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Variation of G Due to the Rotation of Earth

A planet of m...

Question

A planet of mass $m$ moves along an ellipse so that perihelion and aphelion distances are $r_{1} and r_{2}$ respectively. Find the angular momentum of the planet. Assume $M_{s}$ to be the mass of the sun.

A

m \sqrt{\frac{G M_{s} r_{1} r_{2}}{r_{1} + r_{2}}}

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B

m \sqrt{\frac{2 G M_{s} r_{1} r_{2}}{r_{1} + r_{2}}}

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C

\sqrt{G M_{s} (r_{1} + r_{2}})

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D

m \sqrt{2 G M_{s} (r_{1} + r_{2})}

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Solution

The correct option is B $m \sqrt{\frac{2 G M_{s} r_{1} r_{2}}{r_{1} + r_{2}}}$
Since there is no external torque, we can apply conservation of angular momentum at perihelion and aphelion positions.
$m v_{1} r_{1} = m v_{2} r_{2}$
$\Rightarrow v_{1}^{2} = v_{2}^{2} {(\frac{r_{2}}{r_{1}})}^{2}$

From conservation of energy at the two points,
$\frac{- G m M_{s}}{r_{1}} + \frac{m v_{1}^{2}}{2} = \frac{- G m M_{s}}{r_{2}} + \frac{m v_{2}^{2}}{2}$

$\Rightarrow \frac{- G M_{s}}{r_{1}} + \frac{G M_{s}}{r_{2}} = \frac{v_{2}^{2}}{2} - \frac{v_{2}^{2}}{2} {(\frac{r_{2}}{r_{1}})}^{2}$

$\Rightarrow G M_{s} (\frac{r_{1} - r_{2}}{r_{1} r_{2}}) = \frac{v_{2}^{2}}{2} (\frac{r_{1}^{2} - r_{2}^{2}}{r_{1}^{2}})$

$\Rightarrow G M_{s} (\frac{r_{1} - r_{2}}{r_{1} r_{2}}) = \frac{v_{2}^{2}}{2} (\frac{(r_{1} - r_{2}) (r_{1} + r_{2})}{r_{1}^{2}})$

$\Rightarrow v_{2} = \sqrt{\frac{2 G M_{s} r_{1}}{r_{2} (r_{1} + r_{2})}}$

Therefore, angular momentum $= m v_{2} r_{2}$
$= m \sqrt{\frac{2 G M_{s} r_{1}}{r_{2} (r_{1} + r_{2})}} \times r_{2}$
$= m \sqrt{\frac{2 G M_{s} r_{1} r_{2}}{r_{1} + r_{2}}}$

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