JEE Mains · Physics · STD 11 - 7. gravitation
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 \(2\,mv^2\)
- B \(mv^2\)
- C \(\frac{1}{2}\,m{v^2}\)
- D \(\frac{3}{2}\,m{v^2}\)
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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