Bi-elliptic transfer — when it wins over Hohmann: Visual walkthrough

3.2.21 · D2 · Physics › Orbital Mechanics & Astrodynamics › Bi-elliptic transfer — when it wins over Hohmann

Step 1 — Do circles, aur "orbit speed" ka matlab kya hai

KYA. Hum poora problem draw karte hain: ek heavy body (planet ya star) centre mein, ek choti circle radius $r_{1}$ ki jahan se hum start karte hain, aur ek badi circle radius $r_{2}$ ki jahan hum pahunchna chahte hain. Dono circles ek hi flat plane mein hain — orbits coplanar hain, isliye koi tilting nahi chahiye.

KYUN. Kisi bhi burn se pehle, hume ek single fact pe agree karna hoga: radius $r$ ki circle par tum kitni fast move karte ho? Aage jo bhi hoga woh speeds ka comparison hai, isliye pehle speed rule chahiye.

PICTURE. Step 1 dekho. Inner circle par ek fast blue arrow hai (lamba); outer circle par ek slow blue arrow hai (chota). Jitna door = utna slow. Yahi ek sach is poore trick ka engine hai.

Step 2 — Ek single burn tumhe circles ke beech teleport kyun nahi kar sakta

KYA. Hum poochte hain: kya ek engine push tumhe inner circle se directly outer circle par le ja sakta hai? Nahi. Ek burn tumhari speed us spot par change karta hai jahan tum ho; yeh tumhara radius instantly nahi badalta. Toh ek burn tumhari circle ko ek stretched loop mein bend kar deta hai — ek ellipse — jo tumhe time ke saath outward le jaata hai.

KYUN. Yahi reason hai ki transfers ka shape hota hai. Tum curved bridges (ellipses) par travel karne ke liye force ho, aur har ellipse do circles ko do special points par touch karti hai. Un points ko naam dena Step 3 ka kaam hai.

PICTURE. Step 2 mein inner circle par ek single prograde nudge (pink arrow, "speed up") dikhti hai. Circle immediately ek yellow ellipse mein bloom ho jaati hai jiski near end abhi bhi $r_{1}$ ko kiss karti hai lekin jiski far end ek naye, bade radius tak swing karti hai.

Step 3 — Ek ellipse ko naam dene wala ek number: $a$

KYA. Hum ek ellipse ko ek single length se measure karte hain: periapsis distance $r_{peri}$ aur apoapsis distance $r_{apo}$ ko end to end rakh ke average lo.

KYUN. Vis-viva ko ek number chahiye poori orbit describe karne ke liye, aur woh number semi-major axis $a$ hai. Yeh secretly orbit ki total energy store karta hai (dekho Semi-major axis and orbital energy): bada $a$ = higher, zyada energetic orbit. Har burn ki cost really yahi hai: "maine $a$ kitna move kiya?"

PICTURE. Step 3 do apsidal radii ko centre se guzarne wali ek seedhi line par rakhta hai. Unka full span $r_{peri} + r_{apo}$ hai; aadha karo aur tum $a$ par pahunch jaate ho, ellipse ki waistline par marked.

Ab har orbit — circle ya ellipse — ko ek master speed rule mein daala ja sakta hai.

Step 4 — Master ruler: vis-viva kisi bhi orbit par speed padhta hai

KYA. Hum woh ek equation state karte hain jo tumhari speed kisi bhi radius $r$ par, kisi bhi orbit jiska size $a$ hai, deta hai.

KYUN. Har burn hai "new orbit par speed jo main rakhta hoon" minus "old orbit par speed jo main rakhta tha," dono same spot $r$ par measured. Agar hamare paas ek formula hai jo $(r, a)$ se speed read karta hai, toh har $Δ v$ bas do readings subtract karne se milti hai. Yahi poora toolkit hai — hume aur kuch nahi chahiye.

PICTURE. Step 4 ek dial hai: radius $r$ fix karo (jahan tum khade ho), aur vis-viva tumhari speed return karta hai. Same spot par do alag ellipses par do alag needle positions milti hain; ek burn unke beech ka jump hai.

Step 5 — Teen burns, ek map par draw kiye

KYA. Hum ab bi-elliptic route ke teeno impulses ko ek figure par rakhte hain aur do bridging ellipses ko naam dete hain.

Ellipse 1 $r_{1}$ (periapsis) se $r_{b}$ (apoapsis) tak jaati hai, ek intermediate radius jo $r_{2}$ se badi choose ki gayi hai. Uska size: $a_{1} = \frac{r _{1} + r _{b}}{2}$ .
Ellipse 2 $r_{2}$ (periapsis) se usi far point $r_{b}$ (apoapsis) tak jaati hai. Uska size: $a_{2} = \frac{r _{2} + r _{b}}{2}$ .

