Question:
A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P', R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is _________. 'P' has the time period of 24 hours.
A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P', R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is _________. 'P' has the time period of 24 hours.
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Apply Kepler's third law \(T^2 \propto r^3\), with orbit radii 12R and 3R measured from the planet's centre.
Updated On: Nov 4, 2026
- 3
- 5
- \(6\sqrt{2}\)
- \(\frac{6}{\sqrt{2}}\)
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The Correct Option is A
Solution and Explanation
Orbit radius is measured from the centre of the planet: for the geostationary satellite \(r_1 = R + 11R = 12R\), with \(T_1 = 24\ \mathrm{h}\). For the other satellite \(r_2 = R + 2R = 3R\).
Kepler's third law: \(\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2} = \left(\frac{3R}{12R}\right)^{3/2} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}\).
\(T_2 = \frac{24}{8} = 3\ \mathrm{hours}\).
Answer: (1) 3
Kepler's third law: \(\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2} = \left(\frac{3R}{12R}\right)^{3/2} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}\).
\(T_2 = \frac{24}{8} = 3\ \mathrm{hours}\).
Answer: (1) 3
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