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. (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.

### Step-by-Step Solution:

1. **Identify the Setup**:
- We have two masses \( M \) located at points \( A \) and \( B \) with a separation of \( 2L \).
- The midpoint, where the mass \( m \) is projected, is at a distance \( L \) from each mass.

2. **Calculate the Gravitational Potential Energy**:
- The gravitational potential energy \( U \) between two masses is given by:
\[ U = -\frac{G M m}{r} \]
- At the midpoint, the distance from mass \( M \) at \( A \) to mass \( m \) is \( L \) and similarly for mass \( B \).
- Therefore, the total potential energy \( U \) at the midpoint is:
\[ U = -\frac{G M m}{L} - \frac{G M m}{L} = -\frac{2GMm}{L} \]

3. **Set Up the Energy Conservation Equation**:
- The total mechanical energy \( E \) at the starting point (where the particle is projected) must equal the total mechanical energy at infinity (where it escapes the gravitational influence).
- At infinity, the potential energy is \( 0 \) and the kinetic energy is also \( 0 \) (as the particle just escapes). Thus, the total energy at infinity is:
\[ E_{\text{final}} = 0 \]
- At the midpoint, the total energy is the sum of kinetic energy \( K \) and potential energy \( U \):
\[ E_{\text{initial}} = K + U = \frac{1}{2} mv^2 - \frac{2GMm}{L} \]
- Setting the initial energy equal to the final energy:
\[ \frac{1}{2} mv^2 - \frac{2GMm}{L} = 0 \]

4. **Solve for the Initial Velocity \( v \)**:
- Rearranging the equation gives:
\[ \frac{1}{2} mv^2 = \frac{2GMm}{L} \]
- 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:
\[ v = 2\sqrt{\frac{GM}{L}} \]

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