Question:
A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass 'm' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is -
A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass 'm' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is -
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'Just escapes' means total energy is zero, so \(KE = \frac{GMm}{r}\); the orbital speed gives \(\frac{GMm}{r}=mv^2\).
Updated On: Nov 5, 2026
- \(mv^2\)
- \(\frac{1}{2}mv^2\)
- \(\frac{3}{2}mv^2\)
- \(2mv^2\)
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The Correct Option is A
Solution and Explanation
For the circular orbit of radius r: \(\frac{mv^2}{r} = \frac{GMm}{r^2}\), so \(mv^2 = \frac{GMm}{r}\).
To just escape from distance r the total energy must be zero: \(KE - \frac{GMm}{r} = 0\), so \(KE = \frac{GMm}{r}\).
Therefore \(KE = mv^2\).
Answer: (1) \(mv^2\)
To just escape from distance r the total energy must be zero: \(KE - \frac{GMm}{r} = 0\), so \(KE = \frac{GMm}{r}\).
Therefore \(KE = mv^2\).
Answer: (1) \(mv^2\)
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