Exercises — Bi-elliptic transfer — when it wins over Hohmann

• Teen burns, do ellipses.
• Burn 1 $r_1$ par (apoapsis ko $r_b$ tak raise karo), Burn 2 $r_b$ par ($r_1$ waale opposite apsis ko $r_2$ tak raise karo), Burn 3 $r_2$ par (circularize karo).
• Hohmann transfer se comparison: do burns, ek ellipse.

Intermediate radius ko $r_b>r_2$ satisfy karna chahiye (yaad karo $r_b$ upar define kiya gaya overshoot apoapsis hai). Yahan $r_b=15<20=r_2$ hai, jo geometrically impossible hai: pehle ellipse ka apoapsis $r_b$ target se aage hona chahiye taaki doosra ellipse $r_2$ par "coast back down" kar sake. Agar $r_b\le r_2$ ho toh ye bilkul bi-elliptic transfer nahi hai ($r_b=r_2$ par ye Hohmann mein degenerate ho jaata hai).

• $R<11.94$: Hohmann hamesha jeet ta hai — $r_b$ ki koi bhi choice bi-elliptic ko usse beat nahi karne deti.
• $R>15.58$: bi-elliptic jeetta hai agar $r_b$ kaafi bada choose kiya jaye.
• $11.94<R<15.58$: ye $r_b$ par depend karta hai (grey zone).

• $v_{c1}=\sqrt{\mu /r_1}=\sqrt{1/1}=1$.
• $v_{c2}=\sqrt{\mu /r_2}=\sqrt{1/20}=0.223607$. Ye formula kyun? Ye vis-viva hai jab $a=r$ (ek circle), isliye dono energy terms sirf $v^2=\mu /r$ chhod dete hain. Iska matlab: outer orbit slow hai — woh slowness hi woh resource hai jise bi-elliptic exploit karta hai.

• $a_1=\frac{r_1+r_b}{2}=\frac{1+100}{2}=50.5$. Kyun: $r_1$ is ellipse ka periapsis hai, $r_b$ iska apoapsis; inhe average karo.
• Ellipse 1 par $r_1$ par speed: vis-viva $r=r_1,a=a_1$ ke saath: $\sqrt{2-\frac{1}{50.5}}=\sqrt{1.98020}=1.40719$. Kyun yahan: Burn 1 $r_1$ par fire hota hai, isliye hum naye orbit ki speed $r_1$ par evaluate karte hain.
• Circular speed subtract karo: $\Delta v_1=1.40719-1=0.40719$. Subtract kyun? Hum pehle se $v_{c1}=1$ speed se ek circle mein move kar rahe the; burn sirf orbit ko bahar stretch karne ke liye zaroori extra speed ke liye pay karta hai — ek pure prograde kick, same direction, isliye cost sirf gap hai.

• $a_2=60$. Kyun: ellipse 2 ka apoapsis $r_b=100$ aur periapsis $r_2=20$ hai; inhe average karo.
• $r_b$ par ellipse 2 ki speed (vis-viva, $r=100,a=60$): $\sqrt{\frac{2}{100}-\frac{1}{60}}=\sqrt{0.02-0.016667}=\sqrt{0.003333}=0.057735$.
• $r_b$ par ellipse 1 ki speed ($r=100,a=50.5$): $\sqrt{\frac{2}{100}-\frac{1}{50.5}}=\sqrt{0.02-0.019802}=\sqrt{0.000198}=0.014079$.
• $\Delta v_2=0.057735-0.014079=0.043656$. Figure padho: dono speeds far point par prograde arrows hain, aur dono same dominant $\frac{2}{r_b}=0.02$ term share karte hain — isliye arrows almost equal length ke hain aur tum sirf unke beech ka chota orange gap pay karte ho, unka sum nahi. Ye tiny far-out cost hi poori wajah hai ki bi-elliptic jeet sakta hai.

• $a_2=60$ (2.3 se). Burn 3 kahan aur kyun fire hota hai: $r_2=20$ par, jo ellipse 2 ka periapsis hai, isliye craft wahan circle se zyada fast arrive karta hai — vis-viva $r=20,a=60$ par: $\sqrt{\frac{2}{20}-\frac{1}{60}}=\sqrt{0.083333}=0.288675$.
• Target circular speed: $v_{c2}=\sqrt{1/20}=0.223607$.
• $\Delta v_3=0.288675-0.223607=0.065068$ — ek brake (retro): tum circular speed se upar ho, isliye circularize karne ke liye slow down karna padega. Magnitude lo.
• Total: $0.40719+0.043656+0.065068=0.51591$.
• Intuition tie-back: teen burns hain "climb out" (0.407) + "cheap far-out reshape" (0.044) + "gentle brake in" (0.065). Tiny middle term hi poora trick hai. Kyunki $R=20>15.58$, bi-elliptic ko jeetnaa chahiye — aur parent ka Hohmann $=0.5347>0.51591$ confirm karta hai. ✅

