Gravitation

JEE Main level, single correct, medium

194 of 223

Question 194 of 223

A satellite is moving with a constant speed vv 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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Answer: (B) mv2m{v}^{2}

At height rr from the centre of the Earth, orbital velocity is given as:
orbitalvelocity=GMrorbitalvelocity=\sqrt{\dfrac{GM}{r}}
Ve=2v{V}_{e}=\sqrt{2}v

K.E=12m(2v)2=mv2K.E=\dfrac{1}{2}m{\left(\sqrt{2}v\right)}^{2}=m{v}^{2}

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