Two bodies, each of mass M, are kept fixed at 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)

Let the two bodies be at (-L, 0) and (L, 0). The gravitational potential energy of the mass m at (0, y) is given by:

U = -GMm/√(L² + y²) - GMm/√(L² + y²)

U = -2GMm/√(L² + y²)

The minimum initial velocity required to escape is given when the total energy is zero. The total energy is the sum of kinetic energy and potential energy:

1/2 mv² - 2GMm/√(L² + y²) = 0

At the midpoint (0, 0), the potential energy is U = -2GMm/L

The minimum velocity required to escape the gravitational field is:

1/2 mv² = 2GMm/L

v² = 4GM/L

v = 2√(GM/L)

Therefore, the minimum initial velocity of mass m to escape the gravitational field of the two bodies is 2√(GM/L).