The time period of a geo-stationary satellite is 24 ~h , at a...

From Kepler's third law, the time period of revolution of satellite around earth is $\mathrm{T}^2 \propto \mathrm{r}^3 \text{ or } \mathrm{T} \propto \mathrm{r}^{3 / 2}$ where, $\mathrm{r}$ is the radius of satellite's orbit. Here, $\mathrm{r}_1=6 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}, \mathrm{T}_1=24 \mathrm{ h}\mathrm{r}_2=2.5 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}, \mathrm{T}_2=$ ? where $\mathrm{R}_{\mathrm{E}}=$ radius of earth So, from Eq. (i), we get \begin{aligned}\frac{\mathrm{T}_1}{\mathrm{ T}_2}& =\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^{3 / 2}\\ \frac{24}{\mathrm{ T}_2}& =\left(\frac{6 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}}{2.5 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}}\right)^{3 / 2}=\left(\frac{7}{3.5}\right)^{3 / 2}\\ \Rightarrow \quad \mathrm{T} }_{2} & =\frac{24}{(2)^{3 / 2}}=\frac{24}{2 \sqrt{2}}=\frac{12}{\sqrt{2}}=6 \sqrt{2} \mathrm{ h} \end{aligned}