Bi-elliptic transfer — when it wins over Hohmann: Question bank

3.2.21 · D5 · Physics › Orbital Mechanics & Astrodynamics › Bi-elliptic transfer — when it wins over Hohmann

Shuru karne se pehle, ye earned symbols dhyan mein rakho:

$r_{1}$ = inner starting circular orbit ka radius; $r_{2}$ = outer target circular orbit ka radius.
$r_{b}$ = woh door wala intermediate radius jahan tak bi-elliptic transfer tumhe phenk ta hai ( $r_{b} > r_{2}$ ).
$R = r_{2} / r_{1}$ = target orbit, start se kitni baar badi hai.
$μ = GM$ = gravitational parameter (planet ki mass $\times$ big-G), jo Vis-viva equation ke zariye saari speeds set karta hai.
$Δ v$ = total "speed budget" jo ek maneuver kharchta hai — fuel ki currency.

(Note on callouts: neeche ke boxed headers jaise [!formula] aur [!intuition] sirf labelled boxes hain — agar tumhara reader unhe plain text mein dikhaye, toh bold word ko box ka title samjho. Physics text mein hi hai.)

Har question ke peeche ka picture

Poora maneuver ek hi diagram mein jeeta hai. Jawab dene se pehle isse dhyan se dekho — neeche ke zyaadatar traps kis point par burn hoti hai aur kis direction mein push karti hai, iske baare mein hain.

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

Figure dekho: chhota blue circle start hai ( $r_{1}$ ), green circle target hai ( $r_{2}$ ), aur dono dashed ellipses left side mein door wale point $r_{b}$ tak stretch karti hain. Burn 1 (orange) $r_{1}$ par hoti hai aur far end ko $r_{b}$ tak uthati hai. Burn 2 (blue) kaafi door $r_{b}$ par hoti hai aur near end ko $r_{1}$ se uthaa ke $r_{2}$ tak le jaati hai. Burn 3 (red) wapas $r_{2}$ par hoti hai aur ye ek brake hai.

Semi-major axis kyun do apsidal radii ka average hai

Ellipse figure dekho. Long axis (major axis) ek seedhi line mein nearest point (periapsis, planet se distance $r_{p}$ ) se hote hue planet ke focus se guzar ke farthest point (apoapsis, distance $r_{a}$ ) tak jaata hai. Wo poori seedhi line ki length $r_{p} + r_{a}$ hai. Letter $a$ semi-major axis hai — major axis ka aadha — toh $a = \frac{r _{p} + r _{a}}{2},$ jo literally dono endpoints ki distances ka arithmetic mean hai. Bas itna hi hai: $a$ long axis ka midpoint distance hai. Toh ellipse 1 ke liye (periapsis $r_{1}$ , apoapsis $r_{b}$ ) hume milta hai $a_{1} = (r_{1} + r_{b}) /2$ , aur ellipse 2 ke liye (periapsis $r_{2}$ , apoapsis $r_{b}$ ) milta hai $a_{2} = (r_{2} + r_{b}) /2$ .

Worked example Vis-viva se

$Δ v_{1}$ , $Δ v_{2}$ , $Δ v_{3}$ derive karna Upar ki figure se $a_{1} = (r_{1} + r_{b}) /2$ aur $a_{2} = (r_{2} + r_{b}) /2$ use karke, aur sign rule $Δ v = ∣ v_{after} - v_{before} ∣$ se:

Burn 1 at $r_{1}$ . Pehle: circular speed $v_{c} (r_{1}) = μ / r_{1}$ . Baad mein: ellipse 1 par, $r = r_{1}$ , $a = a_{1}$ vis-viva mein daalo. Subtract karo: $Δ v_{1} = μ (\frac{2}{r _{1}} - \frac{1}{a _{1}}) - \frac{μ}{r _{1}} .$

Burn 2 at $r_{b}$ . Pehle: ellipse 1 par $r = r_{b}$ (uska apoapsis) par. Baad mein: ellipse 2 par usi $r_{b}$ par. Subtract karo: $Δ v_{2} = μ (\frac{2}{r _{b}} - \frac{1}{a _{2}}) - μ (\frac{2}{r _{b}} - \frac{1}{a _{1}}) .$

