Kepler's equation

(Redirected from Kepler's Equation)
For specific applications of Kepler's equation, see Kepler's laws of planetary motion.
Kepler's equation solutions for five different eccentricities between 0 and 1

In orbital mechanics, Kepler's equation relates various geometric properties of the orbit of a body subject to a central force.

It was first derived by Johannes Kepler in 1609 in Chapter 60 of his Astronomia nova,[1][2] and in book V of his Epitome of Copernican Astronomy (1621) Kepler proposed an iterative solution to the equation.[3][4] The equation has played an important role in the history of both physics and mathematics, particularly classical celestial mechanics.

Equation

Kepler's equation is

 $M = E -\varepsilon \sin E$

where M is the mean anomaly, E is the eccentric anomaly, and ε is the eccentricity.

The 'eccentric anomaly' E is useful to compute the position of a point moving in a Keplerian orbit. As for instance, if the body passes the periastron is at coordinates x=a(1−e), y=0, at time t=t0, then to find out the position of the body at any time, then you first calculate the mean anomaly M from the time and the mean motion n by the formula M = n (tt0), then solve the Kepler equation above to get E, then get the coordinates from:

 $\begin{array}{lcl} x & = & a (\cos E -\varepsilon ) \\ y & = & b \sin E \end{array}$

Kepler's equation is a transcendental equation because sine is a transcendental function, meaning it cannot be solved for E algebraically. Numerical analysis and series expansions are generally required to evaluate E.

Alternate forms

There are several forms of Kepler's equation. Each form is associated with a specific type of orbit. The standard Kepler equation is used for elliptic orbits (0 ≤ ε < 1). The hyperbolic Kepler equation is used for hyperbolic orbits (ε ≫ 1). The radial Kepler equation is used for linear (radial) orbits (ε = 1). Barker's equation is used for parabolic orbits (ε = 1). When ε = 1, Kepler's equation is not associated with an orbit.

When ε = 0, the orbit is circular. Increasing ε causes the circle to flatten into an ellipse. When ε = 1, the orbit is completely flat, and it appears to be a either a segment if the orbit is closed, or a ray if the orbit is open. An infinitesimal increase to ε results in a hyperbolic orbit with a turning angle of 180 degrees, and the orbit appears to be a ray. Further increases reduce the turning angle, and as ε goes to infinity, the orbit becomes a straight line of infinite length.

Hyperbolic Kepler equation

The Hyperbolic Kepler equation is:

 $M = \varepsilon \sinh H - H$

where H is the hyperbolic eccentric anomaly. This equation is derived by multiplying Kepler's equation by the square root of −1; i = √−1 for imaginary unit, and replacing

$E = i H$

to obtain

$M = i \left( E - \varepsilon \sin E \right)$

 $t(x) = \sin^{-1}( \sqrt{ x } ) - \sqrt{ x ( 1 - x ) }$

where t is time, and x is the distance along an x-axis. This equation is derived by multiplying Kepler's equation by 1/2 making the replacement

$E = 2 \sin^{-1}(\sqrt{ x })$

and setting ε = 1 gives

$t(x) = \frac{1}{2}\left[ E - \sin( E ) \right].$

Inverse problem

Calculating M for a given value of E is straightforward. However, solving for E when M is given can be considerably more challenging.

Kepler's equation can be solved for E analytically by Lagrange inversion. The solution of Kepler's equation given by two Taylor series below.

Confusion over the solvability of Kepler's equation has persisted in the literature for four centuries.[5][6] Kepler himself expressed doubt at the possibility of ﬁnding a general solution.

Inverse Kepler equation

The inverse Kepler equation is the solution of Kepler's equation for all real values of ε:

$E = \begin{cases} \displaystyle \sum_{n=1}^\infty {\frac{M^{\frac{n}{3}}}{n!}} \lim_{\theta \to 0^+} \! \Bigg( \frac{\mathrm{d}^{\,n-1}}{\mathrm{d}\theta^{\,n-1}} \bigg( \bigg( \frac{\theta}{ \sqrt[3]{\theta - \sin(\theta)} } \bigg)^{\!\!\!n} \bigg) \Bigg) , & \varepsilon = 1 \\ \displaystyle \sum_{n=1}^\infty { \frac{ M^n }{ n! } } \lim_{\theta \to 0^+} \! \Bigg( \frac{\mathrm{d}^{\,n-1}}{\mathrm{d}\theta^{\,n-1}} \bigg( \Big( \frac{ \theta }{ \theta - \varepsilon \sin(\theta)} \Big)^{\!n} \bigg) \Bigg) , & \varepsilon \ne 1 \end{cases}$

