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A spherical planet has uniform density p...

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))`).

Text Solution

Verified by Experts

The correct Answer is:
C
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)/R rArr v=sqrt((GM)/R)`
`T_"min"=(2piR)/sqrt((GM)/R) = (2piRsqrtR)/sqrt(GM)` Using `M=4/3piR^3 rho T_"min"=sqrt((3pi)/(G rho))`
Using values `T_"min"` = 3000 s `therefore` P = 300
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