If a particles of mass `m` is projected with minimum velocity form the surface of a star with kinetic energy `(K_(1)GMm)/a` and potential energy at surface of the star `(K_(2)GMm)/a` towards the star of same mass `m` and radius `a` (`K_(1)` and `K_(2)` are constant) to reach the other star. Find the distance between the centre of the two stars:

Options

A. `(2a)/((K_(2)-K_(1)))`
B. `(4a)/((K_(2)-K_(1)))`
C. `(2a)/((K_(1)-K_(2)))`
D. `(a)/((K_(1)-K_(2)))`

Answer

B

Explanation

Since the mass and radius of two stars are same, point `P` at which gravitational force will be equal, will be at equal distance (say `r`) from centre of each star. The particle should be projected to cross `P`, beyond which the particle will be attracted towards the other star.
`KE=PE` of body at `P` -PE at surface of the star
`(K_(1)GMm)/a=[-(GMm)/r-(GMm)/r]-[-K_(2) (GMm)/a]`
On solving, `r=(2a)/((K_(2)-K_(1)))2r=(4a)/((K_(2)-K_(1)))`