Orbit determination — Gauss's method, Gibbs method: Visual walkthrough

We are heading toward the central result of Gibbs's method: a formula for the velocity $v_{2}$ at the middle point, built only from the three position arrows. Rather than print it now (its symbols aren't defined yet), we assemble it live across the steps below.

Prerequisites we lean on: Two-body Problem, Angular Momentum & Eccentricity Vector, and later Classical Orbital Elements.

Step 1 — Three dots in space, and what "position vector" means

WHAT. We are given three arrows $r_{1}, r_{2}, r_{3}$ — three snapshots of the same satellite.

WHY. A single orbit is a fixed curve in space. If three arrows all touch that curve, they are heavily constrained: the curve has to thread all three arrowheads and keep the Earth at its focus. That is enough to reconstruct everything — including the speed. No stopwatch required.

PICTURE. Look at the three coloured arrows below, all sprouting from the same origin (Earth's centre, the black dot). Their tips $P_{1}, P_{2}, P_{3}$ are the three "photos" of the satellite.

Figure — Orbit determination — Gauss's method, Gibbs method

Throughout this page every $r_{i}$ is one of these arrows, and each magnitude $r_{i} = ∣ r_{i} ∣$ is just how long that arrow is.

Step 2 — They must be coplanar: the flat-sheet test

WHAT. Before trusting the data, check that $P_{1}, P_{2}, P_{3}$ really share one plane.

WHY. If they don't, no single Keplerian orbit can pass through all three, and any velocity we compute would be garbage. This is the very first sanity gate.

How to test it with a cross product. The cross product $a \times b$ of two arrows produces a new arrow perpendicular to both. So $r_{2} \times r_{3}$ is perpendicular to the plane containing $r_{2}$ and $r_{3}$ . If $r_{1}$ also lies in that plane, then $r_{1}$ is perpendicular to that perpendicular, so their dot product (a number measuring overlap, zero when perpendicular) is zero: $unit arrow to P_{1} \overset{u}{^}_{r 1} \cdot normal to plane of P_{2}, P_{3} (\overset{u}{^}_{r 2} \times \overset{u}{^}_{r 3}) \approx 0$

Here $\overset{u}{^}_{r i} = r_{i} / r_{i}$ is the unit version of each arrow (length 1, direction only) — we normalise so the test doesn't depend on how long the arrows happen to be.

PICTURE. The green shaded sheet is the orbital plane. The yellow arrow is the normal $r_{2} \times r_{3}$ ; notice $r_{1}$ lies flat in the sheet, so it makes a right angle with the yellow normal — dot product zero.

Step 3 — The one equation every point obeys: the conic

WHAT. Write down the one algebraic law all three points share.

WHY. This is the secret glue. All three arrowheads obey this same equation with the same $p$ and $e$ — only $θ$ changes from point to point. Gibbs's whole trick is to combine the three copies of this equation so that the unknown angles $θ_{i}$ cancel out, leaving only measurable arrows.

PICTURE. The ellipse with focus at Earth's centre; the semi-latus rectum $p$ is the special vertical distance at the focus; $θ$ is measured from perigee (red dot).

With that convention fixed, each point's arrow decomposes cleanly: $r_{i} = along \overset{e}{^} r_{i} cos θ_{i} \overset{e}{^} + along \overset{p}{^} r_{i} sin θ_{i} \overset{p}{^}$ "how far along perigee-direction" plus "how far sideways". The angles $θ_{i}$ are the only unknowns — and they are what the next step kills.

Step 4 — Killing the unknown angles: build $N$ , $D$ , $S$

WHAT. Form these three helper vectors from the data: $N = r_{1} (r_{2} \times r_{3}) + r_{2} (r_{3} \times r_{1}) + r_{3} (r_{1} \times r_{2})$ $D = (r_{1} \times r_{2}) + (r_{2} \times r_{3}) + (r_{3} \times r_{1})$ $S = r_{1} (r_{2} - r_{3}) + r_{2} (r_{3} - r_{1}) + r_{3} (r_{1} - r_{2})$

Reading each symbol where it sits:

In $N$ : each cross product $r_{j} \times r_{k}$ is an area-arrow perpendicular to the plane (it points along $h$ ). Each is weighted by a scalar magnitude $r_{i}$ . So $N$ is a stack of plane-normal arrows — it points along the orbit normal and its length carries the scale.
In $D$ : the same three cross products, but unweighted. $D$ also points along the normal. The ratio of lengths $N / D$ turns out to be exactly the scale $p$ .
In $S$ : notice each arrow $r_{i}$ is weighted by a difference of the other two distances $(r_{j} - r_{k})$ . Differences of distances encode "which way is downhill toward perigee" — so $S$ lies in the plane and points related to $\overset{e}{^}$ .

Formula WHY

$N ∥ h$ , $D ∥ h$ , and $N / D = p$ Why $D$ is along the normal. Each term $r_{j} \times r_{k}$ is, by the meaning of the cross product, perpendicular to the plane — i.e. parallel to $\hat{h}$ . A sum of arrows all pointing the same way (all $\pm \hat{h}$ ) is again along $\hat{h}$ . So $D = D \hat{h}$ . The same argument makes $N = N \hat{h}$ : every term is a scalar times a plane-normal cross product. Hence both $N$ and $D$ are parallel to $h$ , and therefore parallel to each other — their cross product is zero. Why $N / D = p$ . Substitute the conic $r_{i} = p / (1 + e cos θ_{i})$ and the basis decomposition into $N$ and $D$ . Each cross product $r_{j} \times r_{k} = r_{j} r_{k} sin (θ_{k} - θ_{j}) \hat{h}$ . Carrying the algebra through, the sums differ by exactly one factor of $p$ : $N = p D$ . Geometrically: $D$ measures the unweighted signed areas, $N$ the distance-weighted ones, and the conic law forces their ratio to be the semi-latus rectum. The numeric checks in VERIFY confirm both $N \times D = 0$ and $N / D = h^{2} / μ = p$ on the worked example.

