Exercises — Gauss's method for Lambert's problem

Inputs: do position vectors ${\mathbf{r}}_1,{\mathbf{r}}_2$ (ek hi focus = central body se measure kiye gaye) aur unke beech time-of-flight $\Delta t$. Outputs: connecting Kepler orbit, concretely do velocities ${\mathbf{v}}_1$ (point 1 par) aur ${\mathbf{v}}_2$ (point 2 par) ke roop mein diya gaya. ${\mathbf{r}}_1,{\mathbf{v}}_1$ se tum har classical orbital element compute kar sakte ho, isliye orbit poori tarah determined ho jaati hai.

$y=\frac{\text{elliptic sector ka swept area}}{\text{triangle ka area}}$ jo focus aur do position points se banta hai. $y\approx 1$ matlab swept "pie slice" aur points ke beech flat triangle lagbhag same hain — arc barely bahar nikalti hai, jo chhote transfer angles $\theta$ par hota hai. Figure dekho: yellow sector blue triangle ke saath chipka hua hai.

Time-of-flight squared hona chahiye: $m=\frac{\mu \Delta t^2}{(2\sqrt{r_1{r}_2}\cos (\theta /2){)}^3}$. Squared kyun? Yeh Kepler's-third-law scaling hai — parent note time information ko $\propto \Delta t^2$ ke roop mein pack karta hai. $\Delta t$ linearly use karna dimensionally aur physically galat hai.

Hum kya karte hain: focus–point–point triangle par law of cosines apply karo. Kyun: $c$ seedhi-line distance hai; right side par sab kuch already pata hai. $c^2=r_1^2+r_2^2-2r_1{r}_2\cos \theta =4+9-2(6)(0.76604)=13-9.19252=3.80748,$ $c=\sqrt{3.80748}=\mathrm{1.9513.}$

$\cos (\theta /2)=\cos 20^{\circ }=0.93969$, $\sqrt{r_1{r}_2}=\sqrt{6}=2.44949$. $\ell =\frac{r_1+r_2}{4\sqrt{r_1{r}_2}\cos (\theta /2)}-\frac{1}{2}=\frac{5}{4(2.44949)(0.93969)}-\mathrm{0.5.}$ Denominator $=4(2.44949)(0.93969)=9.2065$, isliye $\ell =\frac{5}{9.2065}-0.5=0.54309-0.5=0.04309$.

Cube ka base hai $2\sqrt{r_1{r}_2}\cos (\theta /2)=2(2.44949)(0.93969)=4.60326$. $m=\frac{\mu \Delta t^2}{(4.60326{)}^3}=\frac{(1)(1.8{)}^2}{97.552}=\frac{3.24}{97.552}=\mathrm{0.033213.}$

Step A — $x$ paane ke liye Gauss's first equation ko invert karo. Pehli equation hai $y^2=\frac{m}{\ell +x}$; ise $x$ ke liye solve karne par $\ell +x=\frac{m}{y^2}$ milta hai, yani $x=\frac{m}{y^2}-\ell =\frac{0.033213}{1^2}-0.04309=-\mathrm{0.009877.}$ Step B — series $X(x)$. Do terms rakho: $X(x)=\frac{4}{3}\left(1+\frac{6}{5}x+\cdots \right)=\frac{4}{3}(1+1.2(-0.009877))=\frac{4}{3}(0.988148)=\mathrm{1.31753.}$ Step C — $y$ update karo $y=1+(\ell +x)(x-\frac{1}{2}+X)$ se (yeh Gauss's second equation hai $y^2$ se divide karke, $m/y^2=\ell +x$ use karte hue):

• $\ell +x=0.04309-0.009877=0.033213$
• $x-\frac{1}{2}+X=-0.009877-0.5+1.31753=0.807653$
• $y=1+(0.033213)(0.807653)=1+0.026823=\mathrm{1.02682.}$

Matlab: guess $y=1$ thoda chhota tha; sector triangle se $\sim 2.7\%$ bada hai. Ek aur iteration bahut kam move karti hai.

