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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 some instant...

A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit 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. +GMm2R + \dfrac{{GMm}}{{2R}}
D. −GMm6R - \dfrac{{GMm}}{{6R}}

Explanation

Solution

Recall the formula for the total energy of the orbiting body and express its total energy before it splits. The total energy after it splits into two bodies is the sum of total energy of each body. Subtract the total energy of single initial body from the total energy of system of two bodies to get the required answer.

Formula used:
Kinetic energy, K=GMm2RK = \dfrac{{GMm}}{{2R}}
where, G is the gravitational constant and R is the radius of the circular orbit of the body of mass m.
Potential energy, U=−GMmRU = - \dfrac{{GMm}}{R}

Complete step by step solution:
We know that the total energy of the body around the planet is the sum of its kinetic energy and potential energy. We have the expression for the kinetic energy of the body of mass m around the planet of mass M,
K=GMm2RK = \dfrac{{GMm}}{{2R}} …… (1)
Here, G is the gravitational constant and R is the radius of the circular orbit of the body of mass m.
We also have the expression for the potential energy of the body around the planet of mass M,
U=−GMmRU = - \dfrac{{GMm}}{R} …… (2)
Adding equation (1) and (2) to get the total energy of the body, we get,
E=GMm2R−GMmRE = \dfrac{{GMm}}{{2R}} - \dfrac{{GMm}}{R}
⇒E=−GMm2R\Rightarrow E = - \dfrac{{GMm}}{{2R}} …… (3)
This is the total energy of the body of mass m revolving around the planet of M in the circular orbit of radius R.
Now, we have given that the body is split into two equal masses and two masses are revolving in two different orbits of radius R2\dfrac{R}{2} and 3R2\dfrac{{3R}}{2}respectively. Now, the total energy of these two masses is the sum of the total energy of each mass. Therefore, we can write,
E′=−GM(m/2)2(R/2)−GM(m/2)2(3R/2)E' = - \dfrac{{GM\left( {m/2} \right)}}{{2\left( {R/2} \right)}} - \dfrac{{GM\left( {m/2} \right)}}{{2\left( {3R/2} \right)}}
⇒E′=−GMm2R−GMm6R\Rightarrow E' = - \dfrac{{GMm}}{{2R}} - \dfrac{{GMm}}{{6R}}
⇒E′=−GMmR(12+16)\Rightarrow E' = - \dfrac{{GMm}}{R}\left( {\dfrac{1}{2} + \dfrac{1}{6}} \right)
⇒E′=−2GMm3R\Rightarrow E' = - \dfrac{{2GMm}}{{3R}} …… (4)
Subtracting equation (3) from equation (4) to get the difference in the final and initial total energy, we get,
ΔE=E′−E=−2GMm3R+GMm2R\Delta E = E' - E = - \dfrac{{2GMm}}{{3R}} + \dfrac{{GMm}}{{2R}}
⇒ΔE=GMmR(−23+12)\Rightarrow \Delta E = \dfrac{{GMm}}{R}\left( { - \dfrac{2}{3} + \dfrac{1}{2}} \right)
∴ΔE=−GMm6R\therefore \Delta E = - \dfrac{{GMm}}{{6R}}

So, the correct answer is option D.

Note: The potential energy always bears a negative sign and therefore, the total energy is also negative. You can determine the velocity of the two bodies using the law of conservation of momentum. The two bodies have two different radii of their circular orbits is the consequence of the law of conservation of momentum. The second body which has a larger radius will have lower orbital speed to conserve its angular momentum.