• $a_1=\frac{1+60}{2}=30.5$, $a_2=\frac{20+60}{2}=40$. Kyun: same "apsides average karo" rule naye $r_b=60$ ke saath.
• Burn 1 ($r_1$ par, ellipse 1 enter karo): $\sqrt{2-\frac{1}{30.5}}-1=\sqrt{1.967213}-1=1.402574-1=0.402574$.
• Burn 2 ($r_b=60$ par, ellipse 1→2): $\sqrt{\frac{2}{60}-\frac{1}{40}}-\sqrt{\frac{2}{60}-\frac{1}{30.5}}=\sqrt{0.008333}-\sqrt{0.000546}=0.091287-0.023372=0.067915$.
• Burn 3 ($r_2=20$ par brake): $\sqrt{\frac{2}{20}-\frac{1}{40}}-0.223607=\sqrt{0.075}-0.223607=0.273861-0.223607=0.050254$.
• Total: $0.402574+0.067915+0.050254=0.520743$.
• Interpretation: $r_b=100$ wale $0.51591$ se compare karein, ye thoda worse hai. $r_b=60$ par far point utna slow nahi, isliye Burn 2 bada hai (0.068 vs 0.044). Bada $r_b$ total ko neeche laa raha tha — ye trend bi-parabolic floor ki taraf continue karta hai.

• Burn 2 ↓ (physical reason): ye $r=r_b$ par fire hota hai. Vis-viva mein $\frac{2}{r_b}\to 0$ jab $r_b\to \infty$ dono ellipse speeds ke liye, isliye dono roots zero ki taraf shrink karte hain aur unka difference vanish ho jaata hai. Bahar jaao = sab kuch slower = sasti reshape. Ye parent ke merry-go-round analogy ka far-out "calm zone" hai.
• Burn 1 ↑ (physical reason): $r_1$ par ellipse-1 speed $\sqrt{2/r_1-1/a_1}$ hai jahan $a_1=(r_1+r_b)/2$. Jab $r_b\to \infty$, $1/a_1\to 0$, isliye ye speed $\to \sqrt{2/r_1}=$ escape speed ho jaati hai. Bahar fling karne ke liye bada initial kick chahiye.
• Net: $r_b:60\to 100$ jaane par, Burn-2 saving (aur Burn-3 shift ke saath) ne Burn-1 rise ko outweigh kiya, isliye total $0.520743\to 0.51591$ gira. Dono effects ek tug-of-war hain jo sirf bi-parabolic limit par improve karna band karte hain.

• $a_1=50.5,a_2=(6+100)/2=53$.
• Burn 1: $\sqrt{2-\frac{1}{50.5}}-1=1.407190-1=0.407190$ — Ex 2.2 jaisa hi, kyunki Burn 1 sirf $r_1$ aur $r_b$ par depend karta hai, $r_2$ par nahi.
• Burn 2: $\sqrt{\frac{2}{100}-\frac{1}{53}}-\sqrt{\frac{2}{100}-\frac{1}{50.5}}=\sqrt{0.001132}-\sqrt{0.000198}=0.033651-0.014079=0.019572$.
• Burn 3 ($r_2=6$ par brake): $\sqrt{\frac{2}{6}-\frac{1}{53}}-\sqrt{\frac{1}{6}}=\sqrt{0.314570}-0.408248=0.560866-0.408248=0.152618$.
• Total: $0.407190+0.019572+0.152618=0.579380$.
• Parent ke $0.6034$ par: woh value rounded intermediate roots use karta tha (e.g. $\sqrt{0.3144}$, $a_2$ printed as $53$) aur un roundings ko accumulate kiya; full precision rakhne par honest total $0.5794$ hai. $\approx 0.024$ gap nahi hai final answer ki trivial rounding — ye har ek square root ko subtract se pehle 4 digits tak round karne ka compounded effect hai jo almost-equal numbers par (especially Burn 2 mein catastrophic cancellation). Kisi bhi case mein conclusion same rehta hai: $0.5794>0.4925$, Hohmann jeetta hai.
• Culprit: Burn 1 ($0.407$). Chhote $R$ par tum abhi bhi $r_b=100$ tak fling karne ki full escape-like cost pay karte ho, lekin target sirf $r_2=6$ par hai — woh badi climb mostly waste hai, aur meagre Burn-2 saving use repay nahi kar sakti. $R=6<11.94$ → Hohmann jeetta hai.

Hohmann, $a_H=(1+13)/2=7$:

• Burn A: $\sqrt{2-\frac{1}{7}}-1=\sqrt{1.857143}-1=1.362770-1=0.362770$. Kyun: Hohmann ellipse par $r_1$ par vis-viva, minus $v_{c1}$.
• Burn B: $v_{c2}-\sqrt{\frac{2}{13}-\frac{1}{7}}=\sqrt{1/13}-\sqrt{0.153846-0.142857}=0.277350-0.104828=0.172522$. Kyun: Hohmann $r_2$ par circular speed se neeche arrive karta hai (ye apoapsis hai), isliye ye burn speed add karta hai.
• Hohmann total $=0.362770+0.172522=0.535292$.

