Hohmann transfer — derivation, minimum energy transfer: Visual walkthrough

3.2.19 · D2 · Physics › Orbital Mechanics & Astrodynamics › Hohmann transfer — derivation, minimum energy transfer

Hum ek planet ke orbit mein hain. Planet ki gravitational strength ko $μ = GM$ kehte hain — yeh sirf ek number hai jo batata hai "yeh planet kitna kheenchta hai." $G$ universal gravity constant hai, $M$ planet ki mass hai. Bada $μ$ = zyada strong pull. Bas itna hi matlab hai $μ$ ka; kuch aur nahi chhupaaya gaya hai.

Step 1 — Do circles draw karo jinke beech tum jump karne wale ho

KYA. Ek hi planet ke around do circular orbits, ek hi flat plane mein (jaise ek kaagaz par do rings). Inner wali ka radius $r_{1}$ hai (planet centre se ring tak ki doori). Outer wali ka radius $r_{2}$ hai, jahan $r_{2} > r_{1}$ .

YAHAN SE SHURU KYUN. Koi bhi number calculate karne se pehle tumhe pata hona chahiye kya chhod rahe ho aur kahan pahunch rahe ho. Radius bas "kitna bahar ho" hai. Poora problem hai: chhhoti ring se badi ring par saste mein pahuncho.

PICTURE. Planet centre mein hai (dot). Inner ring wahan hai jahan tum shuru karte ho; outer ring tumhari destination hai. Do arrows dikhate hain tum kis direction mein ghoom rahe ho.

Step 2 — Pata karo tum ABHI kitni tez chal rahe ho (inner circle)

KYA. Kisi bhi circular orbit par, gravity ki andar ki taraf pull exactly woh "turning force" provide karti hai jo tumhe loop mein rakhti hai seedha uda jaane ke bajaye. Unhe balance karne par ek clean speed milti hai: $v_{c 1} = \frac{μ}{r _{1}}$

Term by term:

$v_{c 1}$ — " $c$ " ka matlab hai circular, " $1$ " ka matlab hai inner ring. Yeh tumhari speed hai abhi, engine chhuye bina.
$μ$ upar — zyada strong planet ⇒ andar girne se bachne ke liye zyada tez move karna padega.
$r_{1}$ neeche, square root ke andar — zyada bahar ⇒ dheema. Well mein gehraai mein hona tumhe tez daudaata hai; bahar hona tumhe aaraam se chalane deta hai.

YEH TOOL (square root) KYUN. Hum speed chahte hain, lekin energy balance naturally $v^{2}$ deta hai. Square ko undo karne ke liye aur $v$ padhne ke liye, hum square root lete hain — woh operation jo jawaab deta hai "kaunsa number, khud se multiply hone par, $v^{2}$ deta hai?" Yahi ek wajah hai ki aata hai.

PICTURE. Velocity arrow circle ka tangent hai — ring ke along sideways point karta hai, kabhi andar ya bahar nahi. Yeh sideways direction ek moment mein bahut important hoga.

Step 3 — Transfer ellipse draw karo jo dono rings ko "kiss" kare

KYA. Humein ab ek curve chahiye jo do rings ko bridge kare. Woh curve ek ellipse hai. Ellipse sirf ek "dabaa hua circle" se zyada hai — iska ek precise defining rule hai, aur wahi rule iska natural orbit shape hona make karta hai.

Hum woh ellipse choose karte hain jiska nearest point (periapsis) inner ring ko $r_{1}$ par graze kare aur jiska farthest point (apoapsis) outer ring ko $r_{2}$ par graze kare.

Semi-major axis $a_{t}$ ellipse ke sabse lambe diameter ka aadha hai. Woh lambaida diameter inner grazing point se seedha baahri tak jaata hai: $2 a_{t} = r_{1} + r_{2} ⟹ a_{t} = \frac{r _{1} + r _{2}}{2}$

Term by term:

$r_{1}$ — sabse kareeb ki doori (periapsis).
$r_{2}$ — sabse door ki doori (apoapsis).
Unka sum poora lamba diameter hai; $a_{t}$ paane ke liye aadha karo. Toh $a_{t}$ literally do radii ka average hai.

