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

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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# .

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

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.

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#

#U_f = 0#

#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#

#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)#

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