Gravitation

JEE Main level, single correct, medium

154 of 223

Question 154 of 223

What is the minimum energy required to launch a satellite of mass m\text{m} from the surface of a planet of mass M\text{M} and radius R\text{R} in a circular orbit at an altitude of 2R\text{2R}?

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Answer: (C) 5GmM6R\dfrac{5\text{GmM}}{6\text{R}}

As we know,
Gravitational potential cnergy =−GMmr=\dfrac{-GMm}{r}
and orbital velocity, v0=GM/R+h{v}_{0}=\sqrt{GM/R+h}
Ef=12mv02−GMm3R=12mGM3R−GMm3R{E}_{f}=\dfrac{1}{2}{mv}_{0}^{2}-\dfrac{GMm}{3R}=\dfrac{1}{2}m\dfrac{GM}{3R}-\dfrac{GMm}{3R}
=GMm3R(12−1)=−GMm6R=\dfrac{GMm}{3R}\left(\dfrac{1}{2}-1\right)=\dfrac{-GMm}{6R}
Ei=−GMmR+K{E}_{i}=\dfrac{-GMm}{R}+K
Ei=Ef{E}_{i}={E}_{f}
Therefore minimum required energy,
K=5GMm6RK=\dfrac{5GMm}{6R}

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