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Kepler's Law

A planet of m...

Question

A planet of mass $m$ moves in an elliptical orbit around an unknown star of mass $M$ such that its maximum and minimum distance from the star are equal to $r_{1}$ and $r_{2}$ respectively. The angular momentum of the planet relative to the centre of the star is:

A

m \sqrt{\frac{2 G M r_{1} r_{2}}{r_{1} + r_{2}}}

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B

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C

m \sqrt{\frac{2 G M (r_{1} + r_{2})}{r_{1} r_{2}}}

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D

\sqrt{\frac{2 G M m r_{1}}{(r_{1} + r_{2}) r_{2}}}

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Solution

The correct option is A $m \sqrt{\frac{2 G M r_{1} r_{2}}{r_{1} + r_{2}}}$
From law of conservation of angular momentum
$m v_{1} r_{1} = m v_{2} r_{2}$
$\Rightarrow v_{2} = \frac{v_{1} r_{1}}{r_{2}} \to (1)$
From law of conservation of total mechanical energy
$\frac{- G M m}{r_{1}} + \frac{1}{2} m v_{1}^{2} = \frac{- G M m}{r_{2}} + \frac{1}{2} m v_{2}^{2} \to (2)$
From equation (1) and (2)
$v_{1} = \sqrt{\frac{2 G m r_{2}}{(r_{1} + r_{2}) r_{1}}}$
Angular momentum
$L = m v_{1} r_{1} = m \sqrt{\frac{2 G M r_{1} r_{2}}{r_{1} + r_{2}}}$

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