Two stars each of mass `M` and radius `R` are approaching each other for a head-on collision. They start approaching each other when their separation is `rgt gtR`. If their speed at this separation are negligible, the speed `v` with which they collide would be

Options

A. `sqrt(Gm (1/R - 1/r))`
B. `sqrt(Gm ((1)/(2R) - 1/r)`
C. `sqrt ( Gm (1/R + 1/r))`
D. `sqrt(Gm ( (1)/(2R) + 1/r))`

Answer

B

Explanation

Since the speeds of the stars are negligible when they are at a distance of r, the initial kinetic energy of the system is zero. Therefore, initial total energy of the system is
`E_i= KE + PE = 0 + ( - (Gmm)/( r))` ....(i)
where m represents the mass of each star and r the initial separation between them. When two stars collide, their centres will be at a distance twice the radius of a star i.e. 2R. Let v be the speed with which two stars collide. Then total energy of the system at the instant of their collision is
`E_f = (1/2 mv^2) xx 2 + (-(Gmm)/(2R)) = mv^2 - (GMm)/(2R)`
According to law of conservation of energy, `E_f=E_i`
` therefore mv^2 - (Gmm)/(2R) = - (Gmm)/(r ) " or " v^2 = Gm ((1)/(2R) - 1/r)`
or ` v = sqrt(Gm ((1)/(2R) - 1/r))`