InterviewSolution
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InterviewSolution
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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 beA. `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 )))` |
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Answer» Correct Answer - B (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)))` |
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