Question

Satellite A of massm is orbiting around the earth with time periodT. Another satellite B of mass 2m is orbiting around the earth with half the radius of orbit as that of A. What is the time period of revolution of satellite B around the earth?

Answer 1

The period of revolution of a satellite in a given orbit is independent of its mass. Time period of revolution is given by the formula,
$$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{{\mathrm{r}}^3}{\mathrm{GM}}} \\ \mathrm{where},\mathrm{r}=\mathrm{radius}\mathrm{of}\mathrm{orbit},\mathrm{G}=\mathrm{Gravitational}\mathrm{constant}\mathrm{and}\mathrm{M}=\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{earth} \\ \mathrm{For}\mathrm{satellite}\mathrm{A}, \\ \mathrm{Suppose}\mathrm{radius}\mathrm{of}\mathrm{orbit}\mathrm{is}{\mathrm{r}}_1. \\ \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{{{\mathrm{r}}_1}^3}{\mathrm{GM}}} \\ \mathrm{For}\mathrm{satellite}\mathrm{B}, \\ \mathrm{radius}\mathrm{of}\mathrm{orbit}={\mathrm{r}}_2 \\ \mathrm{T}\text{'}=2\mathrm{\pi }\sqrt{\frac{{{\mathrm{r}}_2}^3}{\mathrm{GM}}} \\ \mathrm{Given},{\mathrm{r}}_2={\mathrm{r}}_1/2 \\ \mathrm{So},\mathrm{T}\text{'}=2\mathrm{\pi }\sqrt{\frac{({\mathrm{r}}_1/2{)}^3}{\mathrm{GM}}}=2\mathrm{\pi }\sqrt{\frac{{{\mathrm{r}}_1}^3}{8\mathrm{GM}}} \\ =\frac{1}{2\sqrt{2}}2\mathrm{\pi }\sqrt{\frac{{{\mathrm{r}}_1}^3}{\mathrm{GM}}}=\frac{1}{2\sqrt{2}}\mathrm{T} \end{gathered}$$