Exercises — Lambert's problem — connecting two positions in given time

Lambert's theorem kehta hai ki time exactly teen scalars par depend karta hai: $a$, $c$, aur $r_1+r_2$. To correct set hai (b), (c), (d). Eccentricity akela (a) aur arc ki individual shape (e) enter nahi karte — yahi to pura miracle hai: $e$ time law se cancel ho jaata hai.

Chord. $\cos 90^{\circ }=0$, isliye $c=\sqrt{1^2+1.5^2-2(1)(1.5)(0)}=\sqrt{1+2.25}=\sqrt{3.25}=\mathrm{1.8028.}$ Semiperimeter. $s=\frac{1}{2}(1+1.5+1.8028)=\frac{1}{2}(4.3028)=\mathrm{2.1514.}$

$a_{\min }=\frac{s}{2}=\frac{2.1514}{2}=\mathrm{1.0757.}$ $a=a_{\min }=s/2$ par, ratio $\frac{s}{2a}=\frac{s}{s}=1$, isliye $\sin \frac{\alpha }{2}=\sqrt{1}=1\Rightarrow \frac{\alpha }{2}=\frac{\pi }{2}\Rightarrow \alpha =\pi .$ $\beta$ ke liye: $\frac{s-c}{2a}=\frac{2.1514-1.8028}{2(1.0757)}=\frac{0.3486}{2.1514}=0.1620$, isliye $\sin \frac{\beta }{2}=\sqrt{0.1620}=0.4025\Rightarrow \frac{\beta }{2}=0.4145\text{ rad}\Rightarrow \beta =0.8290\text{ rad}.$ To minimum-energy ellipse par, $\alpha =\pi$ exactly — yahi parent note ka landmark hai.

Lambert's law mein $\alpha =\pi$, $\beta =0.8290$ plug karo:

• $\alpha -\sin \alpha =\pi -\sin \pi =\pi -0=\mathrm{3.1416.}$
• $\beta -\sin \beta =0.8290-\sin (0.8290)=0.8290-0.7368=\mathrm{0.0922.}$
• Bracket: $(\alpha -\sin \alpha )-(\beta -\sin \beta )=3.1416-0.0922=\mathrm{3.0494.}$
• $a^{3/2}=1.0757^{1.5}=\mathrm{1.1157.}$

$\sqrt{\mu }\Delta t=1.1157\times 3.0494=3.402\Rightarrow \Delta t_{\min }=\mathrm{3.402.}$ Yeh woh sabse fast time hai jis par minimum-energy ellipse yeh arc fly karta hai. Koi bhi aur feasible transfer bada $a$ use karta hai.

Hume $F(a)=\text{(Lambert time)}-4.0$ ki root chahiye. $a_0=1.0757$ par: Lambert time $=3.402$, isliye $F(a_0)=3.402-4.0=-0.598$. Numerical slope. Paas ka point $a=1.20$ lo:

• $\frac{s}{2a}=\frac{2.1514}{2.40}=0.8964\Rightarrow \sin (\alpha /2)=0.9468\Rightarrow \alpha /2=1.2464\Rightarrow \alpha =2.4928$.
• $\alpha -\sin \alpha =2.4928-0.6003=\mathrm{1.8925.}$
• $\frac{s-c}{2a}=\frac{0.3486}{2.40}=0.14525\Rightarrow \sin (\beta /2)=0.38112\Rightarrow \beta /2=0.39125\Rightarrow \beta =0.7825$.
• $\beta -\sin \beta =0.7825-0.7053=\mathrm{0.0772.}$
• Bracket $=1.8925-0.0772=1.8153$; $a^{3/2}=1.3145$; time $=1.3145\times 1.8153=2.386$.
• $F(1.20)=2.386-4.0=-1.614$.

Ruko — $a$ badhne par time ghata! Yahi woh non-monotonic behavior hai jiske baare mein parent note mein warning di gayi thi. Is branch par ($a_{\min }$ se smoothly continue karte hue, jahan $\alpha$, $\pi$ se aage badhta hai), $a$ upar karna is range mein time ko "ulti" direction mein le jaata hai. Newton slope yahan hai: $F^{\prime }\approx \frac{-1.614-(-0.598)}{1.20-1.0757}=\frac{-1.016}{0.1243}=-\mathrm{8.17.}$ Newton step: $a_1=a_0-\frac{F(a_0)}{F^{\prime }}=1.0757-\frac{-0.598}{-8.17}=1.0757-0.0732=\mathrm{1.0025.}$ Yeh $a$ ko $a_{\min }$ se neeche push kar deta hai — jo infeasible hai ($a\ge s/2$). Sabak: $a_{\min }$ ke paas $\alpha$ ki galat branch naive Newton ko diverge kara deti hai. Sahi increasing-time branch ke liye $\alpha \to 2\pi -\alpha$ (woh doosri branch jo $a_{\min }$ ke baad hai) lena padta hai, jise hum aage explore karenge.

