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?

Options

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)`

Answer

A

Explanation

`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)`