Hohmann transfer — derivation, minimum energy transfer: Exercises

3.2.19 · D4 · Physics › Orbital Mechanics & Astrodynamics › Hohmann transfer — derivation, minimum energy transfer

Intuition Is page ko kaise use karein

Upar se neeche karo. Har problem self-contained hai — har symbol dobara introduce kiya gaya hai taaki tumhe kabhi scroll back na karna pade. Pehle solution collapse karke try karo, phir [!recall]- kholke check karo. Levels "kya tum formula spot kar sakte ho" se lekar "kya tum argument invent kar sakte ho" tak jaate hain.

Constants jo poore page mein use hote hain (jab tak problem kuch aur na kahe): $μ_{\oplus} = 3.986 \times 1 0^{5} km^{3} / s^{2} = 398600 km^{3} / s^{2}$ (Earth ka gravitational parameter, matlab Earth ke liye $GM$ — woh number jo gravity ki strength set karta hai; dono forms same number hain, ek scientific notation mein aur ek seedha likha hua, aur har solution mein $398600$ plug in kiya gaya hai), $μ_{⊙} = 1.327 \times 1 0^{11} km^{3} / s^{2}$ (Sun ka). Jab koi formula sirf $μ$ kahe, to jo body tum orbit kar rahe ho uske hisaab se inme se sahi wala use karo.

Core toolkit — neeche words mein dobara bataya gaya hai (pehli teen lines hi sach mein "big three" hain; baaki lines sirf named shorthands hain jo in teeno se bani hain):

Formula Toolkit (

Vis-viva equation aur Orbital energy & semi-major axis se — sab kuch) Big three:

Circular speed radius $r$ par: $v_{circ} = μ / r$ . "Size $r$ ka perfect circle hold karne ke liye kitni fast." Inner circle ke liye $v_{c 1} = μ / r_{1}$ aur outer ke liye $v_{c 2} = μ / r_{2}$ likhte hain.
Vis-viva (kisi bhi orbit par kisi bhi jagah speed): $v = μ (\frac{2}{r} - \frac{1}{a})$ , jahan $r$ tumhari current distance hai aur $a$ semi-major axis hai (ellipse ka sabse lamba diameter ka aadha).
Transfer ellipse size: $a_{t} = \frac{r _{1} + r _{2}}{2}$ — iska low point chhote radius ko touch karta hai, high point bade radius ko touch karta hai.

Named shorthands (big three se derive kiye gaye):

Periapsis speed $v_{p}$ = transfer ellipse ki speed uske closest point par (vis-viva ko chhote radius par evaluate karo, $a = a_{t}$ ke saath). Apoapsis speed $v_{a}$ = transfer ellipse ki speed uske farthest point par (vis-viva bade radius par, $a = a_{t}$ ke saath). $v_{p}$ fast end hai, $v_{a}$ slow end.
Do burns (raising ke liye): $Δ v_{1} = v_{p} - v_{c 1}$ ( $r_{1}$ par, ellipse par chadhne ke liye speed up karo), $Δ v_{2} = v_{c 2} - v_{a}$ ( $r_{2}$ par, circularize karne ke liye speed up karo). Lowering transfer ke liye signs flip ho jaate hain — L2.4 dekho.
Half-orbit coast time: $t = π a_{t}^{3} / μ$ (Kepler's Third Law se).

Raise ka map Figure 1 mein dekho neeche — har problem is ek picture ka variation hai.

Figure 1 — Hohmann geometry (raise).

Figure — Hohmann transfer — derivation, minimum energy transfer

Level 1 — Recognition

L1.1

Ek satellite Earth ke around $r = 8000$ km radius ki circular orbit mein hai. Uski speed kya hai?

Recall Solution

Yeh circular speed formula hai — "circle hold karne ke liye kitni fast" sawaal ka exactly ek hi jawab hota hai. $v_{circ} = \frac{μ}{r} = \frac{398600}{8000} = 49.825 = 7.059 km/s$

L1.2

Ek Hohmann transfer $r_{1} = 7000$ km se $r_{2} = 105000$ km tak jaata hai. Transfer ellipse ka semi-major axis $a_{t}$ kya hai?

Recall Solution

Ellipse ka sabse lamba diameter inner circle se outer circle tak pahunchta hai, isliye uski length $r_{1} + r_{2}$ hai, aur $a_{t}$ uska aadha hai. $a_{t} = \frac{r _{1} + r _{2}}{2} = \frac{7000 + 105000}{2} = 56000 km$

L1.3

L1.2 ke transfer ke liye, ellipse ka kaunsa end periapsis (closest point) hai aur kaunsa apoapsis (farthest point)?

