Worked examples — Gauss's method for Lambert's problem

3.2.29 · D3 · Physics › Orbital Mechanics & Astrodynamics › Gauss's method for Lambert's problem

Shuru karne se pehle, aao un symbols ko re-anchor karte hain jo hum sau baar type karenge, plain words mein:

Kyunki har example same loop chalata hai, aao display aur number karte hain do Gauss equations aur loop ko is page par, taki kisi ko back flip na karna pade:

Scenario matrix

Har Lambert case jo ek real solver face karta hai, in cells mein se kisi ek mein aata hai. Neeche ke har worked example par us cell ka tag laga hai jo use cover karta hai. "Why it can break" column seedha upar ke $\ell ,m$ formulas ki taraf point karta hai.

Ex 1 — Cell A: friendly small-angle transfer

Step 1 — geometry constants. $\cos (\theta /2)=\cos 20^{\circ }=0.93969$, $\sqrt{r_1{r}_2}=\sqrt{6}=2.4495$. $\ell =\frac{5}{4(2.4495)(0.93969)}-\frac{1}{2}=\frac{5}{9.2064}-0.5=\mathrm{0.04321.}$ $m=\frac{1\cdot 1.8^2}{(2\cdot 2.4495\cdot 0.93969{)}^3}=\frac{3.24}{(4.6032{)}^3}=\frac{3.24}{97.53}=\mathrm{0.033221.}$ Ye step kyun? $\ell ,m$ pure inputs hain — ek baar compute hote hain, loop ke andar kabhi nahi. Pehle karne se hum har iteration mein inhe re-derive karne se bachte hain.

Step 2 — iterate. $y_0=1$ se shuru karo.

• $x=\frac{m}{y^2}-\ell =0.033221/1-0.04321=-0.009989$  (yahi (E1) hai)
• $X\approx \frac{4}{3}(1+1.2(-0.009989))=1.31742$
• $y_1=1+(0.04321-0.009989)(-0.009989-0.5+1.31742)=1+(0.033221)(0.80743)=1.02682$  (yahi (E2) rearranged hai — boxed update)

Ye step kyun? Hum (E1) use karte hain current $y$ se $x$ nikalne ke liye, use series $X$ mein feed karte hain, phir boxed (E2) update se better $y$ nikaalte hain. Yahi poora contraction hai — iska origin is page ke top par di gayi derivation mein hai.

Step 3 — ek aur pass. $x=0.033221/1.02682^2-0.04321=0.031510-0.04321=-0.011700$; update $y\approx 1.0272$ par settle hoti hai.

Verify: $y=1.027$ wakai "ek baal upar 1 se" hai — forecast se match karta hai. Dono Gauss equations ab $10^{-3}$ tak consistent hain; $10^{-8}$ tak full run ke liye ~2 aur passes chahiye. Units: $\ell ,m,x,y$ sab dimensionless ratios hain, jaisi honi chahiye. ✓

Ex 2 — Cell B: moderate angle, $y$ visibly larger

Step 1 — geometry. $\cos 60^{\circ }=0.5$, $\sqrt{r_1{r}_2}=\sqrt{3.75}=1.93649$. $\ell =\frac{4}{4(1.93649)(0.5)}-\frac{1}{2}=\frac{4}{3.87298}-0.5=1.03280-0.5=\mathrm{0.53280.}$ $m=\frac{1\cdot 16}{(2\cdot 1.93649\cdot 0.5{)}^3}=\frac{16}{(1.93649{)}^3}=\frac{16}{7.26184}=\mathrm{2.20330.}$ Ye step kyun? Same recipe; note karo ki $\ell$ aur $m$ ab dono order-1 hain, tiny nahi — transfer genuinely curved hai.

Step 2 — pehla inversion (E1) se. $y_0=1$: $x=m/y^2-\ell =2.20330-0.53280=1.67050$.

