Question:
An earth satellite \(S\) has orbit radius which is 4 times that of communication satellite \(C\). The period of revolution of \(S\) will be
An earth satellite \(S\) has orbit radius which is 4 times that of communication satellite \(C\). The period of revolution of \(S\) will be
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Geostationary satellite has period exactly 1 day.
Updated On: Apr 23, 2026
- \(32\ \text{days}\)
- \(18\ \text{days}\)
- \(8\ \text{days}\)
- \(9\ \text{days}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Kepler's third law: \(T^2 \propto R^3\). Communication satellite is geostationary with \(T_C = 1\ \text{day}\).
Step 2: Detailed Explanation:
Given \(R_S = 4R_C\). \(\frac{T_S^2}{T_C^2} = \left(\frac{R_S}{R_C}\right)^3 = (4)^3 = 64\).
\(\frac{T_S}{T_C} = \sqrt{64} = 8 \Rightarrow T_S = 8 \times T_C = 8\ \text{days}\).
Step 3: Final Answer:
Thus, period of S = \(8\ \text{days}\).
Kepler's third law: \(T^2 \propto R^3\). Communication satellite is geostationary with \(T_C = 1\ \text{day}\).
Step 2: Detailed Explanation:
Given \(R_S = 4R_C\). \(\frac{T_S^2}{T_C^2} = \left(\frac{R_S}{R_C}\right)^3 = (4)^3 = 64\).
\(\frac{T_S}{T_C} = \sqrt{64} = 8 \Rightarrow T_S = 8 \times T_C = 8\ \text{days}\).
Step 3: Final Answer:
Thus, period of S = \(8\ \text{days}\).
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