Recall Solution

Periapsis = Earth ke sabse paas = chhota radius $= r_{1} = 7000 km$ . Apoapsis = Earth se sabse door = bada radius $= r_{2} = 105000 km$ . Satellite periapsis par sabse fast aur apoapsis par sabse slow hoti hai (chadhte waqt speed height ke badle mein exchange hoti hai).

Level 2 — Application

L2.1

Ek spacecraft ko Earth ke around $r_{1} = 6800$ km se $r_{2} = 20000$ km tak raise karo. $Δ v_{1}$ , pehla burn, nikalo.

Recall Solution

Step A — current speed (inner circle par): $v_{c 1} = \frac{398600}{6800} = 58.618 = 7.656 km/s$ Step B — transfer ellipse size: $a_{t} = \frac{6800 + 20000}{2} = 13400$ km. Step C — periapsis par zaroori speed (vis-viva at $r = r_{1}$ , $a = a_{t}$ ): $v_{p} = 398600 (\frac{2}{6800} - \frac{1}{13400}) = 398600 (2.9412 \times 1 0^{- 4} - 7.4627 \times 1 0^{- 5})$ $= 398600 \times 2.1949 \times 1 0^{- 4} = 87.489 = 9.354 km/s$ Step D — burn (dono velocities same direction mein point karti hain, isliye magnitudes subtract karo): $Δ v_{1} = v_{p} - v_{c 1} = 9.354 - 7.656 = 1.698 km/s$

L2.2

Same transfer jaise L2.1. $Δ v_{2}$ , doosra burn, aur total $Δ v$ nikalo.

Recall Solution

Step A — apoapsis par speed (vis-viva at $r = r_{2} = 20000$ , abhi bhi $a = a_{t} = 13400$ ): $v_{a} = 398600 (\frac{2}{20000} - \frac{1}{13400}) = 398600 (1.0 \times 1 0^{- 4} - 7.4627 \times 1 0^{- 5})$ $= 398600 \times 2.5373 \times 1 0^{- 5} = 10.114 = 3.180 km/s$ Step B — $r_{2}$ par target circular speed: $v_{c 2} = \frac{398600}{20000} = 19.93 = 4.464 km/s$ Step C — burn (tum bahut slow pahunche, isliye speed up karo): $Δ v_{2} = v_{c 2} - v_{a} = 4.464 - 3.180 = 1.284 km/s$ Total: $Δ v = 1.698 + 1.284 = 2.982 km/s$ .

L2.3

Same transfer ke liye, do burns ke beech coast kitni lambi hai?

Recall Solution

Tum ellipse ka aadha hissa fly karte ho, isliye time orbital period ka aadha hai. Half-period formula: $t = π \frac{a _{t}^{3}}{μ} = π \frac{1340 0 ^{3}}{398600} = π \frac{2.406 \times 1 0 ^{12}}{398600}$ $= π 6.037 \times 1 0^{6} = π \times 2457.0 = 7719 s \approx 2.14 h$

L2.4

Edge case — orbit lower karna. Ek spacecraft ko outer circle $r_{1} = 20000$ km se neeche inner circle $r_{2} = 6800$ km tak jaana hai (L2.1 ka reverse). Dono burns (unke signs/direction ke saath) aur total $Δ v$ nikalo.

Recall Solution

Pehle sign convention. Fuel cost ke roop mein $Δ v$ hamesha positive magnitude $∣ v_{after} - v_{before} ∣$ hoti hai; direction (prograde = speed up, retrograde = brake) woh hoti hai jo lowering mein flip hoti hai. Transfer ellipse L2.1 waali same ellipse hai — sirf travel direction reverse ho jaata hai: ab periapsis target $r_{2}$ hai aur apoapsis start $r_{1}$ hai. $a_{t} = \frac{20000 + 6800}{2} = 13400$ km (unchanged — averaging symmetric hai).

Burn 1 — outer circle $r_{1} = 20000$ par (yeh ellipse ka apoapsis hai). Tum abhi outer circle par $v_{c 1} = 398600/20000 = 4.464$ km/s par ho. Ellipse par drop karne ke liye tumhe apoapsis speed $v_{a} = 3.180$ km/s tak slow karna hoga (same $v_{a}$ jaise L2.2). Kyunki tum slower jaate ho, yeh retrograde (braking) burn hai: $Δ v_{1} = v_{c 1} - v_{a} = 4.464 - 3.180 = 1.284 km/s (brake)$ Burn 2 — inner circle $r_{2} = 6800$ par (ellipse ka periapsis). Tum periapsis speed $v_{p} = 9.354$ km/s par pahunchte ho (same $v_{p}$ jaise L2.1), jo inner circular speed $v_{c 2} = 398600/6800 = 7.656$ km/s se faster hai. Circularize karne ke liye tumhe dobara brake karna hoga: $Δ v_{2} = v_{p} - v_{c 2} = 9.354 - 7.656 = 1.698 km/s (brake)$ Total: $Δ v = 1.284 + 1.698 = 2.982 km/s$ .

