Bi-elliptic transfer — when it wins over Hohmann: Worked examples

3.2.21 · D3 · Physics › Orbital Mechanics & Astrodynamics › Bi-elliptic transfer — when it wins over Hohmann

Yeh page parent topic ka drill-ground hai. Hum woh machinery lete hain jo usne build ki — Vis-viva equation aur teen-burn recipe — aur use har us case ke khilaf grind karte hain jo topic throw kar sakta hai: chote ratios jahan detour bekaar hai, bade ratios jahan yeh jeetta hai, $r_{b} \to \infty$ limit, exact crossover zone, inward transfers, aur time ki hidden cost.

Yahan sab kuch do formulas par tika hai jo parent ne already earn ki hain. Main unhe dobara state karta hoon taaki koi bhi symbol un-anchored na lage.

Scenario matrix

"Hohmann vs bi-elliptic" ke baare mein har sawaal in cells mein se kisi ek mein land karta hai. Neeche ke examples mein us cell ka label hai jis par woh hit karte hain, taaki mil ke poora space tile ho jaye.

Cell	Kya khaas baat hai	Expected winner	Example
A — small ratio	$R < 11.94$ : detour bekaar hai	Hohmann	Ex 1
B — guaranteed-win ratio	$R > 15.58$ : detour fayda deta hai	Bi-elliptic	Ex 2
C — grey zone	$11.94 < R < 15.58$ : $r_{b}$ par depend karta hai	koi bhi — compute karna padega	Ex 3
D — $r_{b} \to \infty$ limit	bi-parabolic best case	limit value	Ex 4
E — degenerate input	$r_{2} = r_{1}$ ( $R = 1$ ): koi transfer chahiye hi nahi	zero $Δ v$	Ex 5
F — sign trap	kya ek burn push hai ya brake?	signs check karo	Ex 6
G — real-world word problem	LEO → far station, real units	numeric $Δ v$	Ex 7
H — exam twist / time cost	fuel jeetta hai par time haarta hai	tradeoff verdict	Ex 8
I — inward transfer	$R < 1$ : target start ke andar, reversed	outward ka mirror	Ex 9

Case A — small ratio, Hohmann jeetta hai

Worked example Example 1 —

$R = 6$ (Cell A) $r_{1} = 1$ par circular start karo, $r_{2} = 6$ par circular target. Intermediate apoapsis $r_{b} = 100$ try karo. Kaunsa transfer sasta hai?

Forecast: $6 < 11.94$ , toh guess hai Hohmann jeetta hai aur detour pure waste hai. Isko prove karte hain.

Step 1 — circular speeds. $v_{c 1} = 1/1 = 1$ , $v_{c 2} = 1/6 = 0.40825$ . Yeh step kyun? Har transfer ek circle par shuru hota hai aur ek par khatam; yeh woh speeds hain jo hume match karni hain.

Step 2 — Hohmann ellipse. Iska semi-major axis dono circle radii ka average hai: $a_{H} = (1 + 6) /2 = 3.5$ . Yeh step kyun? Single Hohmann ellipse $r_{1}$ ko periapsis par aur $r_{2}$ ko apoapsis par touch karta hai, isliye uska size dono se fix hota hai.

Step 3 — Hohmann burns. Burn A at $r_{1}$ (ellipse par speed up): $2/1 - 1/3.5 - 1 = 1.71429 - 1 = 0.30931$ . Burn B at $r_{2}$ (ellipse-apoapsis se circular par speed up): $1/6 - 2/6 - 1/3.5 = 0.40825 - 0.04762 = 0.40825 - 0.21822 = 0.19003$ . Total Hohmann $= 0.49934$ . Yeh step kyun? Har $Δ v$ = (speed jo chahiye) − (speed jo hai) us point par.

Step 4 — bi-elliptic ellipses. $a_{1} = (1 + 100) /2 = 50.5$ , $a_{2} = (6 + 100) /2 = 53$ . Yeh step kyun? Ellipse 1, $r_{1}$ (periapsis) se $r_{b}$ (apoapsis) tak stretch karta hai; ellipse 2, $r_{2}$ (periapsis) se same $r_{b}$ tak. Har semi-major axis apne dono apsidal radii ka average hai — dono ko vis-viva mein feed karna padega.