Evaluating this yields:

$E = \begin{cases} \displaystyle x + \frac{1}{60} x^3 + \frac{1}{1400}x^5 + \frac{1}{25200}x^7 + \frac{43}{17248000}x^9 + \frac{ 1213}{7207200000 }x^{11} + \frac{151439}{12713500800000 }x^{13}+ \cdots \ | \ x = ( 6 M )^\frac{1}{3} , & \varepsilon = 1 \\ \\ \displaystyle \frac{1}{1-\varepsilon} M - \frac{\varepsilon}{( 1-\varepsilon)^4 } \frac{M^3}{3!} + \frac{(9 \varepsilon^2 + \varepsilon)}{(1-\varepsilon)^7 } \frac{M^5}{5!} - \frac{(225 \varepsilon^3 + 54 \varepsilon^2 + \varepsilon ) }{(1-\varepsilon)^{10} } \frac{M^7}{7!} + \frac{ (11025\varepsilon^4 + 4131 \varepsilon^3 + 243 \varepsilon^2 + \varepsilon ) }{(1-\varepsilon)^{13} } \frac{M^9}{9!}+ \cdots , & \varepsilon \ne 1 \end{cases}$

These series can be reproduced in Mathematica with the InverseSeries operation.

InverseSeries[Series[M - Sin[M], {M, 0, 10}]]
InverseSeries[Series[M - e Sin[M], {M, 0, 10}]]

These functions are simple Taylor series. Taylor series representations of transcendental functions are considered to be definitions of those functions. Therefore this solution is a formal definition of the inverse Kepler equation. While this solution is the simplest in a certain mathematical sense, for values of ε near 1 the convergence is very poor, other solutions are preferable for most applications. Alternatively, Kepler's equation can be solved numerically.

The solution for ε ≠ 1 was discovered by Karl Stumpff in 1968,[8] but its significance wasn't recognized.[9]

The inverse radial Kepler equation is:

$x( t ) = \sum_{n=1}^{ \infty } \left[ \lim_{ r \to 0^+ } \left( {\frac{ t^{ \frac{ 2 }{ 3 } n }}{ n! }} \frac{\mathrm{d}^{\,n-1}}{\mathrm{ d } r ^{\,n-1}} \! \left( r^n \left( \frac{ 3 }{ 2 } \Big( \sin^{-1}( \sqrt{ r } ) - \sqrt{ r - r^2 } \Big) \right)^{ \! -\frac{2}{3} n } \right) \right) \right]$

Evaluating this yields:

$x(t) = p - \frac{1}{5} p^2 - \frac{3}{175}p^3 - \frac{23}{7875}p^4 - \frac{1894}{3931875}p^5 - \frac{3293}{21896875}p^6 - \frac{2418092}{62077640625}p^7 - \ \cdots \ \bigg| { p = \left( \tfrac{3}{2} t \right)^{2/3} }$

To obtain this result using Mathematica:

InverseSeries[Series[ArcSin[Sqrt[t]] - Sqrt[(1 - t) t], {t, 0, 15}]]

Numerical approximation of inverse problem

For most applications, the inverse problem can be computed numerically by finding the root of the function:

$f(E) = E - \varepsilon \sin(E) - M(t)$

This can be done iteratively via Newton's method:

$E_{n+1} = E_{n} - \frac{f(E_{n})}{f'(E_{n})} = E_{n} - \frac{ E_{n} - \varepsilon \sin(E_{n}) - M(t) }{ 1 - \varepsilon \cos(E_{n})}$

Note that E and M are in units of radians in this computation. This iteration is repeated until desired accuracy is obtained (e.g. when f(E) < desired accuracy). For most elliptical orbits an initial value of E0 = M(t) is sufficient. For orbits with ε > 0.8, an initial value of E0 = π should be used.[10] A similar approach can be used for the hyperbolic form of Kepler's equation. In the case of a parabolic trajectory, Barker's equation is used.