Q: A particle is projected from the mid-point of the line joining two fixed particles each of mass m. If the separation between the fixed particles is *l*, the minimum velocity of projection of the particle so as to escape is equal to

(A)√(Gm/*l)*

(B)√(Gm/2*l)*

(C )√(2Gm/*l)*

(D) 2√(2Gm/*l)*

Solution :

The gravitational potential at the mid-point P is

V = V_{1} + V_{2}

= -Gm/(l/2)-Gm/(l/2) = -4Gm/l

The gravitational potential energy

= U = -4Gmm_{0}/l

, m_{0} = mass of particle

When it is projected with a speed v, it just escapes to infinity, and the potential & kinetic energy will become zero.

ΔKE + ΔPE = 0

= 0 – ½mv² +{0-(-4Gmm_{0}/*l*)}= 0

v = 2√2Gm/*l*