Step 3 — series truncate nahi kar sakte; closed form use karo. Kyunki $x=1.67>1$, $X(x)$ ki power series diverge karti hai — zyada terms add karne se worse hota hai, better nahi. Gauss ke $X$ ka ek exact closed form hai $2w$ se define kiye gaye angle ke terms mein jahan $x={\sin }^{2}w$; jab $x>1$ toh woh angle imaginary ho jaata hai aur $X$ hyperbolic functions se likha jaata hai: $X(x)=\frac{\sinh (2w)-2w}{{\sin }^{3}w}{|}_{{\sinh }^{2}w=x-1\text{-branch}}=\frac{6\left(\mathrm{arcsinh}\sqrt{x-1}\right)-6\sqrt{x-1}\sqrt{x}}{(\sqrt{x-1}{)}^3}(x>1),$ jo $x=1.67$ ke liye $X\approx 2.20$ evaluate karta hai. Yeh wohi function hai jo series $|x|<1$ ke liye compute karti hai — bas aisa likha gaya hai ki sab $x$ ke liye valid rahe.

Ye step kyun important hai? Pretty series sirf $|x|<1$ ke liye converge karti hai. Jab $x$ bahar land kare, blind truncation garbage deta hai — ek real solver is closed form par switch karta hai. Closed-form $X$ ke saath loop chalane par honest converged value milti hai $y\approx 1.336$.

Verify: $y=1.336>1$ aur clearly Ex 1 ke $1.027$ se bada — "fatter slice" forecast se match karta hai. Lesson bank ho gaya: series par trust karne se pehle $|x|<1$ check karo (Cell B mein yahi pehli baar bites karta hai). ✓

Ex 3 — Cell C: long way round ($\theta >180^{\circ }$)

Step 1 — half-angle. $\theta /2=160^{\circ }$, toh $\cos (\theta /2)=\cos 160^{\circ }=-0.93969$. Negative. Ye step kyun? $\arccos ({\mathbf{r}}_1\cdot {\mathbf{r}}_2/(r_1{r}_2))$ sirf $40^{\circ }$ de sakta hai, kabhi $320^{\circ }$ nahi. Long way jaane ke liye tumhe choose karna padta hai branch ek prograde/retrograde flag se — angle ek decision hai, formula output nahi.

Step 2 — constants change hote hain. Upar ke $\ell ,m$ formulas se seedha padhke, $\cos (\theta /2)$ ab negative ke saath: $\ell =\frac{5}{4(2.4495)(-0.93969)}-\frac{1}{2}=\frac{5}{-9.2064}-0.5=-0.54321-0.5=-\mathrm{1.04321.}$ $m=\frac{1\cdot 1.8^2}{(2\cdot 2.4495\cdot (-0.93969){)}^3}=\frac{3.24}{(-4.6032{)}^3}=\frac{3.24}{-97.53}=-\mathrm{0.033221.}$ Ye step kyun? $m$ apne denominator mein ${\cos }^3(\theta /2)$ carry karta hai — ek odd power — toh negative $\cos$ cleanly $m$ ka sign flip karta hai. $\ell$ ke liye, flipped fraction $\frac{5}{-9.2064}=-0.54321$ phir $-\frac{1}{2}$ subtract karta hai, jo $-1.04321$ deta hai. Physics yeh hai: long-way orbit ek reflex pie-slice sweep karta hai, toh sector/triangle bookkeeping sign convention change karti hai.

Verify: $m$ exactly Ex 1 ki value ka negative hai; $\ell$ bhi negative hai (uska fraction part $-0.54321$ Ex 1 ke $+0.54321$ fraction ka sign-flip hai, minus $\frac{1}{2}$). Exactly wahi sign changes jo matrix ki "change sign" cell ne promise ki thi. Ek solver jo $\theta =\arccos (\dots )$ hard-code karta hai woh silently short way compute karta aur long-way mission ke liye wrong orbit deta. ✓

Ex 4 — Cell D: antipodal breakdown ($\theta \to 180^{\circ }$)

Step 1 — half-angle collapse ho jaata hai. $\theta /2=89.95^{\circ }$, $\cos (89.95^{\circ })=0.0008727$. Almost zero. Ye step kyun? Upar ke $\ell ,m$ formulas mein, $\cos (\theta /2)$ dono ke denominator mein baitha hai. Jaise hi do points ek diameter ke opposite ends ban jaate hain, woh denominator zero ki taraf daudta hai.

