Physics - Gravitation Question with Solution | TestHub

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 $$\begin{array}{rl}\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}& ={\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^{3/2}\\ \frac{24}{{\mathrm{T}}_2}& ={\left(\frac{6{\mathrm{R}}_{\mathrm{E}}+{\mathrm{R}}_{\mathrm{E}}}{2.5{\mathrm{R}}_{\mathrm{E}}+{\mathrm{R}}_{\mathrm{E}}}\right)}^{3/2}={\left(\frac{7}{3.5}\right)}^{3/2}\\ \Rightarrow {\mathrm{T}}_2& =\frac{24}{(2{)}^{3/2}}=\frac{24}{2\sqrt{2}}=\frac{12}{\sqrt{2}}=6\sqrt{2}\mathrm{h}\end{array}$$