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Newton's Second Law: Rotatory Version

A planet of m...

Question

A planet of mass $m$ moves around the sun of mass $M$ in an elliptical orbit. The maximum and minimum distances of the planet from the sun are $r_{1}$ and $r_{2}$ respectively. The time period of revolution of planet is proportional to

A

r_{1}^{3 / 2}

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B

r_{2}^{3 / 2}

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C

(r_{1} + r_{2})^{3 / 2}

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D

(r_{1} - r_{2})^{3 / 2}

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Solution

The correct option is C $(r_{1} + r_{2})^{3 / 2}$
From Kepler's law of time period,

The square of the time period of revolution of a planet is proportional to the cube of the semi- major axis of the ellipse traced by the planet.

$∴ T^{2} \propto a^{3}$

From the geometry,

$r_{1} = a + a e$ and $r_{2} = a - a e$

Position of sun is at $s (- a e, 0)$

$∴ r_{1} + r_{2} = 2 a$

Therefore, $T^{2} \propto [\frac{(r_{1} + r_{2})}{2}]^{3}$

$\Rightarrow T \propto {(r_{1} + r_{2})}^{3 / 2}$

Hence, option $(c)$ is correct.

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Newton's Second Law: Rotatory Version

Standard XII Physics

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