A spherical planet has uniform density `pi/2xx10^(4)kg//m^(3)`. Find out the minimum period for a satellite in a circular orbit around it in seconds (Use `G=20/3xx10^(-11) (N-m^(2))/(kg^(2))`).

Options

A. `7500`
B. `3000`
C. `4500`
D. `6000`

Answer

B

Explanation

Time period is minimum for the satellites with minimum radius of the orbit i.e. equal to the radius of the planet. Therefore,
`(GMm)/(R^(2))=(mv^(2))/Rimplies V=sqrt((GM)/R)`
`T_(min)=(2piR)/(sqrt((GM)/R))=(2piRsqrt(R))/(sqrt(GM))`
Using `M=4/3pR^(3). rho T_(min)=sqrt((3pi)/(grho))`
Using values `T_(min)=3000s`