A satellite revolves in an orbit close to the surface of a planet of mean density `5.51 xx 10^(3) kg m^(-3)`. Calculate the time period of satellite.
Given `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`.

Options

A. `562.7s`
B. `5062.7 s`
C. `506.27 s`
D. `56.27 s`

Answer

B

Explanation

Here `rho=5.51xx10^(3) kgm^(-3)`
`G=6.67xx10^(-11) Nm^(2)kg^(-2)` Time period of the satellite near the surface of the plate `(h=0)` is
`T=sqrt((3pi)/(Grho))`
`=sqrt((3xx3.14)/(6.67xx10^(-11)xx5.51xx10^(3)))=5062.7 s`