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

Method: Gibbs. Gibbs ka input exactly teen position vectors hain, aur isko koi time information nahi chahiye (parent-note mistake box "I need the times for Gibbs" dekho). Ek orbiting body ki full state chhe numbers hoti hai: position $\vec{r}$ (jo tumhare paas middle point par already hai) plus velocity $\vec{v}$. Toh jo ek vector missing hai woh middle point par velocity ${\vec{v}}_2$ hai. Gibbs ka poora kaam ${\vec{v}}_2$ produce karna hai.

• (a) → Gauss. Angles akele direction ${\hat{\rho }}_i$ dete hain lekin distance nahi (slant range ${\rho }_i$). Gauss ki cleverness un unknown ranges ko dhundhne mein hai taaki ${\vec{r}}_i={\vec{R}}_i+{\rho }_i{\hat{\rho }}_i$ fully known ho jaye.
• (b) → Gibbs. Positions already known hain, toh seedha velocity-recovery step par jao. Pipeline mnemonic: Angles → Gauss → Positions → Gibbs → Velocity.

Root $r_2$ hai, middle position vector ki magnitude (Earth ke centre se uski distance). Woh root choose karo jo (1) positive aur real ho (distance negative ya imaginary nahi ho sakti) aur (2) central body ki radius se bada ho — Earth ke liye, $R_{\oplus }\approx 6378\text{km}$ se bada — aur jo positive slant ranges ${\rho }_i>0$ de (satellite surface ke upar aur telescope ke saamne hona chahiye, peechhe nahi). Complex, negative, ya sub-surface roots (koi bhi $r_2<6378\text{km}$) unphysical hain aur discard kar do.

Magnitude $r_i=\sqrt{r_{ix}^2+r_{iy}^2+r_{iz}^2}$ hai (ek scalar length). $r_1=\sqrt{294.32^2+4265.1^2+5986.7^2}=7356.5\text{km}$ $r_2=\sqrt{1365.5^2+3637.6^2+6346.8^2}=7440.3\text{km}$ $r_3=\sqrt{2940.3^2+2473.7^2+6555.8^2}=7598.9\text{km}$ Teeno $R_{\oplus }=6378$ km ✓ se zyada hain (yeh genuine above-surface positions hain). Yeh scalars weights hain jo $\vec{N}$ aur $\vec{S}$ ke andar appear hote hain, isliye pehle yahi aane chahiye.

Yeh test kyun: ${\vec{r}}_2\times {\vec{r}}_3$ ek vector hai jo ${\vec{r}}_2$ aur ${\vec{r}}_3$ wali plane ke perpendicular hai. Agar ${\vec{r}}_1$ usi plane mein hai, toh woh us normal ke perpendicular hai, isliye dot product (near) zero hoga. Neeche figure dekho.

Figure s01 — coplanarity test. Axes arbitrary in-plane coordinates $x,y$ hain. Mint parallelogram orbit plane hai; teen coloured arrows (lavender ${\vec{r}}_1$, coral ${\vec{r}}_2$, butter ${\vec{r}}_3$) sab focus se shuru hote hain aur us plane mein hain; slate arrow normal ${\vec{r}}_2\times {\vec{r}}_3$ hai. Coplanarity = teeno coloured arrows slate wale ke perpendicular hain. Numbers ko seedha karna (units matter karte hain). km mein do vectors ka cross product km·km = km² units mein hota hai. ${\vec{r}}_2\times {\vec{r}}_3$ ko component by component karte hain: ${\vec{r}}_2\times {\vec{r}}_3\approx (8.1454\times 10^6,1.9095\times 10^6,7.3908\times 10^6){\text{km}}^2.$ ${\vec{r}}_1$ (km) se dot karne par triple product km³ mein milta hai. Uske teen product-terms mein har ek $r_1\cdot |{\vec{r}}_2\times {\vec{r}}_3|\sim 7.4\times 10^3\cdot 1.1\times 10^7\approx 8\times 10^{10}{\text{km}}^3$ order ka hai, phir bhi nearly cancel ho jaate hain: ${\vec{r}}_1\cdot ({\vec{r}}_2\times {\vec{r}}_3)\approx -8.0\times 10^4{\text{km}}^3.$ Toh leftover ka relative size $\frac{8\times 10^4}{8\times 10^{10}}\approx 1\times 10^{-6}$ hai — ek million mein ek hissa. Ise padhne ka saaf tarika unit vectors par angle test hai: ${\hat{u}}_{r1}\cdot ({\hat{u}}_{r2}\times {\hat{u}}_{r3})\approx 1.3\times 10^{-6}$, jiska arcsine out-of-plane angle $\ll 1^{\circ }$ hai. Coplanar ✓ — ek single orbit teeno ko thread kar sakta hai.

