Question

A geostationary satellite is orbiting the earth at a height of 5R above the 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. $\frac{6}{\sqrt{2}}$
• B. $5$
• C. $10$
• D. $6\sqrt{2}$

Answer 1

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

From Kepler's $3^{rd}$ Law,

$T^2\propto r^3$

Hence, $T_1^2{\propto r}_1^3$ and $T_2^2{\propto r}_2^3$

So, $\frac{T_2^2}{T_1^2}=\frac{r_2^3}{r_1^3}={\left(\frac{3R}{6R}\right)}^3=\frac{1}{8}{\therefore T}_2^2=\frac{1}{8}{T}_1^2\therefore T=\frac{24}{2\sqrt{2}}=6\sqrt{2}hrs$