Exercises
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?
Given: The planet's orbital speed is twice the Earth's, i.e., the planet completes one revolution in half the time of the Earth.
So T_planet = T_earth / 2.
By Kepler's Third Law: T² ∝ a³ (where a = semi-major axis).
(Tₚ / Tₑ)² = (aₚ / aₑ)³
(1/2)² = (aₚ / aₑ)³
1/4 = (aₚ / aₑ)³
aₚ / aₑ = (1/4)^(1/3) = (1/2)^(2/3)
aₚ / aₑ = (0.5)^(2/3) = 0.63
The orbital size of the planet is 0.63 times (about 63%) that of Earth's orbit.
Answer: aₚ / aₑ = (1/2)^(2/3) ≈ 0.63. The planet's orbital radius is about 0.63 times Earth's orbital radius.
