Orbit determination — Gauss's method, Gibbs method: Worked examples

This page is the drill hall for the parent topic. The parent told you the machinery; here we run it on every kind of input you can meet: clean data, degenerate data, coplanarity-failing data, near-circular vs eccentric orbits, and an exam-style twist. Every symbol below was already earned in the parent note; where a step needs a fact from Angular Momentum & Eccentricity Vector, Lagrange Coefficients (f and g), Classical Orbital Elements or Coordinate Frames (ECI, topocentric) we point there.

The scenario matrix

Every cell below is hit by at least one worked example. "Cell" = the code in the last column.

#	Scenario class	What is special about the input	Expected behaviour	Cell
1	Baseline Gibbs	3 clean coplanar points, moderate eccentricity	clean $v_{2}$	`G-base`
2	Coplanarity PASS check	verify the triple product is tiny	angle $< 1^{\circ}$	`G-coplanar-ok`
3	Coplanarity FAIL	one point nudged out of plane	machine must refuse	`G-coplanar-bad`
4	Near-circular limit	$e \to 0$	radial term $S \to 0$	`G-circular`
5	Sign / quadrant of result	velocity components with mixed signs	direction sanity	`G-signs`
6	Degenerate: collinear points	3 points on a straight line	$D = 0$ , blow-up	`G-degenerate`
7	Gauss root selection	the 8th-degree polynomial	pick the ONE physical root	`Ga-root`
8	Word problem (radar pass)	real numbers, full pipeline	element sanity	`word`
9	Exam twist	"which $v$ ? which root?" traps	avoid the trap	`twist`

We use Earth throughout: $μ = 398600 km^{3} / s^{2}$ .

Definition Notation recap (so line one needs nothing external)

$r_{i}$ = position vector of the body at observation $i$ , tail at Earth's centre, tip at the body. $r_{i} = ∣ r_{i} ∣$ is its length (the distance).
$\times$ = cross product: $a \times b$ is a new vector perpendicular to both, with length = area of the parallelogram they span. It is zero when $a, b$ are parallel.
$\cdot$ = dot product: $a \cdot b = a b cos (angle)$ ; it is zero when the two are perpendicular.
$\overset{u}{^} = u /∣ u ∣$ is the unit vector (same direction, length 1).
$N, D, S$ are the three Gibbs helper vectors; their lengths are $N = ∣ N ∣$ and $D = ∣ D ∣$ . The semi-latus rectum is a scalar, defined as the ratio of two lengths: $p \equiv ∣ N ∣/∣ D ∣$ . We write it shorthand as $p = N / D$ , but read this as "length of $N$ divided by length of $D$ " — it is never a vector-divided-by-vector (that operation does not exist).

Example 1 — Baseline Gibbs `G-base`

Forecast: Guess before computing — is $∣ v_{2} ∣$ nearer $3$ , $5$ , or $8$ km/s for a low-Earth orbit? (Low-Earth circular speed is about $μ / r_{2} \approx 7.3$ km/s, so expect a few km/s per component.)

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

Magnitudes. $r_{1} = 7359.6, r_{2} = 7433.5, r_{3} = 7598.0$ km. Why this step? The scalars $r_{i}$ are the weights inside $N$ and $S$ ; nothing downstream works without them. (Check the third: $2940. 3^{2} + 2473. 7^{2} + 6555. 8^{2} \approx 7598.0$ km.)
The three helper vectors (definitions straight from the parent): $N = r_{1} (r_{2} \times r_{3}) + r_{2} (r_{3} \times r_{1}) + r_{3} (r_{1} \times r_{2})$ $D = (r_{1} \times r_{2}) + (r_{2} \times r_{3}) + (r_{3} \times r_{1})$ $S = r_{1} (r_{2} - r_{3}) + r_{2} (r_{3} - r_{1}) + r_{3} (r_{1} - r_{2})$ Why this step? $N$ and $D$ carry the orbit's scale (their length ratio gives the semi-latus rectum $p = ∣ N ∣/∣ D ∣$ ); $S$ carries the direction toward perigee. In figure s01 the three black arrows are $r_{1}, r_{2}, r_{3}$ from the focus, and the red arrow is $S$ — notice it points along the eccentricity direction $e$ toward perigee.
Velocity formula. $v_{2} = \frac{μ}{N D} (\frac{D \times r _{2}}{r _{2}} + S)$ Why this step? The bracket splits velocity into its only two possible directions on a conic: the transverse part (from $D \times r_{2}$ , perpendicular to $r_{2}$ ) and the radial part (from $S$ , along/against $r_{2}$ ).
Result: $v_{2} \approx (- 6.218, - 4.014, 1.599)$ km/s.

