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

3.2.18 · D2 · Physics › Orbital Mechanics & Astrodynamics › Orbit determination — Gauss's method, Gibbs method

Hum Gibbs's method ke central result ki taraf ja rahe hain: $v_{2}$ ke liye ek formula — middle point ki velocity — sirf teen position arrows se bana. Ise abhi print nahi karte (uske symbols abhi define nahi hue), balki neeche steps mein live assemble karte hain.

Prerequisites jinpar hum rely karte hain: Two-body Problem, Angular Momentum & Eccentricity Vector, aur baad mein Classical Orbital Elements.

Step 1 — Space mein teen dots, aur "position vector" ka matlab

KYA HAI. Humein teen arrows $r_{1}, r_{2}, r_{3}$ diye gaye hain — ek hi satellite ke teen snapshots.

KYUN. Ek single orbit space mein ek fixed curve hoti hai. Agar teen arrows sab us curve ko touch karte hain, toh ve heavily constrained hain: curve ko teenon arrowheads se guzarna hai aur Earth ko apne focus par rakhna hai. Yahi sab kuch reconstruct karne ke liye kaafi hai — speed bhi include karke. Koi stopwatch ki zaroorat nahi.

PICTURE. Neeche teen coloured arrows dekho, sab ek hi origin (Earth ka centre, black dot) se nikal rahe hain. Unke tips $P_{1}, P_{2}, P_{3}$ satellite ke teen "photos" hain.

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

Is poore page mein har $r_{i}$ inhi arrows mein se ek hai, aur har magnitude $r_{i} = ∣ r_{i} ∣$ bas yeh batati hai ki woh arrow kitna lamba hai.

Step 2 — Ye sab coplanar hone chahiye: flat-sheet test

KYA HAI. Data par trust karne se pehle check karo ki $P_{1}, P_{2}, P_{3}$ sach mein ek plane share karte hain.

KYUN. Agar nahi karte, toh koi single Keplerian orbit teeno se nahi guzar sakti, aur jo bhi velocity hum compute karein woh bakwaas hogi. Yeh pehla sanity gate hai.

Cross product se test kaise karein. Do arrows ka cross product $a \times b$ ek naya arrow produce karta hai jo dono ke perpendicular hota hai. Toh $r_{2} \times r_{3}$ , $r_{2}$ aur $r_{3}$ wale plane ke perpendicular hai. Agar $r_{1}$ bhi usi plane mein hai, toh $r_{1}$ us perpendicular ke perpendicular hai, isliye unka dot product (ek number jo overlap measure karta hai, zero jab perpendicular ho) zero hoga: $unit arrow to P_{1} \overset{u}{^}_{r 1} \cdot normal to plane of P_{2}, P_{3} (\overset{u}{^}_{r 2} \times \overset{u}{^}_{r 3}) \approx 0$

Yahan $\overset{u}{^}_{r i} = r_{i} / r_{i}$ har arrow ka unit version hai (length 1, sirf direction) — hum normalize isliye karte hain taaki test arrows ki lengths par depend na kare.

PICTURE. Green shaded sheet orbital plane hai. Yellow arrow normal $r_{2} \times r_{3}$ hai; notice karo $r_{1}$ sheet mein flat pada hai, isliye yellow normal se right angle banta hai — dot product zero.

Step 3 — Woh ek equation jise har point follow karta hai: conic

KYA HAI. Woh ek algebraic law likho jo teenon points share karte hain.

KYUN. Yeh secret glue hai. Teenon arrowheads is same equation ko usi same $p$ aur $e$ ke saath follow karte hain — sirf $θ$ ek point se doosre par change hota hai. Gibbs ki poori trick yeh hai ki is equation ki teen copies ko combine karo taaki unknown angles $θ_{i}$ cancel ho jaayein, sirf measurable arrows bachein.

PICTURE. Ellipse jiska focus Earth's centre par hai; semi-latus rectum $p$ focus par special vertical distance hai; $θ$ perigee (red dot) se measure hota hai.

Us convention ke fix hote hi, har point ka arrow cleanly decompose hota hai: $r_{i} = along \overset{e}{^} r_{i} cos θ_{i} \overset{e}{^} + along \overset{p}{^} r_{i} sin θ_{i} \overset{p}{^}$ "perigee-direction mein kitna aage" plus "sideways kitna". $θ_{i}$ angles hi sirf unknowns hain — aur yahi agla step khatam karta hai.

