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(C) Time taken by the Earth to complete one revolution around the Sun,$T_{e} = 1 	ext{ year}$.Orbital radius of the Earth,$R_{e} = 1 	ext{ AU}$.Give

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Lesser by a factor of $0.63$

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(C) Time taken by the Earth to complete one revolution around the Sun,$T_{e} = 1 	ext{ year}$.Orbital radius of the Earth,$R_{e} = 1 	ext{ AU}$.Given that the planet goes around the Sun twice as fast as the Earth,its orbital period $T_{p}$ is half that of the Earth:$T_{p} = \frac{1}{2} T_{e} = 0.5 	ext{ year}$.According to Kepler's third law of planetary motion,the square of the orbital period is proportional to the cube of the semi-major axis (orbital radius):$\left(\frac{R_{p}}{R_{e}}ight)^{3} = \left(\frac{T_{p}}{T_{e}}ight)^{2}$.Substituting the values:$\frac{R_{p}}{R_{e}} = \left(\frac{T_{p}}{T_{e}}ight)^{2/3} = \left(\frac{0.5}{1}ight)^{2/3} = (0.5)^{0.666} \approx 0.63$.Thus,the orbital size of the planet is lesser by a factor of $0.63$ compared to that of the Earth.

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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