Ek fixed $a>a_{\min }$ ke liye, number $x=\sqrt{s/2a}$ satisfy karta hai $0<x<1$. Equation $\sin (\alpha /2)=x$ ke $(0,\pi )$ mein $\alpha /2$ ke do solutions hain: $\frac{\alpha }{2}=\arcsin x\text{ya}\frac{\alpha }{2}=\pi -\arcsin x,$ jisse milta hai $\alpha =2\arcsin x$ (woh "$\alpha <\pi$" branch) ya $\alpha =2\pi -2\arcsin x$ (woh "$\alpha >\pi$" branch).

• $\alpha <\pi$ branch: swept eccentric-anomaly arc chhota hota hai → chhota time.
• $\alpha >\pi$ branch: arc door wali side wrap karta hai → zyada time.

Physically yeh do genuinely alag ellipses hain (chord ke relative swept region ki doosri taraf focus hota hai) jo dono same $a$ ke saath ${\vec{r}}_1,{\vec{r}}_2$ connect karte hain. Exactly $a=a_{\min }$ par dono branches $\alpha =\pi$ par merge ho jaati hain. Isliye "ek time → ek orbit" galat hai, aur isliye ek solver ko iterate karne se pehle pata hona chahiye ki use kaunsi branch chahiye.

(i) Chord. $\cos 270^{\circ }=0$, bilkul $\cos 90^{\circ }$ jaisa, isliye $c=\sqrt{1+2.25-2(1.5)(0)}=\sqrt{3.25}=\mathrm{1.8028.}$ Same chord, same $s=2.1514$, same $a_{\min }=1.0757$. Chord kisi bhi taraf ghoomne mein symmetric hota hai — sirf arc alag hota hai. (ii) Modification. Long way ke liye hum $\beta$ ka sign flip karte hain: $\beta \to -\beta$ use karo, isliye bracket ban jaata hai $(\alpha -\sin \alpha )-(-\beta -\sin (-\beta ))=(\alpha -\sin \alpha )+(\beta -\sin \beta ).$ (iii) Long-way min-energy time. $\alpha =\pi$ aur $\beta =0.8290$ ke saath (2.1 se):

• $\alpha -\sin \alpha =3.1416$.
• $+(\beta -\sin \beta )=+0.0922$.
• Bracket $=3.1416+0.0922=3.2338$.
• $\times a^{3/2}=1.1157$: $\Delta t_{\text{long,min}}=1.1157\times 3.2338=3.608$.

Comparison: short way ne $3.402$ diya, long way $3.608$ deta hai. Long way slower hai same $a$ par — samajh mein aata hai, kyunki yeh ek bada arc sweep karta hai. $\beta$ ka sign exactly yahi difference encode karta hai.

(i) Hohmann $a$. Ek Hohmann ellipse dono circles ko touch karta hai, isliye uska major axis $r_1+r_2$ hai: $a_H=\frac{r_1+r_2}{2}=\frac{1+1.5}{2}=\mathrm{1.25.}$ (ii) $\Delta \theta =180^{\circ }$ par Geometry. $\cos 180^{\circ }=-1$: $c=\sqrt{1+2.25-2(1)(1.5)(-1)}=\sqrt{1+2.25+3}=\sqrt{6.25}=\mathrm{2.5.}$ Yeh bas $r_1+r_2$ hai — samajh mein aata hai, do points diametrically opposite hain isliye chord poora diameter hai. $s=\frac{1}{2}(1+1.5+2.5)=\frac{1}{2}(5)=2.5,a_{\min }=\frac{s}{2}=\mathrm{1.25.}$ To $a_H=1.25=a_{\min }$: Hohmann transfer exactly in radii ke beech half-revolution ke liye minimum-energy Lambert solution hai. Yeh ek beautiful unification hai — Hohmann ek alag theory nahi hai, yeh Lambert ka woh $\Delta \theta =\pi$, $a=a_{\min }$ corner hai. Yeh recognize karna (jaise humne abhi prove kiya $a_H=a_{\min }=1.25$) kisi bhi Lambert solver ko sanity-check karne deta hai: use 180° half-orbit do aur usse Hohmann reproduce karna chahiye. Inhe unrelated treat karna is powerful cross-check ko chhupa deta hai.

