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, Te = 1 year
Orbital radius of the Earth in its orbit, Re = 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 = Rp

From Kepler’s third law of planetary motion, we can write:
$(\frac{R_p}{R_e})^3 = (\frac{T_p }{ T_e})^2$
$\frac{R_p}{R_e} = (\frac{T_p}{T_e})^{\frac{2}{3}}$
$=(\frac{\frac{1}{2}}{1} )^{\frac{2}{3}}= (0.5)^{\frac{2}{3} }= 0.63$

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