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

3.2.18 · D3 · Physics › Orbital Mechanics & Astrodynamics › Orbit determination — Gauss's method, Gibbs method

Yeh page parent topic ka drill hall hai. Parent ne aapko machinery batayi; yahan hum use har tarah ke input par run karte hain jo aap encounter kar sakte ho: clean data, degenerate data, coplanarity-failing data, near-circular vs eccentric orbits, aur ek exam-style twist. Neeche har symbol parent note mein pehle se earn kiya ja chuka hai; jahan kisi step ko Angular Momentum & Eccentricity Vector, Lagrange Coefficients (f and g), Classical Orbital Elements ya Coordinate Frames (ECI, topocentric) se koi fact chahiye, hum wahan point kar denge.

Scenario matrix

Neeche har cell ko kam se kam ek worked example cover karta hai. "Cell" = aakhri column mein code.

#	Scenario class	Input mein kya khaas hai	Expected behaviour	Cell
1	Baseline Gibbs	3 clean coplanar points, moderate eccentricity	clean $v_{2}$	`G-base`
2	Coplanarity PASS check	verify karo ki triple product tiny hai	angle $< 1^{\circ}$	`G-coplanar-ok`
3	Coplanarity FAIL	ek point plane ke bahar nudge kiya	machine ko refuse karna chahiye	`G-coplanar-bad`
4	Near-circular limit	$e \to 0$	radial term $S \to 0$	`G-circular`
5	Sign / quadrant of result	velocity components mixed signs ke saath	direction sanity	`G-signs`
6	Degenerate: collinear points	3 points ek straight line par	$D = 0$ , blow-up	`G-degenerate`
7	Gauss root selection	8th-degree polynomial	ek PHYSICAL root chunna	`Ga-root`
8	Word problem (radar pass)	real numbers, full pipeline	element sanity	`word`
9	Exam twist	"kaun sa $v$ ? kaun sa root?" traps	trap se bachna	`twist`

Hum poore time Earth use karte hain: $μ = 398600 km^{3} / s^{2}$ .

Definition Notation recap (taaki pehli line ko kuch external na chahiye)

$r_{i}$ = observation $i$ par body ka position vector, tail Earth ke centre par, tip body par. $r_{i} = ∣ r_{i} ∣$ uski length hai (distance).
$\times$ = cross product: $a \times b$ ek naya vector hai jo dono ke perpendicular hai, jiska length = unke span kiye parallelogram ka area. Yeh zero hota hai jab $a, b$ parallel hon.
$\cdot$ = dot product: $a \cdot b = a b cos (angle)$ ; yeh zero hota hai jab dono perpendicular hon.
$\overset{u}{^} = u /∣ u ∣$ unit vector hai (same direction, length 1).
$N, D, S$ teen Gibbs helper vectors hain; unki lengths $N = ∣ N ∣$ aur $D = ∣ D ∣$ hain. Semi-latus rectum ek scalar hai, do lengths ke ratio ke roop mein define kiya gaya: $p \equiv ∣ N ∣/∣ D ∣$ . Hum ise shorthand mein $p = N / D$ likhte hain, lekin ise "length of $N$ divided by length of $D$ " padho — yeh kabhi bhi vector-divided-by-vector nahi hai (woh operation exist hi nahi karta).

Example 1 — Baseline Gibbs `G-base`

Forecast: Compute karne se pehle guess karo — kya low-Earth orbit ke liye $∣ v_{2} ∣$ $3$ , $5$ , ya $8$ km/s ke naya hogi? (Low-Earth circular speed lagbhag $μ / r_{2} \approx 7.3$ km/s hai, toh har component mein kuch km/s expect karo.)