YEH ELLIPSE KYUN, KOI AUR NAHI. Sab ellipses mein se jo dono rings ko bridge kar sakti hain, yeh bi-tangent wali har ring ko exactly ek baar touch karti hai aur wahan har ring ki velocity direction share karti hai (tangent to tangent). Woh shared direction hi burns ko sasta banata hai — hum Step 5 mein iska fayda uthayenge.

PICTURE. Planet ek focus par hai (filled dot); doosra focus ellipse ke andar hollow dot hai. String-sum property ek sample point se dono foci tak do lines dikhake show ki gayi hai. Periapsis inner ring par low point hai; apoapsis outer ring par high point hai; dashed line lamba diameter $2 a_{t}$ hai.

Step 4 — Ellipse ke START par chahiye speed pata karo

KYA. Is ellipse par ride karne ke liye tumhe inner ring se jaate waqt ek specific speed par move karna hoga. Vis-viva equation deta hai: $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$

Intuition Vis-viva aisa KYUN dikhta hai (energy bookkeeping)

Ek chalte spacecraft ke paas per unit mass do tarah ki energy hoti hai: motion energy $\frac{1}{2} v^{2}$ (tez hone par bada) aur height energy $- μ / r$ (ek debt jo door jaane par zero ki taraf shrink hoti hai, aur kareeb hone par deeply negative hoti hai). Unka sum coasting orbit par kabhi nahi badalta — yeh energy conservation hai: $\frac{1}{2} v^{2} - \frac{μ}{r} = ε (constant) .$ Ab, ek alag geometric fact us constant ko orbit ki size se jodata hai: kisi bhi ellipse ke liye total energy sirf semi-major axis par depend karti hai, $ε = - μ /2 a$ (bada orbit ek shallow, low-energy orbit hai; tight orbit deeply bound hai). $ε$ ke dono expressions ko equal set karo aur $v^{2}$ ke liye solve karo: $\frac{1}{2} v^{2} - \frac{μ}{r} = - \frac{μ}{2 a} ⟹ v^{2} = μ (\frac{2}{r} - \frac{1}{a}) .$ Toh vis-viva bas "kinetic + potential = constant, aur woh constant $a$ se fix hai." (Poora algebra Vis-viva equation mein hai.)

Ise radius $r = r_{1}$ par ellipse ke $a = a_{t}$ use karke evaluate karo: $v_{p} = μ (\frac{2}{r _{1}} - \frac{1}{a _{t}}) = μ (\frac{2}{r _{1}} - \frac{2}{r _{1} + r _{2}})$

Term by term:

$v_{p}$ — ellipse ke periapsis par speed.
$\frac{2}{r _{1}}$ — "main kitna kareeb hun" term ( $r_{1}$ par evaluate ki gayi height-energy $- μ / r$ se); gehraai mein hona zyada speed deta hai.
$\frac{1}{a _{t}}$ — "meri poori orbit kitni badi hai" term ( $ε = - μ /2 a_{t}$ se); badi orbit zyada energy store karti hai aur is subtraction ko thoda reduce karti hai. Kyunki $a_{t} = (r_{1} + r_{2}) /2$ , $\frac{1}{a _{t}} = \frac{2}{r _{1} + r _{2}}$ .

$v_{p} > v_{c 1}$ KYUN. Step 4 ko Step 2 se compare karo. Circle ne $a = r_{1}$ use kiya, toh uska term tha $\frac{2}{r _{1}} - \frac{1}{r _{1}} = \frac{1}{r _{1}}$ . Ellipse $a_{t} > r_{1}$ use karta hai, toh $\frac{1}{a _{t}} < \frac{1}{r _{1}}$ — hum kam subtract karte hain, root ke neeche bada number chodke. Physically: $r_{2}$ tak poori taraf swing karne ke liye extra energy chahiye, aur periapsis par woh extra speed ke roop mein dikhta hai.

