Q. A geostationary satellite is orbiting the earth at the height of 6 R above the surface of earth, $R$ being radius of earth. The time period of another satellite at a height of 2.5 R from the surface of earth, is
A) 10 h
B) $\frac{6}{\sqrt{2}}h$
C) 6 h
D) $6\sqrt{2}\ h$
Explanation
(d) By Kepler’s law $T^{2} \propto R^{3}$
$Hence,\ \ \ \ {\ \left( \frac{T_{2}}{T_{1}} \right)}^{2} = {\ \left( \frac{R_{2}}{R_{1}} \right)}^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left( \frac{2.5R + R}{6R + R} \right)^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left( \frac{1}{2} \right)^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{2} = \frac{T_{1}}{(2)^{3/2}}$
For a geostationary satellite
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{1} = 24\ h$
$So,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{2} = \frac{24}{2\sqrt{2}} = 6\sqrt{2}\ h$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.