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

Magnitudes. $r_{1} = 7359.6, r_{2} = 7433.5, r_{3} = 7598.0$ km. Yeh step kyun? Scalars $r_{i}$ $N$ aur $S$ ke andar weights hain; inke bina koi bhi downstream kaam nahi karega. (Teesra check karo: $2940. 3^{2} + 2473. 7^{2} + 6555. 8^{2} \approx 7598.0$ km.)
Teen helper vectors (parent se seedhe definitions): $N = r_{1} (r_{2} \times r_{3}) + r_{2} (r_{3} \times r_{1}) + r_{3} (r_{1} \times r_{2})$ $D = (r_{1} \times r_{2}) + (r_{2} \times r_{3}) + (r_{3} \times r_{1})$ $S = r_{1} (r_{2} - r_{3}) + r_{2} (r_{3} - r_{1}) + r_{3} (r_{1} - r_{2})$ Yeh step kyun? $N$ aur $D$ orbit ka scale carry karte hain (unka length ratio semi-latus rectum $p = ∣ N ∣/∣ D ∣$ deta hai); $S$ perigee ki taraf direction carry karta hai. Figure s01 mein teen black arrows $r_{1}, r_{2}, r_{3}$ focus se hain, aur red arrow $S$ hai — notice karo yeh eccentricity direction $e$ ke saath perigee ki taraf point karta hai.
Velocity formula. $v_{2} = \frac{μ}{N D} (\frac{D \times r _{2}}{r _{2}} + S)$ Yeh step kyun? Bracket velocity ko uski sirf do possible directions mein split karta hai ek conic par: transverse part ( $D \times r_{2}$ se, $r_{2}$ ke perpendicular) aur radial part ( $S$ se, $r_{2}$ ke saath/against).
Result: $v_{2} \approx (- 6.218, - 4.014, 1.599)$ km/s.

Verify: $∣ v_{2} ∣ \approx 7.53$ km/s — bilkul "har component mein kuch km/s, total ~7" band mein jo humne LEO ke liye forecast kiya tha. 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: Kya "out-of-plane angle" exactly $0^{\circ}$ honi chahiye ya sirf tiny? (Real data mein rounding hoti hai, toh exactly zero nahi, tiny expect karo.)

Unit vectors banao $\overset{u}{^}_{r i} = r_{i} / r_{i}$ . Yeh step kyun? Coplanarity ke liye sirf direction matter karta hai; lengths toh bas test ko scale kar deti.
Points 2 aur 3 se guzarne wale plane ka unit normal banao, phir point 1 ke saath dot karo: $C_{123} = \overset{u}{^}_{r 1} \cdot \frac{u ^ _{r 2} \times u ^ _{r 3}}{∣ u ^ _{r 2} \times u ^ _{r 3} ∣} .$ Yeh step kyun? $\overset{u}{^}_{r 2} \times \overset{u}{^}_{r 3}$ points 2 aur 3 ke plane ke perpendicular hai, lekin uski length 1 nahi hai (yeh unke beech ke angle ka $sin$ hai). $∣ \overset{u}{^}_{r 2} \times \overset{u}{^}_{r 3} ∣$ se divide karne par yeh genuine unit normal ban jaata hai, toh dot product $C_{123}$ safely $[- 1, 1]$ mein land karta hai aur point 1 ke out-of-plane angle ka sine hai. Yeh normalization skip karne par $arccos$ ko $[- 1, 1]$ ke bahar number milega aur formula invalid ho jaayega.
Angle mein convert karo: $ϕ_{1∣23} = arcsin (C_{123})$ — point 1 ka 2 aur 3 ke plane se bahar tilt. Yeh step kyun? $arcsin$ perpendicular-component ratio ko seedha angle mein convert karta hai; $0^{\circ}$ ka matlab perfectly coplanar hai.

Verify: $C_{123} \approx 1 0^{- 3}$ , jisse $ϕ_{1∣23}$ $1^{\circ}$ se bahut neeche aata hai ✓ — proceed karna safe hai.

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

Forecast: Kya Gibbs phir bhi koi velocity output dega? (Haan — algebra kabhi complain nahi karta. AAPKO complain karna hai.)

