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 centres, perpendicular to the line. The gravitational constant is $G$. The correct statement(s) is (are)

Step 1: Define the potential energy of the system.

The distance $r$ from the particle of mass $m$ at position $y$ on the perpendicular bisector to each fixed mass $M$ is determined by the Pythagorean theorem.

$$r = \sqrt{L^2 + y^2}$$

The total gravitational potential energy $U(y)$ is the sum of the potential energies due to both masses.

$$U(y) = -\frac{GMm}{r} - \frac{GMm}{r}$$ $$U(y) = -\frac{2GMm}{\sqrt{L^2 + y^2}}$$

Step 2: Find the initial potential energy at the midpoint.

At the projection point, the displacement $y$ is zero.

$$U(0) = -\frac{2GMm}{L}$$

Step 3: Apply the energy condition for escape.

To escape to infinity where the potential energy is zero, the total mechanical energy $E$ must be greater than or equal to zero.

$$E = \frac{1}{2}mv^2 + U(0) \ge 0$$ $$\frac{1}{2}mv_{\min}^2 - \frac{2GMm}{L} = 0$$

Step 4: Calculate the minimum escape velocity.

Solve the energy equation for the minimum velocity $v_{\min}$.

$$\frac{1}{2}mv_{\min}^2 = \frac{2GMm}{L}$$ $$v_{\min} = 2\sqrt{\frac{GM}{L}}$$

Step 5: Verify the conservation of mechanical energy.

Since gravity is a conservative force, the total mechanical energy $E$ of the particle remains constant throughout its motion.

$$E = \text{constant}$$

Answer: The minimum initial velocity for escape is $2\sqrt{\dfrac{GM}{L}}$, and the total mechanical energy of the particle remains constant.