Naya $x$: $x=\frac{0.033213}{1.02682^2}-0.04309=\frac{0.033213}{1.054360}-0.04309=0.031501-0.04309=-0.011589$. Naya $X$: $X=\frac{4}{3}(1+1.2(-0.011589))=\frac{4}{3}(0.986093)=1.314791$. Naya $y$:

• $\ell +x=0.04309-0.011589=0.031501$
• $x-\frac{1}{2}+X=-0.011589-0.5+1.314791=0.803202$
• $y=1+(0.031501)(0.803202)=1+0.025302=\mathrm{1.025302.}$

$y$ $1.02682$ se $1.02530$ ho gaya — sirf $0.0015$ ka change. Yeh map $y\approx 1$ ke paas ek contraction hai; ek aur step $y\approx 1.0253$ par settle kar deta hai. Yeh "3–5 iterations" waala claim concretely prove ho gaya — figure par trace karo: blue markers $1$ se green dashed fixed line ki taraf step karte hain, har jump pehle se chhota.

Semi-latus rectum. Parent ki extraction formula use karte hue $p=\frac{r_1{r}_2(1-\cos \theta )}{r_1+r_2-2\sqrt{r_1{r}_2}\cos (\theta /2)(1-2x)}.$ Numerator: $r_1{r}_2(1-\cos \theta )=6(1-0.76604)=6(0.23396)=1.40376$. Denominator: $2\sqrt{r_1{r}_2}\cos (\theta /2)=4.60326$; $(1-2x)=1-2(-0.011589)=1.023178$; product $=4.71028$; toh denominator $=5-4.71028=0.28972$. $p=\frac{1.40376}{0.28972}=\mathrm{4.8452.}$ Lagrange $f,g,\dot{g}$ (parent formulas):

• $f=1-\frac{r_2}{p}(1-\cos \theta )=1-\frac{3}{4.8452}(0.23396)=1-0.14487=\mathrm{0.85513.}$
• $g=\frac{r_1{r}_2\sin \theta }{\sqrt{\mu p}}=\frac{6\sin 40^{\circ }}{\sqrt{4.8452}}=\frac{6(0.64279)}{2.20118}=\frac{3.85674}{2.20118}=\mathrm{1.75212.}$
• $\dot{g}=1-\frac{r_1}{p}(1-\cos \theta )=1-\frac{2}{4.8452}(0.23396)=1-0.09658=\mathrm{0.90342.}$

Yeh kyun: $f,g$ closed-form propagator coefficients hain (Lagrange f and g functions), isliye ${\mathbf{r}}_2=f{\mathbf{r}}_1+g{\mathbf{v}}_1$ — invert karne par velocity milti hai bina Kepler's equation dobara solve kiye.

${\mathbf{v}}_1=\frac{{\mathbf{r}}_2-f{\mathbf{r}}_1}{g}.$ $f{\mathbf{r}}_1=(0.85513\cdot 2,0)=(1.71026,0)$. ${\mathbf{r}}_2-f{\mathbf{r}}_1=(2.29813-1.71026,1.92836-0)=(0.58787,1.92836)$. ${\mathbf{v}}_1=\frac{1}{1.75212}(0.58787,1.92836)=(0.33552,1.10059).$ ${\mathbf{v}}_2=\frac{\dot{g}{\mathbf{r}}_2-{\mathbf{r}}_1}{g}.$ $\dot{g}{\mathbf{r}}_2=(0.90342\cdot 2.29813,0.90342\cdot 1.92836)=(2.07623,1.74212)$. ${\mathbf{r}}_1=(2,0)$ subtract karo: $(0.07623,1.74212)$. ${\mathbf{v}}_2=\frac{1}{1.75212}(0.07623,1.74212)=(0.04351,0.99429).$ Yeh connecting orbit ki velocities hain — Lambert's problem ka output.

