Question

A geostationary satellite is revolving at a height $6R$ above the earth's surface, where $R$ is the radius of earth. The period of revolution of satellite orbiting at a height $2.5R$ above the earth's surface will be.

• A. $\text{24 hour}$
• B. $\text{12 hour}$
• C. $\text{6 hour}$
• D. $6\sqrt{2}\text{hour}$

Answer 1

The correct option is D $6\sqrt{2}\text{hour}$

$\text{The time period of satellite orbiting at a distance from the centre of the earth is given by,}$

$T^2=\frac{4{\pi }^2{r}^3}{G{M}^2}$

$\text{where M is the mass of the earth.}$

$\text{Therefore,the ratio of the time periods of two satellites at distance}{r}_1\text{and}{r}_2$

$\text{respectively from the centre of the earth is given by,}$

$\frac{T_1}{T_2}=(\frac{r_1}{r_2}{)}^{3/2}$

$\text{or}{T}_2=T_1{\left(\frac{r_2}{r_1}\right)}^{3/2}$

$\text{For the geostationary satellite}{T}_1=1\text{day}=24\text{hours and}{r}_1=6R+R=7R$

$\text{For the other satellite}{r}_2=2.5R+R=3.5R$

Therefore $T_2=24\times (\frac{3.5R}{7R}{)}^{3/2}=24\times (\frac{1}{2}{)}^{3/2}=6\sqrt{2}$ hours