Verified Solution • 2025–26 Session moderate

QJEEM18Ph4657: JEE Main 2018 Physics - Class 12 JEEM18Ph4657 Solution: A body of mass m is moving in a circular orbit of radius R…

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}, and the other mass, in a circular orbit of radius 3R2{3R \over 2} . The difference between the final and initial total energies is :

Multiple Choice Options

  • A −GMm2R- {{GMm} \over {2R}}
  • B +GMm6R+ {{GMm} \over {6R}}
  • C GMm2R{{GMm} \over {2R}}
  • D −GMm6R- {{GMm} \over {6R}}

Step-by-Step Verified Solution

Initially gravitational potenrial energy

Ei = −- GMm2R{{GMm} \over {2R}}

Final gravitational potential energy

Ef = −- GM(m2)2(R2){{GM\left( {{m \over 2}} \right)} \over {2\left( {{R \over 2}} \right)}} −- GM(m2)2(3R2){{GM\left( {{m \over 2}} \right)} \over {2\left( {{{3R} \over 2}} \right)}}

= −- GMm2R−GMm6R{{GMm} \over {2R}} - {{GMm} \over {6R}}

= −- 4GMm6R{{4GMm} \over {6R}}

= −- 2GMm3R{{2GMm} \over {3R}}

∴    \therefore\,\,\,\, Required difference in energies

= Ef−Ei{E_f} - {E_i}

= −- GMmR(23−12){{GMm} \over R}\left( {{2 \over 3} - {1 \over 2}} \right)

= −- GMm6R{{GMm} \over {6R}}

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