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Gravitational Force

A planet of m...

Question

A planet of mass $m$ is moving in an elliptical orbit round the sun of mass $M$ . If the maximum and minimum distances of the planet from the sun be $l_{1}$ and $l_{2}$ , the angular momentum of the planet about the sun will be

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

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

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

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Solution

The correct option is C $m \sqrt{\frac{2 G M l_{1} l_{2}}{(l_{1} + l_{2})}}$
For a planet moving around the sun in an orbit, angular momentum( $L$ ) is constant.
$L = m v_{1} l_{1} = m v_{2} l_{2} \Rightarrow v_{1} = v_{2} \frac{l_{2}}{l_{1}}$
By conserving energy between the two points, farthest and nearest.
$\frac{- G M m}{l_{2}} + \frac{1}{2} m v_{2}^{2} = \frac{- G M m}{l_{1}} + \frac{1}{2} m v_{1}^{2}$
$\Rightarrow G M (\frac{1}{l_{1}} - \frac{1}{l_{2}}) = \frac{1}{2} (v_{1}^{2} - v_{2}^{2})$
Substituting $v_{1}$ from momentum equation.
$\Rightarrow G M (\frac{1}{l_{1}} - \frac{1}{l_{2}}) = \frac{1}{2} (v_{2}^{2} (\frac{l_{2}^{2}}{l_{1}^{2}}) - v_{2}^{2})$
$\Rightarrow G M (\frac{l_{2} - l_{1}}{l_{1} l_{2}}) = \frac{v_{2}^{2}}{2} (\frac{(l_{2} - l_{1}) (l_{2} + l_{1})}{l_{1}^{2}})$
$\Rightarrow v_{2}^{2} = \frac{2 G M l_{1}}{(l_{1} + l_{2}) l_{2}}$
$∴ L = m v_{2} l_{2} = m \sqrt{\frac{2 G M l_{1}}{(l_{1} + l_{2}) l_{2}}} l_{2} = m \sqrt{\frac{2 G M l_{1} l_{2}}{l_{1} + l_{2}}}$

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