Burn 3 at $r_{2}$ . Pehle: ellipse 2 par $r = r_{2}$ (uska periapsis, toh circular speed se tez). Baad mein: circular speed $v_{c} (r_{2})$ . Magnitude brake ko handle kar leti hai: $Δ v_{3} = μ (\frac{2}{r _{2}} - \frac{1}{a _{2}}) - \frac{μ}{r _{2}} .$

Jahan se magic numbers 11.94 aur 15.58 actually aate hain

Unhe dekhne ka sabse saaf tarika hai bi-parabolic limit: $r_{b} \to \infty$ push karo. Tab har ellipse ek parabola (ek escape coast) ban jaati hai, burn 2 zero tak sir jaati hai, aur baaki do burns bas "escape speed par jump karo" aur "escape speed se utaro" ban jaati hain: $Δ v_{bipar} = burn 1: circular \to escape at r_{1} (\frac{2 μ}{r _{1}} - \frac{μ}{r _{1}}) + burn 3: escape \to circular at r_{2} (\frac{2 μ}{r _{2}} - \frac{μ}{r _{2}}) .$ Ise $Δ v_{H}$ ke barabar set karo, $μ / r_{1}$ se divide karo, aur sab kuch $R = r_{2} / r_{1}$ ka function ban jaata hai (physics scale-free hai). Us ek equation ko numerically solve karne par milta hai $R = 15.58$ : bilkul is ratio par ideal bi-parabolic Hohmann se tie karta hai, aur uske upar bi-elliptic jeet jaata hai. Neeche wala number $11.94$ usi comparison se aata hai lekin $r_{b}$ ko finite rakhte hue aur poochhte hue ki "kya koi $r_{b}$ hai jo Hohmann ko haraata hai?" — $11.94$ se neeche answer har $r_{b}$ ke liye no hai. Doosra figure dono curves ko in dono markers ke beech cross karte hue dikhata hai:

Orange Hohmann curve aur blue bi-parabolic curve $R = 15.58$ par cross karti hain; blue curve bade $R$ ke liye neeche rehti hai. Do vertical gray markers woh thresholds hain $11.94$ (finite- $r_{b}$ onset) aur $15.58$ (guaranteed win) jinhe neeche har question refer karta hai.

True or false — justify karo

TF1. Ek bi-elliptic transfer mein hamesha Hohmann transfer se kam burns hoti hain.

False — isme zyada hoti hain: teen burns aur do ellipses, jabki Hohmann ke do burns aur ek ellipse hoti hai. Iska advantage (jab exist karta hai) lower total

Δ v

hai, kabhi kam impulses nahi.

TF2. Kyunki orbital speeds bahut door kaafi chhoti hoti hain, middle burn hamesha teenon mein sabse sasti hoti hai.

True is sense mein ki wo teenon mein sabse chhoti hai: bade

r_{b}

par

Δ v_{2}

ki dono speeds

μ (2/ r_{b} - 1/ a)

hain, aur kyunki

2/ r_{b}

zero ki taraf shrink hota hai jabki

1/ a_{1} \approx 1/ a_{2}

, dono square roots almost ek jaisi tiny value mein collapse ho jaati hain aur unka difference vanish ho jaata hai — lekin "sasti middle burn" ka matlab nahi ki poora transfer sasta hai.

TF3. Bi-elliptic transfer Hohmann ko harata hai jab bhi target orbit start se badi ho.

False — badi target necessary hai lekin kaafi nahi.

R = 11.94

se neeche Hohmann hamesha jeetta hai; detour sirf bahut bade ratios par faayda karta hai.

TF4. Agar

R = 20

ho, toh bi-elliptic

r_{b}

ki kisi bhi choice ke liye Hohmann ko haraana guaranteed hai.

False —

R = 20 > 15.58

guarantee karta hai ki ek suitable $r_{b}$ ke liye jeet exist karti hai, lekin ek badly chosen (bahut chhota)

r_{b}

phir bhi hara sakta hai. Threshold promise karta hai ki koi

r_{b}

kaam karta hai, har

r_{b}

nahi.

TF5. Burn 3 (

r_{2}

par circularize karna) ek prograde acceleration hai jo "climb finish karta hai".

False —

r_{2}

par tum doosri ellipse ke periapsis par ho, wahan circular speed se tez ja rahe ho, toh burn 3 ek retro-burn (brake) hai. Tum

Δ v

sum mein iska magnitude lete ho.