Key symmetry: L2.1–L2.2 ke raise se identical total (dono burns ke same magnitudes hain, sirf swap ho gaye hain aur ab retrograde hain). Same do circles ke beech lower karna aur raise karna exactly same fuel cost karta hai — tum sirf boost ki jagah brake karte ho.

Level 3 — Analysis

L3.1

Ek transfer $r_{1} = 7000$ km se $r_{2} = 42164$ km (LEO→GEO) tak jaata hai. Koi claim karta hai "kyunki hum sirf $6$ ke factor se raise kar rahe hain, total $Δ v$ kisi base value ka $6 \times$ hona chahiye." Actual total $Δ v$ compute karo aur explain karo kyun raise ratio cost ko linearly scale nahi karta.

Recall Solution

$a_{t} = (7000 + 42164) /2 = 24582$ km. $v_{c 1} = 398600/7000 = 7.546 km/s$ $v_{p} = 398600 (2/7000 - 1/24582) = 398600 (2.85714 \times 1 0^{- 4} - 4.06802 \times 1 0^{- 5}) = 97.616 = 9.880 km/s$ $Δ v_{1} = 9.880 - 7.546 = 2.334 km/s$ $v_{a} = 398600 (2/42164 - 1/24582) = 398600 (4.74338 \times 1 0^{- 5} - 4.06802 \times 1 0^{- 5}) = 2.6920 = 1.641 km/s$ $v_{c 2} = 398600/42164 = 3.075 km/s$ $Δ v_{2} = 3.075 - 1.641 = 1.434 km/s$ $Δ v_{total} = 2.334 + 1.434 = 3.769 km/s$ Comment: har burn ki size radii ke square roots ke differences par depend karti hai, raise ratio par nahi. $r_{2}$ double karne se $Δ v$ double nahi hota; asliyat mein jab $r_{2} \to \infty$ total $Δ v$ ek finite limit ki taraf jaata hai (L4.1 dekho). Cost ratio mein sub-linear hai — orbits ko bada banana sasta hai, deep banana mehenga.

L3.2

Do spacecraft dono apni orbit ko same height $Δ r = 3000$ km se raise karte hain. Ek neeche se shuru karta hai ( $r_{1} = 7000 \to r_{2} = 10000$ km), doosra upar se ( $r_{1} = 40000 \to r_{2} = 43000$ km). Kaunsa transfer kam total $Δ v$ cost karta hai, aur kyun?

Recall Solution

Low starter ( $7000 \to 10000$ , $a_{t} = 8500$ ): $v_{c 1} = 398600/7000 = 7.546, v_{p} = 398600 (2/7000 - 1/8500) = 8.183$ $Δ v_{1} = 8.183 - 7.546 = 0.637$ $v_{a} = 398600 (2/10000 - 1/8500) = 5.729, v_{c 2} = 398600/10000 = 6.313$ $Δ v_{2} = 6.313 - 5.729 = 0.584, total = 1.221 km/s$ High starter ( $40000 \to 43000$ , $a_{t} = 41500$ ): $v_{c 1} = 398600/40000 = 3.157, v_{p} = 398600 (2/40000 - 1/41500) = 3.212$ $Δ v_{1} = 3.212 - 3.157 = 0.0546$ $v_{a} = 398600 (2/43000 - 1/41500) = 2.988, v_{c 2} = 398600/43000 = 3.045$ $Δ v_{2} = 3.045 - 2.988 = 0.0568, total = 0.111 km/s$ High starter 10× se zyada sasta hai same height gain ke liye. Kyun: gravity well mein deep speeds badi hoti hain aur gravity steep hai, isliye orbit ki energy badalne ke liye bade velocity changes chahiye. Door bahar, sab kuch slow hai aur gravity gentle hai, isliye same height almost kuch cost nahi karti. Yeh Oberth effect ka geometric chehra hai ulta chal raha hai — deep hona energy per $Δ v$ amplify karta hai lekin saath hi baseline speeds bhi badhata hai jo tumhe orbit reshape karne ke liye overcome karni padti hain.