PICTURE. $N$ and $D$ (both along the yellow normal, out of the plane), and $S$ lying flat in the green plane pointing toward perigee.

Step 5 — Recovering the two velocity directions

WHAT. The transverse direction is $\frac{D \times r _{2}}{r _{2}}$ ; the radial content comes from $S$ .

Reading the transverse term. $D$ points along the normal (out of the plane). Crossing it with $r_{2}$ gives an arrow that is in the plane and perpendicular to $r_{2}$ — precisely the "around the orbit" direction. Dividing by $r_{2}$ keeps its length calibrated.

Reading the radial term — with the projection check. The radial unit direction is $\overset{r}{^}_{2} = r_{2} / r_{2}$ . To show $S$ genuinely supplies in/out motion, project it onto $\overset{r}{^}_{2}$ . Substituting the conic and basis into $S$ (same algebra as Step 4), one finds $S = e (sin θ_{2} \overset{e}{^} - cos θ_{2} \overset{p}{^}) \cdot (scale) \Rightarrow S \cdot \overset{r}{^}_{2} \propto e sin θ_{2},$ which is nonzero exactly when the satellite is climbing or falling (true anomaly $θ_{2}$ away from $0$ or $180°$ ) and zero at perigee/apogee, where radial speed must vanish. That is the fingerprint of a radial velocity: it tracks $e sin θ$ . So $S$ carries the radial part; the transverse term carries the rest. A numeric projection ( $S \cdot \overset{r}{^}_{2}$ having the same sign as the true radial speed $v_{2} \cdot \overset{r}{^}_{2}$ ) is verified in VERIFY.

PICTURE. At $P_{2}$ , the blue transverse arrow $D \times r_{2} / r_{2}$ (perpendicular to $r_{2}$ ) and the red radial arrow (along $r_{2}$ ) add up to the true velocity $v_{2}$ (yellow).

Step 6 — The magnitude, and the finished formula

WHY a square root and not $μ / N D$ ? Because $h \propto μ p$ — angular momentum scales like the square root of both the gravity strength and the orbit size. A plain ratio would give the wrong units (an acceleration, not a speed).

Now — and only now, with every symbol defined — we can write the finished result: $v_{2} = \frac{μ}{N D} (\frac{D \times r _{2}}{r _{2}} + S)$

$μ / N D$ — the speed scale (Step 6).
$\frac{D \times r _{2}}{r _{2}}$ — the transverse direction (Step 5).
$S$ — the radial contribution (Step 5).

Feed $(r_{2}, v_{2})$ into the state-to-elements conversion of Classical Orbital Elements and the orbit is fully known.

Step 7 — Degenerate & edge cases (never leave the reader stranded)

PICTURE. Left: collinear (no curve possible). Middle: circular (equal radii, $S = 0$ ). Right: hyperbolic ( $e > 1$ , large $S$ , open arc).

The one-picture summary

Everything at once: three coplanar dots → the shared conic → helper vectors $N, D, S$ → the two velocity directions at $P_{2}$ → the assembled $v_{2}$ .

Recall Feynman retelling of the whole walkthrough

You have three photos of a ball flying in a big loop, each marked with an arrow from the centre out to the ball. First you check the three arrowheads lie flat on one tilted sheet — because gravity keeps the loop in a single plane (that's the cross-product test). Then you remember all three obey the same simple curve-law, differing only by an unknown angle. Gibbs's cleverness is three special recipes ( $N$ , $D$ , $S$ ) that mix the arrows so those unknown angles cancel — $N$ and $D$ point out of the sheet and their length-ratio tells you how big the orbit is, while $S$ lies in the sheet and points toward the closest approach. Finally, velocity in a plane can only go "around" or "in-and-out": the around-part is $D$ crossed with the middle arrow, the in-out part comes from $S$ , and a factor $μ / N D$ gives them the right speed. Add them and you have the missing velocity — no stopwatch ever needed.

Recall Quick checks

What makes $D \times r_{2}$ point around the orbit? ::: $D$ is along the normal, so crossing with $r_{2}$ gives an in-plane arrow perpendicular to $r_{2}$ — the transverse direction. Why does a circular orbit give $S = 0$ ? ::: All radii equal, so every $(r_{j} - r_{k}) = 0$ ; a circle has no radial velocity, correctly. What does $N / D$ equal? ::: The semi-latus rectum $p$ , the orbit's scale. Why a square root in the prefactor? ::: Because $h = μ p$ — angular momentum scales as the square root of gravity strength times orbit size. Does Gibbs work for a hyperbola? ::: Yes — the conic law and $h = μ p$ hold for all conics; only collinear/coincident points break it.

Flashcards

Gibbs coplanarity test in one dot product?

\overset{u}{^}_{r 1} \cdot (\overset{u}{^}_{r 2} \times \overset{u}{^}_{r 3}) \approx 0

What are the two directions velocity can have in the orbital plane?

Transverse (around) and radial (in/out)

In Gibbs, which vector points along the orbit normal

h

Both

N

and

D