# How to find eccentric anomaly by mean anomaly?

Wikipedia gives me the equation

M = E - e*sinE

I have M and need to find E.

How to do it?

As the wikipedia article states, there is no closed form way to express $$E$$ in terms of $$M$$.

You would have to approximate it, for example using a series expansion:

$$E \approx M + \left(e - \frac{1}{8} e^3\right) \sin{(M)} + \left(\frac{1}{2}e^2 \right)\sin{(2M)} + \left(\frac{3}{8} e^3\right) \sin{(3M)}$$

• How accurate that approxymation will be? What is the allowed range of e value? Will it work with any e? Oct 13 at 14:17
• @Robotex The omitted series terms would be the error term. The approximation would be more accurate for low e values than high e values. Oct 13 at 14:25
• What do you mean by high? Is 1.0 high? Or higher? How big this error can be? I need it for calculation true anomaly. Do I understand right that M(t) is calculated with enough accuracy and calculating the true anomaly from it will not collect error with time flow since initial position/velocity measurements? Oct 13 at 15:42
• In the circular case $(e=0)$, you don't have to do anything. For a circular orbit, Mean Anomaly $(M)$ = Eccentric Anomaly $(E)$ = True Anomaly $(\theta)$ Oct 13 at 19:34
• I downvoted because this is not the best approach. Newton-Raphson works quite nicely, even for highly eccentric elliptical orbits. Oct 14 at 3:44

You can use a root solving method to calculate eccentric anomaly

As you stated, Kepler's equation for eccentric / mean anomaly and eccentricity is:

$$M=E-e sin(E)$$

and there is no closed form solution for E as a function of M but you can still iteratively calculate E.

You can set this up as a root solving problem by subtracting M from both sides of the equation to create a function that you want to find the root of (or when it equals 0 as a function of E):

$$f(E)=E-esin(E)-M=0$$

The root solver's solution will then calculate a value of E such that this equation equals 0, thus telling you the eccentric anomaly that satisfies Kepler's equation (the solution).

For a Newton root solver you need to calculate the derivative of this function (with respect to E), which is:

$$\dfrac{d(f(E))}{dE}=1-ecos(E)$$

With those two equations, values for mean anomaly (M), eccentricity (e), and an initial guess for eccentric anomaly (E) (this can be 0), the Newton root solver will calculate the solution.

As other comments stated, this only works for closed orbits (circular and elliptical), because by definition parabolic and hyperbolic orbits don't have a period. The hyperbolic eccentric anomaly is used for parabolic / hyperbolic orbits: What is hyperbolic eccentric anomaly F?

Here's a good diagram explaining mean anomaly ( https://en.wikipedia.org/wiki/Mean_anomaly ). To summarize it, mean anomaly is the "true anomaly" of an orbit with the same period as the orbit you are modeling, except it is circular, thus has a constant ("mean") angular velocity.

Here is the Python script used for making this plot. The Newton root solver can be found here: https://github.com/alfonsogonzalez/AWP/blob/main/src/python_tools/numerical_tools.py

'''
Create visualizations for Kepler's equation
of Mean / Eccentric anomalies and eccentricity
'''

from numerical_tools import newton_root_single_args
from numerical_tools import d2r, r2d

import numpy             as np
import matplotlib.pyplot as plt
plt.style.use( 'dark_background' )

def keplers_eq( E, args ):
return E - args[ 'e' ] * np.sin( E ) - args[ 'M' ]

def dkep_dE( E, args ):
return 1.0 - args[ 'e' ] * np.cos( E )

if __name__ == '__main__':
args0   = { 'M': 100 * d2r, 'e': 0.0 }
args1   = { 'M': 100 * d2r, 'e': 0.5 }
args2   = { 'M': 100 * d2r, 'e': 0.9 }
args3   = { 'M': 200 * d2r, 'e': 0.1 }
args4   = { 'M': 200 * d2r, 'e': 0.9 }
Es      = np.arange( 0, 2 * np.pi, 0.01 )
Es_deg  = Es * r2d
args    = [ args0, args1, args2, args3, args4 ]
colors  = [ 'r', 'g', 'b', 'c', 'm' ]
labels  = [ '0 | $$M=100,e=0$$', '1 | $$M=100,e=0.5$$', '2 | $$M=100,e=0.9$$' ]
labels += [ '3 | $$M=200,e=0.1$$', '4 | $$M=200,e=0.9$$' ]

plt.figure( figsize = ( 12, 8 ) )
for n in range( len( args ) ):
fs   = keplers_eq( Es, args[ n ] )
root = newton_root_single_args( keplers_eq, dkep_dE, 0.0, args[ n ] )
plt.plot( Es_deg, fs, colors[ n ], label = labels[ n ] )
plt.plot( root[ 0 ] * r2d, 0, colors[ n ] + 'o' )

plt.ylabel( 'f( E )' )
plt.xlabel( 'E $$(degrees)$$' )
plt.grid( linestyle = 'dotted' )
plt.legend()
plt.show()

• Is the Hyperbolic Anomaly used for parabolic orbits? Parabolas have that annoying property where they have an infinite semi-major axis so neither an osculating circle (for the Elliptic anomaly) nor an appropriate equilateral hyperbola (for the hyperbolic anomaly) exists., and you have to pin the Mean Anomaly to something else. That said, the equations of motion are directly solvable for parabolas, because they are a very special case, similar to the way circles are. Oct 16 at 12:51
• When plotting an elliptical orbit you can start at $E=M=0$, and then for subsequent points you can use the $E$ of the previous point as the initial approximation. 2 days ago
• @AlfonsoGonzalez Does anyone take spacecraft interplanetary with C3=0 on Earth departure? It requires both ridiculous levels of precision in your burns, and wastes delta-V. If you're going anywhere interplanetary that's not exactly on the Earth's orbit, your earth departure should be hyperbolic, 2 days ago
• @notovny Not in reality of course, but for preliminary analysis it can be used. I was in a situation in an internship where we were doing interplanetary trajectory design assuming the spacecraft getting dropped off with C3 > 0, but we also did analysis with C3 = 0 to see how that would affect mission design (knowing that this exact number wouldn't apply in real life) 2 days ago
• @Alfonso No, although $M$ is also a reasonable first approximation. Another option is $E=M+e\sin M$. My previous suggestion is that when calculating the $E$ at time $t$, use the $E$ at time $(t-\Delta t)$ as your 1st approximation. That is, the $E$ of the previous point in the orbit you're plotting. There's not much difference between those 3 options when $e$ is small, but for eccentric orbits, it can make a difference, especially if $\Delta M$ is small, so each new point is close to the previous one. I'll post a small demo in the next comment. 2 days ago