(i) $a$. Kyunki $\Delta t=\Delta t_{\min }$, iska sirf wahi ellipse achieve karta hai jo $a=a_{\min }=1.0757$ hai. (ii) Solutions ki sankhya. $\Delta t>\Delta t_{\min }$ ke liye do branches hoti hain (short-time aur long-time $\alpha$). Lekin $\Delta t=\Delta t_{\min }$ par dono branches merge ho gayi hain — yeh time-vs-$a$ curve ka tangent bottom hai. Isliye exactly ek zero-rev short-way solution hai. (iii) Planner insight. Ek merged (double) root matlab solution $\Delta t(a)$ curve ka minimum hai: $\frac{d\Delta t}{da}=0$ wahan. Required $\Delta t$ mein chhote changes $a$ ko thoda hi move karte hain, lekin yeh ek knife-edge hai — is geometry ke kisi bhi ellipse par $\Delta t_{\min }$ se tez nahi ja sakte. Ek porkchop plot mein yeh feasible $\Delta v$ region ki boundary ke roop mein dikhta hai: minimum-energy contour se neeche simply koi solution nahi hai. Planners $\Delta t$ ko $\Delta t_{\min }$ se comfortably upar rakhte hain taaki do branches (isliye flexibility) available rahe.

Bracket. $\Delta t=4.5>\Delta t_{\min }=3.402$, isliye $a>a_{\min }=1.0757$ ke saath ek solution exist karta hai. Kyunki $4.5$ minimum time se zyada hai, hume long-time behavior chahiye — woh $\alpha >\pi$ branch, jahan arc badhane se flight lambi hoti hai. $\alpha >\pi$ branch par hum $\alpha =2\pi -2\arcsin \sqrt{s/2a}$ use karte hain. Trial $a=1.30$:

• $\sqrt{s/2a}=\sqrt{2.1514/2.60}=\sqrt{0.8275}=0.9097\Rightarrow \arcsin =1.1441\Rightarrow \alpha =2\pi -2.2882=3.9950$.
• $\alpha -\sin \alpha =3.9950-(-0.7469)=4.7419$.
• $\sqrt{(s-c)/2a}=\sqrt{0.3486/2.60}=\sqrt{0.13408}=0.36617\Rightarrow \beta /2=0.37504\Rightarrow \beta =0.75008$; $\beta -\sin \beta =0.75008-0.68164=0.06844$.
• Bracket $=4.7419-0.0684=4.6735$; $a^{3/2}=1.4822$; time $=1.4822\times 4.6735=6.928$. Bahut zyada.

Trial $a=1.10$:

• $\sqrt{s/2a}=\sqrt{2.1514/2.20}=\sqrt{0.97791}=0.98890\Rightarrow \arcsin =1.42280\Rightarrow \alpha =2\pi -2.84560=3.4376$.
• $\alpha -\sin \alpha =3.4376-(-0.29327)=3.7309$.
• $\sqrt{(s-c)/2a}=\sqrt{0.3486/2.20}=\sqrt{0.15845}=0.39806\Rightarrow \beta /2=0.40952\Rightarrow \beta =0.81904$; $\beta -\sin \beta =0.81904-0.73053=0.08851$.
• Bracket $=3.7309-0.0885=3.6424$; $a^{3/2}=1.15369$; time $=1.15369\times 3.6424=4.202$. Thoda chhota.

Target $4.5$, $a=1.10$ (time $4.202$) aur $a=1.30$ (time $6.928$) ke beech hai. Interpolating: $a\approx 1.10+\frac{4.5-4.202}{6.928-4.202}(1.30-1.10)=1.10+\frac{0.298}{2.726}(0.20)=1.10+0.0219=\mathrm{1.122.}$ $a=1.122$ par refine karo:

• $\sqrt{s/2a}=\sqrt{2.1514/2.244}=\sqrt{0.95873}=0.97914\Rightarrow \arcsin =1.36494\Rightarrow \alpha =2\pi -2.72988=3.5533$.
• $\alpha -\sin \alpha =3.5533-(-0.40241)=3.9557$.
• $\sqrt{(s-c)/2a}=\sqrt{0.3486/2.244}=\sqrt{0.15535}=0.39415\Rightarrow \beta /2=0.40527\Rightarrow \beta =0.81054$; $\beta -\sin \beta =0.81054-0.72469=0.08585$.
• Bracket $=3.9557-0.0859=3.8698$; $a^{3/2}=1.18845$; time $=1.18845\times 3.8698=4.599$. $4.5$ ke bahut paas.

Ek aur thoda sa neeche nudge karne se $a\approx 1.117$ milta hai. 3 sig figs tak, $a\approx 1.12$ $\alpha >\pi$ (long-time) branch par.