Suppose a comet orbits the Sun every 68 years and has an eccentricity of 0.95. What is the length of its semi-major axis in AU?

Answer 1

So, we have the period T = 68 years, the eccentricity e = 0.95, and want to find the semi-major axis, a.

One thing to note from Kepler's Third Law, is that the period depends only on the semi-major axis, so we don't need the eccentricity to answer this. The mathematical form of Kepler's Third Law is T²=a³. Considering we have the period, we just need to take the cube-root of both sides of the equation, giving T^(2/3)=a. Remember this is the period in years and the semi-major axis in AU, so we can use our 68 years to get a semi-major axis of 16.66 AU or 17 AU if we round.