$\vec{N}=r_1({\vec{r}}_2\times {\vec{r}}_3)+r_2({\vec{r}}_3\times {\vec{r}}_1)+r_3({\vec{r}}_1\times {\vec{r}}_2)$ $\vec{D}=({\vec{r}}_1\times {\vec{r}}_2)+({\vec{r}}_2\times {\vec{r}}_3)+({\vec{r}}_3\times {\vec{r}}_1)$ $\vec{S}={\vec{r}}_1(r_2-r_3)+{\vec{r}}_2(r_3-r_1)+{\vec{r}}_3(r_1-r_2)$ (Yahan bina arrow ke har $r_i$ scalar magnitude weight hai.) Arithmetic carry out karte hain: $\vec{N}\approx (6.0983\times 10^{10},1.4262\times 10^{10},5.5333\times 10^{10}){\text{km}}^3,N\approx 8.3465\times 10^{10}$ $\vec{D}\approx (8.1971\times 10^6,1.9169\times 10^6,7.4370\times 10^6){\text{km}}^2,D\approx 1.1215\times 10^7$ $\vec{S}\approx (6708.3,-10516,-2516.2){\text{km}}^2$ Semi-latus rectum $p=N/D\approx 7442\text{km}$. (Yeh $p$ ka $r_2$ ke close hona batata hai ki point 2 semi-latus-rectum crossing ke paas hai — ek achha sanity feel hai, requirement nahi.)

Prefactor: $\sqrt{\mu /(ND)}=\sqrt{398600/(8.3465\times 10^{10}\cdot 1.1215\times 10^7)}\approx 2.066\times 10^{-7}$. Bracket ke do pieces hain — transverse part $\vec{D}\times {\vec{r}}_2/r_2$ (motion ke saath velocity) aur radial part $\vec{S}$ (distance change karne wali velocity). Add karke scale karte hain: $\boxed{{\vec{v}}_2\approx (-6.218,-4.014,1.599)\text{km/s}}$ parent note se match karta hai. Ab $({\vec{r}}_2,{\vec{v}}_2)$ complete state hai.