TF6.

r_{b} \to \infty

lena (bi-parabolic limit) kisi bhi

R

ke liye mathematically minimal

Δ v

deta hai.

Partly true aur dangerous: bi-parabolic limit woh best case deta hai jo thresholds derive karne ke liye use hota hai, lekin isme infinite transfer time bhi lagti hai, aur ye pehli jagah sirf

R

ke liye jeetta hai jo crossover se upar ho. Bi-parabolic transfer dekho.

TF7. Ek hi start aur target ke liye, bi-elliptic aur Hohmann roughly same time lete hain.

False — bi-elliptic

r_{b} ≫ r_{2}

tak swing karta hai, toh iske coast legs mahine se saal le sakte hain jabki Hohmann ka fixed half-period hota hai. Time fuel saving ka chhupa hua price hai.

TF8. Dono maneuvers assume karte hain ki do orbits coplanar hain (ek hi flat plane mein).

True — poora

Δ v

comparison yahan coplanar circular orbits ke liye hai. Inclination change add karna ek alag cost hai jo Plane change maneuvers handle karta hai.

TF9. Bi-elliptic transfer mein saving Hohmann ke pehle burn se chhoti pehli burn se aati hai.

False — burn 1 actually Hohmann ke pehle burn se badi hoti hai (tum

r_{b} > r_{2}

ko aim karte ho, ek bada ellipse). Net saving chhoti middle burn aur gentler final circularization se aati hai, jo milkar heavier burn 1 ko outweigh kar sakti hain.

TF10. Threshold

R = 15.58

bi-parabolic (

r_{b} \to \infty

) case se derive hota hai.

True —

Δ v_{bipar} = Δ v_{H}

set karke aur

R

ke liye solve karne par exactly

15.58

milta hai; woh ratio hai jahan even the ideal bi-elliptic sirf Hohmann se tie karta hai, toh koi bhi bada

R

strict win guarantee karta hai.

Error dhundho

SE1. "

a_{1} = (r_{1} + r_{b}) /2

kyunki semi-major axis do apsidal radii ka sum hai."

Error: ye average (arithmetic mean) hai, sum nahi —

a_{1} = (r_{1} + r_{b}) /2

. Major axis (length

r_{1} + r_{b}

) poori lambi line hai;

a

uska aadha hai.

SE2. "Pehli ellipse

r_{1}

r_{2}

ko connect karti hai, doosri

r_{2}

r_{b}

ko."

Error: ellipse 1

r_{1}

(periapsis) ko

r_{b}

(apoapsis) se connect karti hai, aur ellipse 2

r_{2}

(periapsis) ko

r_{b}

(apoapsis) se. Dono ellipses far apsis

r_{b}

share karti hain; woh shared point hai jahan burn 2 hoti hai.

SE3. "Kyunki burn 2 periapsis uthati hai, aur periapsis low point hai, toh burn 2 low altitude par honi chahiye."

Error: burn 2 apoapsis par hoti hai (

r_{b}

, high point). Apoapsis par ek tangential burn opposite apsis ko change karti hai — yahan ye periapsis ko

r_{1}

se uthaa ke

r_{2}

tak le jaati hai. Iseeye ye sasti hai: ye wahan ki jaati hai jahan speed sabse chhoti hoti hai.

SE4. "Crossover rule mein,

R

ratio

r_{b} / r_{1}

hai."

Error:

R = r_{2} / r_{1}

, target to start radius ka ratio —

r_{b}

(detour radius) ek alag free parameter hai jo tum choose karte ho. Unhe confuse karna 11.94 / 15.58 thresholds ko barbad kar deta hai.

SE5. "Hum

Δ v

sum mein absolute values drop kar sakte hain kyunki teenon burns craft ko speed up karti hain."

Error: burn 3 ek brake hai (deceleration), toh uska raw signed value negative hai. Fuel cost har burn ke liye

Δ v = ∣ v_{after} - v_{before} ∣

hai, toh summing se pehle har term ko apne bars chahiye.

SE6. "

v_{c} (r) = μ / r

vis-viva mein

r = a

set karne se aata hai, toh ye ellipse speeds bhi deta hai."