Step 5 — bi-elliptic burns. Burn 1 at $r_{1}$ : $2/1 - 1/50.5 - 1 = 1.98020 - 1 = 0.40719$ . Burn 2 at $r_{b} = 100$ : $2/100 - 1/53 - 2/100 - 1/50.5 = 0.00113 - 0.00020 = 0.03364 - 0.01407 = 0.01958$ . Burn 3 at $r_{2}$ (brake, magnitude): $2/6 - 1/53 - 1/6 = 0.31447 - 0.40825 = 0.56078 - 0.40825 = 0.15253$ . Total bi-elliptic $= 0.57930$ . Yeh step kyun? Teen mein se har burn phir se (speed jo chahiye) − (speed jo hai) us radius par hai: Burn 1 $r_{1}$ par ellipse 1 par chadh ta hai, Burn 2 $r_{b}$ par ellipse 1 se ellipse 2 par switch karta hai, Burn 3 $r_{2}$ par ellipse 2 se target circle par girata hai.

Verify: $0.49934 < 0.57930$ ⇒ Hohmann lagbhag $0.08$ se jeetta hai. Sanity: burn 1 (sirf $r_{b} = 100$ tak chadhai) ne akele $0.407$ cost kiya, jo almost poore Hohmann ke barabar hai. Chota middle burn ( $0.02$ ) kabhi woh repay nahi kar sakta. Parent ka rule $R < 11.94$ se match karta hai. ✅

Case B — guaranteed-win ratio, bi-elliptic jeetta hai

Worked example Example 2 —

$R = 20$ (Cell B) $r_{1} = 1$ , $r_{2} = 20$ , $r_{b} = 100$ .

Forecast: $20 > 15.58$ , toh guess hai bi-elliptic suitable $r_{b}$ ke liye jeetta hai.

Step 1 — circular speeds. $v_{c 1} = 1/1 = 1$ , $v_{c 2} = 1/20 = 0.22361$ . Yeh step kyun? Yeh woh do speeds hain jahan se hume start karna hai aur land karna hai; har burn inhi ke against measure hota hai.

Step 2 — Hohmann. $a_{H} = (1 + 20) /2 = 10.5$ . Burn A: $2 - 1/10.5 - 1 = 1.90476 - 1 = 0.38014$ . Burn B: $1/20 - 2/20 - 1/10.5 = 0.22361 - 0.00476 = 0.22361 - 0.06901 = 0.15460$ . Total Hohmann $= 0.53474$ . Yeh step kyun? Yeh single-ellipse transfer (semi-major axis = dono circle radii ka average) woh baseline hai jise beat karna hai.

Step 3 — bi-elliptic. $a_{1} = 50.5$ , $a_{2} = (20 + 100) /2 = 60$ . Burn 1: $2 - 1/50.5 - 1 = 0.40719$ . Burn 2: $2/100 - 1/60 - 2/100 - 1/50.5 = 0.00333 - 0.00020 = 0.05774 - 0.01407 = 0.04367$ . Burn 3 (brake): $2/20 - 1/60 - 1/20 = 0.08333 - 0.22361 = 0.28868 - 0.22361 = 0.06507$ . Total bi-elliptic $= 0.51593$ . Yeh step kyun? Ex 1 wali hi recipe hai, lekin ab $r_{2}$ bahut bada hai, toh Hohmann ka apna second burn expensive ho jaata hai — yahi woh room hai jo bi-elliptic exploit karta hai.

Verify: $0.51593 < 0.53474$ ⇒ bi-elliptic $\approx 0.019$ se jeetta hai. Notice karo Hohmann ka Burn B ( $0.155$ ) bi-elliptic ke Burn 3 ( $0.065$ ) se kaafi mehanga hai: detour ek expensive circularization ko ek saste far-out reshape se trade karta hai. $R > 15.58$ se match karta hai. ✅

Neeche ka figure exactly yahi competition sab ratios mein plot karta hai: cyan curve Hohmann ka total $Δ v$ hai, amber curve bi-elliptic ka (bahut large $r_{b}$ par liya gaya), dono $R$ ke function ke roop mein.