Setup — sab kuch orbit plane mein likhो. Orbit plane mein ek 2-D polar frame rakho jisme focus origin par hai aur polar axis perigee ke along hai. True anomaly ${\theta }_i$ par ek point ka vector phir hoga ${\vec{r}}_i=r_i(\cos {\theta }_i\hat{x}+\sin {\theta }_i\hat{y}),\frac{1}{r_i}=\frac{1+e\cos {\theta }_i}{p}.$ Cross products sine ban jaate hain. In vectors mein se koi bhi do satisfy karte hain ${\vec{r}}_i\times {\vec{r}}_j=r_i{r}_j\sin ({\theta }_j-{\theta }_i)\hat{z},$ kyunki do planar vectors ka cross product (lengths ka product)×(unke beech angle ka sine) normal $\hat{z}$ ke along hota hai. Teen anomaly gaps likhte hain ${\theta }_{21}={\theta }_2-{\theta }_1$, ${\theta }_{32}={\theta }_3-{\theta }_2$, ${\theta }_{31}={\theta }_3-{\theta }_1$. Index 2 par do bracket pieces. Upar se, transverse term hai $\frac{\vec{D}\times {\vec{r}}_2}{r_2}=\frac{1}{r_2}[({\vec{r}}_1\times {\vec{r}}_2)+({\vec{r}}_2\times {\vec{r}}_3)+({\vec{r}}_3\times {\vec{r}}_1)]\times {\vec{r}}_2.$ Har inner cross product ek $\hat{z}$ vector hai $\propto \sin (\cdot )$; $\hat{z}$ ko ${\vec{r}}_2$ se cross karna ${\vec{r}}_2$ ko $90^{\circ }$ plane ke andar rotate karta hai, toh yeh term ek planar vector hai jiske $\cos {\theta }_2$ aur $\sin {\theta }_2$ components par "${\theta }_2$" tag hai. Iske saath $\vec{S}={\vec{r}}_1(r_2-r_3)+{\vec{r}}_2(r_3-r_1)+{\vec{r}}_3(r_1-r_2)$ bhi $\cos {\theta }_i,\sin {\theta }_i$ pieces mein expand hota hai. Cancellation: $r_i=p/(1+e\cos {\theta }_i)$ dono mein substitute karo. ${\theta }_1$ tags $\vec{S}$ mein $\propto \sin ({\theta }_{32})$ factors se multiply hoke appear karte hain aur ${\theta }_3$ tags $\propto \sin ({\theta }_{21})$ factors se — jo exactly wahi sine factors hain jo transverse term index 2 par produce karta hai — lekin opposite sign ke saath (ek ${\vec{r}}_1\times {\vec{r}}_2$ se aata hai, doosra ${\vec{r}}_2\times {\vec{r}}_3$ se, jo orientation mein alag hain). Dono brackets add karne par, ${\theta }_1$-tagged pieces ${\theta }_1$-tagged pieces ko annihilate karte hain aur ${\theta }_3$ ke liye bhi aise hi; sirf middle anomaly ${\theta }_2$ tagged terms, saath mein pure constants $p$ aur $e$, bachte hain. Result transverse aur radial directions mein $\sqrt{\mu p}$ ka ek clean multiple hai — matlab exactly ${\vec{v}}_2$. Index 1 kyun fail karta hai. $\vec{D}\times {\vec{r}}_1/r_1$ se dobara karo. Ab transverse term "${\theta }_1$" tagged hai, lekin $\vec{S}$ abhi bhi ${\theta }_1,{\theta }_2,{\theta }_3$ pieces supply karta hai. ${\theta }_2$ tag ka would-be partner missing hai (koi $\vec{D}\times {\vec{r}}_1$ term nahi hai jo matching $\sin ({\theta }_2-{\theta }_2)=0$ produce kare), toh $\propto \sin ({\theta }_2-{\theta }_1)\ne 0$ ka ek residual bachta hai aur answer ko pollute karta hai. Symmetric sums literally centred hain toh books sirf middle index par balance hoti hain.

Figure s02 — velocity decomposition. Axes arbitrary in-plane $x,y$ hain. Lavender curve orbit hai, slate dot focus hai, coral arrow ${\vec{r}}_2$ middle point ka radius hai. Tip par, mint arrow transverse direction hai (${\vec{r}}_2$ ke perpendicular, motion ke along) $\vec{D}\times {\vec{r}}_2/r_2$ se; butter arrow radial direction hai (${\vec{r}}_2$ ke along) $\vec{S}$ se; slate arrow unka sum hai, true velocity ${\vec{v}}_2$. Kisi bhi conic par, ek point ki velocity do perpendicular directions mein decompose hoti hai:

• Transverse (radius ke perpendicular, motion ke along) — $\frac{\vec{D}\times {\vec{r}}_2}{r_2}$ se supply hota hai. Kyunki $\vec{D}$ (near) orbit-normal direction hai, isse ${\vec{r}}_2$ se cross karna ${\vec{r}}_2$ ko $90^{\circ }$ plane ke andar rotate karta hai, along-track direction deta hai.
• Radial (${\vec{r}}_2$ ke along, distance change karta hai) — $\vec{S}$ se supply hota hai, jo $\dot{r}$ part carry karta hai. Ek plane mein velocity ko exactly do independent directions chahiye; radial + transverse orbits ke liye natural pair hai, toh sum complete hai — kuch missing nahi.

