Foundations — Gauss's method for Lambert's problem

This page is the toolbox. Before you can read Lambert's problem or the parent Gauss's method note, you need every symbol below. We build each from nothing: plain words → a picture → why the topic needs it. Do not skip; every later symbol leans on an earlier one.

1. A position vector $\mathbf{r}$ — "an arrow to where something is"

The central body sits at a special point called the focus. Every orbit we discuss has the central body at one focus of its ellipse — that is Kepler's first law, and we simply take it as given here.

• The picture: look at figure s01. The orange dot is the focus. Two magenta arrows, ${\mathbf{r}}_1$ and ${\mathbf{r}}_2$, reach out to two positions of the spacecraft — an earlier one and a later one.
• Why the topic needs it: Lambert's problem is stated in terms of these two arrows. They are the known data. Everything else is squeezed out of them.

2. The angle between the two arrows — the transfer angle $\theta$

The symbol $\nu$ (Greek "nu") is the true anomaly — the angle of the spacecraft measured around its orbit. So $\Delta \nu$ (the change in true anomaly) and $\theta$ are the same thing: the swept angle.

• The picture: in figure s01 the violet wedge between the two magenta arrows is $\theta$.
• Why the topic needs it: $\theta$ tells you how "spread out" the two points are around the central body. A tiny $\theta$ means they are almost in line; a large $\theta$ means the spacecraft went most of the way round.

3. The dot product — how we compute $\cos \theta$ from the arrows

We know the two arrows, but not yet the angle between them. The tool that extracts an angle from two arrows is the dot product.

The $\cos$ (cosine) here is the cosine function: for an angle, it is a number between $-1$ and $+1$ that says how much two arrows "agree" in direction. $\cos 0^{\circ }=1$ (same direction), $\cos 90^{\circ }=0$ (perpendicular), $\cos 180^{\circ }=-1$ (opposite).

4. The chord $c$ and the triangle

Draw a straight line directly connecting the two spacecraft positions. That straight segment is the chord $c$.

• The picture: figure s02. The focus, tip of ${\mathbf{r}}_1$, and tip of ${\mathbf{r}}_2$ form a triangle. The two arrows are two sides; the chord $c$ is the third side (the orange straight line).
• Why we need $c$: it is a pure-geometry summary of "how far apart the two endpoints are." The chord appears in every serious Lambert solver.

• Why the topic needs the triangle: it is the flat stand-in for the curved region the planet actually sweeps. Gauss compares the curved region to this flat one.

5. Sector vs triangle — the picture behind the ratio $y$

The planet does not travel along the straight chord. It travels along the curved orbit. So the region it truly sweeps out is bounded by the two arrows and the curve — a "pie slice." That is the elliptic sector.

• The picture: figure s03. The violet region is the true elliptic sector (bounded by the curved orbit). The magenta dashed region is the flat triangle (bounded by the chord). The sector is always a little fatter than the triangle because the curve bulges outward.
• Why we need both: Gauss's whole trick is one number, $y\equiv \frac{\text{area of sector}}{\text{area of triangle}}.$ When the orbit barely bends (small $\theta$), the curve is almost the chord, sector $\approx$ triangle, so $y\approx 1$. That is the sensible first guess the algorithm starts from.

6. Kepler's second law — why area equals time

Why does area matter at all? Because of Kepler's second law:

Here $h$ is the specific angular momentum — a fixed number for a given orbit measuring how fast area is swept. It relates to orbit shape through $h=\sqrt{\mu p}$ (both symbols defined next).

7. $\mu$ — the strength of gravity

• Why we need it: the shape of an orbit and the speed along it both depend on how hard the central body pulls. In canonical units the parent note sets $\mu =1$ to keep the arithmetic clean.
• Where it enters: inside the time constant $m\propto \mu \Delta t^2$ — bigger gravity pulls the planet round faster, so it changes the time budget.

8. $p$ — the semi-latus rectum (the orbit's "width" number)

• Why we need it: once Gauss's iteration lands on the right geometry, $p$ is the number we extract, and from $p$ come the Lagrange f and g functions that hand back the velocities. It is the "answer knob" of the whole method.

9. Eccentric anomaly $E$ and the auxiliary variables $x$, $y$

The true anomaly $\nu$ from section 2 is the real, physical angle. But solving the timing cleanly needs a different bookkeeping angle, the eccentric anomaly $E$ — the angle you'd measure if you stretched the ellipse into a circle. It appears in Kepler's equation, the equation that links angle to time on an orbit.

• The picture: back in figure s03, a bigger bend of the curve = a bigger $x$ = a fatter sector = a bigger $y$. They rise together.

10. The known geometry constants $\ell$ and $m$

These two are just pre-computed numbers built from the data, so the iteration doesn't recompute them each loop.

Notice $\cos (\theta /2)$ in both denominators. When $\theta \to 180^{\circ }$ (the two points nearly opposite the focus), $\theta /2\to 90^{\circ }$ and $\cos (\theta /2)\to 0$ — so $\ell$ and $m$ blow up to infinity. This is exactly where Gauss's method fails, and why near-antipodal transfers need the Universal variable formulation or the Izzo Lambert solver instead.

Prerequisite map

Equipment checklist

Test yourself — cover the right side and see if each item is second nature before you read the parent note.