Solutions To Kepler Equation Numerical Analysis Problems

 1) The mean anomaly  (M) of Mercury is 179 ^o .796  and its eccentricity is e= 0.20563.

Use this to do a Newton-Raphson iteration to obtain the eccentric anomaly. 

Solution - From Mathcad

Thus E = 3. 139

2) The mean anomaly of Earth is 198 ^o .115  and its eccentricity is e = 0.01672.

a) Use this data  to obtain the eccentric anomaly E using a Newton-Raphson iteration.

Solution - From Mathcad

Thus E = 3.45268

b) Compare the preceding result with E obtained from applying the Lagrange  expansion theorem, viz.

E = M + e sin M +  ( e ²/ 2) sin 2M +  e³/3! 2e ²  [3 ²sin 3M - 3 sin 3M) + ...

Solution - From Mathcad

Both answers 2(a) and 2(b) arrive at the textbook solution (3.45268)

c) How do the above results compare with E obtained from the simplest formula:

tan E  = sin M/ (cos M - e)

Basic computation shows the answer is way off, i.e.

sin M = sin (3.4578) = - 0.31096  

sin M/ (cos M - e)  =  sin (3.4578)/ [cos (3.4578) - 0. 01672] = 0.32153

tan E = 0.33309

E = arctan (0.33309) = 0.31109

We can conclude this (early) approximation formula is off,  because even with e = 0 the result is nowhere close to the actual value for E.