Question:
A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:
A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:
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Orbital speed at radius R is \(\sqrt{GM/R}\) and escape speed is \(\sqrt{2GM/R}\).
Updated On: Nov 5, 2026
- \(2\)
- \(1\)
- \(\sqrt{2}\)
- \(\frac{1}{\sqrt{2}}\)
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The Correct Option is D
Solution and Explanation
Gravity provides the centripetal force: \(\frac{mv_o^2}{R} = \frac{GMm}{R^2} \Rightarrow v_o = \sqrt{\frac{GM}{R}}\).
Escape velocity: \(v_e = \sqrt{\frac{2GM}{R}}\).
Ratio \(\frac{v_o}{v_e} = \frac{1}{\sqrt{2}}\).
Answer: (4) \(\frac{1}{\sqrt{2}}\)
Escape velocity: \(v_e = \sqrt{\frac{2GM}{R}}\).
Ratio \(\frac{v_o}{v_e} = \frac{1}{\sqrt{2}}\).
Answer: (4) \(\frac{1}{\sqrt{2}}\)
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