Step 4 — Unknown angles ko khatam karna: $N$ , $D$ , $S$ banao

KYA HAI. Data se yeh teen helper vectors banao: $N = r_{1} (r_{2} \times r_{3}) + r_{2} (r_{3} \times r_{1}) + r_{3} (r_{1} \times r_{2})$ $D = (r_{1} \times r_{2}) + (r_{2} \times r_{3}) + (r_{3} \times r_{1})$ $S = r_{1} (r_{2} - r_{3}) + r_{2} (r_{3} - r_{1}) + r_{3} (r_{1} - r_{2})$

Har symbol ko uski jagah par padhna:

$N$ mein: har cross product $r_{j} \times r_{k}$ ek area-arrow hai jo plane ke perpendicular hai (yeh $h$ ke along point karta hai). Har ek scalar magnitude $r_{i}$ se weighted hai. Toh $N$ plane-normal arrows ka stack hai — yeh orbit normal ke along point karta hai aur uski length scale carry karti hai.
$D$ mein: wahi teen cross products, lekin unweighted. $D$ bhi normal ke along point karta hai. Lengths ka ratio $N / D$ exactly scale $p$ nikalta hai.
$S$ mein: notice karo har arrow $r_{i}$ , doosre do distances ke difference $(r_{j} - r_{k})$ se weighted hai. Distances ke differences encode karte hain "perigee ki taraf downhill kidhar hai" — toh $S$ plane mein lie karta hai aur $\overset{e}{^}$ se related direction mein point karta hai.

Formula

$N ∥ h$ , $D ∥ h$ , aur $N / D = p$ kyun $D$ normal ke along kyun hai. Har term $r_{j} \times r_{k}$ cross product ke meaning se, plane ke perpendicular hai — yaani $\hat{h}$ ke parallel hai. Ek hi direction (sab $\pm \hat{h}$ ) mein point karne wale arrows ka sum phir se $\hat{h}$ ke along hoga. Toh $D = D \hat{h}$ . Yahi argument $N = N \hat{h}$ banata hai: har term ek scalar times ek plane-normal cross product hai. Isliye dono $N$ aur $D$ , $h$ ke parallel hain, aur therefore ek doosre ke parallel bhi — unka cross product zero hai. $N / D = p$ kyun. Conic $r_{i} = p / (1 + e cos θ_{i})$ aur basis decomposition ko $N$ aur $D$ mein substitute karo. Har cross product $r_{j} \times r_{k} = r_{j} r_{k} sin (θ_{k} - θ_{j}) \hat{h}$ . Algebra carry karne par, sums exactly $p$ ke ek factor se differ karte hain: $N = p D$ . Geometrically: $D$ unweighted signed areas measure karta hai, $N$ distance-weighted ones, aur conic law force karti hai ki unka ratio semi-latus rectum ho. VERIFY mein numeric checks confirm karte hain ki worked example par dono $N \times D = 0$ aur $N / D = h^{2} / μ = p$ hain.

PICTURE. $N$ aur $D$ (dono yellow normal ke along, plane se bahar), aur $S$ green plane mein flat rehkar perigee ki taraf point karta hua.

Step 5 — Do velocity directions recover karna

KYA HAI. Transverse direction hai $\frac{D \times r _{2}}{r _{2}}$ ; radial content $S$ se aata hai.

Transverse term padhna. $D$ normal ke along point karta hai (plane se bahar). Ise $r_{2}$ se cross karne par ek ऐसा arrow milta hai jo plane mein hai aur $r_{2}$ ke perpendicular hai — exactly "orbit ke around" direction. $r_{2}$ se divide karna uski length calibrated rakhta hai.

Radial term padhna — projection check ke saath. Radial unit direction hai $\overset{r}{^}_{2} = r_{2} / r_{2}$ . Yeh dikhane ke liye ki $S$ genuinely in/out motion supply karta hai, ise $\overset{r}{^}_{2}$ par project karo. Conic aur basis ko $S$ mein substitute karne par (Step 4 jaisi algebra), milta hai $S = e (sin θ_{2} \overset{e}{^} - cos θ_{2} \overset{p}{^}) \cdot (scale) \Rightarrow S \cdot \overset{r}{^}_{2} \propto e sin θ_{2},$ jo nonzero exactly tab hota hai jab satellite chadh ya utar raha ho (true anomaly $θ_{2}$ , $0$ ya $180°$ se alag) aur perigee/apogee par zero hota hai, jahan radial speed vanish honi chahiye. Yahi radial velocity ki fingerprint hai: yeh $e sin θ$ track karta hai. Toh $S$ radial part carry karta hai; transverse term baaki carry karta hai. VERIFY mein ek numeric projection ( $S \cdot \overset{r}{^}_{2}$ ka true radial speed $v_{2} \cdot \overset{r}{^}_{2}$ ke saath same sign hona) verify kiya gaya hai.