Verify: $∣ v_{2} ∣ \approx 7.53$ km/s — squarely in the "few km/s per component, ~7 total" band we forecast for LEO. Units: $km^{3} s^{- 2} / km^{4} \cdot km = km^{- 1} s^{- 2} \cdot km = km/s$ ✓.

Example 2 — Coplanarity PASS `G-coplanar-ok`

Forecast: Should the "out-of-plane angle" be exactly $0^{\circ}$ or just tiny? (Real data has rounding, so expect tiny, not exactly zero.)

Form the unit vectors $\overset{u}{^}_{r i} = r_{i} / r_{i}$ . Why this step? Direction is all that matters for coplanarity; lengths would just scale the test.
Build the unit normal of the plane through points 2 and 3, then dot with point 1: $C_{123} = \overset{u}{^}_{r 1} \cdot \frac{u ^ _{r 2} \times u ^ _{r 3}}{∣ u ^ _{r 2} \times u ^ _{r 3} ∣} .$ Why this step? $\overset{u}{^}_{r 2} \times \overset{u}{^}_{r 3}$ is perpendicular to the plane of points 2 and 3, but its length is not 1 (it equals $sin$ of the angle between them). Dividing by $∣ \overset{u}{^}_{r 2} \times \overset{u}{^}_{r 3} ∣$ makes it a genuine unit normal, so the dot product $C_{123}$ lands safely in $[- 1, 1]$ and is the sine of point 1's out-of-plane angle. Skipping this normalization would feed $arccos$ a number outside $[- 1, 1]$ and the formula would be invalid.
Convert to an angle: $ϕ_{1∣23} = arcsin (C_{123})$ — the tilt of point 1 out of the plane of 2 and 3. Why this step? $arcsin$ turns the perpendicular-component ratio directly into an angle; $0^{\circ}$ means perfectly coplanar.

Verify: $C_{123} \approx 1 0^{- 3}$ , giving $ϕ_{1∣23}$ well under $1^{\circ}$ ✓ — safe to proceed.

Example 3 — Coplanarity FAIL `G-coplanar-bad`

Forecast: Will Gibbs still spit out some velocity? (Yes — algebra never complains. YOU must complain.)

Recompute the normalized coplanarity number $C_{123} = \overset{u}{^}_{r 1} \cdot \frac{u ^ _{r 2} \times u ^ _{r 3^{'}}}{∣ u ^ _{r 2} \times u ^ _{r 3^{'}} ∣}$ with $r_{3}^{'}$ . Why this step? This is the guard clause, and using the unit normal keeps $C_{123} \in [- 1, 1]$ so $arcsin$ is legal. If the resulting angle is large, no Keplerian orbit threads all three points, so any $v_{2}$ is meaningless.
Convert to an angle $ϕ_{1∣23} = arcsin (C_{123})$ .

Verify: The out-of-plane angle jumps to tens of degrees — far above the $1^{\circ}$ threshold. Reject the data — do not run Steps 2–4 of Gibbs. (This is the [!mistake] the parent warns about: geometry that cannot be one orbit.)

Example 4 — Near-circular limit `G-circular` (figure)

Forecast: For a perfect circle, is there any radial velocity? (No — on a circle you never move toward or away from the focus, so the radial term should vanish.)