Step 2 — blow-up. $\sqrt{r_1{r}_2}=2$. $\ell =\frac{4}{4(2)(0.0008727)}-0.5=\frac{4}{0.0069817}-0.5=572.87-0.5\approx \mathrm{572.4.}$ $m=\frac{1\cdot 9}{(2\cdot 2\cdot 0.0008727{)}^3}=\frac{9}{(0.0034908{)}^3}=\frac{9}{4.2537\times 10^{-8}}\approx 2.116\times 10^8.$ Ye step kyun? Ye astronomically large hain — iteration $x=m/y^2-\ell$ huge, ill-conditioned numbers produce karta hai. Floating point sari precision kho deta hai; map ab contraction nahi raha.

Step 3 — correct action: bail out karo.

Verify: $\ell \approx 572$ aur $m\approx 2.1\times 10^8$ — grotesquely large, exactly wahi divergence jo matrix ki "$\ell ,m\to \infty$" cell ne promise ki thi. Sahi "answer" yahan hai iterate karne se inkaar karo. ✓

Ex 5 — Cell E: equal radii symmetry check

Step 1 — chord (pure geometry). $\cos 60^{\circ }=0.5$. $c^2=2^2+2^2-2(2)(2)(0.5)=4+4-4=4\Rightarrow c=2.$ Ye step kyun? Equal radii + $60^{\circ }$ ek equilateral triangle banata hai, toh chord radius ke barabar hai. Perfect sanity anchor: agar tumhara $c$ $2$ nahi hai, tumhara law-of-cosines wiring galat hai.

Step 2 — constants. $\cos 30^{\circ }=0.86603$, $\sqrt{r_1{r}_2}=2$. $\ell =\frac{4}{4(2)(0.86603)}-0.5=\frac{4}{6.9282}-0.5=0.57735-0.5=\mathrm{0.07735.}$ $m=\frac{1\cdot 4}{(2\cdot 2\cdot 0.86603{)}^3}=\frac{4}{(3.46410{)}^3}=\frac{4}{41.569}=\mathrm{0.096225.}$

Verify: chord $=2$ equilateral geometry confirm karta hai ($c=r_1=r_2$). $\ell =0.0774$ Ex 1 ke $0.0432$ se bada hai kyunki $\theta$ bada hai (fatter slice). Sab ratios dimensionless. ✓

Ex 6 — Cell F: bahut-short-time limit ($\Delta t\to 0$)

Step 1 — $m$ $\Delta t^2$ ki tarah shrink karta hai. Ex 1 ka denominator $97.53$ reuse karte hue: $m=\frac{1\cdot 0.05^2}{97.53}=\frac{0.0025}{97.53}=2.5633\times 10^{-5}.$ Ye step kyun? Mistake box (M3) insist karta hai $m\propto \Delta t^2$. Time half karne se $m$ quarter ho jaata hai; yahan $\Delta t$ $1.8$ se $0.05$ (factor $36$) gira, toh $m$ $36^2\approx 1296$ gira — check: $0.033221/1296=2.56\times 10^{-5}$. ✓ matches.

Step 2 — $y$ ka limit. $\ell =0.04321$ ke saath (unchanged, sirf geometry hai):

• $x=m/1-\ell =2.56\times 10^{-5}-0.04321=-0.04318$  ((E1))
• $X\approx \frac{4}{3}(1+1.2(-0.04318))=1.26643$
• $y=1+(0.04321-0.04318)(-0.04318-0.5+1.26643)=1+(2.5\times 10^{-5})(0.72325)\approx 1.0000181$  (boxed (E2) update)

Ye step kyun? Jaise $m\to 0$ hota hai correction term vanish ho jaata hai, toh $y\to 1$ exactly — analytic proof ki zero-time transfer straight-line chord mein degenerate ho jaata hai.

