JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 22)
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 $${R \over 2},$$ and the other mass, in a circular orbit of radius $${3R \over 2}$$. The difference between the final and initial total energies is :
$$ - {{GMm} \over {2R}}$$
$$ + {{GMm} \over {6R}}$$
$${{GMm} \over {2R}}$$
$$ - {{GMm} \over {6R}}$$
Explanation
Initially gravitational potenrial energy
Ei = $$-$$ $${{GMm} \over {2R}}$$
Final gravitational potential energy
Ef = $$-$$ $${{GM\left( {{m \over 2}} \right)} \over {2\left( {{R \over 2}} \right)}}$$ $$-$$ $${{GM\left( {{m \over 2}} \right)} \over {2\left( {{{3R} \over 2}} \right)}}$$
= $$-$$ $${{GMm} \over {2R}} - {{GMm} \over {6R}}$$
= $$-$$ $${{4GMm} \over {6R}}$$
= $$-$$ $${{2GMm} \over {3R}}$$
$$\therefore\,\,\,\,$$ Required difference in energies
= $${E_f} - {E_i}$$
= $$-$$ $${{GMm} \over R}\left( {{2 \over 3} - {1 \over 2}} \right)$$
= $$-$$ $${{GMm} \over {6R}}$$
Ei = $$-$$ $${{GMm} \over {2R}}$$
Final gravitational potential energy
Ef = $$-$$ $${{GM\left( {{m \over 2}} \right)} \over {2\left( {{R \over 2}} \right)}}$$ $$-$$ $${{GM\left( {{m \over 2}} \right)} \over {2\left( {{{3R} \over 2}} \right)}}$$
= $$-$$ $${{GMm} \over {2R}} - {{GMm} \over {6R}}$$
= $$-$$ $${{4GMm} \over {6R}}$$
= $$-$$ $${{2GMm} \over {3R}}$$
$$\therefore\,\,\,\,$$ Required difference in energies
= $${E_f} - {E_i}$$
= $$-$$ $${{GMm} \over R}\left( {{2 \over 3} - {1 \over 2}} \right)$$
= $$-$$ $${{GMm} \over {6R}}$$
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