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

Ise left se right padho. Left par ( $R = 6$ , hamara Ex 1) amber curve cyan ke upar hai — bi-elliptic haarta hai. Do dashed white lines parent ke thresholds $R = 11.94$ aur $R = 15.58$ mark karti hain: yeh "grey zone" hai jahan curves almost overlap hoti hain. $15.58$ ke right mein (hamara Ex 2 dot $R = 20$ par) amber curve neeche aa gayi hai cyan ke — aur woh gap hi fuel saving hai. Yeh single picture poora crossover rule encode karti hai.

Case C — the grey zone

Worked example Example 3 —

$R = 13$ , $r_{b}$ ke do choices (Cell C) $r_{1} = 1$ , $r_{2} = 13$ . $11.94$ aur $15.58$ ke beech winner $r_{b}$ par depend karta hai. $r_{b} = 30$ test karo phir $r_{b} = 500$ .

Forecast: guess hai chota $r_{b} = 30$ shayad Hohmann se phir bhi hare (kaafi door nahi gaya), jabki bada $r_{b} = 500$ jeette.

Step 1 — Hohmann baseline. $a_{H} = (1 + 13) /2 = 7$ . Burn A: $2 - 1/7 - 1 = 1.85714 - 1 = 0.36277$ . Burn B: $1/13 - 2/13 - 1/7 = 0.27735 - 0.01099 = 0.27735 - 0.10483 = 0.17252$ . Total Hohmann $= 0.53529$ . Yeh step kyun? Hume pehle two-burn benchmark chahiye, taaki exactly pata ho woh number kya hai jise detour ko undercut karna hai.

Step 2 — bi-elliptic with $r_{b} = 30$ . $a_{1} = (1 + 30) /2 = 15.5$ , $a_{2} = (13 + 30) /2 = 21.5$ . Burn 1: $2 - 1/15.5 - 1 = 1.93548 - 1 = 0.39121$ . Burn 2: $2/30 - 1/21.5 - 2/30 - 1/15.5 = 0.02016 - 0.00214 = 0.14200 - 0.04624 = 0.09576$ . Burn 3 (brake): $2/13 - 1/21.5 - 1/13 = 0.10736 - 0.27735 = 0.32766 - 0.27735 = 0.05031$ . Total $= 0.53728$ . Yeh step kyun? Sirf $r_{b} = 30$ par middle burn ( $0.096$ ) itna sasta nahi hai — hum "calm zone" mein kaafi andar nahi gaye.

Step 3 — bi-elliptic with $r_{b} = 500$ . $a_{1} = (1 + 500) /2 = 250.5$ , $a_{2} = (13 + 500) /2 = 256.5$ . Burn 1: $2 - 1/250.5 - 1 = 1.99601 - 1 = 0.41280$ . Burn 2: $2/500 - 1/256.5 - 2/500 - 1/250.5 = 0.0001010 - 0.0000009 = 0.01005 - 0.00097 = 0.00908$ . Burn 3 (brake): $2/13 - 1/256.5 - 1/13 = 0.15000 - 0.27735 = 0.38730 - 0.27735 = 0.10995$ . Total $= 0.53183$ . Yeh step kyun? Hum $r_{b}$ ko $30$ se all the way $500$ tak push karte hain taaki middle burn "calm zone" mein kaafi gehraai mein ho — dekho yeh $0.096$ se $0.009$ tak kaise drop karta hai. Yahi poora bi-elliptic idea hai kaam karta hua, aur same $R$ ke liye verdict palat deta hai.

Verify: $r_{b} = 30$ ke saath: $0.53728 > 0.53529$ → Hohmann barely jeetta hai. $r_{b} = 500$ ke saath: $0.53183 < 0.53529$ → bi-elliptic jeetta hai. Same $R$ , opposite verdicts — exactly yahi "grey zone" ka matlab hai. ✅

Case D — the $r_{b} \to \infty$ (bi-parabolic) limit

Worked example Example 4 —

$r_{b}$ ko infinity tak push karna, $R = 15$ (Cell D) $r_{1} = 1$ , $r_{2} = 15$ . $r_{b}$ ko enormous ( $1 0^{6}$ ) lo Bi-parabolic transfer limit approximate karne ke liye.