Build the three points: $r_{i} = 7000 (cos θ_{i}, sin θ_{i}, 0)$ , so $r_{1} = r_{2} = r_{3} = 7000$ km. Why this step? Equal magnitudes are the algebraic signature of a circle; they will make the $S$ sum collapse.
Look at $S = r_{1} (r_{2} - r_{3}) + r_{2} (r_{3} - r_{1}) + r_{3} (r_{1} - r_{2})$ . With $r_{1} = r_{2} = r_{3}$ , every bracket is zero, so $S = 0$ . Why this step? $S$ was the radial/perigee term. A circle has $e = 0$ : no perigee direction, no radial motion — exactly why $S$ dies. In figure s02 the red arrow is $v_{2}$ ; it lies exactly tangent to the circle (perpendicular to $r_{2}$ ), showing zero radial component.
Velocity reduces to the transverse term only: $v_{2} = μ / (N D) (D \times r_{2}) / r_{2}$ . Why this step? In the Gibbs formula the radial direction enters only through $S$ . Once $S = 0$ (Step 2), the bracket keeps just $D \times r_{2}$ , which is perpendicular to $r_{2}$ — the transverse direction. So the drop-out of the radial piece is a direct algebraic consequence of $S = 0$ , not a coincidence.

Verify: Numerically $∣ v_{2} ∣ \approx 7.546$ km/s, matching circular speed $μ /7000 \approx 7.546$ km/s ✓, and the radial component $v_{2} \cdot \overset{r}{^}_{2} \approx 0$ ✓.

Example 5 — Signs of the result `G-signs`

Forecast: "Climbing" here means altitude — the radial speed $v_{2} \cdot \overset{r}{^}_{2}$ . Positive ⇒ moving away from Earth's centre (altitude rising); negative ⇒ falling. Guess the sign before computing.

Radial speed $= v_{2} \cdot \overset{r}{^}_{2}$ . Sign $> 0$ ⇒ climbing (altitude increasing), $< 0$ ⇒ falling. Why this step? The sign of the radial component is the ONE number that tells you pre- vs post-perigee, independent of quadrant.
Check near-perpendicularity: for a nearly circular case $v_{2} \cdot \overset{r}{^}_{2} \approx 0$ ; here the orbit has some eccentricity so a small nonzero value is expected. Why this step? If $∣ v_{2} \cdot \overset{r}{^}_{2} ∣$ is a small fraction of $∣ v_{2} ∣$ , the motion is mostly transverse, confirming a low-eccentricity orbit.

Verify: $v_{2} \cdot \overset{r}{^}_{2} \approx + 0.44$ km/s (small, positive) ⇒ the body is climbing — its altitude is increasing, so it is on the post-perigee side heading toward apogee; and $∣ v_{2} \cdot \overset{r}{^}_{2} ∣ ≪ ∣ v_{2} ∣$ confirms motion is mostly transverse ✓ — consistent with a low-eccentricity orbit.

Example 6 — Degenerate: collinear points `G-degenerate`

Forecast: A straight radial line cannot be a closed orbit around a focus — so something must divide by zero. What?

Every cross product of two parallel vectors is $0$ : $r_{1} \times r_{2} = r_{2} \times r_{3} = r_{3} \times r_{1} = 0$ . Why this step? $D$ is a sum of exactly these cross products, so $D = 0$ and $D = 0$ .
The prefactor $μ / (N D)$ then has $D = 0$ in the denominator → undefined. Why this step? This is the machine's honest way of saying "collinear points define no orbital plane." Also $N = 0$ , so $p = N / D$ is $0/0$ — doubly degenerate.

Verify: $D = ∣ D ∣ = 0$ exactly ✓ ⇒ Gibbs is undefined. Interpretation: collinear points give no unique conic, so reject — you need genuinely spread-out geometry.

Example 7 — Gauss root selection `Ga-root`

Forecast: An 8th-degree polynomial has up to 8 roots. How many are physically legal for a satellite?

Solve numerically; keep only real roots. Why this step? $r_{2}$ is a distance — complex roots are non-physical artefacts.
Discard non-positive roots and any with $r_{2} <$ Earth radius ( $6378$ km). Why this step? A satellite cannot sit inside the Earth; negative distances are meaningless. This is the parent's [!mistake] "take any root."
The surviving positive real root near $\approx 7500$ km is $r_{2}$ .

Verify: The polynomial has exactly one positive real root $\approx 7548$ km $> 6378$ km ✓; all others are negative or complex and are discarded.

Example 8 — Word problem: full radar pass `word`

Forecast: Radar caught a LEO satellite. Do you expect $e$ near $0$ (nearly circular) or near $1$ (highly elliptical)?

