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Escape Velocity

A planet of m...

Question

A planet of mass $m$ revolves in elliptical orbit around the sun so that its maximum and minimum distances from the sun are equal to $r_{a}$ and $r_{p}$ respectively. Find the angular momentum of this planet relative to the sun.

L = m \sqrt{\frac{G M r_{p} r_{a}}{(r_{p} + r_{a})}}

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L = m \sqrt{\frac{2 G M r_{p} r_{a}}{(r_{p} + r_{a})}}

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L = M \sqrt{\frac{2 G M r_{p} r_{a}}{(r_{p} + r_{a})}}

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L = M \sqrt{\frac{(r_{p} + r_{a})}{G m r_{p} r_{a}}}

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Solution

The correct option is B $L = m \sqrt{\frac{2 G M r_{p} r_{a}}{(r_{p} + r_{a})}}$
For a planet moving around the sun in an orbit, angular momentum $(L)$ is constant.

$L = m v_{a} r_{a} = m v_{b} r_{b} \Rightarrow v_{a} = v_{b} \frac{r_{b}}{r_{a}}$

By conserving energy between the two points, farthest and nearest.

$\frac{- G M m}{r_{b}} + \frac{1}{2} m v_{b}^{2} = \frac{- G M m}{r_{a}} + \frac{1}{2} m v_{a}^{2}$

$\Rightarrow G M (\frac{1}{r_{a}} - \frac{1}{r_{b}}) = \frac{1}{2} (v_{a}^{2} - v_{b}^{2})$

Substituting $v_{a}$ from momentum equation.

$\Rightarrow G M (\frac{1}{r_{a}} - \frac{1}{r_{b}}) = \frac{1}{2} [v_{b}^{2} (\frac{r_{b}^{2}}{r_{a}^{2}}) - v_{b}^{2}]$
$\Rightarrow G M [\frac{r_{b} - r_{a}}{r_{a} r_{b}}] = \frac{v_{b}^{2}}{2} [\frac{(r_{b} - r_{a}) (r_{b} + r_{a})}{r_{a}^{2}}]$
$\Rightarrow v_{b}^{2} = \frac{2 G M r_{a}}{(r_{a} + r_{b}) r_{b}}$
$∴ L = m v_{b} r_{b} = m \sqrt{\frac{2 G M r_{a}}{(r_{a} + r_{b}) r_{b}}} r_{b} = m \sqrt{\frac{2 G M r_{a} r_{b}}{r_{a} + r_{b}}}$

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