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

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/2l)
(C )√(2Gm/l)
(D) 2√(2Gm/l)

Solution :

The gravitational potential at the mid-point P is
V = V1 + V2
= -Gm/(l/2)-Gm/(l/2) = -4Gm/l
 The gravitational potential energy
= U = -4Gmm0/l
 , m0 = 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-(-4Gmm0/l)}= 0
v = 2√2Gm/l

Author: Rajesh Jha

QuantumStudy.com for CBSE|IIT-JEE|NEET