[Mathematical Notation]

Kepler's Equation

The mean anomaly M, eccentric anomaly E, and true anomaly nu of an object in a Keplerian orbit are determined by

(Eq 1) M = 2 pi * t / P

(Eq 2) E = M + e sin E

(Eq 3) tan (nu/2) = sqrt((1 + e)/(1 - e)) tan (E/2)

where P is the object's orbital period, e is the eccentricity of the orbit, and all angles are counted in radians (2 pi radians equals 360 degrees). M advances at a constant rate, and nu is the angular distance of the object from its periapsis. Equation 2 is called Kepler's equation and is generally not directly solvable for E. Various iterative schemes exist for finding a solution for E. The easiest one is to take M as an initial guess for E and to use the right-hand side of Equation 2 to get successively more accurate estimates, i.e.,

E1 = M + e sin M

E2 = M + e sin E1

E3 = M + e sin E2

etcetera. We determine how the accuracy of successive estimates increases by denoting the solution of the equation E_f and writing

(Eq 4) E_i = E_f + d_i.

After one iteration we find

(Eq 5) d2 = e(sin(E_f + d1) - sin E_f)

which shows that the error in a result of this iteration is at most 2e times as large as the error in the result of the previous iteration. When the current error d1 is much smaller than 1 radian (i.e., a few degrees or less), then the next error d2 is approximately equal to e*d1, so then the error reduces by a factor of approximately e during every iteration.

True Anomaly; Equation of Center

Once you have solved Kepler's equation for your values of M and e, you can calculate the true anomaly nu. The equation of center C is defined as

(Eq 6) C = nu - M,

i.e. it is the correction to be applied to the mean anomaly M to get the true anomaly nu. When the eccentricity e is much smaller than 1 (i.e., for nearly circular orbits), then we can combine equations 1 through 3 and determine an approximation for the equation of center:

(Eq 7) C = 2 e sin M

+ (5/4) e^2 sin (2M)

+ (1/12) e^3 (13 sin 3M - 3 sin M)

+ (1/96) e^4 (103 sin 4M - 44 sin 2M)

+ (1/960) e^5 (1097 sin 5M - 645 sin 3M + 50 sin M)

+ (1/960) e^6 (1223 sin 6M - 902 sin 4M + 85 sin M)

+ seventh-order terms

Rearranged to combine similar terms of M we find

(Eq 8) C = (2e - (1/4) e^3 + (5/96) e^5) sin M

+ ((5/4) e^2 - (11/24) e^4 + (17/192) e^6) sin 2M

+ ((13/12) e^3 - (43/64) e^5) sin 3M

+ ((103/96) e^4 - (451/480) e^6) sin 4M

+ (1097/960) e^5 sin 5M

+ (1223/960) e^6 sin 6M

+ seventh-order terms

For the Earth's orbit, with a current eccentricity of e = 0.01671, we find, transformed to degrees, to third order in e,

(Eq 9) C_Earth = 1.9148 sin M + 0.0200 sin 2M + 0.0003 sin 3M.

[LS 16 November 1996]

From true to mean anomaly

In the opposite direction, from true to mean anomaly, we find

(Eq 10) M = nu - 2 e sin nu

+ (3/4) e^2 sin 2nu

- (1/3) e^3 sin 3nu

+ (1/16) e^4 ((2 + 5 cos 2nu) sin 2nu)

- (1/40) e^5 (5 sin 3nu + 3 sin 5nu)

+ (1/96) e^6 ((8 + 18 cos 2nu + 7 cos 4nu) sin 2nu)

+ seventh-order terms in e

or

(Eq 11) M = nu

- 2 e sin 2nu

+ ((3/4) e^2 + (1/8) e^4 + (1/12) e^6 + ((5/16) e^4 + (3/16) e^6) cos 2nu) sin 2nu

- ((1/3) e^3 + (1/8) e^5) sin 3nu

+ (7/96) e^6 sin 2nu cos 4nu

- (3/40) e^5 sin 5nu

+ seventh-order terms in e

For the Earth's orbit, to third order in e, in degrees,

(Eq 12) M = nu - 1.9148 sin nu + 0.0120 sin 2nu - 0.0001 sin 3nu

[LS 29 August 1997]