Question

2 point masses each having mass `m` are placed at a distance `2a`.Escape velocity for a mass `m` projected from position P?

all the three masses lie in the same straight line and the point P is the midpoint of the distance separating the 2 masses,i.e,P is situated at a distance `a` from each of the masses `m`.

Answer 1

`sqrt((4GM)/a)`, if the two masses at distance `2a` are held fixed by other forces.

If all three bodies are free to move, then this is a special case of the three body problem, where predicting the final outcome is extremely difficult.

Explanation:

Let us assume that the two masses at distance `2a` are held fixed by other forces.

The potential energy of the mass `m` at the point P is

`U_i = -2 times (Gm^2)/a = -(2Gm^2)/a`

If this mass can escape, i.e. reach an infinitely distant point, its final potential energy is

`U_f = 0`

Conservation of energy tells us that

`K_i+U_i = K_f+U_f implies`

`K_i = (U_f-U_i)+K_f = (2Gm^2)/a+K_f >= (2Gm^2)/a`

Thus

`1/2mv_i^2 >= (2Gm^2)/a implies v_i^2 >= (4GM)/a`

Thus, under these conditions, the third body will escape if its velocity is at least `sqrt((4GM)/a)`