PICTURE. Usi periapsis point se do velocity arrows: chhhota wala circular speed $v_{c 1}$ hai jo tumhare paas hai, lamba wala $v_{p}$ hai jo chahiye. Dono same direction mein point karte hain (tangent).

Step 5 — PEHLA burn (sirf subtraction kyun hai)

KYA. Engine prograde (jis direction mein already ja rahe ho) fire karo speed gap pura karne ke liye: $Δ v_{1} = v_{p} - v_{c 1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1)$

Term by term:

$Δ v_{1}$ — "delta-vee," speed mein woh change jo engine ko deliver karna hai. Yeh spaceflight ka fuel cost currency hai (dekho Delta-v budget).
$v_{p} - v_{c 1}$ — chahiye speed minus current speed.
Bracket wali form: $μ / r_{1} = v_{c 1}$ factor out karo; bacha hua $(\frac{2 r _{2}}{r _{1} + r _{2}} - 1)$ hamesha positive hai kyunki $r_{2} > r_{1}$ , toh burn hamesha tumhe tez karta hai.

YEH PLAIN SUBTRACTION KYUN HAI, VECTOR MATH NAHI. Step 4 mein dono arrows same direction mein point karte the. Jab do velocities collinear hoon, vector mein change bas length mein change hai: $Δ v = ∣ v_{p} ∣ - ∣ v_{c 1} ∣$ . Yahi poori wajah hai ki hum tangent burns choose karte hain — yeh hard vector subtraction ko easy arithmetic mein badal deta hai.

PICTURE. Green $Δ v_{1}$ arrow exactly Step 4 ke dono arrows ke beech ka gap hai — usi line par end to tip rakhaa gaya.

Step 6 — Aadha ellipse coast karo; upar SLOW pahuncho

KYA. Engine off. Tum $r_{1}$ se $r_{2}$ tak ellipse ke aadhe par climb karte ho. Energy conserved hai, toh jaise $r$ badhta hai tumhari speed drop hoti hai. Apoapsis par vis-viva $r = r_{2}$ par (same ellipse par, toh abhi bhi $a = a_{t}$ ) deta hai: $v_{a} = μ (\frac{2}{r _{2}} - \frac{2}{r _{1} + r _{2}})$

Term by term:

$v_{a}$ — apoapsis par speed (climb ka top).
$\frac{2}{r _{2}}$ — "main kitna kareeb hun" term, ab $r_{2}$ par bahar evaluate kiya gaya; yeh chhhota hai, toh akele yeh thodi speed deta hai. Tumhari "steam khatam ho gayi."
$\frac{2}{r _{1} + r _{2}}$ — yeh exactly $\frac{1}{a _{t}}$ hai, Step 4 ka same orbit-size term (kyunki tumne transfer ellipse kabhi nahi chhoda — uska semi-major axis $a_{t}$ poore coast mein unchanged hai). Yeh ellipse ki size se set hona fixed energy debt hai; ise subtract karna encode karta hai "yeh ek ellipse hai, $r_{2}$ par circle nahi."

TUM SLOW KYUN HOTE HO. Gravity well se bahar climb karna kinetic energy (motion) ko potential energy (height) mein trade karta hai, bilkul waisi tarah jaise ek upar phenki gayi ball dheemi hoti jaati hai. Same total energy, redistribute. (Yeh Orbital energy & semi-major axis action mein hai.)

PICTURE. Coasting arc, jisme velocity arrow shrink hota hai jaise craft periapsis se apoapsis tak climb karta hai. Is arc par kahi bhi engine nahi.

Step 7 — DOOSRA burn (BRAKE mat karo, tez CHALO!)