Non-coplanar case. Gibbs assume karta hai ki teeno points se ek single planar conic guzarti hai. Agar woh genuinely non-coplanar hain, aisa koi orbit exist nahi karta, toh formula phir bhi ${\vec{v}}_2$ ke liye ek number return karta hai lekin woh number meaningless hai: reconstructed $\vec{h}$ aur $\vec{e}$ kisi real orbit ke corresponding nahi hain jo points se guzre. Symptoms: resulting orbit points 1 aur 3 ko badly miss karta hai, aur derived elements (eccentricity, inclination) ek re-propagation ke saath inconsistent aate hain. Yahi reason hai ki $<1^{\circ }$ gate ek hard prerequisite hai. Collinear (singular) case — doosra failure mode. $\vec{D}=({\vec{r}}_1\times {\vec{r}}_2)+({\vec{r}}_2\times {\vec{r}}_3)+({\vec{r}}_3\times {\vec{r}}_1)$ dekho. Har cross product do position vectors ke beech angle ke sine ke proportional hai. Agar teen points nearly collinear hain (sab ek line ke along bunched — bahut chhota arc jisme radius vectors ke beech tiny angular separation hai), toh har cross product zero ki taraf shrink karta hai, toh $\vec{D}\to \vec{0}$ aur $\vec{N}\to \vec{0}$. Prefactor $\sqrt{\mu /(ND)}$ phir blow up karta hai ($1/\sqrt{ND}\to \infty$) jabki bracket $\to \vec{0}$ — ek $0\times \infty$ indeterminate form jo catastrophically noise-sensitive hai: tiny cross products mein koi bhi measurement jitter enormously amplify ho jaata hai, toh recovered ${\vec{v}}_2$ signal ki jagah noise se dominated hai. Rule of thumb: Gibbs ko teen points chahiye jo true anomaly mein spread out hon (well-separated angles). Chhote, tightly-spaced arcs ke liye, Gauss's series treatment prefer karo ya zyada widely-spaced observations ikkathe karo.

$\vec{h}={\vec{r}}_2\times {\vec{v}}_2\approx (3.1310\times 10^4,-3.7660\times 10^4,2.7106\times 10^4){\text{km}}^2/\text{s}$ $h=|\vec{h}|\approx 5.5757\times 10^4{\text{km}}^2/\text{s}$ Conservation check sahi tarike se karna. Authoritative semi-latus rectum state se $p=h^2/\mu$ ke zariye aata hai: $p=\frac{h^2}{\mu }=\frac{(5.5757\times 10^4{)}^2}{398600}\approx 7800\text{km}.$ Phir $\sqrt{\mu p}=\sqrt{398600\cdot 7800}=\sqrt{3.109\times 10^9}\approx 5.576\times 10^4=h$ ✓ — $h=\sqrt{\mu p}$ relation last digit tak close ho jaata hai kyunki dono sides ab same $p$ use karti hain. Pehle wala $N/D\approx 7442$ kyun alag tha: woh $p$ hand-carried, rounded helper vectors $\vec{N},\vec{D}$ se assemble hua tha; har component mein chhoti rounding $N/D$ mein propagate hoti hai. Lesson: ek baar $({\vec{r}}_2,{\vec{v}}_2)$ ho jaaye, conserved quantities state se derive karo, jo internally consistent hai, intermediate scratch values se nahi.

${\hat{r}}_2={\vec{r}}_2/7440.3\approx (-0.18354,0.48889,0.85304)$ aur 4.1 se $\vec{h}$ ke saath: $\vec{e}=\frac{{\vec{v}}_2\times \vec{h}}{\mu }-{\hat{r}}_2\approx (-0.0868,-0.0663,0.0195)$ $e=|\vec{e}|\approx 0.111$ Kyunki $0<e<1$, orbit ek ellipse hai (bound). $e\approx 0.11$ mildly elliptical hai — nearly circular lekin clearly perfect circle nahi. $\vec{e}$ ki direction perigee (closest approach) ki taraf point karti hai, orbit ki orientation uske plane ke andar fix karti hai.

flowchart TD
  A["Three angle pairs alpha delta plus observer R_i"] --> B["Line of sight unit vectors rho hat"]
  B --> C["Gauss f and g series about middle time"]
  C --> D["Eighth degree polynomial in r_2"]
  D --> E["Pick positive real root above Earth radius"]
  E --> F["Back out slant ranges rho_i positive"]
  F --> G["Positions r_i equals R_i plus rho_i rho hat"]
  G --> H["Gibbs helper vectors N D S"]
  H --> I["Velocity v_2 from state"]
  I --> J["State r_2 and v_2"]
  J --> K["Compute h and e vectors"]
  K --> L["Six classical elements a e i RAAN arg perigee"]
  L --> M["True anomaly then eccentric then mean anomaly"]
  M --> N["Kepler equation for time since perigee"]
  E --> P["Refine f and g exact then re solve"]
  P --> D

Quadrant / sign care (woh part jo beginners skip karte hain):

