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
$$
\begin{aligned}
T & =\sqrt{\frac{3 \pi R_\rho^3}{G \rho R_\rho^3}} \\
\Rightarrow \quad T & =\sqrt{\frac{3 \pi}{G \rho}}
\end{aligned}
$$