KYA. Apoapsis par tum outer ring ke radius par ho lekin wahan circle hold karne ke liye bahut dheeme ho. Outer circular speed hai $v_{c 2} = \frac{μ}{r _{2}},$ aur kyunki $v_{a} < v_{c 2}$ , tumhe phir prograde fire karna hoga pakad ne ke liye: $Δ v_{2} = v_{c 2} - v_{a} = \frac{μ}{r _{2}} (1 - \frac{2 r _{1}}{r _{1} + r _{2}})$

Term by term:

$v_{c 2}$ — outer ring par circular speed (Step 2 ka formula $r_{2}$ ke saath).
$v_{a}$ — Step 6 se tumhari slow arrival speed.
Bracket $(1 - \frac{2 r _{1}}{r _{1} + r _{2}})$ positive hai kyunki $r_{1} < r_{2}$ , toh phir burn prograde hai (speed-up).

BRAKE KYUN NAHI. Car-wala intuition ("upar ja rahe ho ⇒ slow karo") ek trap hai. Tum $r_{2}$ par ruk nahi rahe; tum wahan circular motion mein rehne ki koshish kar rahe ho, jisme ek definite speed $v_{c 2}$ chahiye jo tumhari arrival speed $v_{a}$ se zyada hai. Energy do, mat lo.

PICTURE. Chhota arrival arrow $v_{a}$ , lamba circular arrow $v_{c 2}$ , aur orange $Δ v_{2}$ gap unke beech — sab outer ring ke tangent.

Step 8 — Edge cases: check karo formula kabhi break nahi karta

KYA. Inputs ko unki limits tak push karo aur confirm karo picture abhi bhi sense banata hai.

Case A — $r_{2} = r_{1}$ (rings coincide). Tab $a_{t} = r_{1}$ , aur dono brackets collapse hote hain: $\frac{2 r _{1}}{2 r _{1}} - 1 = 0$ . Toh $Δ v_{1} = Δ v_{2} = 0$ . Picture: ellipse degenerate hokar us circle mein wapas aa jaata hai jis par tum already ho — koi burn nahi chahiye, bilkul sahi.

Case B — $r_{2} ≫ r_{1}$ (bahut bade orbit par jump). Periapsis factor $\frac{2 r _{2}}{r _{1} + r _{2}} \to 2$ , toh $Δ v_{1} \to v_{c 1} (2 - 1) \approx 0.414 v_{c 1}$ — pehla burn saturate ho jaata hai. Meanwhile $v_{c 2} \to 0$ , toh $Δ v_{2} \to 0$ : ek huge, faint orbit par pahunchna upar lagbhag kuch nahi karta. Picture: itna stretched ellipse ki seedha bahar ki tarah lagta hai.

Case C — ANDAR jaana ( $r_{2} < r_{1}$ ). Re-label karo taki bada radius start ho. Ab dono burns retrograde hain (tum backward fire karte ho speed subtract karne ke liye): tum high circle par brake lagate ho ellipse par drop karne ke liye, speed gain karte hue neeche coast karte ho, phir neeche phir brake lagate ho. Formulas same hain; sirf burns ki direction prograde se retrograde flip hoti hai. Picture: same diagram reverse mein, arrows backward pointing.

Yeh minimum-energy do-burn transfer KYUN hai

Intuition Claim, seedhe kaha gaya

Do coplanar circular orbits ke beech sab do-burn transfers mein se, Hohmann ellipse woh hai jiska sabse chhhota total $Δ v$ hai — matlab sabse kam fuel. Yahi wajah hai ki yeh page matter karta hai: humne sirf orbit change karne ka ek tarika nahi dhoondha, humne sabse sasta two-impulse tarika dhoondha.

Yeh KYUN jeet ta hai (ek saanth mein): har burn ek tangent point par rakhaa gaya hai, toh koi bhi thrust kabhi velocity ko sideways ghoomane mein waste nahi hota — fuel ki har boond speed change karne mein jaati hai, jo orbital energy change karta hai. Koi bhi non-tangent transfer apna kuch burn velocity vector ko re-point karne mein kharch karta hai, jo pure waste hai. (Extreme ratios $r_{2} / r_{1} ≳ 11.94$ ke liye ek three-burn Bi-elliptic transfer ise beat kar sakta hai, lekin do-burn transfers mein Hohmann sab realistic ratios ke liye optimal hai.)

