Two particles of masses m and M are initially at rest at an infinite distance apart. They move towards each other and gain speeds due to gravitational attraction. Find their speeds when the separation between the masses becomes equal to d.

Explanation

To solve the problem of finding the speeds of two particles of masses \( m \) and \( M \) when the separation between them is \( d \), we will use the principles of conservation of energy and conservation of momentum.

### Step-by-Step Solution:

1. **Initial Conditions**:
- The two particles are initially at rest at an infinite distance apart. Therefore, their initial kinetic energy (KE) and gravitational potential energy (PE) are both zero.
- Initial KE: \( KE_i = 0 \)
- Initial PE: \( PE_i = 0 \)

2. **Final Conditions**:
- When the particles are at a separation \( d \), they will have gained some speeds \( V_1 \) (for mass \( m \)) and \( V_2 \) (for mass \( M \)).
- Final KE:
\[ KE_f = \frac{1}{2} m V_1^2 + \frac{1}{2} M V_2^2 \]
- Final PE (using the formula for gravitational potential energy):
\[ PE_f = -\frac{G m M}{d} \]
- Here, \( G \) is the gravitational constant.

3. **Conservation of Energy**:
- The total mechanical energy is conserved, so we set the initial total energy equal to the final total energy:
\[ KE_i + PE_i = KE_f + PE_f \]
\[ 0 + 0 = \frac{1}{2} m V_1^2 + \frac{1}{2} M V_2^2 - \frac{G m M}{d} \]
- Rearranging gives us:
\[ \frac{1}{2} m V_1^2 + \frac{1}{2} M V_2^2 = \frac{G m M}{d} \quad \text{(Equation 1)} \]

4. **Conservation of Momentum**:
- Since there are no external forces acting on the system, the momentum is conserved:
\[ 0 = m V_1 - M V_2 \]
- Rearranging gives:
\[ m V_1 = M V_2 \quad \Rightarrow \quad V_1 = \frac{M}{m} V_2 \quad \text{(Equation 2)} \]

5. **Substituting Equation 2 into Equation 1**:
- Substitute \( V_1 \) from Equation 2 into Equation 1:
\[ \frac{1}{2} m \left(\frac{M}{m} V_2\right)^2 + \frac{1}{2} M V_2^2 = \frac{G m M}{d} \]
- Simplifying gives:
\[ \frac{1}{2} \frac{M^2}{m} V_2^2 + \frac{1}{2} M V_2^2 = \frac{G m M}{d} \]
- Factoring out \( V_2^2 \):
\[ \frac{1}{2} V_2^2 \left(\frac{M^2}{m} + M\right) = \frac{G m M}{d} \]
- Solving for \( V_2^2 \):
\[ V_2^2 = \frac{2 G m M / d}{\frac{M^2}{m} + M} = \frac{2 G m M}{d \left(\frac{M^2 + mM}{m}\right)} = \frac{2 G m^2 M}{d (M + m)} \]
- Therefore, \( V_2 \) is:
\[ V_2 = \sqrt{\frac{2 G m^2 M}{d (M + m)}} \]

6. **Finding \( V_1 \)**:
- Substitute \( V_2 \) back into Equation 2:
\[ V_1 = \frac{M}{m} V_2 = \frac{M}{m} \sqrt{\frac{2 G m^2 M}{d (M + m)}} = \sqrt{\frac{2 G M^2}{d (M + m)}} \]

### Final Answers:
- The speeds of the particles when the separation is \( d \) are:
\[ V_1 = \sqrt{\frac{2 G M^2}{d (M + m)}} \]
\[ V_2 = \sqrt{\frac{2 G m^2 M}{d (M + m)}} \]