JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Gravitation

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

(A) $\frac{1}{2} m V^{2}$
(B) $m V^{2}$ ✓ Correct
(C) $\frac{3}{2} m V^{2}$
(D) $2 m V^{2}$

Answer: (B)

Solution

For a satellite in circular orbit, the centripetal force equals gravitational force:

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

, giving

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

. For the object to just escape, its total energy must be zero. The gravitational potential energy at radius r is

- \frac{GM m}{r}

. Therefore, the required kinetic energy is

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

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Single correct option

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