Level 4 — Synthesis

L4.1

Dikhao ki jab outer radius without bound badhta hai ( $r_{2} \to \infty$ ) fixed inner radius $r_{1}$ ke liye, pehla burn $Δ v_{1}$ escape-boost ki taraf approach karta hai: $Δ v_{1} \to (2 - 1) v_{c 1}$ . Phir ise $r_{1} = 7000$ km ke liye evaluate karo.

Recall Solution

Pehle, pehle-burn ka formula toolkit se redrive karo. $Δ v_{1} = v_{p} - v_{c 1}$ jahan $v_{c 1} = μ / r_{1}$ aur $v_{p} = μ (2/ r_{1} - 1/ a_{t})$ , $a_{t} = (r_{1} + r_{2}) /2$ . $1/ a_{t} = 2/ (r_{1} + r_{2})$ substitute karo: $v_{p} = μ (\frac{2}{r _{1}} - \frac{2}{r _{1} + r _{2}}) = \frac{μ}{r _{1}} \frac{2 r _{2}}{r _{1} + r _{2}}$ (aakhri step bracket se $μ / r_{1}$ factor out karta hai). Isliye working form: $Δ v_{1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) = v_{c 1} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1)$ Limit lo. Root ke andar top aur bottom ko $r_{2}$ se divide karo: $\frac{2 r _{2}}{r _{1} + r _{2}} = \frac{2}{r _{1} / r _{2} + 1} r_{2} \to \infty \frac{2}{0 + 1} = 2$ Toh $Δ v_{1} \to v_{c 1} (2 - 1)$ Yeh exactly woh burn hai jo circle ko parabola (escape orbit) mein convert karta hai: $2 v_{c 1}$ escape speed hai. Interpretation: infinitely-large Hohmann ka periapsis burn escape burn hai — samajh mein aata hai, kyunki infinite ellipse parabola hi hoti hai. Numbers: $v_{c 1} = 398600/7000 = 7.546$ km/s, toh $Δ v_{1} \to (2 - 1) (7.546) = 0.41421 \times 7.546 = 3.126 km/s$

L4.2

Sun ke around planet-to-planet Hohmann ke liye, Earth ki orbit $r_{1} = 1.496 \times 1 0^{8}$ km aur Mars ki $r_{2} = 2.279 \times 1 0^{8}$ km lo, $μ_{⊙} = 1.327 \times 1 0^{11} km^{3} / s^{2}$ ke saath. Dono heliocentric burns aur transfer time days mein nikalo. (Yeh woh Delta-v budget hai jo ek probe ko Earth chhodne ke upar chahiye.)

Recall Solution

$a_{t} = (1.496 + 2.279) \times 1 0^{8} /2 = 1.8875 \times 1 0^{8}$ km. $v_{c 1} = \frac{1.327 \times 1 0 ^{11}}{1.496 \times 1 0 ^{8}} = 887.03 = 29.783 km/s$ $v_{p} = 1.327 \times 1 0^{11} (\frac{2}{1.496 \times 1 0 ^{8}} - \frac{1}{1.8875 \times 1 0 ^{8}}) = 1.327 \times 1 0^{11} \times (1.33690 \times 1 0^{- 8} - 5.2980 \times 1 0^{- 9})$ $= 1.327 \times 1 0^{11} \times 8.0710 \times 1 0^{- 9} = 1071.0 = 32.726 km/s$ $Δ v_{1} = 32.726 - 29.783 = 2.943 km/s$ $v_{c 2} = \frac{1.327 \times 1 0 ^{11}}{2.279 \times 1 0 ^{8}} = 582.27 = 24.130 km/s$ $v_{a} = 1.327 \times 1 0^{11} (\frac{2}{2.279 \times 1 0 ^{8}} - \frac{1}{1.8875 \times 1 0 ^{8}}) = 1.327 \times 1 0^{11} \times (8.7758 \times 1 0^{- 9} - 5.2980 \times 1 0^{- 9})$ $= 1.327 \times 1 0^{11} \times 3.4778 \times 1 0^{- 9} = 461.50 = 21.482 km/s$ $Δ v_{2} = 24.130 - 21.482 = 2.648 km/s$ Transfer time: $t = π \frac{a _{t}^{3}}{μ _{⊙}} = π \frac{( 1.8875 \times 1 0 ^{8} ) ^{3}}{1.327 \times 1 0 ^{11}} = π \frac{6.7255 \times 1 0 ^{24}}{1.327 \times 1 0 ^{11}} = π 5.0682 \times 1 0^{13}$ $= π \times 7.1191 \times 1 0^{6} = 2.2366 \times 1 0^{7} s$ Days mein convert karo ek din ke seconds se divide karke, $86400 s/day$ (kyunki $60 s \times 60 min \times 24 h = 86400$ ): $t = \frac{2.2366 \times 1 0 ^{7} s}{86400 s/day} = 258.9 days$

