Between the Particles
The gravitational force (\( F \)) between the two particles when they are at a distance \( d \) is given by Newton's law of gravitation:
\[ F = \frac{G m M}{d^2} \]
where \( G \) is the gravitational constant.

### Step 3: Use the Concept of Energy
The gravitational potential energy (\( U \)) at a distance \( d \) is given by:
\[ U = -\frac{G m M}{d} \]
When the particles are at rest at an infinite distance, the potential energy is zero. As they move towards each other, the potential energy decreases, and this change in potential energy is converted into kinetic energy.

### Step 4: Relate Kinetic Energy to Potential Energy
The change in potential energy as the particles approach each other from infinity to a distance \( d \) is equal to the kinetic energy gained by the system:
\[ \Delta U = K \]
The total kinetic energy for two particles can be expressed as:
\[ K = \frac{1}{2} \mu v^2 \]
where \( v \) is the relative velocity of approach.

### Step 5: Set Up the Equation
Setting the change in potential energy equal to the kinetic energy gives:
\[ -\frac{G m M}{d} = \frac{1}{2} \mu v^2 \]

### Step 6: Substitute Reduced Mass
Substituting the expression for reduced mass:
\[ -\frac{G m M}{d} = \frac{1}{2} \left(\frac{mM}{m + M}\right) v^2 \]

### Step 7: Solve for Relative Velocity
Rearranging the equation to solve for \( v \):
\[ v^2 = -\frac{2G m M}{d} \cdot \frac{m + M}{mM} \]
\[ v^2 = \frac{2G(m + M)}{d} \]
Taking the square root gives:
\[ v = \sqrt{\frac{2G(m + M)}{d}} \]

### Final Answer
Thus, the relative velocity of approach of the two particles due to gravitational attraction when their separation is \( d \) is:
\[ v = \sqrt{\frac{2G(m + M)}{d}} \]