• Inclination $i=\arccos (h_z/h)$ $[0^{\circ },180^{\circ }]$ mein cleanly land karta hai — $\arccos$ already cover karta hai, koi ambiguity nahi.
• RAAN $\Omega$ aur argument of perigee $\omega$ ek ratio ke $\arccos$ se aate hain, jo sirf $[0^{\circ },180^{\circ }]$ return karta hai. Doosra half sign test se resolve karo: agar node vector ka $y$-component $n_y<0$ hai toh $\Omega =360^{\circ }-\arccos (\dots )$; agar $e_z<0$ hai toh $\omega =360^{\circ }-\arccos (\dots )$.
• True anomaly $\theta$: agar $\vec{r}\cdot \vec{v}<0$ hai toh body inward move kar rahi hai (apogee ke baad, perigee ki taraf), toh $\theta =360^{\circ }-\arccos (\dots )$.
• Kepler's equation $M=E-e\sin E$ phir eccentric anomaly $E$ ko mean anomaly $M$ (isliye time) se relate karta hai. $M$ se $E$ numerically solve karo (Newton's method), $E$ ko $\theta$ ke same revolution mein rakhte hue. Iterate loop (right branch) key subtlety hai: Gauss ke pehle ${\rho }_i$ truncated Lagrange series se aate hain, toh exact Lagrange Coefficients (f and g) se refine karo aur polynomial re-solve karo jab tak ${\rho }_2$ move karna band na ho jaaye.

Scalar triple product hi $3\times 3$ determinant hai teen vectors ko rows ki tarah stack karke — ek standard identity. ${\vec{r}}_2\times {\vec{r}}_3$ ke components expand karo aur ${\vec{r}}_1$ se dot karo; resulting sum of $3!=6$ signed products exactly us determinant ka cofactor expansion hai. Geometry: $|{\vec{r}}_1\cdot ({\vec{r}}_2\times {\vec{r}}_3)|$ teen vectors se spanned parallelepiped ka volume hai. Teen vectors ek common plane mein hain iff woh box flat hai — zero volume. Isliye determinant $=0\iff$ coplanar. Yahi reason hai ki test principle mein exact hai aur noisy practice mein (term size ke relative) near-zero hota hai.

Units: $\vec{N}$ (km)(km²)=km³ hai, toh $N$ km³ hai. $\vec{D}$ km² hai, toh $D$ km² hai. Isliye $p=N/D$ ke units ${\text{km}}^3/{\text{km}}^2=\text{km}$ ✓ hain, aur $p\approx 8.35\times 10^{10}/1.12\times 10^7\approx 7.4\times 10^3$ km. Prefactor: $\mu /(ND)$ ke units $\frac{{\text{km}}^3{\text{s}}^{-2}}{{\text{km}}^3\cdot {\text{km}}^2}={\text{km}}^{-2}{\text{s}}^{-2}$ hain, toh $\sqrt{\mu /(ND)}$ ${\text{km}}^{-1}{\text{s}}^{-1}$ ✓ hai. Bracket $\frac{\vec{D}\times {\vec{r}}_2}{r_2}+\vec{S}$ ke units $\frac{{\text{km}}^2\cdot \text{km}}{\text{km}}={\text{km}}^2$ hain. Multiply karo: ${\text{km}}^{-1}{\text{s}}^{-1}\cdot {\text{km}}^2=\text{km}{\text{s}}^{-1}$ ✓. Dimensional analysis akela formula ki shape force karta hai — ek powerful sanity net.

Agar ${\vec{r}}_1={\vec{r}}_2$, toh kai building blocks collapse ho jaate hain: ${\vec{r}}_1\times {\vec{r}}_2=\vec{0}$, aur pair $\vec{D}$ ya $\vec{N}$ mein kuch naya contribute nahi karta; sums degenerate ho jaate hain jaise sirf do distinct points hon. Conic par do points use uniquely determine nahi karte (infinitely many conics alag-alag eccentricities wali do given points se do ek shared focus ke saath guzar sakti hain), toh reconstructed ${\vec{v}}_2$ undetermined/ill-conditioned hai. Physically: tumhe curve ke genuinely alag teen samples chahiye taaki scale ($p$), plane ($\vec{h}$ direction), aur shape/orientation ($\vec{e}$) pin ho sake. Yeh wahi $\vec{D}\to \vec{0}$ pathology hai jaise Exercise 3.3 mein collinear case hai — coincident points sirf "not spread out" ka extreme hain: jab samples ke beech angular spread vanish hota hai, cross-product sums vanish hote hain aur $\sqrt{\mu /(ND)}$ diverge karta hai.