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. `v=sqrt(GM((1)/( R)-(1)/(r )))`
B. `v=sqrt(GM((1)/( 2R)-(1)/(r )))`
C. `v=sqrt(GM((1)/( R)+(1)/(r )))`
D. `v=sqrt(GM((1)/( 2R)+(1)/(r )))`

Answer

B

Explanation

(b) Since the speeds of the stars are negligible when they are at a distance r, hence the initial kinetic energy of the system is zero. Therefore, the initial total energy of the systme is
`E_(i)=KE+PE=0+(-(GMM)/( r))=-(GM^(2))/( r)`
where M represent the mass of each star and r is initial seperation 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 given by.
`E_(i)=KE+PE=0+(-(GMM)/(r ))=-(GM^(2))/(r )`
where M represents the mass of each star and r is initial separation between them.
When two stars collide their centres will be at a distancee 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 given by
`E_(f)=2xx((1)/(2)Mv^(2))+(-(GMM)/(2R))=Mv^(2)-(GM^(2))/(2R)`
According to law of conservation of mechanical energy
`E_(f)=E_(i)`
`Mv^(2)-(GM^(2))/(2R)=-(GM^(2))/(r ) or v^(2)=GM((1)/(2R)-(1)/( r)) or v=sqrt(GM((1)/(2R)-(1)/( r)))`