A satellite is revolving very close to a planet of density $\rho$. The period of revolution of satellite is

We know that, Period of revolution Let, $T$ be the time of one revolution of the satellite. Then, $$ \begin{aligned} T & =\frac{2 \pi r}{v_0}=\frac{2 \pi\left(R_p+h\right)}{v_0} \\ T & =\frac{2 \pi\left(R_p+h\right)}{\left[G M_{\ominus} /\left(R_\rho+h\right)\right]^{1 / 2}} \\ \because \quad v_0 & =\sqrt{\frac{G M_\rho}{\left(R_p+h\right)}} \\ T & =2 \pi \sqrt{\frac{\left(R_p+h\right)^3}{G M_p}} \end{aligned} $$ Again, $\quad G M_e=g R_e^2$ $\therefore \quad T=2 \pi \sqrt{\frac{\left(R_p+h\right)^3}{g R_\rho^2}}$ If the planet is supposed to be a sphere of mean density $\rho$, then the mass of the planet is given by $$ \begin{aligned} M_\rho & =\text { volume } \times \text { density } \\ & =\frac{4}{3} \pi r_\rho^3 \cdot \rho \end{aligned} $$ So, $$ T=\sqrt{\frac{3 \pi\left(R_p+h\right)^3}{G \rho R_p^3}} $$ If a satellite is orbiting very close to the planet's surface $\left(h \ll R_p\right)$, then putting $h=0$ in eq. (iii), we have $$ \begin{aligned} T & =\sqrt{\frac{3 \pi R_\rho^3}{G \rho R_\rho^3}} \\ \Rightarrow \quad T & =\sqrt{\frac{3 \pi}{G \rho}} \end{aligned} $$