Jab $\theta \to 180^{\circ }$, $\theta /2\to 90^{\circ }$, isliye $\cos (\theta /2)\to 0$.

• $\ell =\frac{r_1+r_2}{4\sqrt{r_1{r}_2}\cos (\theta /2)}-\frac{1}{2}\to \frac{2}{4\cdot 1\cdot 0^{+}}-\frac{1}{2}\to +\infty .$
• $m=\frac{\mu \Delta t^2}{(2\sqrt{r_1{r}_2}\cos (\theta /2){)}^3}\to \frac{\mu \Delta t^2}{(0^{+}{)}^3}\to +\infty .$

Physical reason: do points antipodal ho rahe hain (focus ke opposite sides par). Transfer plane ab ${\mathbf{r}}_1,{\mathbf{r}}_2$ se uniquely define nahi hota — infinitely many orbital planes dono points contain karti hain — isliye geometry degenerate ho jaati hai. Gauss ki half-angle parameterization exactly is singularity par baith ti hai. Fix: kisi aisa solver use karo jiske denominator mein half-angle na ho — Universal variable formulation Lambert solvers (Battin's method, Izzo Lambert solver) $180^{\circ }$ ke through well-behaved rehte hain. Rule of thumb: Gauss $\theta \lesssim 90^{\circ }$–$120^{\circ }$ ke liye best hai.

• $\cos 320^{\circ }=\cos 40^{\circ }=0.76604$ — unchanged (cosine $360^{\circ }$ ke around even hai). Isliye law of cosines se chord $c$ identical hai! Yahi reason hai ki dot product akela branches distinguish nahi kar sakta.
• $\theta /2=160^{\circ }$, $\cos 160^{\circ }=-0.93969$ — ab negative.

Phir $2\sqrt{r_1{r}_2}\cos (\theta /2)=-4.60326$ (negative). Consequences:

• $\ell =\frac{5}{4(2.44949)(-0.93969)}-0.5=\frac{5}{-9.2065}-0.5=-0.54309-0.5=-\mathrm{1.04309.}$
• $m=\frac{3.24}{(-4.60326{)}^3}=\frac{3.24}{-97.552}=-0.033213$ (negative).

Matlab: sign flip long-way geometry encode karta hai. Iteration aur extraction formulas in signed constants ke saath consistently run honi chahiye — short-way $\cos (\theta /2)$ ko long-way $\cos \theta$ ke saath mix nahi kar sakte. Yeh L1 "short vs long" trap ka mathematical face hai.

Ek porkchop plot (departure date, arrival date) pairs ki ek grid fix karta hai. Har pair ${\mathbf{r}}_1$ (departure par planet A), ${\mathbf{r}}_2$ (arrival par planet B), aur $\Delta t$ (unka date difference) fix karta hai — exactly teen Lambert inputs. Har grid cell par Lambert solve karne par ${\mathbf{v}}_1,{\mathbf{v}}_2$ milta hai; planets ki apni velocities subtract karne par hyperbolic excess speeds milti hain, aur $C_3=v_{\infty }^2$ (departure energy) woh hai jo contours color karti hain. Toh yeh plot literally Lambert's problem ka millions-fold sweep hai, aur Gauss's method (ya $180^{\circ }$ ke paas uske robust cousins) neeche ka engine hai.

• Inputs/outputs? ::: In: ${\mathbf{r}}_1,{\mathbf{r}}_2,\Delta t$. Out: ${\mathbf{v}}_1,{\mathbf{v}}_2$.
• $y$ kya hai? ::: sector-area ÷ triangle-area, chhote $\theta$ ke liye $\approx 1$.
• $\Delta t$ square kyun? ::: $m$ mein Kepler-third-law scaling.
• Gauss kahan khatam hota hai? ::: $\theta \to 180^{\circ }$, $\cos (\theta /2)\to 0$, $\ell ,m\to \infty$.
• Kitni iterations? ::: $10^{-8}$ accuracy tak 3–5 iterations, kyunki $y\approx 1$ ke paas update map ek contraction hai.