A further refinement to an approximate solution to Kepler's Equation

In my previous post I described the importance of Kepler's Equation in orbital dynamics and came up with an approximate solution which covers the cases where the eccentricity of the orbit is less than or approximately equal to 0.2. I wondered if we could do slightly better than this and come up with a solution for larger values of the eccentricity.

If we go back to equation (4) of that last post and consider a second order approximation to sin Δ and cos Δ of the form sin Δ ≈ Δ and cos Δ ≈ 1 - Δ²/2 then equation (4) becomes

Δ ≈  e (sin M (1 - Δ²/2) + Δ cos M )

Rearranging this we obtain a quadratic in Δ

½ (e sin M) Δ² + (1 - e cos M) Δ - e sin M = 0                . . . (6)

(I have assumed equality for now). If we write this in the form of

a Δ² + b Δ + c = 0

with a = ½ (e sin M), b = (1 - e cos M) and c =  - e sin M then the two solutions are 

Δ = (- b ± √(b² - 4ac))/2a

that is

Δ = (- (1 - e cos M) ± √((1 - e cos M)² + 2e² sin² M))/e sin M                . . . (7)

Note that (1 - e cos M)² + 2e² sin² M is the sum of two squares and so is always greater than or equal to zero. This means that the solutions in (7) are real. Unfortunately, the solutions are not defined when M = 0, ±π, ±2π as then sin M = 0 and they may become unstable as M tends to any of these values.

A bit of further analysis shows that only one of these two solutions is viable and that is

Δ ≈ (- (1 - e cos M) + √((1 - e cos M)² + 2e² sin² M))/e sin M                . . . (8)

I have compared this solution to the true values of Δ over all values of M and for e = 0.2 the maximum deviation is less than 0.1%. For e = 0.4 the deviation is less than 0.8% and e = 0.6 the deviation is less than 3.1%. This shows that this approximation can probably be successfully used where e ≤ 0.5.

All text and images © Duncan Hale-Sutton 2025