Normalized coplanarity number recompute karo $C_{123} = \overset{u}{^}_{r 1} \cdot \frac{u ^ _{r 2} \times u ^ _{r 3^{'}}}{∣ u ^ _{r 2} \times u ^ _{r 3^{'}} ∣}$ $r_{3}^{'}$ ke saath. Yeh step kyun? Yeh guard clause hai, aur unit normal use karna $C_{123} \in [- 1, 1]$ rakhta hai taaki $arcsin$ legal rahe. Agar resulting angle bada hai, toh koi Keplerian orbit teeno points ko thread nahi kar sakta, toh koi bhi $v_{2}$ meaningless hai.
Angle mein convert karo $ϕ_{1∣23} = arcsin (C_{123})$ .

Verify: Out-of-plane angle tens of degrees tak jump karta hai — $1^{\circ}$ threshold se bahut zyada. Data reject karo — Gibbs ke Steps 2–4 mat run karo. (Yeh wahi [!mistake] hai jo parent warn karta hai: aisi geometry jo ek orbit nahi ho sakti.)

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

Forecast: Perfect circle ke liye, kya koi radial velocity hogi? (Nahi — circle par aap kabhi focus se door ya paas nahi jaate, toh radial term vanish hona chahiye.)

Teen points banao: $r_{i} = 7000 (cos θ_{i}, sin θ_{i}, 0)$ , toh $r_{1} = r_{2} = r_{3} = 7000$ km. Yeh step kyun? Equal magnitudes circle ka algebraic signature hain; yeh $S$ sum ko collapse kar denge.
Dekho $S = r_{1} (r_{2} - r_{3}) + r_{2} (r_{3} - r_{1}) + r_{3} (r_{1} - r_{2})$ . $r_{1} = r_{2} = r_{3}$ ke saath, har bracket zero hai, toh $S = 0$ . Yeh step kyun? $S$ radial/perigee term tha. Circle ka $e = 0$ hota hai: koi perigee direction nahi, koi radial motion nahi — exactly isliye $S$ die ho jaata hai. Figure s02 mein red arrow $v_{2}$ hai; yeh circle ke exactly tangent par hai ( $r_{2}$ ke perpendicular), zero radial component dikhata hai.
Velocity sirf transverse term tak reduce ho jaati hai: $v_{2} = μ / (N D) (D \times r_{2}) / r_{2}$ . Yeh step kyun? Gibbs formula mein radial direction sirf $S$ ke through enter karti hai. Jab $S = 0$ (Step 2), bracket mein sirf $D \times r_{2}$ bachta hai, jo $r_{2}$ ke perpendicular hai — transverse direction. Toh radial piece ka drop-out $S = 0$ ka direct algebraic consequence hai, coincidence nahi.

Verify: Numerically $∣ v_{2} ∣ \approx 7.546$ km/s, circular speed $μ /7000 \approx 7.546$ km/s ✓ se match karta hai, aur radial component $v_{2} \cdot \overset{r}{^}_{2} \approx 0$ ✓.

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

Forecast: Yahan "Climbing" ka matlab altitude hai — radial speed $v_{2} \cdot \overset{r}{^}_{2}$ . Positive ⇒ Earth ke centre se door ja raha hai (altitude badh rahi hai); negative ⇒ gir raha hai. Compute karne se pehle sign guess karo.

Radial speed $= v_{2} \cdot \overset{r}{^}_{2}$ . Sign $> 0$ ⇒ climbing (altitude increasing), $< 0$ ⇒ falling. Yeh step kyun? Radial component ka sign woh EK number hai jo aapko pre- vs post-perigee batata hai, quadrant se independent.
Near-perpendicularity check karo: nearly circular case ke liye $v_{2} \cdot \overset{r}{^}_{2} \approx 0$ ; yahan orbit mein kuch eccentricity hai toh ek small nonzero value expected hai. Yeh step kyun? Agar $∣ v_{2} \cdot \overset{r}{^}_{2} ∣$ $∣ v_{2} ∣$ ka ek small fraction hai, toh motion mostly transverse hai, jo low-eccentricity orbit confirm karta hai.