Ek-picture summary

Upar sab kuch, ek frame mein compress kiya gaya — dono directions dikhate hue. Raise ke liye left-to-right padho (dono burns prograde, green), ya lower ke liye right-to-left (dono burns retrograde/braking, slate): same ellipse, same magnitudes, opposite burn directions.

$Δ v_{total} = burn 1: inner ring chhodo \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) + burn 2: outer ring hold karo \frac{μ}{r _{2}} (1 - \frac{2 r _{1}}{r _{1} + r _{2}})$

Raise ke liye ( $r_{2} > r_{1}$ ) dono terms positive hain aur dono burns prograde hain. Lower ke liye ( $r_{2} < r_{1}$ ) same do magnitudes retrograde (backward) burn ki jaati hain, tumhe badi ring se ellipse par aur phir chhhoti ring par neeche braaking karte hue.

Recall Poore walkthrough ki Feynman retelling

Tum ek planet ke around ek chhhoti ring par circle kar rahe ho, kisi fixed speed par sideways move karte hue (Step 2). Tum bahar wali badi ring chahte ho. Tum seedha bahar drive nahi kar sakte — gravity sab kuch curve karta hai. Toh tum ek dabaa hua loop draw karte ho, ek ellipse (ek curve jahan do fixed pins tak do doriyan hamesha ek hi total mein add hoti hain), jo tumhari chhhoti ring ko apne low point par aur badi ring ko apne high point par bas touch kare (Step 3). Ise ride karne ke liye tumhe thodi tez move karna hoga jitne tum ho (Step 4), toh tum ek prograde shove dete ho — aur kyunki tum pehle se jis direction mein ja rahe ho usi taraf point kar rahe ho, cost bas speed difference hai, ek simple subtraction (Step 5). Phir tum engine band karo aur coast karo. Bahar climb karte hue, tum dheeme hote ho, motion ko height ke liye trade karte hue, aur badi ring par bahut dheeme pahunchte ho wahan rehne ke liye (Step 6). Ek aur prograde shove tumhe badi ring ki circular speed tak tez karta hai aur tum ho gaye (Step 7). Do pushes, dono forward, sabse sasta ride. Agar rings same size ki hoti toh koi push nahi chahiye tha; agar tum andar drop karna chahte instead, toh same do pushes gentle brakes ban jaati hain jo peeche fire hoti hain — retrograde (Step 8). Yahi poora Hohmann transfer hai.

Recall Active recall — answers cover karo

$Δ v$ plain subtraction kyun hai aur vector math nahi? ::: Kyunki burns tangent hain, toh do velocities collinear hain; ek vector ki length mein change uski magnitude mein change ke barabar hai. Transfer ellipse ka semi-major axis kya hai? ::: $a_{t} = (r_{1} + r_{2}) /2$ , do radii ka average. Apoapsis par tum brake karne ki jagah TEZ kyun karte ho? ::: Tum outer circular speed se dheeme pahunchte ho ( $v_{a} < v_{c 2}$ ), toh bade circle ko hold karne ke liye energy add karni padti hai. Ellipse ki defining property kya hai? ::: Uske kisi bhi point se do foci tak doriyon ka sum constant hai; planet ek focus par baitha hai. Jab $r_{2} = r_{1}$ ho toh burns ka kya hota hai? ::: Dono zero ho jaate hain — ellipse us circle mein collapse ho jaata hai jis par tum already ho. Ek huge outward jump ke liye, $Δ v_{1}$ kya approach karta hai? ::: $v_{c 1} (2 - 1) \approx 0.414 v_{c 1}$ — yeh saturate ho jaata hai. Kis sense mein Hohmann optimal hai? ::: Yeh coplanar circular orbits ke beech sab do-burn transfers mein minimum-total- $Δ v$ transfer hai ( $r_{2} / r_{1} \approx 11.94$ tak).

Dekho bhi: Vis-viva equation · Orbital energy & semi-major axis · Kepler's Third Law · Oberth effect · Bi-elliptic transfer · Delta-v budget · Two-body problem