InterviewSolution
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InterviewSolution
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If a spherically symmetric star of radius `R` collapsed under its own weight, neglecting any forces other than gravitational ones, what is the time required for collapse?A. `(8pi^(2)R^(3))/((GM)^(0.5))`B. `(2pi^(2)R^(3))/((3GM)^(0.5))`C. `(pi^(2)R^(3))/((8GM)^(0.5))`D. `((2R^(3))/(GM))^(0.5)` |
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Answer» Correct Answer - A `4pir^(2)dr`=volume of an element of thickness `dr`. The number of particles, `n_(i)=4pir^(2)dr.rho`, assumming `rho` is the number of the particles per unit volume. The acceleration due to gravity `=g/Rxxr` for each particle `n_(j)g=4pir^(2)drrho.g/Rr=4pig/Rrhor^(3)dr` `:.` mean acceleration of the particle due to gravity, `g=(int_(O)^(R)((4pi)/Rr^(3)dr)rho)/((4//3)piR^(3)rho)` Average distance travelled by the particle is `=(sumn_(i)r)/(sumn_(i))rArrbar(r)=3/4R` `:. bar(r)=1/2bar(at^(2))` where `bar(a)` average acceleration `rArr 3/4R=1/2.3/4g.t^(2)rArr t=sqrt((2R)/g)` `g=(GM)/(R^(2))rArr ((2R.R^(2))/(GM))^(1//2)=((2R^(3))/(GM))^(0.5)` |
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