Question:
Radius of orbit of satellite of earth is \(R\). Its kinetic energy is proportional to
Radius of orbit of satellite of earth is \(R\). Its kinetic energy is proportional to
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Total energy of satellite = \(-KE\) for circular orbits.
Updated On: Apr 23, 2026
- \(\frac{1}{R}\)
- \(\frac{1}{\sqrt{R}}\)
- \(R\)
- \(\frac{1}{R^{3/2}}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
For a satellite in circular orbit, gravitational force provides centripetal force.
Step 2: Detailed Explanation:
\(\frac{GMm}{R^2} = \frac{mv^2}{R} \Rightarrow v^2 = \frac{GM}{R}\).
KE = \(\frac{1}{2}mv^2 = \frac{1}{2}m \cdot \frac{GM}{R} \propto \frac{1}{R}\).
Step 3: Final Answer:
Thus, KE \(\propto \frac{1}{R}\).
For a satellite in circular orbit, gravitational force provides centripetal force.
Step 2: Detailed Explanation:
\(\frac{GMm}{R^2} = \frac{mv^2}{R} \Rightarrow v^2 = \frac{GM}{R}\).
KE = \(\frac{1}{2}mv^2 = \frac{1}{2}m \cdot \frac{GM}{R} \propto \frac{1}{R}\).
Step 3: Final Answer:
Thus, KE \(\propto \frac{1}{R}\).
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