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Gravitation question

2018 · 15 Apr · Shift 1 · Q67
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Gravitation question

2018 · 15 Apr · Shift 1 · Q67

JEE MainPhysicsGravitationMCQ+4 / −1
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 of radius R2,{R \over 2},2R​, and the other mass, in a circular orbit of radius 3R2{3R \over 2}23R​. The difference between the final and initial total energies is :
  1. A
    −GMm2R- {{GMm} \over {2R}}−2RGMm​
  2. B
    +GMm6R+ {{GMm} \over {6R}}+6RGMm​
  3. C
    GMm2R{{GMm} \over {2R}}2RGMm​
  4. D
    −GMm6R- {{GMm} \over {6R}}−6RGMm​
View written solutionFree

Correct answer: D

  1. Initial total energy of the body

For a satellite of mass mmm in a circular orbit of radius RRR around a planet of mass MMM,

E=−GMm2RE=-\frac{GMm}{2R}E=−2RGMm​

So initially,

Ei=−GMm2RE_i=-\frac{GMm}{2R}Ei​=−2RGMm​

  1. After splitting: two equal masses

The body splits into two equal parts, so each has mass

m2\frac{m}{2}2m​

For a mass m′m'm′ in a circular orbit of radius rrr, total energy is

E=−GMm′2rE=-\frac{GMm'}{2r}E=−2rGMm′​

  1. Energy of first fragment

First fragment has mass m2\frac{m}{2}2m​ and orbit radius R2\frac{R}{2}2R​.

Thus,

E1=−GM(m2)2(R2)E_1=-\frac{GM\left(\frac{m}{2}\right)}{2\left(\frac{R}{2}\right)}E1​=−2(2R​)GM(2m​)​

Simplify:

E1=−GM(m/2)R=−GMm2RE_1=-\frac{GM(m/2)}{R}=-\frac{GMm}{2R}E1​=−RGM(m/2)​=−2RGMm​

  1. Energy of second fragment

Second fragment has mass m2\frac{m}{2}2m​ and orbit radius 3R2\frac{3R}{2}23R​.

Thus,

E2=−GM(m2)2(3R2)E_2=-\frac{GM\left(\frac{m}{2}\right)}{2\left(\frac{3R}{2}\right)}E2​=−2(23R​)GM(2m​)​

Since

2⋅3R2=3R2\cdot \frac{3R}{2}=3R2⋅23R​=3R

we get

E2=−GM(m/2)3R=−GMm6RE_2=-\frac{GM(m/2)}{3R}=-\frac{GMm}{6R}E2​=−3RGM(m/2)​=−6RGMm​

  1. Final total energy

Ef=E1+E2=−GMm2R−GMm6RE_f=E_1+E_2=-\frac{GMm}{2R}-\frac{GMm}{6R}Ef​=E1​+E2​=−2RGMm​−6RGMm​

Taking LCM 6R6R6R,

Ef=−3GMm6R−GMm6R=−4GMm6R=−2GMm3RE_f=-\frac{3GMm}{6R}-\frac{GMm}{6R}=-\frac{4GMm}{6R}=-\frac{2GMm}{3R}Ef​=−6R3GMm​−6RGMm​=−6R4GMm​=−3R2GMm​

  1. Difference between final and initial energies

The question asks for

Ef−EiE_f-E_iEf​−Ei​

So,

Ef−Ei=−2GMm3R−(−GMm2R)E_f-E_i=-\frac{2GMm}{3R}-\left(-\frac{GMm}{2R}\right)Ef​−Ei​=−3R2GMm​−(−2RGMm​)

=−2GMm3R+GMm2R= -\frac{2GMm}{3R}+\frac{GMm}{2R}=−3R2GMm​+2RGMm​

Taking LCM 6R6R6R,

Ef−Ei=−4GMm6R+3GMm6R=−GMm6RE_f-E_i=-\frac{4GMm}{6R}+\frac{3GMm}{6R}=-\frac{GMm}{6R}Ef​−Ei​=−6R4GMm​+6R3GMm​=−6RGMm​

  1. Option check

Thus the required difference is

−GMm6R-\frac{GMm}{6R}−6RGMm​

So the correct option is D.

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