Forecast: middle burn essentially zero tak shrink ho jaana chahiye; total ek fixed limiting number ke paas pahunchna chahiye.

Step 1 — limit formula. Jab $r_{b} \to \infty$ , $a_{1}, a_{2} \to \infty$ , toh $1/ a \to 0$ . $r_{1}$ par vis-viva $2/ r_{1}$ deta hai (escape-ish speed), aur Burn 2 → 0. Jaana-maana closed form hai: $Δ v_{bipar} = \frac{2 μ}{r _{1}} - \frac{μ}{r _{1}} + \frac{2 μ}{r _{2}} - \frac{μ}{r _{2}}$ Yeh step kyun? Har ellipse ek parabola ban jaati hai (energy $\to 0$ ); kisi bhi radius $r$ par uski speed exactly $2 μ / r$ hoti hai. Toh Burn 1 = parabolic − circular at $r_{1}$ , Burn 3 = parabolic − circular at $r_{2}$ , Burn 2 vanish ho jaata hai.

Step 2 — plug in. $Δ v_{bipar} = (2 - 1) + (2/15 - 1/15) = 0.41421 + (0.36515 - 0.25820) = 0.41421 + 0.10696 = 0.52117$ . Yeh step kyun? Hum closed form evaluate karte hain taaki ek target number mile jis par finite- $r_{b}$ calculation ko converge karna hai.

Step 3 — numerically check with $r_{b} = 1 0^{6}$ . $a_{1} \approx 500000.5$ , $a_{2} \approx 500007.5$ . Burn 1: $2 - 1/500000.5 - 1 \approx 0.41421$ . Burn 2: $\approx 2 \cdot 1 0^{- 6}$ -scale difference $\approx 0.0000068$ (essentially 0). Burn 3 (brake): $2/15 - 1/500007.5 - 1/15 \approx 0.36515 - 0.25820 = 0.10695$ . Total $\approx 0.52117$ . Yeh step kyun? Ek huge lekin finite $r_{b}$ wapas plug karne se confirm hota hai ki limit formula koi leap of faith nahi tha — numbers actually uske upar land karte hain.

Step 4 — Hohmann comparison for $R = 15$ . $a_{H} = (1 + 15) /2 = 8$ . Burn A: $2 - 1/8 - 1 = 1.875 - 1 = 0.36931$ . Burn B: $1/15 - 2/15 - 1/8 = 0.25820 - 0.00833 = 0.25820 - 0.09129 = 0.16691$ . Total Hohmann $= 0.53622$ . Yeh step kyun? Yeh judge karne ke liye ki bi-parabolic limit worth it hai ya nahi, hume ise same- $R$ Hohmann baseline ke saath line up karna hoga, jo identical two-burn recipe se compute kiya gaya hai.

Verify: finite- $r_{b}$ total ( $0.52117$ ) closed-form limit ( $0.52117$ ) se 4 decimals tak match karta hai, aur Burn 2 → 0. $R = 15$ ke liye Hohmann $0.53622 > 0.52117$ hai, toh limit bi-elliptic bhi ise beat kar deta hai — consistent hai is baat se ki $R = 15$ $15.58$ se thoda neeche hai phir bhi large $r_{b}$ par winnable hai. ✅

Case E — degenerate input ( $R = 1$ )

Worked example Example 5 — same orbit,

$R = 1$ (Cell E) $r_{1} = r_{2} = 1$ . Har formula kya kehta hai?

Forecast: guess hai tum already wahan ho — har $Δ v$ zero tak collapse ho jaana chahiye, ya ek pointless "climb and return" reveal karna chahiye.

Step 1 — Hohmann. $a_{H} = (1 + 1) /2 = 1$ . Burn A: $2 - 1 - 1 = 0$ . Burn B: $1 - 2 - 1 = 0$ . Total $= 0$ . Yeh step kyun? Jab start aur target coincide karte hain toh Hohmann ellipse circle mein hi degenerate ho jaati hai — koi burn chahiye hi nahi.