PICTURE. $P_{2}$ par, blue transverse arrow $D \times r_{2} / r_{2}$ ( $r_{2}$ ke perpendicular) aur red radial arrow ( $r_{2}$ ke along) milke true velocity $v_{2}$ (yellow) banate hain.

Step 6 — Magnitude, aur completed formula

Square root kyun, $μ / N D$ nahi? Kyunki $h \propto μ p$ — angular momentum gravity strength aur orbit size dono ke square root ke jaisa scale hota hai. Ek plain ratio galat units deta (ek acceleration, speed nahi).

Ab — aur sirf ab, har symbol define hone ke baad — hum finished result likh sakte hain: $v_{2} = \frac{μ}{N D} (\frac{D \times r _{2}}{r _{2}} + S)$

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

$(r_{2}, v_{2})$ ko Classical Orbital Elements ke state-to-elements conversion mein feed karo aur orbit poori tarah jaani jaayegi.

Step 7 — Degenerate & edge cases (reader ko kabhi stranded mat chhoddo)

PICTURE. Left: collinear (koi curve possible nahi). Middle: circular (equal radii, $S = 0$ ). Right: hyperbolic ( $e > 1$ , bada $S$ , open arc).

Ek-picture summary

Sab ek saath: teen coplanar dots → shared conic → helper vectors $N, D, S$ → $P_{2}$ par do velocity directions → assembled $v_{2}$ .

Recall Poore walkthrough ki Feynman-style retelling

Tumhare paas ek ball ki teen photos hain jo ek bade loop mein ud rahi hai, har ek mein centre se ball tak ek arrow marked hai. Pehle check karo ki teenon arrowheads ek tilted sheet par flat hain — kyunki gravity loop ko ek single plane mein rakhti hai (yahi cross-product test hai). Phir yaad karo ki teeno usi simple curve-law ko follow karte hain, sirf ek unknown angle se alag hote hain. Gibbs ki cleverness yeh hai ki teen special recipes ( $N$ , $D$ , $S$ ) arrows ko aisi tarah mix karti hain ki woh unknown angles cancel ho jaayein — $N$ aur $D$ sheet se bahar point karte hain aur unka length-ratio batata hai orbit kitni badi hai, jabki $S$ sheet mein rehta hai aur closest approach ki taraf point karta hai. Aakhir mein, ek plane mein velocity sirf "around" ya "andar-bahar" ja sakti hai: around-part hai $D$ crossed with middle arrow, in-out part $S$ se aata hai, aur ek factor $μ / N D$ unhe sahi speed deta hai. Inhe add karo aur tumhare paas missing velocity hai — koi stopwatch kabhi nahi chahiye.

Recall Quick checks

$D \times r_{2}$ orbit ke around kyun point karta hai? ::: $D$ normal ke along hai, toh $r_{2}$ se cross karne par ek in-plane arrow milta hai jo $r_{2}$ ke perpendicular hai — transverse direction. Circular orbit kyun $S = 0$ deti hai? ::: Saari radii equal hain, toh har $(r_{j} - r_{k}) = 0$ ; circle mein radial velocity nahi hoti, correctly. $N / D$ kiske barabar hai? ::: Semi-latus rectum $p$ , orbit ka scale. Prefactor mein square root kyun? ::: Kyunki $h = μ p$ — angular momentum gravity strength times orbit size ke square root ke jaisa scale karta hai. Kya Gibbs hyperbola ke liye kaam karta hai? ::: Haan — conic law aur $h = μ p$ sab conics ke liye hold karte hain; sirf collinear/coincident points ise todta hai.

Flashcards

Gibbs coplanarity test ek dot product mein?

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

Orbital plane mein velocity ki do directions kya hain?

Transverse (around) aur radial (in/out)

Gibbs mein kaun sa vector orbit normal

h

ke along point karta hai?

Dono

N

aur

D