JEE Main · 2013 · 07 Apr 2013 · Offline · Gravitation

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

(A) $\frac{2 GmM}{3 R}$
(B) $\frac{G m M}{2 R}$
(C) $\frac{GmM}{3 R}$
(D) $\frac{5 GmM}{6 R}$ ✓ Correct

Answer: (D)

Solution

The minimum energy is simply the difference between the satellite's total mechanical energy in its final orbit and its energy sitting on the surface. On the surface it has only potential energy,

E_{i} = - \frac{GM m}{R}

. In a circular orbit at altitude

2 R

the orbital radius is

3 R

, and the total energy of a circular orbit is half its potential energy,

E_{f} = - \frac{GM m}{2 ( 3 R )} = - \frac{GM m}{6 R}

. The energy that must be supplied is

E_{f} - E_{i} = - \frac{GM m}{6 R} + \frac{GM m}{R} = \frac{5 GM m}{6 R}

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option

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JEE Main · 2013 · 07 Apr 2013 · Offline · Gravitation

Solution

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