8.3 Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

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Vishal Baghel | Contributor-Level 10

4 months ago

Time taken by the Earth to complete one revolution around the Sun, $T_{e}$ = 1 year

Orbital radius of the Earth in its orbit, $R_{e}$ = 1 AU

Time taken by the planet to complete one revolution around the Sun, $T_{p}$ = $\frac{1}{2} T_{e}$ = $\frac{1}{2}$ year

Orbital radius of the planet = $R_{p}$

From Kepler's 3rd law of planetary motion, we can write:

( ${\frac{R_{p}}{R_{e}})}^{3}$ = ( ${\frac{T_{p}}{T_{e}})}^{2}$

$\frac{R_{p}}{R_{e}}$ = ( ${\frac{T_{p}}{T_{e}})}^{2 / 3}$ = ( ${\frac{1}{2})}^{2 / 3}$ = 0.63

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth

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