KYUN. Hum deliberately target overshoot karte hain ( $r_{b} > r_{2}$ ) taaki reshaping burn bahut door ho jahan Step 1 ne promise kiya tha ki speeds tiny hoti hain. Us middle burn ki cost ko hum starve karna chahte hain.

PICTURE. Step 5 full trajectory dikhata hai: burn 1 (pink, speed up) $r_{1}$ par yellow ellipse 1 launch karta hai; burn 2 (yellow, gentle) far point $r_{b}$ par blue ellipse 2 par lift karta hai; burn 3 (pink, brake) $r_{2}$ par outer circle par settle karta hai.

Har burn ab bas same $r$ par do vis-viva readings, subtracted hai.

Step 6 — Middle burn ko vanish hote dekho (the far-out limit)

KYA. Hum $r_{b}$ ko bada se bada karte hain aur teeno burns ko respond hote dekhte hain.

KYUN. Trick feel karne ke liye tum trend dekhna chahte ho: burn 2 zero ki taraf collapse karta hai, burn 3 apni limit ki taraf flatten hota hai, jabki burn 1 climb karta rehta hai. Unka tug-of-war sab decide karta hai.

PICTURE. Step 6 har burn ka size $r_{b}$ ke against plot karta hai. Burn 2 (blue) zero ki taraf dives; burn 3 (pink) flatten hota hai; burn 1 (yellow) rise karta hai. Sum (white) bi-parabolic limit $r_{b} \to \infty$ ke paas ek minimum ki taraf sag karta hai.

Step 7 — Crossover: detour kab finally pay karta hai? (11.94 aur 15.58 derive karna)

KYA. Hum total bi-elliptic $Δ v$ ko do-burn Hohmann transfer cost ke saath overlay karte hain, ratio $R = r_{2} / r_{1}$ sweep karte hue, aur exactly woh point dhundte hain jahan dono equal hain.

KYUN. Yahi payoff picture hai — yeh sari geometry ko ek single yes/no rule mein convert karti hai. Famous numbers paane ke liye hume $Δ v_{H} (R) = Δ v_{bi} (R)$ solve karna hoga.

Numbers kaise arise karte hain. Normalized units mein kaam karo $μ = 1, r_{1} = 1$ , toh $r_{2} = R$ . Do natural comparisons do thresholds dete hain:

Threshold $R \approx 15.58$ (guaranteed-win number). Hohmann ko best-case bi-parabolic limit ( $r_{b} \to \infty$ , Step 6 se) ke against compare karo. Do $Δ v$ expressions ko equal set karne se $R$ mein akela ek transcendental equation milta hai; use numerically solve karne par $R^{⋆} \approx 15.58$ milta hai. Iske upar, best Hohmann bhi best bi-elliptic se haarta hai, toh bi-elliptic kisi finite $r_{b}$ ke liye win karta hai.
Threshold $R \approx 11.94$ (never-below number). Ab opposite poochho: kis $R$ ke neeche koi finite $r_{b}$ kabhi Hohmann ko beat nahi kar sakta? Yeh sabse bada $R$ hai jis par $\frac{d}{d r _{b}} Δ v_{bi}_{r_{b} = r_{2}} \geq 0$ — yani overshoot shuru karna cost raise karta hai. Woh boundary condition $R \approx 11.94$ solve hota hai.
Unke beech ( $11.94 < R < 15.58$ ): overshoot karna help karta hai, lekin sirf itna ki win ho agar $r_{b}$ kaafi bada ho — winner tumhare chosen $r_{b}$ par depend karta hai (aur tumhare time budget par).

PICTURE. Step 7 dono curves $R$ ke against plot karta hai. $R \approx 11.94$ ke neeche Hohmann ki line neeche hai; $R \approx 15.58$ ke upar bi-elliptic line neeche dip karti hai. Beech wala grey band "yeh $r_{b}$ par depend karta hai" zone hai.

Step 8 — Edge & degenerate cases (kabhi mat pakde jaao)

KYA + PICTURE. Step 8 corner cases ko side by side collect karta hai; left panel algebraically aur visually key one dikhata hai.

