Kepler's equation

Where on its orbit is a body at time $t$? On a circle that's trivial: angle increases at a constant rate. On an ellipse, motion isn't uniform — the body sprints near periapsis, crawls past apoapsis. Converting elapsed time to actual position requires solving Kepler's equation, the most famous transcendental equation in celestial mechanics.

Three angles, one ellipse

To talk about position on an elliptical orbit we need to be careful which angle we mean — there are three different ones in standard use, and they're easily confused.

• True anomaly $\nu$ — the actual angle, as seen from the central body (focus), between periapsis and the orbiting body. This is what you'd measure with a protractor on the screen. It's nonlinear in time: it accelerates near periapsis and slows near apoapsis.
• Eccentric anomaly $E$ — a clever auxiliary angle, defined geometrically by drawing a circle of radius $a$ around the ellipse center and projecting the orbiting body up onto that circle. The angle from the ellipse center to the projection is $E$. It's an intermediate quantity with no direct physical interpretation.
• Mean anomaly $M$ — the angle the body would have swept out by now if it were moving at constant angular speed. It's linear in time by construction: $M = n(t - t_p)$, where $n = 2\pi/T$ is the mean motion and $t_p$ is the time of periapsis passage.

Time gives you $M$ for free. The end goal is $\nu$ (where the body actually is). Bridging them requires going through $E$, and that's where Kepler's equation comes in.

The equation itself

Kepler discovered, around 1620, that for any elliptical orbit:

$E$ is what we want; $M$ and $e$ are known. The equation is transcendental — there's no algebraic rearrangement that gives $E$ explicitly in terms of $M$. You can't write $E = (\text{something})$. The answer exists, but only as the solution of an equation, and you have to find it numerically.

Solving it: Newton–Raphson

Rewrite Kepler's equation as $f(E) = 0$ and apply Newton's method. Define

and iterate

Starting from $E_0 = M + e\sin M$ (a sharp first guess that already gets you close), the iteration converges quadratically — meaning the number of correct digits roughly doubles with each step. For any reasonable orbit ($e < 0.95$, say), three or four iterations gets you to floating-point precision.

Watch it converge

Pick a mean anomaly and an eccentricity. The table shows the Newton iterations explicitly: at each step, the current $E$, the residual $f(E)$, and the correction $\Delta E$. Watch the residual drop by orders of magnitude per step.

Even at $e = 0.97$ — close to a parabolic escape — the iteration still converges in about half a dozen steps. Halley's Comet has $e \approx 0.967$, and its position-finding code uses exactly this loop, run a few times per orbit.

From $E$ to $\nu$

Once you have the eccentric anomaly, the true anomaly comes from a half-angle relation:

And the focus-distance is $r = a(1 - e\cos E)$, the form used throughout this curriculum. With $\nu$ and $r$ in hand, you can convert to Cartesian coordinates relative to the orbital plane and you're done — that's how every Kepler-propagation animation in modules 3, 4, 5, 7, and 8 actually computes the body's position. Each frame: take the elapsed time, compute $M$, run the loop above for $E$, project onto the ellipse.

Getting back