Kepler's Equation

In my previous description of a program that I wrote to calculate binary star orbits I mentioned Kepler's Equation which can be stated as:-

E - e sin E = M                . . . (1)

(see section 43 on page 84 of Practical Astronomy with your Calculator by Peter Duffett-Smith). E is the eccentric anomaly, M is the mean anomaly and e is the eccentricity of the orbit. From E we can calculate the true anomaly ν. For a pair of binary stars, the mean anomaly is essentially how far round star B has gone round star A assuming that it orbited at a constant rate. If we count the number of years since the epoch of periastron (the closest approach for which data is recorded) and divide this by the orbital period then the remainder is the fraction of an orbit that has been completed since the last periastron. Multiply this by 2π and you get the mean anomaly in radians. The true anomaly is the angle taking into consideration that star B moves in an ellipse about star A, A is at a focus of this ellipse and star B does not move at a constant rate.

As you can see equation (1) is not easy to solve for E if you know M and e and that is why people resort to numerical solutions (see page 85 for Duffett-Smith's book). I wondered if there was an approximate solution which didn't rely on a numerical solution. Firstly, if we write 

Δ = E - M                        . . . (2)

Then equation (1) becomes

Δ = e sin E                     . . . (3)

If you look at page 122 of Duffett-Smith's book you can see plots Δ of versus M for various values of e. Note that the modulus of Δ is always less than e which is less than 1. This comes from equation (3); since -1 ≤ sin E ≤ 1 this implies -e ≤ e sin E ≤ e (e > 0), so from (3) -e ≤ Δ ≤ e. 

This lead me to an idea for an approximate solution to (1). Substituting for E = M + Δ in equation (3) we get 

Δ = e sin (M + Δ) = e (sin M cos Δ + cos M sin Δ)                . . . (4)

If Δ is much less than 1 then sin Δ ≈ Δ and cos Δ ≈ 1 (to first order). Thus (4) becomes

Δ ≈ e (sin M + Δ cos M)

Rearranging we get

Δ ≈ e sin M / (1 - e cos M)            . . . (5)

I have found that this solution does pretty well if e ≤ 0.2 (the maximum difference between this solution and the true estimate of Δ is less than 2.1% over all values of M).

All text and images © Duncan Hale-Sutton 2025