There’s no elementary formula for the circumference of an ellipse, but there is an elementary approximation that is extremely accurate.

An ellipse has equation (*x*/*a*)² + (*y*/*b*)² = 1. If *a* = *b*, the ellipse reduces to a circle and the circumference is simply 2π*a*. But the general formula for circumference requires the hypergeometric function _{2}*F*_{1}:

where λ = (*a* – *b*)/(*a* + *b*).

However, if the ellipse is anywhere near circular, the following approximation due to Ramanujan is extremely good:

To quantify what we mean by extremely good, the error is O(λ^{10}). When an ellipse is roughly circular, λ is fairly small, and the error is on the order of λ to the 10th power.

To illustrate the accuracy of the approximation, I tried the formula out on some planets. The error increases with ellipticity, so I took the most most elliptical orbit of a planet or object formerly known as a planet. That distinction belongs to Pluto, in which case λ = 0.016. If Pluto’s orbit were exactly elliptical, you could use Ramanujan’s approximation to find the circumference of its orbit with an error less than one micrometer.

Next I tried it on something with a much more elliptical orbit: Halley’s comet. Its orbit is nearly four times longer than it is wide. For Halley’s comet, λ = 0.59 and Ramanujan’s approximation agrees with the exact result to seven significant figures. The exact result is 11,464,319,022 km and the approximation is 11,464,316,437 km.

Here’s a video showing how elliptical the comet’s orbit is.

If you’d like to experiment with the approximation, here’s some Python code:

from scipy import pi, sqrt from scipy.special import hyp2f1 def exact(a, b): t = ((a-b)/(a+b))**2 return pi*(a+b)*hyp2f1(-0.5, -0.5, 1, t) def approx(a, b): t = ((a-b)/(a+b))**2 return pi*(a+b)*(1 + 3*t/(10 + sqrt(4 - 3*t))) # Semimajor and semiminor axes for Halley's comet orbit a = 2.667950e9 # km b = 6.782819e8 # km print exact(a, b) print approx(a, b)