JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2019
- Medium
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(mv^2\)
Step-by-step Solution
Detailed explanation
\(Initially,\,kinetic\,energy = \frac{1}{2}m{v^2} = \frac{1}{2}m\frac{{G{M_e}}}{r}\)
By conservation of \(M.E\).,
\(\frac{{ - G{M_e}m}}{r} + KE = 0\)
\(KE = \frac{{G{M_e}m}}{r} = m{v^2}\)
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