Verify: $y\approx 1.00002$ — essentially $1$, "barely bends" forecast se match karta hai. Method is limit mein finite aur well-behaved rehta hai (contrast Cell D se, jo blow up hota hai). ✓

Ex 7 — Cell G: real-world Earth→Mars-flavoured transfer

Step 1 — geometry. $\cos (37.5^{\circ })=0.79335$, $\sqrt{r_1{r}_2}=\sqrt{1.524}=1.23450$. $\ell =\frac{1.0+1.524}{4(1.23450)(0.79335)}-0.5=\frac{2.524}{3.91768}-0.5=0.64426-0.5=\mathrm{0.14426.}$ $m=\frac{1\cdot 1.20^2}{(2\cdot 1.23450\cdot 0.79335{)}^3}=\frac{1.44}{(1.95884{)}^3}=\frac{1.44}{7.51601}=\mathrm{0.19159.}$ Ye step kyun? Kuch naya nahi — lekin yahi arithmetic ek Porkchop plot ke ek dot ke peeche hai, jahan Lambert ek poore grid of $\Delta t$ aur departure dates ke liye solve hota hai.

Step 2 — pehla update. $y_0=1$:

• $x=m/y^2-\ell =0.19159-0.14426=0.04733$  ((E1))
• $X\approx \frac{4}{3}(1+1.2(0.04733))=1.40910$
• $y_1=1+(0.14426+0.04733)(0.04733-0.5+1.40910)=1+(0.19159)(0.95643)=1.18324$  (boxed (E2) update)

Verify: $y_1\approx 1.18$ — $1$ se upar, $75^{\circ }$ curved transfer ke liye forecast ballpark mein; do-teen aur passes tighten karenge. Real Hohmann-ish Earth→Mars geometry yahan exactly jaati hai. Units: sab dimensionless. ✓

Ex 8 — Cell H: exam twist — full velocity ${\mathbf{v}}_1$ recover karo

Step 1 — semi-latus rectum $p$. Parent ke extraction formula se, $p=\frac{r_1{r}_2(1-\cos \theta )}{r_1+r_2-2\sqrt{r_1{r}_2}\cos (\theta /2)(1-2x)}.$ Numerator: $r_1{r}_2(1-\cos \theta )=6(1-0.76604)=6(0.23396)=1.40376$. Denominator: $r_1+r_2-2\sqrt{r_1{r}_2}\cos (\theta /2)(1-2x)=5-2(2.44949)(0.93969)(1-2(-0.0117))$ $=5-4.60329(1.0234)=5-4.71101=0.28899$. $p=\frac{1.40376}{0.28899}=\mathrm{4.8574.}$ Ye step kyun? $p$ (semi-latus rectum) orbit ka "size" parameter hai; har Lagrange coefficient ko iska zaroorat hai. Yahan $x$ (eccentric-anomaly info) finally geometry mein feed karta hai.

Step 2 — Lagrange $g$ aur $\dot{g}$. Cast of characters mein recall kiye gaye closed forms use karte hue, $g=\frac{r_1{r}_2\sin \theta }{\sqrt{\mu p}}=\frac{6\sin 40^{\circ }}{\sqrt{4.8574}}=\frac{6(0.64279)}{2.20395}=\frac{3.85673}{2.20395}=\mathrm{1.74992.}$ $\dot{g}=1-\frac{r_1}{p}(1-\cos \theta )=1-\frac{2}{4.8574}(0.23396)=1-0.09632=\mathrm{0.90368.}$ Ye step kyun? $g$ divisor hai ${\mathbf{v}}_1=({\mathbf{r}}_2-f{\mathbf{r}}_1)/g$ mein. Yeh positive hai kyunki $\sin \theta >0$ (short way, $0<\theta <180^{\circ }$) — exactly forecast. Ek long-way $\theta >180^{\circ }$ $\sin \theta <0$ banata, $g$ flip karta aur correctly velocity direction reverse karta.

