A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, R being the radius of the earth. The time period of another satellite at a height 2.5 R from the surface of the earth is

given that a geostationary satellites revolving at a distance from centre ${\text{r}}_{\text{1}}=\text{R}+6\text{R}=7\text{R}$

Time period ${\text{T}}_1=24\text{hours}$

another planet is revolving at height 2.5 R

its distance from centre ${\text{r}}_2=3.5\text{R}=\frac{7}{2}\text{R}$

Time period ${\text{T}}_2=?$

We know $\frac{{\text{T}}_{\text{1}}^{\text{2}}}{{\text{T}}_{\text{2}}^{\text{2}}}=\frac{{\text{r}}_{\text{1}}^{\text{3}}}{{\text{r}}_{\text{2}}^{\text{3}}}$

$\Rightarrow \frac{{\text{24}}^{\text{2}}}{{\text{T}}_{\text{2}}^{\text{2}}}={\left(\frac{\text{7R}}{\text{7/2R}}\right)}^3$

$\frac{{\text{24}}^{\text{2}}}{{\text{T}}_{\text{2}}^{\text{2}}}=8$

$\Rightarrow \therefore {\text{T}}_{\text{2}}^{\text{2}}=\frac{24\times 24}{8}$

${\text{T}}_{\text{2}}^{\text{2}}=72$

$\therefore$ Time period of satellite ${\text{T}}_2=6\sqrt{2}\text{hours}$