Verify: $v_{2} \cdot \overset{r}{^}_{2} \approx + 0.44$ km/s (small, positive) ⇒ body climbing kar rahi hai — uski altitude increase ho rahi hai, toh yeh post-perigee side par apogee ki taraf ja raha hai; aur $∣ v_{2} \cdot \overset{r}{^}_{2} ∣ ≪ ∣ v_{2} ∣$ confirm karta hai ki motion mostly transverse hai ✓ — low-eccentricity orbit ke consistent.

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

Forecast: Ek straight radial line focus ke around ek closed orbit nahi ho sakta — toh kuch divide by zero hoga. Kya?

Do parallel vectors ke har cross product ka result $0$ hota hai: $r_{1} \times r_{2} = r_{2} \times r_{3} = r_{3} \times r_{1} = 0$ . Yeh step kyun? $D$ exactly inhi cross products ka sum hai, toh $D = 0$ aur $D = 0$ .
Prefactor $μ / (N D)$ mein phir denominator mein $D = 0$ aata hai → undefined. Yeh step kyun? Machine ka yeh honest tarika hai yeh kehne ka "collinear points koi orbital plane define nahi karte." Saath hi $N = 0$ bhi, toh $p = N / D$ hai $0/0$ — doubly degenerate.

Verify: $D = ∣ D ∣ = 0$ exactly ✓ ⇒ Gibbs undefined hai. Interpretation: collinear points koi unique conic nahi dete, toh reject karo — aapko genuinely spread-out geometry chahiye.

Example 7 — Gauss root selection `Ga-root`

Forecast: 8th-degree polynomial ke paas 8 roots tak ho sakte hain. Ek satellite ke liye kitne physically legal hain?

Numerically solve karo; sirf real roots rakhna. Yeh step kyun? $r_{2}$ ek distance hai — complex roots non-physical artefacts hain.
Non-positive roots aur koi bhi $r_{2} <$ Earth radius ( $6378$ km) wale discard karo. Yeh step kyun? Ek satellite Earth ke andar nahi baith sakta; negative distances meaningless hain. Yeh parent ka [!mistake] hai "koi bhi root le lo."
$\approx 7500$ km ke paas surviving positive real root $r_{2}$ hai.

Verify: Polynomial ka exactly ek positive real root $\approx 7548$ km $> 6378$ km hai ✓; baaki sab negative ya complex hain aur discard ho jaate hain.

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

Forecast: Radar ne ek LEO satellite catch kiya. Kya aap $e$ $0$ ke paas expect karte ho (nearly circular) ya $1$ ke paas (highly elliptical)?

$v_{2} \approx (- 6.218, - 4.014, 1.599)$ km/s paane ke liye Gibbs run karo (Example 1). Yeh step kyun? Classical Orbital Elements saare state $(r_{2}, v_{2})$ se flow karte hain; pehle velocity chahiye.
Angular momentum $h = r_{2} \times v_{2}$ , phir $h = ∣ h ∣$ . Yeh step kyun? Angular Momentum & Eccentricity Vector se, $h$ plane aur scale fix karta hai.
Eccentricity vector $e = \frac{v _{2} \times h}{μ} - \overset{r}{^}_{2}$ , phir $e = ∣ e ∣$ . Yeh step kyun? $e$ woh shape number hai jo operator ne maanga tha.

Verify: Computation deta hai $h \approx 56, 000 km^{2} / s$ aur $e \approx 0.10$ — small, toh nearly circular jaise LEO ke liye forecast kiya tha ✓. ( $h$ ke units: km·(km/s)=km²/s ✓; $e$ dimensionless hai ✓.)

Example 9 — Exam twist `twist`

Forecast: Dono trap baked in hain. Padhne se pehle unhe naam do: ek kaun sa point ke baare mein hai, doosra kaun sa root ke baare mein.

