Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining their cehntres, perpendicualr to the line. The gravitational constant is G. The correct statement (s) is (are)

Options

A. The minimum initial velocity of the mass `m` to escape the gravitational field of the two bodies is `4sqrt((GM)/L)`
B. The minimum initial velocity of the mass `m` to escape the gravitational field of the two bodies is `2sqrt((GM)/L)`
C. The minimum initial velocity of the mass `m` to escape the gravitational field of the two bodies is `sqrt((2GM)/L)`
D. The energy of the mass `m` remains constant.

Explanation

To solve the problem, we need to determine the minimum initial velocity required for a particle of mass \( m \) to escape the gravitational influence of two fixed bodies, each of mass \( M \), separated by a distance of \( 2L \). The particle is projected from the midpoint of the line joining the centers of the two masses, perpendicular to that line.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The two bodies of mass \( M \) are fixed at points \( A \) and \( B \) with a separation of \( 2L \).
- The midpoint \( O \) is at a distance \( L \) from both bodies.
- The particle of mass \( m \) is projected from point \( O \) perpendicular to the line joining the two masses.

2. **Gravitational Potential Energy Calculation**:
- The gravitational potential energy \( U \) of the particle \( m \) at point \( O \) due to both masses can be calculated using the formula:
\[ U = -\frac{G M m}{r_1} - \frac{G M m}{r_2} \]
- Here, \( r_1 = L \) (distance from mass at \( A \)) and \( r_2 = L \) (distance from mass at \( B \)).
- Thus, the total potential energy at point \( O \) is:
\[ U = -\frac{G M m}{L} - \frac{G M m}{L} = -\frac{2GMm}{L} \]

3. **Applying the Conservation of Energy**:
- The total mechanical energy \( E \) of the system when the particle is projected is given by the sum of kinetic energy \( K \) and potential energy \( U \):
\[ E = K + U \]
- At the point of projection, if the particle has an initial velocity \( v \), its kinetic energy is:
\[ K = \frac{1}{2} m v^2 \]
- Therefore, the total energy at the moment of projection is:
\[ E = \frac{1}{2} m v^2 - \frac{2GMm}{L} \]

4. **Condition for Escape Velocity**:
- For the particle to escape the gravitational field of the two masses, the total energy must be zero or greater:
\[ E \geq 0 \]
- Setting \( E = 0 \):
\[ \frac{1}{2} m v^2 - \frac{2GMm}{L} = 0 \]

5. **Solving for Minimum Initial Velocity \( v \)**:
- Rearranging the equation gives:
\[ \frac{1}{2} m v^2 = \frac{2GMm}{L} \]
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = \frac{2GM}{L} \]
- Multiplying both sides by 2:
\[ v^2 = \frac{4GM}{L} \]
- Taking the square root:
\[ v = 2\sqrt{\frac{GM}{L}} \]

6. **Conclusion**:
- The minimum initial velocity \( v \) required for the particle of mass \( m \) to escape the gravitational field of the two bodies is:
\[ v = 2\sqrt{\frac{GM}{L}} \]
- Therefore, the correct statement is that the minimum initial velocity of mass \( m \) to escape the gravitational field of the two bodies is \( 2\sqrt{\frac{GM}{L}} \).