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Question: A satellite of mass *m* is revolving in a circular orbit of radius 2*R*, around the earth of mass *M...

A satellite of mass m is revolving in a circular orbit of radius 2R, around the earth of mass M. The work done to shift the satellite to an orbit of radius 3R is

A

GMm2R\frac{GMm}{2R}

B

GMm3R\frac{GMm}{3R}

C

GMm9R\frac{GMm}{9R}

D

GMm12R\frac{GMm}{12R}

Answer

GMm12R\frac{GMm}{12R}

Explanation

Solution

The total mechanical energy of a satellite of mass mm in a circular orbit of radius rr around a planet of mass MM is given by E=−GMm2rE = -\frac{GMm}{2r}. The work done to shift the satellite from an initial orbit of radius r1r_1 to a final orbit of radius r2r_2 is the difference in their total energies: W=E2−E1W = E_2 - E_1.

Given: Initial orbit radius, r1=2Rr_1 = 2R Final orbit radius, r2=3Rr_2 = 3R

Initial energy, E1=−GMm2r1=−GMm2(2R)=−GMm4RE_1 = -\frac{GMm}{2r_1} = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R} Final energy, E2=−GMm2r2=−GMm2(3R)=−GMm6RE_2 = -\frac{GMm}{2r_2} = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}

Work done, W=E2−E1=(−GMm6R)−(−GMm4R)W = E_2 - E_1 = \left(-\frac{GMm}{6R}\right) - \left(-\frac{GMm}{4R}\right) W=−GMm6R+GMm4RW = -\frac{GMm}{6R} + \frac{GMm}{4R} To combine these terms, we find a common denominator, which is 12R12R: W=−2GMm12R+3GMm12RW = -\frac{2GMm}{12R} + \frac{3GMm}{12R} W=GMm12RW = \frac{GMm}{12R}