If a planet existed that had a period twice that of Earth, what would be the length of the semi-major axis as it orbits the Sun?
A. 4 AU
B. 1.59 AU
C. 2 AU
D. 2.83 AU
The period of an object in orbit can be related to the semi-major axis of its orbit by Kepler's third law:
(P₁/P₂)² = (a₁/a₂)³
Where P₁ is the period of the first object (Earth), P₂ is the period of the second object (unknown planet), a₁ is the semi-major axis of the first object's orbit (1 AU for Earth), and a₂ is the semi-major axis of the second object's orbit (unknown).
We can rearrange this equation to solve for a₂:
(a₂/a₁)³ = (P₁/P₂)²
(a₂/1 AU)³ = (1 year/P₂)²
a₂³ = (1 AU)² * P₁² / P₂²
a₂ = (1 AU)^(2/3) * (P₁/P₂)^(2/3)
Since we are looking for the semi-major axis of the planet that has a period twice that of Earth, P₁/P₂ = 2/1 = 2. Plugging this into the equation:
a₂ = (1 AU)^(2/3) * (2/1)^(2/3)
a₂ = (1 AU)^(2/3) * (8)^(1/3)
a₂ = (1 AU)^(2/3) * 2
a₂ = (1 AU)^(2/3) * (2 AU)
a₂ = 2^(2/3) AU
Calculating this:
a₂ ≈ 1.59 AU
Therefore, the length of the semi-major axis as the planet orbits the Sun would be approximately 1.59 AU.
The correct answer is B. 1.59 AU.