Question

A geostationary satellite is orbiting the earth at a height of $5\,R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2\,R$ from the surface of the earth is

• A. 5
• B. 10
• C. $6\sqrt2$
• D. $\frac {6}{\sqrt2}$

Answer 1

Answer: (C)

According to Kepler's third law $T \propto r^{3 / 2}$
$\therefore \frac{T_{2}}{T_{1}}=\left(\frac{r_{2}}{r_{1}}\right)^{3 / 2}=\left(\frac{R+2 R}{R+5 R}\right)^{3 / 2}=\frac{1}{2^{3 / 2}}$
Since $T_{1}=24$ hours
So, $\frac{T_{2}}{24}=\frac{1}{2^{3 / 2}}$
or $T_{2}=\frac{24}{2^{3 / 2}}=\frac{24}{2 \sqrt{2}}=6 \sqrt{2}$ hours