Question

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. What will be the time period of another satellite at a height 2.5 R from the surface of the earth?

• A. $6\sqrt{2}$ hours
• B. $6\sqrt{2.5}$ hours
• C. $6\sqrt{3}$ hours
• D. $12$ hours

Answer 1

The correct option is A

$6\sqrt{2}$ hours

The time period of satellite orbiting at a distance r from the centre of the earth is given by $T^2=\frac{4{\pi }^2{r}^3}{GM}$ where M is the mass of the earth. Therefore, the ratio of the time periods of two satellites at distance $r_1$ and $r_2$ respectively from the centre of the earth is
$\frac{T_1}{T_2}={\left(\frac{r_1}{r_2}\right)}^{\frac{3}{2}}\Rightarrow T_2=T_1{\left(\frac{r_2}{r_1}\right)}^{\frac{3}{2}}$
For the geostationary satellite, $T_1=1$ day = 24 hours and $r_1=6R+R=7R$.
For the other satellite, $r_2=2.5R+R=3.5R$
Therefore, $T_2=24\times {\left(\frac{3.5R}{7R}\right)}^{\frac{3}{2}}=24\times {\left(\frac{1}{2}\right)}^{\frac{3}{2}}=6\sqrt{2}$ hours
Hence, the correct choice is (a).