Question

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?

Answer 1

Lesser by a factor of 0.63
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: 
\((\frac{R_p}{R_e})^3 = (\frac{T_p }{ T_e})^2\)
\(\frac{R_p}{R_e} = (\frac{T_p}{T_e})^{\frac{2}{3}}\)
\(=(\frac{\frac{1}{2}}{1} )^{\frac{2}{3}}= (0.5)^{\frac{2}{3} }= 0.63\)

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