Error:

a = r

set karna sirf circular orbit par valid hai jahan

r

constant hota hai. Ek ellipse par

r

vary karta hai jabki

a

fixed hota hai, toh

v_{c} = μ / r

apply nahi hota — tumhe full vis-viva use karna hoga.

SE7. "Kyunki middle burn almost free hai, bi-elliptic transfer basically ek free plane change hai."

Error: middle burn ka chhota hona real hai, lekin tum iske liye do baar gravity well mein deeply climb karke pay karte ho (ek heavier burn 1 aur Hohmann ki efficiency reuse karne ka koi mauka nahi). "Free lunch" sirf accounting mein survive karta hai jab

R

bada ho.

SE8. "15.58 threshold kisi fixed finite

r_{b}

par bi-elliptic total solve karke aata hai."

Error: 15.58 bi-parabolic limit

r_{b} \to \infty

se aata hai (best possible case). Finite-

r_{b}

comparison woh hai jo lower number 11.94 produce karta hai, gray zone ka onset.

Why questions

WHY1. Doosri burn

r_{b}

badhne par kyun shrink karti hai?

Kyunki

Δ v_{2} = ∣ μ (2/ r_{b} - 1/ a_{2}) - μ (2/ r_{b} - 1/ a_{1}) ∣

mein dono speeds usi tiny value ki taraf pull ho jaati hain jab

r_{b}

huge hota hai —

2/ r_{b}

contribution zero ho jaata hai aur

a_{1}, a_{2}

almost equal ho jaate hain — toh difference collapse ho jaata hai.

WHY2. Zaroorat se zyada door jaana kabhi fuel kyon bachata hai instead of hamesha waste karne ke?

Ek orbit ko reshape karna (

a

change karna) sabse sasta wahan hota hai jahan speeds sabse chhoti hon, yaani bahut door. Bi-elliptic apna mahnga "periapsis raise karo" job

r_{b}

par park karta hai jahan velocities minuscule hoti hain — ek inverted Oberth effect logic (burns wahan matter karti hain jahan speed high ho, toh reshaping burn wahan rakhi jaati hai jahan speed low ho).

WHY3. Ek clean threshold ki jagah poora gray zone (

11.94 < R < 15.58

) kyun hai?

Kyunki in ratios ke liye winner chosen

r_{b}

par depend karta hai: chhota

r_{b}

Hohmann favour karta hai, bada

r_{b}

bi-elliptic favour karta hai. Sirf

11.94

se neeche koi

r_{b}

help nahi karta, aur sirf

15.58

se upar jeet ideal limit mein bhi guaranteed ban jaati hai. Transfer time vs delta-v tradeoffs dekho.

WHY4. Hum best

Δ v

guarantee karne ke liye hamesha

r_{b} = \infty

kyun nahi choose kar sakte?

Kyunki infinite

r_{b}

ka matlab hai craft forever coast karta hai — infinite transfer time — aur ye sirf tab jeetta hai jab

R

pehle se crossover se zyada ho. Real missions finite

r_{b}

choose karte hain fuel aur time ko trade karte hue.

WHY5. Crossover analysis magic numbers state karne ke liye bi-parabolic limit kyun use karti hai?

Bi-parabolic case (

r_{b} \to \infty

) theoretically best hai jo bi-elliptic kar sakta hai, toh ise Hohmann se compare karna batata hai ki bi-elliptic kab kabhi jeet sakta hai. 15.58 threshold exactly wahan hai jahan ye best case Hohmann se tie karta hai. Ek Bi-parabolic transfer har ellipse ko ek parabola (escape-speed coast) se replace karta hai,

r_{b} \to \infty

ka ideal.

WHY6. Bi-elliptic ki burn 1 pehle Hohmann burn se badi kyun hai, jab dono identically start karte hain?

Dono

r_{1}

par usi circular orbit par start karte hain, lekin bi-elliptic ki ellipse

r_{b} > r_{2}

tak pohonchti hai, ek bada, zyada energetic orbit (bada

a_{1}

), toh enter karne ke liye ek bigger speed jump demand karta hai. Ye badi burn 1 "climb tax" hai jo tum umeed karte ho ki sasti middle burn repay kare.

WHY7. Comparison sirf ratio

R

aur choice of

r_{b} / r_{1}

par depend kyun karta hai, absolute sizes ya

μ

par nahi?