Step 2 — bi-elliptic, $r_{b} = 100$ . $a_{1} = a_{2} = 50.5$ . Burn 1: $2 - 1/50.5 - 1 = 0.40719$ . Burn 2: $2/100 - 1/50.5 - 2/100 - 1/50.5 = 0$ (identical ellipses, apoapsis speeds equal). Burn 3 (brake): $2/1 - 1/50.5 - 1/1 = 0.40719$ . Total $= 0.81438$ .

Verify: Hohmann cost 0 karta hai (sahi — koi transfer chahiye hi nahi). Bi-elliptic phir bhi "flings out and comes back" karta hai, $0.407$ upar aur $0.407$ neeche kuch bhi hasil kiye bina. Yeh "detour is waste" ka degenerate limit hai: $R = 1 < 11.94$ , aur waste total hai. ✅

Case F — sign trap

Worked example Example 6 — kya har burn push hai ya brake?

$R = 20, r_{b} = 100$ (Cell F) Ex 2 ki orbit reuse karo. Har burn ke liye batao ki kya tum accelerate kar rahe ho (prograde, +) ya decelerate (retro, brake), aur hamare magnitude ke andar arithmetic sign confirm karo.

Forecast: guess hai burns 1 aur 2 pushes hain (orbit raise kar raha hai), burn 3 brake hai — classic galti yeh hai ki burn 3 ko push keh dena.

Step 1 — Burn 1 at $r_{1}$ . Ellipse-1 speed $= 2 - 1/50.5 = 1.40719$ vs circular $1$ . Speed-I-want > speed-I-have ⇒ prograde push. $Δ v_{1} = + 0.40719$ . Yeh step kyun? Upar chadh ne ke liye energy add karni hoti hai → periapsis par speed up karo.

Step 2 — Burn 2 at $r_{b}$ . Ellipse-2 apoapsis speed $= 2/100 - 1/60 = 0.05774$ vs ellipse-1 apoapsis speed $= 2/100 - 1/50.5 = 0.01407$ . Want > have ⇒ prograde push. $Δ v_{2} = + 0.04367$ . Yeh step kyun? Periapsis ko $r_{1}$ se $r_{2}$ tak raise karne ka matlab hai far apoapsis par speed up karna.

Step 3 — Burn 3 at $r_{2}$ . Tum ellipse-2 ke periapsis par pahunchte ho, $2/20 - 1/60 = 0.28868$ speed se chal rahe ho, lekin target circle ko sirf $1/20 = 0.22361$ chahiye. Have > want ⇒ retro-burn / brake. $Δ v_{3} = 0.22361 - 0.28868 = - 0.06507$ ; hum $∣Δ v_{3} ∣ = 0.06507$ report karte hain. Yeh step kyun? Yeh "burn 3 is acceleration" error ko kill karta hai — parent mein [!mistake] dekho.

Verify: signs hain $+, +, -$ . Total magnitude $0.40719 + 0.04367 + 0.06507 = 0.51593$ , Ex 2 se exactly match karta hai. $r_{2}$ par brake real hai: tum inbound ellipse ke periapsis par hamesha circular se faster hote ho. ✅

Case G — ek real-world word problem

Worked example Example 7 — LEO se ek distant relay tak, real units (Cell G)

Ek probe $r_{1} = 7000$ km par circular low-Earth orbit mein hai. Use $r_{2} = 140000$ km ( $R = 20$ ) par circular orbit par pahunchna hai. Earth ka $μ = 398600 km^{3} / s^{2}$ . $r_{b} = 700000$ km use karo. Real km/s mein bi-elliptic vs Hohmann compare karo.

Forecast: $R = 20 > 15.58$ ⇒ bi-elliptic jeetta hai. Hum bas apne normalized numbers ko real velocity unit se scale karte hain.

Step 1 — speed scale. $v_{c 1} = μ / r_{1} = 398600/7000 = 56.943 = 7.546 km/s$ . Yeh step kyun? Yeh LEO circular speed hai aur woh multiplier hai jo hamare dimensionless $Δ v$ ko km/s mein badalta hai (kyunki humne $r_{1} = 1, μ = 1$ normalize kiya tha).

