Question

49.The altitude of geostationary satellite is nearly 6times the radius of the earth.The period of revolution of an identical satellite revolving at an altitude 0.75 times the radius of the earth will be -------

Answer 1

As,
$$\begin{gathered} {\mathrm{T}}^2\mathrm{\alpha }{\mathrm{r}}^3 \\ \mathrm{or}\mathrm{T}\mathrm{\alpha }{\mathrm{r}}^{3/2} \\ \mathrm{For}\mathrm{the}\mathrm{first}\mathrm{case}, \\ {\mathrm{r}}_1=6\mathrm{r}+\mathrm{r}=7\mathrm{r} \\ \mathrm{Thus}, \\ {\mathrm{T}}_1\mathrm{\alpha }(7\mathrm{r}{)}^{3/2}---(\mathrm{i}) \\ \mathrm{For}\mathrm{the}\mathrm{second}\mathrm{case}, \\ {\mathrm{r}}_2=0.75\mathrm{r}+\mathrm{r}=1.75\mathrm{r} \\ \mathrm{So},{\mathrm{T}}_2\mathrm{\alpha }(1.75\mathrm{r}{)}^{3/2}--\left(\mathrm{ii}\right) \\ \mathrm{Now},\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=\frac{(7\mathrm{r}{)}^{3/2}}{(1.75\mathrm{r}{)}^{3/2}} \\ \mathrm{Or}\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=4^{3/2}=8 \\ \mathrm{Or}{\mathrm{T}}_2={\mathrm{T}}_1/8 \\ \mathrm{For}\mathrm{geostationary}\mathrm{satellite},{\mathrm{T}}_1=1\mathrm{day}=24\mathrm{hours} \\ \mathrm{Thus}\mathrm{the}\mathrm{time}\mathrm{period},{\mathrm{T}}_2=24/8=3\mathrm{hours} \end{gathered}$$