A satellite is moving with a constant speed ‘V’ in a circular orbit about 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 its ejection, the kinetic energy of the object is

Gravitation PYQ — JEE Advanced 2011 · Shift 2 · Q45 with Solution

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Correct answer: B

For a satellite moving in a circular orbit of radius $r$ about the earth,

\frac{GMm}{r^2} = \frac{mV^2}{r}

So,

V^2 = \frac{GM}{r}

Thus the orbital speed is

V = \sqrt{\frac{GM}{r}}

If the object is ejected so that it just escapes, then its total mechanical energy at the instant of ejection must be zero.

That is,

K + U = 0

where

U = -\frac{GMm}{r}

Hence,

K = \frac{GMm}{r}

Using $V^2 = \dfrac{GM}{r}$ ,

K = mV^2

Therefore, the kinetic energy of the object at the time of ejection is

mV^2

So the correct option is:

B. $mV^2$

Stored correct answer: B

Derived answer: B

They match.