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 be |
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Answer» Here, mass of each star, `M = 2 xx 10^(30) kg`. Initial distance between two stars, `r = 10^(9) km = 10^(12) m`. Initial potential energy of the systam `= (GM m)/(r )` Total K.E. of the mass `= (1)/(2) M upsilon^(2) + (1)/(2) M upsilon^(2) = M upsilon^(2)` where `upsilon` is the speed of stars with which they collide. when the stars are about to collide, the distance between their centres, `r = 2R`. `:.` Final potential energy of two stars `= - (GM M)/(2R)`. Since, gain in `K.E.` is at the cost of loss in `P.E.` `:. M upsilon^(2) = - (GM M)/(r ) - (-(GM M)/(2R)) = (-GM M)/(r) + (GM M)/(2R)` or `2 xx 10^(30) upsilon^(2) = - (6.67 xx 10^(-11) xx (2 xx 10^(30))^(2))/(10^(12)) + (6.67 xx 10^(-11) (2 xx 10^(30))^(2))/(2 xx 10^(7))` `= - 2.668 xx 10^(38) + 1.334 xx 10^(43) = 1.334 xx 10^(43) J` `:. upsilon = sqrt((1.334 xx 10^(43))/(2 xx 10^(30))) = 2.583 xx 10^(6) ms^(-1)` |
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