A spherical planet has uniform density $\frac{\mathrm{\pi }}{2}\times {10}^4 \mathrm{kg}/{\mathrm{m}}^3$. Find out the minimum period for a satellite in a circular orbit around it in seconds. (Use G = $\frac{20}{3}\times {10}^{-11} \frac{\mathrm{N}-{\mathrm{m}}^2}{{\mathrm{kg}}^2}$)

Time period is minimum for the satellite with minimum radius of the orbit i.e. equal to the radius of the planet. Therefore.

       $\frac{\mathrm{GMm}}{{\mathrm{R}}^2}= \frac{{\mathrm{mv}}^2}{\mathrm{R}} \Rightarrow \mathrm{V} = \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$

        ${\mathrm{T}}_{\min } = \frac{2\mathrm{\pi R}}{\sqrt{{\displaystyle \frac{\mathrm{GM}}{\mathrm{R}}}}}= \frac{2\mathrm{\pi R}\sqrt{\mathrm{R}}}{\sqrt{\mathrm{GM}}}$

using M = $\frac{4}{3}{\mathrm{pR}}^3.\mathrm{\rho } {\mathrm{T}}_{\min } = \sqrt{\frac{3\mathrm{\pi }}{\mathrm{G\rho }}}$

using  values  ${\mathrm{T}}_{\min } = 3000 \mathrm{s}$