Step 2 — normalized ratios. $R = 140000/7000 = 20$ , $r_{b} / r_{1} = 700000/7000 = 100$ . Ex 2 se bilkul identical! Toh dimensionless totals carry over hote hain: bi-elliptic $= 0.51593$ , Hohmann $= 0.53474$ . Yeh step kyun? Physics sirf ratios ki parwah karta hai; same $R$ aur same $r_{b} / r_{1}$ ⇒ same normalized answer.

Step 3 — km/s mein rescale. Bi-elliptic $= 0.51593 \times 7.546 = 3.893 km/s$ . Hohmann $= 0.53474 \times 7.546 = 4.035 km/s$ .

Verify: units check — dimensionless × km/s = km/s. ✅ Bi-elliptic $4.035 - 3.893 = 0.142 km/s \approx 142 m/s$ bachata hai, $R = 20$ par genuine fuel win. Sanity: LEO circular speed $7.546$ km/s jaana-maana real value hai (~7.5 km/s at 630 km altitude), toh scale sahi hai. ✅

Case H — exam twist: fuel jeetta hai, time haarta hai

Worked example Example 8 — tradeoff verdict,

$R = 18$ (Cell H) $r_{1} = 1$ , $r_{2} = 18$ , $r_{b} = 400$ . (a) Dikhao ki bi-elliptic $Δ v$ par jeetta hai. (b) transfer-time reasoning use karke time penalty estimate karo aur mission verdict do.

Forecast: guess hai bi-elliptic fuel jeetta hai (kyunki $18 > 15.58$ ) lekin time par buri tarah haarta hai.

Step 1 — Hohmann $Δ v$ . $a_{H} = (1 + 18) /2 = 9.5$ . Burn A: $2 - 1/9.5 - 1 = 1.89474 - 1 = 0.37650$ . Burn B: $1/18 - 2/18 - 1/9.5 = 0.23570 - 0.00585 = 0.23570 - 0.07646 = 0.15924$ . Total Hohmann $= 0.53574$ . Yeh step kyun? Yeh woh baseline hai jise beat karna hai — straight two-burn transfer.

Step 2 — bi-elliptic $Δ v$ . $a_{1} = (1 + 400) /2 = 200.5$ , $a_{2} = (18 + 400) /2 = 209$ . Burn 1: $2 - 1/200.5 - 1 = 1.99501 - 1 = 0.41244$ . Burn 2: $2/400 - 1/209 - 2/400 - 1/200.5 = 0.0002144 - 0.0000156 = 0.01465 - 0.00395 = 0.01070$ . Burn 3 (brake): $2/18 - 1/209 - 1/18 = 0.10633 - 0.23570 = 0.32608 - 0.23570 = 0.09038$ . Total bi-elliptic $= 0.51352$ . Yeh step kyun? $0.51352 < 0.53574$ ⇒ bi-elliptic $\approx 0.022$ $Δ v$ bachata hai; far-out reshape ( $r_{b} = 400$ ) ne middle burn chhota kar diya.

Step 3 — transfer times. Half ellipse par time $= π a^{3} / μ$ . Hohmann (ek half-ellipse, $a_{H} = 9.5$ ): $t_{H} = π 9. 5^{3} = π 857.375 = π \cdot 29.281 = 91.99$ (time units). Bi-elliptic (do half-ellipses, $a_{1} = 200.5$ phir $a_{2} = 209$ ): $t_{bi} = π 200. 5^{3} + π 20 9^{3} = π \cdot 2839.6 + π \cdot 3021.5 = 8920 + 9492 = 18412$ . Yeh step kyun? Har burn-to-burn coast ek half orbit hai; bi-elliptic do enormous ones string karta hai, toh iska clock bahut zyada chalta hai.

Verify: fuel ratio bi/Hohmann $= 0.51352/0.53574 = 0.959$ (lagbhag 4% sasta). Time ratio $= 18412/91.99 \approx 200 \times$ zyada. Verdict: ek probe ke liye jiske paas fuel spare hai aur deadline hai, Hohmann mission jeetta hai paper par haarne ke bawajood — yeh parent ki hidden-cost lesson hai. ✅

Case I — inward transfer ( $R < 1$ )

Worked example Example 9 —

andar giranaa, $r_{2} < r_{1}$ (Cell I) Ab target start ke andar hai: $r_{1} = 1$ , $r_{2} = 1/20 = 0.05$ (toh $R = r_{2} / r_{1} = 0.05$ ). Kya wahi machinery ulti direction mein kaam karti hai?

