Question

A geostationary satellite is orbiting the earth at a height of $5R$ 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 $2R$ from the surface of the earth is

• A. 5
• B. 10
• C. $6\sqrt{2}$
• D. $\frac{6}{\sqrt{2}}$

Answer 1

Answer: (C)

According to Kepler's third law $T\propto r^{3/2}$
$\therefore \frac{T_2}{T_1}={\left(\frac{r_2}{r_1}\right)}^{3/2}={\left(\frac{R+2R}{R+5R}\right)}^{3/2}=\frac{1}{2^{3/2}}$
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