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Question: A body of mass \(m\) is moving in a circular orbit of radius \(R\) about a planet of mass \(M\). At ...

A body of mass mm is moving in a circular orbit of radius RR about a planet of mass MM. At some instant, it splits into two equal masses. The first mass moves into a circular orbit of radius R2\dfrac{R}{2}, and the other mass, in a circular orbit of radius 3R2\dfrac{{3R}}{2}. The difference between the final and initial total energies is:
A. −GMm2R - \dfrac{{GMm}}{{2R}}
B. +GMm6R + \dfrac{{GMm}}{{6R}}
C. −GMm6R - \dfrac{{GMm}}{{6R}}
D. +GMm2R + \dfrac{{GMm}}{{2R}}

Explanation

Solution

We know that energy in this case is dependent on the mass of both the rotating body and the planet as well as on the radius of a circular orbit. In final condition, we are told that the rotating body splits into two equal masses. Therefore the final mass of each rotating body will be half of its original initial mass. We are also given the final radius of the circular orbit of both the masses. So we will find final and initial total energies first and then determine their difference.

Formula used:
E=−Gm1m22rE = - \dfrac{{G{m_1}{m_2}}}{{2r}},where, EE is energy, GG is the gravitational constant, m1{m_1} is the mass of rotating body, m2{m_2} is the mass of the body about which the other body is rotating and rr is the radius of the circular orbit of the rotating mass.

Complete step by step solution:
Here, in the initial condition is given that a body of mass mmis moving in a circular orbit of radius RR about a planet of mass MM.Therefore initial energy is given by,
Ei=−GmM2R{E_i} = - \dfrac{{GmM}}{{2R}}
Now at some instant, the rotating body splits into two equal masses. Therefore, the mass of each rotating body will be m2\dfrac{m}{2}. It is also given that The first mass moves into a circular orbit of radius R2\dfrac{R}{2}, and the other mass, in a circular orbit of radius 3R2\dfrac{{3R}}{2}.
We will now find the individual final energies for both rotating bodies.
E1=−G(m2)M2(R2)=−GmM2R{E_1} = - \dfrac{{G\left( {\dfrac{m}{2}} \right)M}}{{2\left( {\dfrac{R}{2}} \right)}} = - \dfrac{{GmM}}{{2R}}
E2=−G(m2)M2(3R2)=−GmM6R{E_2} = - \dfrac{{G\left( {\dfrac{m}{2}} \right)M}}{{2\left( {\dfrac{{3R}}{2}} \right)}} = - \dfrac{{GmM}}{{6R}}
The total final energy,
Ef=E1+E2=−GmM2R−GmM6R{E_f} = {E_1} + {E_2} = - \dfrac{{GmM}}{{2R}} - \dfrac{{GmM}}{{6R}}
Now, the difference between final and initial energy
Ef−Ei=−GmM2R−GmM6R−(−GmM2R) ⇒Ef−Ei=−GmM2R−GmM6R+GmM2R ∴Ef−Ei=−GmM6R {E_f} - {E_i} = - \dfrac{{GmM}}{{2R}} - \dfrac{{GmM}}{{6R}} - \left( { - \dfrac{{GmM}}{{2R}}} \right) \\\ \Rightarrow{E_f} - {E_i}= - \dfrac{{GmM}}{{2R}} - \dfrac{{GmM}}{{6R}} + \dfrac{{GmM}}{{2R}} \\\ \therefore {E_f} - {E_i}= - \dfrac{{GmM}}{{6R}}

Hence, option C is the right answer.

Note: It is important to know that an isolated system of particles will have the total potential energy that equals the sum of energies for all possible pairs of its constituent particles. This is an example of the application of the superposition principle.