Two planets revolve with same angular velocity about a star. The radius of orbit of outer planet is twice the radius of orbit of the inner planet. If `T` is time period of the revolution of outer planet, find the time in which inner planet will fall into the star. If it was suddenly stopped.

Options

A. `sqrt((23gR)/11)`
B. `sqrt((25gR)/11)`
C. `(v_(a)^(2))/(gR^(2))-R`
D. `(v_(a)^(2))/(2gR^(2))-R`

Answer

B

Explanation

`T_(o)=(2pir)/(omega), T_(1)=(2pir)/(2omega)=(T_(o))/2`
Consider an imaginary comet moving along an ellipse. The extreme points of this ellipse are located on orbit of inner planet and the star, sem-major axis of orbit of such comet will be half of the semi-major axis of the inner planet's orbit. According to Kepler's law. if `T'` is the time period of the comet.
`(T'^(2))/((r//4)^(3))=(T_(1)^(2))/((r//2)^(3))`
`T'^(2)=8/64T_(1)^(2)=(T_(0)^(2))/32 ( :' T_(1)=T_(0)//2)`
`T'=T/(4sqrt(2))`
`(T'//2)` represents time in which inner planet will fall into star.
`((T')/2)=T/(8sqrt(2))=(Tsqrt(2))/16`