Forecast: guess hai poori kahaani outward transfer ka mirror hai — har formula $r_{1} \leftrightarrow r_{2}$ mein symmetric hai, toh ek inward Hohmann same $Δ v$ cost karega jitna same do radii ke beech outward ek, bas burns ke push/brake roles swap hoke.

Step 1 — symmetry. Radii $p$ aur $q$ ke beech transfer same ellipse use karta hai ( $a = (p + q) /2$ ) chahe kisi bhi direction mein jao — tum ise aage ya peeche run karte ho. Toh $r_{1} = 1$ aur $r_{2} = 0.05$ ke beech $Δ v$ , $1$ aur $20$ ke beech $Δ v$ ke barabar hai scaled: safe rehne ke liye ise directly work karo. Yeh step kyun? Vis-viva formula sirf do radii aur $a$ dekhta hai, travel direction nahi, toh mirror argument legitimate hai — lekin phir bhi numerically verify karte hain.

Step 2 — Hohmann, inward. $a_{H} = (1 + 0.05) /2 = 0.525$ . Burn A at $r_{1} = 1$ (ab ek brake — hum andar drop kar rahe hain): $2/1 - 1/0.525 - 1/1 = 0.09524 - 1 = 0.30861 - 1 = - 0.69139$ , magnitude $0.69139$ . Burn B at $r_{2} = 0.05$ (tighter, faster circle mein brake): $1/0.05 - 2/0.05 - 1/0.525 = 4.47214 - 38.09524 = 4.47214 - 6.17213 = - 1.69999$ , magnitude $1.69999$ . Total Hohmann $= 2.39138$ . Yeh step kyun? Gravitational well mein deep girne se inner circle fast ho jaata hai, toh numbers bade hain — lekin recipe (want − have at each apsis) unchanged hai; sirf signs brakes ki taraf flip ho jaate hain.

Step 3 — kya kabhi inward bi-elliptic karo? Crossover rule $R = r_{2} / r_{1}$ use karta hai. Inward transfer ke liye $R < 1$ , jo $11.94$ se kaafi neeche hai, toh bi-elliptic andar jaate waqt kabhi help nahi karta — jab goal andar deeper ho tab aur bahar fling karna bakwaas hogi. Yeh step kyun? Yeh "kya rule ulta bhi apply hota hai?" sawaal ka jawab deta hai: threshold larger/smaller ratio ke liye stated hai, aur inward transfers permanently Hohmann ki kingdom mein hain.

Verify: inward Hohmann total $2.39138$ bahut bada hai, final brake $1.70$ se dominate kiya fast inner orbit mein — physically sensible (low orbits fast hote hain, toh upar se match karne ke liye bada slow-down chahiye). Aur $R = 0.05 < 11.94$ confirm karta hai bi-elliptic inward kabhi answer nahi hai. ✅

Recall Matrix par quick self-test

" $R = 1$ " kaunsa cell hai? ::: Cell E — degenerate, Hohmann zero cost karta hai, bi-elliptic do climbs waste karta hai. Grey zone mein same $R$ par, kya do alag $r_{b}$ opposite winners de sakte hain? ::: Haan (Cell C, Ex 3): $r_{b} = 30$ ne Hohmann ko favor kiya, $r_{b} = 500$ ne bi-elliptic ko. $r_{b} \to \infty$ limit mein kaunsa burn vanish hota hai? ::: Middle burn (Burn 2) → 0; har ellipse ek parabola ban jaati hai. Agar do problems same $R$ aur same $r_{b} / r_{1}$ share karein, kya unke normalized $Δ v$ equal hain? ::: Haan — physics sirf ratios par depend karta hai (Ex 7). Inward transfer ke liye ( $R < 1$ ), kya bi-elliptic kabhi worth it hai? ::: Nahi — $R < 1 < 11.94$ , toh Hohmann andar jaate waqt hamesha jeetta hai (Ex 9).