Run Gibbs (Example 1) to get $v_{2} \approx (- 6.218, - 4.014, 1.599)$ km/s. Why this step? Classical Orbital Elements all flow from the state $(r_{2}, v_{2})$ ; you need velocity first.
Angular momentum $h = r_{2} \times v_{2}$ , then $h = ∣ h ∣$ . Why this step? From Angular Momentum & Eccentricity Vector, $h$ fixes the plane and scale.
Eccentricity vector $e = \frac{v _{2} \times h}{μ} - \overset{r}{^}_{2}$ , then $e = ∣ e ∣$ . Why this step? $e$ is the shape number the operator asked for.

Verify: Computation gives $h \approx 56, 000 km^{2} / s$ and $e \approx 0.10$ — small, so nearly circular as forecast for a LEO ✓. (Units of $h$ : km·(km/s)=km²/s ✓; $e$ is dimensionless ✓.)

Example 9 — Exam twist `twist`

Forecast: Two traps are baked in. Name them before reading on: one is about which point, the other about which root.

Trap 1 — wrong point. The symmetric sums $N, D, S$ are built centred on the middle point, so the boxed formula yields $v_{2}$ only. Feeding it as " $v_{1}$ " is invalid — you would have to re-centre the whole construction on point 1, which the standard formula does not do. Why this step? Using $v_{1}$ or $v_{3}$ is exactly the parent's [!mistake]; the algebra is not symmetric under relabelling, so the student's $v_{1} = (- 3.2, 8.1, 1.0)$ is a mislabelled/invalid output. Fix: report the velocity only at the middle point, $v_{2} \approx (- 6.218, - 4.014, 1.599)$ km/s from Example 1.
Trap 2 — root choice. Both roots are positive and above Earth's radius, so you cannot reject on radius alone. Back-substitute each candidate $r_{2}$ into the slant-range equations and keep the one giving all positive $ρ_{i}$ . Why this step? Recall from Coordinate Frames (ECI, topocentric) that each satellite position is $r_{i} = R_{i} + ρ_{i} \overset{ρ}{^}_{i}$ , where $R_{i}$ is the known observer position, $\overset{ρ}{^}_{i}$ the measured line-of-sight unit vector, and $ρ_{i}$ the slant range (distance to the target). A physical range must satisfy $ρ_{i} > 0$ — you cannot look behind yourself to see the satellite.
Break the tie by orbital speed. Convert each root to the orbital speed it implies via the near-circular estimate $v \approx μ / r_{2}$ . For $r_{2} = 7548$ km, $v \approx 7.27$ km/s (sensible LEO). For $r_{2} = 650000$ km, $v \approx 0.78$ km/s — far too slow for the tightly-spaced short arc that produced these observations. Why this step? A short observed arc means the body swept a large angle quickly, which is impossible at $0.78$ km/s. This rules out the huge root physically, independent of the range test.

Answer: Correct velocity is $v_{2} \approx (- 6.218, - 4.014, 1.599)$ km/s (middle point, not $v_{1}$ ); correct range solution is $r_{2} = 7548$ km (positive $ρ_{i}$ , sensible speed), and $r_{2} = 650000$ km is discarded.

Verify (numeric portion): For the near-circular data, the correct centred computation reproduces $∣ v_{2} ∣ \approx 7.53$ km/s (Example 1), while any relabelling changes the answer — confirming the formula is point-specific ✓. Root $650000$ km implies orbital speed $μ /650000 \approx 0.78$ km/s $< 1$ km/s, absurd for the observed short arc, so it is discarded ✓.

Recall Quick self-test on the matrix

When does $D = 0$ ? ::: When the three points are collinear (all cross products vanish). What makes $S = 0$ ? ::: A circular orbit, $r_{1} = r_{2} = r_{3}$ ⇒ every bracket is zero, no radial motion. Which velocity does Gibbs give? ::: Only $v_{2}$ , at the middle point. How do you break a tie between two positive real Gauss roots? ::: Keep the root giving all positive slant ranges $ρ_{i}$ (and a sensible orbital speed). What is the coplanarity red flag threshold? ::: Out-of-plane angle above about $1^{\circ}$ . Why must $\overset{u}{^}_{r 2} \times \overset{u}{^}_{r 3}$ be normalized before $arcsin$ ? ::: Its length is $sin$ (angle), not 1, so without dividing by it the coplanarity number can fall outside $[- 1, 1]$ .