Astronomy Answers
Kepler's Equation

1. Kepler's Equation

The equation that expresses the relationship (for elliptical and hyperbolic orbits) between the mean anomaly and the eccentric anomaly is called Kepler's Equation, and this equation is nearly but not completely the same for elliptical orbits and for hyperbolic orbits. (Sometimes "Kepler's Equation" is used only for the elliptical variant.)

For elliptical orbits and hyperbolic orbits the eccentric anomaly cannot be calculated directly from the mean anomaly (but the other way around does work), so a special method must be used to yet find a solution.

2. Elliptical Orbits

2.1. Kepler's Equation for Elliptical Orbits

The mean anomaly M, eccentric anomaly E, and true anomaly ν (nu) of an object in an elliptical orbit are determined by

(Eq. 1) M = 2 π t / P

(Eq. 2) E = M + e sin E

(Eq. 3) tan(ν/2) = √((1 + e)/(1 − e)) tan(E/2)

where P is the object's orbital period ("year"), t is the time since the last periapsis passage, e is the eccentricity of the orbit, and all angles are counted in radians (2π radians equal 360 degrees). M advances at a constant rate, and ν is the angular distance of the object from its periapsis (as seen from a focus of the orbit).

Equation 2 is called Kepler's Equation and assumes that the angles are measured in radians and not in degrees. Kepler's Equation 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.,

(Eq. 4) E1 = M

(Eq. 5) E2 = M + e sin E1 = M + e sin M

(Eq. 6) E3 = M + e sin E2

(Eq. 7) E4 = M + e sin E3

etcetera. We determine how the accuracy of successive estimates increases by defining Ei = E + di and performing one iteration of the approximation method. We find

(Eq. 8) d(i+1) = e ( sin(E + di) − sin E )

which shows that the error in a result of this iteration is at most 2 e times as large as the error in the result of the previous iteration. When the current error di is much smaller than 1 radian (i.e., a few degrees or less), then the next error is approximately e times as large. For very elongated elliptical orbits (e barely smaller than 1) and parabolic orbits (e equal to 1) this method is not very suitable.

2.2. The True Anomaly and the Equation of Center

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

(Eq. 11) C = ν − M

i.e., it is the correction to be applied to the mean anomaly M to get the true anomaly nu. If 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. 12) C = 2 e sin M + (5⁄4) e² sin(2 M) + (1⁄12) e³ (13 sin(3 M) − 3 sin M) + (1⁄96) e⁴ (103 sin(4 M) − 44 sin(2 M)) + (1⁄960) e⁵ (1097 sin(5 M) − 645 sin(3 M) + 50 sin M) + (1⁄960) e⁶ (1223 sin(6 M) − 902 sin(4 M) + 85 sin(2 M))

+ smaller terms

Rearranged to combine similar terms of M we find

(Eq. 13) C = (2 e − (¼) e³ + (5⁄96) e⁵) sin M + ((5⁄4) e² − (11⁄24) e⁴ + (17⁄192) e⁶) sin(2 M) + ((13⁄12) e³ − (43⁄64) e⁵) sin(3 M) + ((103⁄96) e⁴ − (451⁄480) e⁶) sin(4 M) + (1097⁄960) e⁵ sin(5 M) + (1223⁄960) e⁶ sin(6 M)

+ smaller 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. 14) C = 1.9148° sin M + 0.0200° sin (2 M) + 0.0003° sin (3 M)

2.3. From True Anomaly to Time

If we know the true anomaly ν, then we can calculate the correspoding time without needing to use iterations. We first calculate the eccentric anomaly E from the true anomaly:

(Eq. 15) E = 2 arctan(√((1 − e)/(1 + e)) tan(ν/2))

then the mean anomaly M from the eccentric anomaly E:

(Eq. 16) M = E − e sin E

(where the angles must again be measured in radians and not in degrees) and then the time t since the last periapsis, from the mean anomaly M:

(Eq. 17) t = P M/(2 π)

If you prefer to use an approximation as before, then you can use the following one:

(Eq. 18) M = ν − 2 e sin ν + (¾) e² sin(2 ν) − (1⁄3) e³ sin(3 ν) + (1⁄16) e⁴ ((2 + 5 cos(2 ν)) sin(2 ν)) − (1⁄40) e⁵ (5 sin(3 ν) + 3 sin(5 ν)) + (1⁄96) e⁶ ((8 + 18 cos(2 ν) + 7 cos(4 ν)) sin(2 ν))

+ smaller terms

or

(Eq. 19) M = ν − 2 e sin(2 ν) + ((¾) e² + (1⁄8) e⁴ + (1⁄12) e⁶ + ((5⁄16) e⁴ + (3⁄16) e⁶) cos(2 ν)) sin(2 ν) − ((1⁄3) e³ + (1⁄8) e⁵) sin(3 ν) + (7⁄96) e⁶ sin(2 ν) cos(4 ν) − (3⁄40) e⁵ sin(5 ν)

+ smaller terms

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

(Eq. 20) M = ν − 1.9148° sin(ν) + 0.0120° sin(2 ν) − 0.0001° sin(3 ν)

3. Hyperbolic Orbits

3.1. Kepler's Equation for Hyperbolic Orbits

The mean anomaly M, eccentric anomaly H, and true anomaly ν (nu) of an object in a hyperbolic orbit are determined by

(Eq. 21) M = 2 π t / P

(Eq. 22) H = M + e sinh H

(Eq. 23) tan(ν/2) = √((1 + e)/(1 − e)) tanh(H/2)

where P is the orbital period ("year") that belongs to an elliptical orbit of the same (absolute) semimajor axis, t is the time since the periapsis passage, e is the eccentricity of the orbit (greater than 1), and all angles are counted in radians (2π radians equal 360 degrees). M advances at a constant rate, and ν is the angular distance of the object from its periapsis (as seen from a focus of the orbit).

Equation 22 is Kepler's Equation for hyperbolic orbits. Note that this equation does not use the ordinary sine function sin but instead the hyperbolic sine function sinh. Equation 23 uses the ordinary tangent function tan on the left and the hyperbolic tangent function tanh on the right. Just like for an elliptical orbit, Kepler's equation for hyperbolic orbits cannot be solved directly for H. The simplest method for finding a solution for H is (just like it was for elliptical orbits) to transform the equation into an approximation formula, take M for its first approximation, and get ever closer to the correct solution by repeatedly applying the approximation formula. The appropriate approximation formula is:

(Eq. 24) H(i+1) = arcsinh((Hi + M)/e)

3.2. The True Anomaly and the Equation of Center

With e = 1 + δ we find approximately for the Equation of Center C = ν − M for orbits with eccentricity e only a little bit larger than 1:

(Eq. 25) C ≈ −(1⁄3) √2 sec³(ν/2) sin(3ν/2) δ(3⁄2) − (sec5(ν/2) (5 sin (ν/2) − 10 sin(3ν/2) + sin(5ν/2)) δ(5⁄2))/(40√2) + ((−181 cos ν + 29 cos(2ν) + cos(3ν) − 17) sec6(ν/2) tan(ν/2) δ(7⁄2))/(896√2) − (sec9(ν/2) (1701 sin(ν/2) − 2097 sin(3ν/2) + 315 sin(5ν/2) + 18 sin(7ν/2) + sin(9ν/2)) δ(9⁄2))/(36864√2) + (sec11(ν/2) (71148 sin(ν/2) − 80586 sin(3ν/2) + 12947 sin(5 ν/2) + 924 sin(7ν/2) + 88 sin(9ν/2) + 5 sin(11ν/2)) δ(11⁄2))/(2883584√2)

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Last updated: 2012-01-13