Bi-elliptic, $r_b=100$: $a_1=50.5,a_2=(13+100)/2=56.5$:

• Burn 1: $0.407190$.
• Burn 2: $\sqrt{\frac{2}{100}-\frac{1}{56.5}}-\sqrt{\frac{2}{100}-\frac{1}{50.5}}=\sqrt{0.002301}-\sqrt{0.000198}=0.047968-0.014079=0.033889$.
• Burn 3 (brake): $\sqrt{\frac{2}{13}-\frac{1}{56.5}}-\sqrt{1/13}=\sqrt{0.136147}-0.277350=0.368981-0.277350=0.091631$.
• Bi-elliptic total $=0.407190+0.033889+0.091631=0.532710$.

Result: $0.532710<0.535292$ → bi-elliptic jeetta hai, lekin barely ($\approx 0.5\%$). Ye grey zone hai: outcome $r_b$ par hinge karta hai aur margin razor-thin hai. Ek planner us sliver ko bahut lambi flight time ke against weigh karta hai.

Limits term by term lo (jahan $r_1=1,r_2=R$), har ek apni physical meaning ke saath.

• Burn 1 (climb): $a_1=\frac{1+r_b}{2}\to \infty$, isliye $\sqrt{2-1/a_1}\to \sqrt{2}$. Iss tarah $\Delta v_1\to \sqrt{2}-1$ — tum $r_1$ par escape speed tak pahuncho aur aur nahi.
• Burn 2 (far-out reshape): $r_b$ par dono speeds $\sqrt{2/r_b}\to 0$ ki tarah scale karte hain; unka difference $\to 0$. Isliye $\Delta v_2\to 0$ — infinity par free plane/apsis reshape.
• Burn 3 (brake in): $a_2=\frac{R+r_b}{2}\to \infty$, isliye $\sqrt{2/R-1/a_2}\to \sqrt{2/R}$. Iss tarah $\Delta v_3\to \sqrt{2/R}-\sqrt{1/R}=\frac{\sqrt{2}-1}{\sqrt{R}}$ — same escape-vs-circular gap, $r_2$ par slower speeds ke liye scaled down. $\boxed{\Delta v_{\text{bipar}}=(\sqrt{2}-1)\left(1+\frac{1}{\sqrt{R}}\right)}$ $R=20$ par: $(\sqrt{2}-1)(1+1/\sqrt{20})=0.414214\times 1.223607=0.506841$. Ex 3.1 ke finite value $0.51591$ se compare karo — bi-parabolic limit lower hai (ye floor hai), confirming karta hai ki bada $r_b$ is bound ki taraf improve karta rehta hai.

Hohmann normalized units mein, $a_H=\frac{1+R}{2}$ (same tarah derive kiya Ex 4.2 ki tarah, ab symbolic): $\Delta v_H=\left(\sqrt{2-\frac{2}{1+R}}-1\right)+\left(\frac{1}{\sqrt{R}}-\sqrt{\frac{2}{R}-\frac{2}{1+R}}\right)$ $R=11.94$ par: $a_H=6.47$.

• Burn A: $\sqrt{2-2/12.94}-1=\sqrt{1.845440}-1=0.358470$.
• Burn B: $1/\sqrt{11.94}-\sqrt{2/11.94-2/12.94}=0.289396-\sqrt{0.012944}=0.289396-0.113772=0.175624$.
• $\Delta v_H=0.534094$.
• Bi-parabolic: $(\sqrt{2}-1)(1+1/\sqrt{11.94})=0.414214\times 1.289396=0.534089$.
• $|0.534094-0.534089|\approx 5\times 10^{-6}$ → rounding tak equal. ✅ Ye equality precisely isliye hai kyunki $11.94$ lower magic number hai: iske neeche bi-parabolic (best-possible bi-elliptic) Hohmann ko match nahi kar sakti.

$R=10$ par (threshold se neeche):

• Hohmann: $a_H=5.5$; Burn A $=\sqrt{2-2/11}-1=0.348469$; Burn B $=1/\sqrt{10}-\sqrt{0.2-2/11}=0.316228-0.134840=0.181388$; total $=0.529857$.
• Bi-parabolic: $0.414214\times (1+1/\sqrt{10})=0.414214\times 1.316228=0.545225$.
• $0.529857<0.545225$ → Hohmann strictly jeetta hai, jaisa $R<11.94$ ke liye promise kiya gaya tha. ✅

Angle $\Delta i$ ka pure plane change, speed $v$ par, cost karta hai $\Delta v\approx 2v\sin (\Delta i/2)$ — ye burn point par speed ke proportional hai (dekho Plane change maneuvers). $r_1$ par spacecraft fast hai ($v_{c1}$ bada), isliye orbit plane rotate karna mehenga hai. $r_b$ par speed tiny hai (vis-viva: $v\to 0$ jab $r_b\to \infty$), isliye same angular rotation almost kuch nahi cost karta.

• Rule: rotate karo jahan slow ho. $r_b$ tak fling karo, us lazy far point par plane change (aur periapsis raise) karo, phir waapis coast karo. Jab required plane change bada ho, ye extra climbs se zyada save kar sakta hai — size-change crossover ka direct cousin, aur Oberth effect ka ulta (jo kehta hai speed changes saste hain jahan tum fast ho).