In this example we will show how to solve Kepler’s equation M = E-e*sin(E)
clear all; clc; M = 0.3458; % Mean anomaly [rad] eps = 1E-9; % Tolerance e = 0.1; % Eccentricity En = M; Ens = En - (En-e*sin(En)- M)/(1 - e*cos(En)); while ( abs(Ens-En) > eps ) En = Ens; Ens = En - (En - e*sin(En) - M)/(1 - e*cos(En)); end; E = Ens; % Eccentric anomaly E [rad]. fprintf('Eccentric anomaly E = %4.4f [rad]\n',E);
Eccentric anomaly E = 0.3832 [rad]
