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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 ?
more by a factor of $2.67$
more by a factor of $1.24$
Lesser by a factor of $0.63$
Lesser by a factor of $0.33$
Solution
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 third law of planetary motion, we can write:
$\left(\frac{R_{p}}{R_{e}}\right)^{3}=\left(\frac{T_{r}}{T_{e}}\right)^{2}$
$\frac{R_{p}}{R_{e}}=\left(\frac{T_{p}}{T_{e}}\right)^{\frac{2}{3}}$
$=\left(\frac{\frac{1}{2}}{1}\right)^{\frac{2}{3}}=(0.5)^{\frac{2}{3}}=0.63$
