The masses and radii of the Earth and the Moon are `M_1, R_1 and M_2,R_2` respectively. Their centres are at a distance d apart. The minimum speed with which a particle of mass m should be projected from a point midway between the two centres so as to escape to infinity is ........

Explanation

To solve the problem of finding the minimum speed with which a particle of mass \( m \) should be projected from a point midway between the Earth and the Moon so as to escape to infinity, we can use the principle of conservation of mechanical energy.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The particle is projected from the midpoint between the Earth and the Moon. The distance between the centers of the Earth and Moon is \( d \).
- Therefore, the distance from the midpoint to the center of the Earth is \( \frac{d}{2} \) and the same applies to the distance to the center of the Moon.

2. **Calculate the Initial Potential Energy**:
- The gravitational potential energy \( U \) between two masses \( M \) and \( m \) separated by a distance \( r \) is given by:
\[ U = -\frac{G M m}{r} \]
- For the particle at the midpoint, the potential energy due to the Earth is:
\[ U_1 = -\frac{G M_1 m}{\frac{d}{2}} = -\frac{2 G M_1 m}{d} \]
- The potential energy due to the Moon is:
\[ U_2 = -\frac{G M_2 m}{\frac{d}{2}} = -\frac{2 G M_2 m}{d} \]
- Therefore, the total initial potential energy \( U_i \) is:
\[ U_i = U_1 + U_2 = -\frac{2 G M_1 m}{d} - \frac{2 G M_2 m}{d} = -\frac{2 G (M_1 + M_2) m}{d} \]

3. **Final Conditions at Infinity**:
- At infinity, the potential energy \( U_f \) is zero, and we assume the kinetic energy \( K_f \) is also zero for the minimum speed case (the particle just escapes).
- Thus, the total mechanical energy at infinity is:
\[ E_f = K_f + U_f = 0 + 0 = 0 \]

4. **Apply Conservation of Mechanical Energy**:
- According to the conservation of mechanical energy:
\[ E_i = E_f \]
- Therefore:
\[ K_i + U_i = 0 \]
- The initial kinetic energy \( K_i \) is given by:
\[ K_i = \frac{1}{2} m v_0^2 \]
- Substituting the initial potential energy:
\[ \frac{1}{2} m v_0^2 - \frac{2 G (M_1 + M_2) m}{d} = 0 \]

5. **Solve for Minimum Speed \( v_0 \)**:
- Rearranging the equation gives:
\[ \frac{1}{2} m v_0^2 = \frac{2 G (M_1 + M_2) m}{d} \]
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_0^2 = \frac{2 G (M_1 + M_2)}{d} \]
- Multiplying both sides by 2:
\[ v_0^2 = \frac{4 G (M_1 + M_2)}{d} \]
- Taking the square root:
\[ v_0 = \sqrt{\frac{4 G (M_1 + M_2)}{d}} = \frac{2 \sqrt{G (M_1 + M_2)}}{\sqrt{d}} \]

### Final Answer:
The minimum speed with which a particle of mass \( m \) should be projected from the midpoint between the Earth and the Moon to escape to infinity is:
\[ v_0 = \frac{2 \sqrt{G (M_1 + M_2)}}{\sqrt{d}} \]