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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Solution

Time taken by the Earth to complete one revolution around the Sun,
$T_{e} = 1$ year
Time taken by the planet to complete one revolution around the Sun, $T_{P} = \frac{1}{2} T_{e} = \frac{1}{2} y e a r$
Orbital radius of the planet = Rp
From Kepler’s third law of planetary motion, we can write:
${(\frac{R_{P}}{R_{e}})}^{3} = {(\frac{T_{P}}{T_{e}})}^{2}$

$(\frac{R_{P}}{R_{e}}) = {(\frac{T_{P}}{T_{e}})}^{2 / 3}$
$= {0.5}^{(2 / 3)} = 0.63 R_{e}$
Hence, the orbital radius of the planet will be 0.63 times of the Earth.

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