Question

an earth satellite X is revolving around earth is an orbit whose radius is one-fourth the radius of orbit of a communication satellite. time period for revolution of X is

Answer 1

Let
Radius of the orbit of communication satellite, $R_1=r$
Radius of the orbit of satellite X, $R_2=\frac{r}{4}$
Time period of revolution of communication satellite, $T_1=24hours$
Time period of revolution of satellite X, $T_2=?$

From Kepler's law of planetary motion,

$$\begin{gathered} T^2\alpha R^3 \\ {\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{R_2}{R_1}\right)}^3 \\ {\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{{\displaystyle \frac{r}{4}}}{r}\right)}^3 \\ {\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{1}{4}\right)}^3 \\ {\left(\frac{T_2}{T_1}\right)}^2=\left(\frac{1}{64}\right) \\ \left(\frac{T_2}{T_1}\right)=\left(\frac{1}{8}\right) \\ {T}_2=\frac{T_1}{8} \\ =\frac{24}{8} \\ =3\mathrm{hours} \end{gathered}$$