Real use of approximate solution to Kepler's Equation

I wanted to try out the 2nd order approximate solution to Kepler's Equation in my binary star program to see how well it compared to the numerical solution. To this end I have replaced the code in the program from the line 

X = M + Zsin M/(1 - Zcos M)

to the line

Lbl 2

(inclusive) with the following code

J = Z sin M

K = 1 - Z cos M

X = M + (-K + √(K² + 2J²))/J

Previously I had a prediction for the binary star Xi Bootis using data from my Norton's 2000.0 Star Atlas. I now want to use the same data to compare the results I had before with the results from the modified program. I get the following:-

 

Numerical                    2nd order solution

Year : Sep. : PA            Sep. : PA

2024.0 : 4.85 : 290.4    4.84 : 290.3

2034.0 : 3.90 : 270.5    3.91 : 270.7

2044.0 : 3.08 : 238.7    3.09 : 239.3

2054.0 : 2.63 : 191.3    2.63 : 191.6

2064.0 : 2.16 : 125.5    2.16 : 125.5

2074.0 : 2.76 : 53.5      2.78 : 52.9

2084.0 : 4.23 : 21.3      4.24 : 21.0

2094.0 : 5.47 : 5.2        5.47 : 5.1

2104.0 : 6.36 : 354.4    6.36 : 354.5

2114.0 : 6.92 : 346.0    6.92 : 346.0

2124.0 : 7.19 : 338.5    7.19 : 338.5

2134.0 : 7.20 : 331.2    7.20 : 331.2

2144.0 : 6.96 : 323.8    6.96 : 323.8

2154.0 : 6.50 : 315.5    6.49 : 315.5

2164.0 : 5.83 : 305.6    5.82 : 305.5

2174.0 : 4.98 : 292.7    4.98 : 292.7

I think the approximate solution does pretty well considering that the eccentricity of these stars orbit is 0.512 which is probably at the limit of what this solution can cater for. The largest error in separation is 0.02 arc seconds but mostly they are within 0.01. The largest error in PA is 0.6 degrees and this occurs in 2044 and 2074. 

If I had the situation where it wasn't practicable to seek a numerical solution then I think this approximate one would do very well.

 

All text and images © Duncan Hale-Sutton 2025