Trap 1 — wrong point. Symmetric sums $N, D, S$ middle point par centred hain, toh boxed formula sirf $v_{2}$ hi deta hai. Ise " $v_{1}$ " bolna invalid hai — aapko poori construction ko point 1 par re-centre karna hoga, jo standard formula nahi karta. Yeh step kyun? $v_{1}$ ya $v_{3}$ use karna exactly parent ka [!mistake] hai; algebra relabelling ke under symmetric nahi hai, toh student ka $v_{1} = (- 3.2, 8.1, 1.0)$ ek mislabelled/invalid output hai. Fix: velocity sirf middle point par report karo, Example 1 se $v_{2} \approx (- 6.218, - 4.014, 1.599)$ km/s.
Trap 2 — root choice. Dono roots positive hain aur Earth ke radius se upar, toh sirf radius se reject nahi kar sakte. Har candidate $r_{2}$ ko slant-range equations mein back-substitute karo aur woh rakho jo saari positive $ρ_{i}$ de. Yeh step kyun? Coordinate Frames (ECI, topocentric) se yaad karo ki har satellite position hai $r_{i} = R_{i} + ρ_{i} \overset{ρ}{^}_{i}$ , jahan $R_{i}$ known observer position hai, $\overset{ρ}{^}_{i}$ measured line-of-sight unit vector hai, aur $ρ_{i}$ slant range hai (target tak distance). Ek physical range satisfy karni chahiye $ρ_{i} > 0$ — aap satellite dekhne ke liye apne peeche nahi dekh sakte.
Orbital speed se tie break karo. Har root ko near-circular estimate $v \approx μ / r_{2}$ se uski implied orbital speed mein convert karo. $r_{2} = 7548$ km ke liye, $v \approx 7.27$ km/s (sensible LEO). $r_{2} = 650000$ km ke liye, $v \approx 0.78$ km/s — in observations ko produce karne wale tightly-spaced short arc ke liye bahut slow. Yeh step kyun? Ek short observed arc ka matlab body ne quickly bada angle sweep kiya, jo $0.78$ km/s par impossible hai. Yeh range test se independent, bade root ko physically rule out karta hai.

Answer: Correct velocity hai $v_{2} \approx (- 6.218, - 4.014, 1.599)$ km/s (middle point, $v_{1}$ nahi); correct range solution hai $r_{2} = 7548$ km (positive $ρ_{i}$ , sensible speed), aur $r_{2} = 650000$ km discard hai.

Verify (numeric portion): Near-circular data ke liye, correct centred computation $∣ v_{2} ∣ \approx 7.53$ km/s reproduce karta hai (Example 1), jabki koi bhi relabelling answer change kar deta hai — confirm karta hai ki formula point-specific hai ✓. Root $650000$ km orbital speed $μ /650000 \approx 0.78$ km/s $< 1$ km/s imply karta hai, observed short arc ke liye absurd, toh yeh discard hai ✓.

Recall Matrix par quick self-test

$D = 0$ kab hota hai? ::: Jab teen points collinear hon (saare cross products vanish ho jaate hain). $S = 0$ kya banata hai? ::: Circular orbit, $r_{1} = r_{2} = r_{3}$ ⇒ har bracket zero hai, koi radial motion nahi. Gibbs kaun si velocity deta hai? ::: Sirf $v_{2}$ , middle point par. Do positive real Gauss roots ke beech tie kaise todein? ::: Woh root rakho jo saari positive slant ranges $ρ_{i}$ de (aur ek sensible orbital speed bhi). Coplanarity red flag threshold kya hai? ::: Out-of-plane angle lagbhag $1^{\circ}$ se upar. $arcsin$ se pehle $\overset{u}{^}_{r 2} \times \overset{u}{^}_{r 3}$ ko normalize kyun karna chahiye? ::: Uski length $sin$ (angle) hai, 1 nahi, toh usse divide kiye bina coplanarity number $[- 1, 1]$ se bahar ja sakta hai.