Q. 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) 5
B) 10
C) $6\sqrt{2}$
D) $\frac{6}{\sqrt{2}}$
Explanation
(c) $\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{R_{1}^{3}}{R_{2}^{3}} = \frac{(6R)^{3}}{(3R)^{3}} = 8$
$\frac{24 \times 24}{T_{2}^{2}} = 8$
$T_{2}^{2} = \frac{24 \times 24}{8}$
$T_{2}^{2} = 72$
$T_{2}^{2} = 36 \times 2$
$T_{2} = 6\sqrt{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.