$r_{b} = r_{2}$ (no overshoot). Tab $a_{2} = \frac{r _{2} + r _{b}}{2} = \frac{r _{2} + r _{2}}{2} = r_{2}$ — ellipse 2 ka size outer circle ke radius ke equal hai, toh ellipse 2 outer circle hai. Uski periapsis speed already circular speed ke equal hai, toh

$Δ v_{2} = μ (\frac{2}{r _{b}} - \frac{1}{a _{2}}) - μ (\frac{2}{r _{b}} - \frac{1}{a _{1}}), aur burn 3 vanish: \frac{μ}{r _{2}} - μ (\frac{2}{r _{2}} - \frac{1}{r _{2}}) = 0.$ Transfer exactly ek Hohmann mein collapse ho jaata hai (ek bridging ellipse, do burns). Toh Hohmann usi family ka " $r_{b} = r_{2}$ " member hai. Step 8 ka left panel yeh collapse draw karta hai.

$r_{b} \to \infty$ (bi-parabolic): burn 2 $\to 0$ , transfer time $\to \infty$ . Given $R$ ke liye lowest possible $Δ v$ , lekin kabhi physically reach nahi hota.
$r_{1} = r_{2}$ ( $R = 1$ ): start aur end circles coincide karte hain, koi transfer nahi chahiye, saare burns → 0.
Chota $R$ (e.g. $R = 6$ ): burn 1 ki extra climb kisi bhi burn-2 saving se zyada hai — detour pure waste hai (parent Example 2).

Ek-picture summary

Ek frame poori story hold karta hai: do circles, do bridging ellipses far point $r_{b}$ par milti hain, teen labelled burns (bada prograde, tiny reshape, prograde-then-brake), aur crossover strip jo batati hai kab yeh Hohmann ko beat karta hai.

Recall Feynman retelling — walkthrough simple words mein

Humne ek cheez notice karke start kiya: jitna door orbit karo, utna slow drift (Step 1) — aur hume uska speed rule mila orbit ka "size" uske radius ke equal set karke. Rocket burn sirf tumhari speed change karta hai, toh yeh tumhari circle ko ek stretched loop mein bend karta hai, ek ellipse, jo tumhe outward le jaati hai (Step 2). Humne har ellipse ko ek number se measure kiya — uski near aur far distances ka average, jise $a$ kehte hain (Step 3); far end ko infinity tak push karo aur woh loop ek parabola ban jaati hai. Ek master rule, vis-viva, tumhari speed padhta hai jahan tum ho aur tumhari orbit kitni badi hai (Step 4). Phir humne bi-elliptic route draw ki: target se bahut aage ek point $r_{b}$ tak fling karo, near-side ko target tak gently nudge karo, phir coast back karo aur circle par brake karo (Step 5) — har $Δ v$ ka sign batata hai kis taraf fire karo, uski size batati hai fuel. Kyunki $r_{b}$ par sab kuch sluggish hai, woh middle nudge almost kuch nahi cost karta, aur jitna zyada door fling karo utna sasta hota hai — lekin burn 1 grow karta hai iske liye pay karne, isliye tum $r_{b}$ choose karte ho ki tum kitna time bardaasht kar sakte ho (Step 6). Simple do-burn Hohmann ke against total race karo aur solve karo jahan woh tie ho — famous numbers milte hain: $R \approx 12$ ke neeche Hohmann always win karta hai, $R \approx 16$ ke upar bi-elliptic always win karta hai (Step 7). Aur humne corners tidy kiye: fling ko target tak shrink karo aur, kyunki $a_{2} = r_{2}$ , burn 2 aur burn 3 collapse ho jaate hain — tum literally Hohmann ban jaate ho (Step 8). Same physics, ek family, ek dial $r_{b}$ kehlata hai.

Recall Quick self-test

Burn 2 almost free kyun hai? ::: Yeh bahut bade radius $r_{b}$ par hota hai, jahan shared $2/ r_{b}$ term dono vis-viva speeds mein dominate karta hai, toh woh almost cancel ho jaate hain. $r_{b} = r_{2}$ bi-elliptic transfer ko kya reduce karta hai? ::: Ek ordinary Hohmann transfer — kyunki $a_{2} = r_{2}$ , toh ellipse 2 outer circle hai aur burns 2, 3 collapse ho jaate hain. Number 15.58 kahan se aata hai? ::: Hohmann ka $Δ v$ best-case bi-parabolic ( $r_{b} \to \infty$ ) $Δ v$ ke equal set karke aur $R$ ke liye solve karke. Kis $R$ ke upar bi-elliptic guaranteed win karta hai? ::: $R > 15.58$ . Burn 3 prograde hai ya retrograde? ::: Retrograde — tum periapsis par already bahut fast pahunchte ho aur brake karna padta hai.