Solution of Kepler's equation

by Newton-Raphson Method

Here you can solve the Kepler's equation M = E-eSin(E) for an elliptical orbit. Given mean anomaly M and eccentricity e , you can solve for eccentric anomaly E. The ranges for e and M are [0,1] and [0,PI]. If M<0 , you can solve for E with |M| and associate a negative sign to the resulting solution.

Eccentricity 'e' of Orbit(in the range [0,1])

Mean anomaly M in degrees(in the range [0,180])

Epsilon

Eccentric Anomaly E in degrees:

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Report all bugs to Sheela V.Belur at sbelur@capaccess.org

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