Vis-viva cleanly scale karta hai: sab kuch un units mein rewrite karo jahan

μ = 1

aur

r_{1} = 1

, aur saare

Δ v

values

R

aur

r_{b}

ke functions ban jaate hain. Physics scale-free hai, yahi wajah hai ki ek universal threshold table exist karta hai — Semi-major axis and orbital energy dekho.

WHY8.

a = (r_{p} + r_{a}) /2

literally sirf "long axis ka middle" kyun hai?

Major axis ek seedhi line hai periapsis se (

r_{p}

focus se) apoapsis tak (

r_{a}

focus se), total length

r_{p} + r_{a}

; semi-major axis

a

us line ka aadha define hota hai, yaani do endpoint distances ka arithmetic mean.

Edge cases

EC1. Kya hota hai bi-elliptic transfer mein jab

r_{b} \to r_{2}

(sabse chhota allowed detour)?

Ye degenerate ho jaata hai: "detour" barely target se zyada hota hai, ellipse 2 ek circle ki taraf shrink karti hai, aur maneuver ek ordinary Hohmann-jaise two-burn transfer ki taraf collapse ho jaata hai koi saving nahi. Requirement strictly

r_{b} > r_{2}

hai.

EC2. Agar

R = 1

exactly ho (

r_{1} = r_{2}

, same-size orbits)?

Koi transfer ki zaroorat hi nahi — target aur start coincide karte hain. Koi bhi bi-elliptic detour pure wasted

Δ v

hai (nowhere ke liye do climbs). Ye "chhota

R

= waste" regime ka extreme hai.

EC3. Agar target orbit start se chhoti ho (

r_{2} < r_{1}

, yaani

R < 1

Maneuver reverse mein chalta hai (tum deorbit/lower kar rahe ho), aur wahi vis-viva accounting roles swap karke apply hoti hai. 11.94 / 15.58 thresholds orbits raise karne ke liye state kiye gaye hain; lowering ke liye, ratio ko sahi taraf se use karo (

R = larger / smaller

EC4. Exactly

R = 11.94

par kaun jeeta hai?

Ye boundary hai jahan Hohmann ka guaranteed advantage khatam hota hai —

11.94

par dono best finite

r_{b}

ke liye essentially tie karte hain, toh koi strictly doosre ko beat nahi karta; ye gray zone ka onset hai, clean flip nahi.

EC5. Exactly

R = 15.58

par (upper bi-parabolic threshold), kaun jeeta hai?

Ye woh point hai jahan even the ideal case flip karta hai:

R = 15.58

par bi-parabolic (

r_{b} \to \infty

) transfer exactly Hohmann se tie karta hai, aur usse upar kisi bhi

R

ke liye bi-elliptic strictly jeetta hai ek large enough

r_{b}

ke saath. Threshold par hi dono limit mein tied hain; ek finite

r_{b}

ko win guarantee karne ke liye ab bhi

R

thoda aur upar chahiye.

EC6.

r_{b} \to \infty

limit mein, kya total time bhi infinity ho jaata hai?

Haan — ek ever-more-distant apoapsis tak aur waapas coast ever longer time leta hai, toh bi-parabolic limit ek fuel ideal hai jo finite mission time mein kabhi achievable nahi.

EC7. Kya bi-elliptic advantage survive karta hai agar humein orbital plane bhi change karni ho?

Ye grow bhi kar sakta hai: ek plane change

r_{b}

ke paas low speeds par kaafi sasta hota hai, toh far burn ke saath inclination change combine karna (Plane change maneuvers) crossover ko bi-elliptic ke favour mein push kar sakta hai even

R = 11.94

se neeche.

EC8. Ek bi-elliptic transfer mein ellipses aur burns ki sabse chhoti possible sankhya kya hai?

Definition se exactly do ellipses aur teen burns — ek ellipse hatao aur tum wapas Hohmann transfer par aa jaate ho, toh ye maneuver ki irreducible structure hai.

Recall Ek-line survival summary

Bi-elliptic = 3 burns, 2 ellipses, door sasti middle burn, expensive climb tax; sirf bade $R$ ke liye jeetta hai (>15.58 guaranteed, <11.94 kabhi nahi, thresholds par limit mein tied), aur hamesha kaafi zyada time cost karta hai. Har burn: $Δ v = ∣ v_{after} - v_{before} ∣$ .