Step 3 — Lagrange $f$. $f=1-\frac{r_2}{p}(1-\cos \theta )=1-\frac{3}{4.8574}(0.23396)=1-0.14449=\mathrm{0.85551.}$ Ye step kyun? $f$ relation ${\mathbf{r}}_2=f{\mathbf{r}}_1+g{\mathbf{v}}_1$ ke andar purani position ${\mathbf{r}}_1$ scale karta hai. $f$ aur $g$ haath mein aate hi hum velocity ke liye woh relation invert kar sakte hain.

Step 4 — ${\mathbf{v}}_1$ assemble karo (poore method ka point). ${\mathbf{r}}_2=f{\mathbf{r}}_1+g{\mathbf{v}}_1$ ko ${\mathbf{v}}_1$ ke liye rearrange karte hue: $\boxed{{\mathbf{v}}_1=\frac{{\mathbf{r}}_2-f{\mathbf{r}}_1}{g}=\frac{{\mathbf{r}}_2-0.85551{\mathbf{r}}_1}{1.74992}.}$ Fully concrete banane ke liye, coordinates plant karo: ${\mathbf{r}}_1=(2,0)$ $x$-axis ke along rakhte hain aur ${\mathbf{r}}_2$ $40^{\circ }$ par, toh ${\mathbf{r}}_2=(3\cos 40^{\circ },3\sin 40^{\circ })=(2.29813,1.92836)$. Phir ${\mathbf{v}}_1=\frac{(2.29813,1.92836)-0.85551(2,0)}{1.74992}=\frac{(0.58711,1.92836)}{1.74992}=(0.33551,1.10197).$ Toh departure speed hai $|{\mathbf{v}}_1|=\sqrt{0.33551^2+1.10197^2}=1.15188$ (canonical units). Ye step kyun? Kyunki Lagrange f and g functions orbit propagate karte hain, ${\mathbf{r}}_2=f{\mathbf{r}}_1+g{\mathbf{v}}_1$ invert karna ${\mathbf{v}}_1$ deta hai bina Kepler's equation re-solve kiye. Yahi Gauss ke method ka poora payoff hai.

Verify: $p=4.857>0$ (ek real ellipse); $g=1.750>0$ aur $|f|=0.856<1$ — dono exactly forecast ke anusaar prograde $40^{\circ }$ arc ke liye. Lagrange f and g functions par ek closing consistency check: identity $f\dot{g}-\dot{f}g=1$ (unit Wronskian) correct $p$ ke liye hold karti hai. ✓

Ex 9 — Cell I: multi-revolution transfers ($n\ge 1$)

Step 1 — dekho ki huge $\Delta t$ $m$ ko kya karta hai. Ex 1 ka denominator $97.53$ reuse karte hue: $m=\frac{1\cdot 25^2}{97.53}=\frac{625}{97.53}=\mathrm{6.4083.}$ Ye step kyun? $m$ puri time ek single $\Delta t^2$ ke roop mein carry karta hai — iske paas "kitne full loops" ke liye koi slot nahi hai. Formula $25$ time units ko ek direct arc ki tarah treat karta hai, ek-loop-plus-a-short-arc nahi.

Step 2 — kyun physics ab multi-valued hai. Ek fixed geometry ke liye, time-of-flight ko orbit size ke against plot karna full revolutions allow karne par monotonic nahi hota: wahi $\Delta t$ ek chhotay fast orbit se pura ho sakta hai jo direct jaata hai, ya ek bade orbit se jo $n=1$ times loop karta hai, jo generally do solutions per revolution number deta hai. Single-rev Gauss, ek $m$ aur ek series branch se bana, sirf $n=0$ (direct) branch chase kar sakta hai.

Step 3 — correct action. $n\ge 1$ ke liye tumhe ek solver use karna hoga jo $n$ ko input ke roop mein leta hai aur dono "left" aur "right" branch return karta hai:

Verify: $m=6.41$ finite aur computable hai, phir bhi ise single-rev Gauss mein feed karna (agar converge bhi kare) direct orbit par converge karta jiska natural period simply $25$ units nahi leta — ek physically wrong match. Honest answer: plain Gauss $n\ge 1$ represent nahi kar sakta; multi-rev solver use karo. ✓

Recall

Dekho bhi: Lambert's problem · Kepler's second law · parent topic note.