Level 5 — Mastery

L5.1

Classic threshold prove karo: ek bi-elliptic transfer Hohmann transfer ko tab beat karna shuru karta hai jab ratio $R = r_{2} / r_{1}$ ≈ 11.94 se zyada ho jaata hai (ek bi-elliptic ke liye jiska intermediate apoapsis $r_{b} \to \infty$ ho). Yeh karo total $Δ v$ 's ko $R$ ke functions ke roop mein compare karke aur dhundho kahaan woh cross karte hain.

Recall Solution

Pehle bi-elliptic aur uska symbol define karo. Ek Bi-elliptic transfer teen burns use karta hai: ek bade ellipse par burn out karo jiska apoapsis ek ==intermediate radius $r_{b}$ == tak pahunchta hai jo $r_{2}$ se bada choose kiya gaya ho; $r_{b}$ tak coast karo aur dobara burn karo periapsis ko $r_{2}$ tak raise karne ke liye; $r_{2}$ tak waapis coast karo aur circularize karne ke liye teesra burn karo. Yahan $r_{b}$ woh intermediate apoapsis radius hai — "detour distance." Hum limiting case $r_{b} \to \infty$ analyze karte hain (infinity tak fling out), jo best deta hai jo bi-elliptic kabhi kar sakta hai.

Units mein kaam karo jahan $r_{1} = 1$ aur $v_{c 1} = μ / r_{1} = 1$ . $R = r_{2} / r_{1}$ lo.

Hohmann total (do burns ka sum, $v_{c 1} = 1$ , $v_{c 2} = 1/ R$ , $a_{t} = (1 + R) /2$ ke saath): $Δ v_{H} = Δ v_{1} (\frac{2 R}{1 + R} - 1) + Δ v_{2} \frac{1}{R} (1 - \frac{2}{1 + R})$

Bi-elliptic with infinite intermediate apoapsis ( $r_{b} \to \infty$ ). Burn 1 circle ko $r_{1} = 1$ par ek escape-like ellipse par raise karta hai: L4.1 se woh $2 - 1$ cost karta hai. Jab apoapsis infinity tak jaata hai beech ka burn cost $0$ ki taraf jaata hai. Final burn tumhe ek infinite-apoapsis ellipse se $r_{2}$ par circle mein drop karta hai: same escape argument apply karo radius $R$ par, jahan circular speed $1/ R$ hai, $\frac{1}{R} (2 - 1)$ cost deta hai. Total: $Δ v_{B E}^{\infty} = (2 - 1) + \frac{1}{R} (2 - 1) = (2 - 1) (1 + \frac{1}{R})$

Crossover dhundho $Δ v_{H} (R) = Δ v_{B E}^{\infty} (R)$ . Yeh transcendental hai, isliye numerically solve karo. Dono expressions tabulate karo:

$R = 10$ : $Δ v_{H} = 0.3944$ , $Δ v_{B E}^{\infty} = 0.4550$ → Hohmann jeet ta hai.
$R = 11.94$ : $Δ v_{H} = 0.3785$ , $Δ v_{B E}^{\infty} = 0.3785$ → tie (yeh root hai).
$R = 15$ : $Δ v_{H} = 0.3606$ , $Δ v_{B E}^{\infty} = 0.3211$ → bi-elliptic (infinite) jeetta hai.

Root yahan hai: $R^{⋆} \approx 11.94$ Padhna: $R < 11.94$ ke liye koi bi-elliptic Hohmann ko beat nahi kar sakta; $R > 11.94$ ke liye ek infinite (aur isliye kafi bada finite) $r_{b}$ kar sakta hai. $R \approx 11.94$ aur $R \approx 15.58$ ke beech sirf ek finite $r_{b}$ jeetta hai; $\approx 15.58$ se upar infinite case bhi outright jeetta hai. Bi-elliptic transfer dekho. Figure dono curves ko cross karte hua dikhata hai.

Figure 2 — Hohmann vs bi-elliptic total $Δ v$ vs ratio $R$ .

L5.2

Design check. Tumhe ek satellite ko $r_{1} = 6600$ km se $r_{2} = 90000$ km ( $R \approx 13.6$ ) tak raise karna hai. Hohmann total $Δ v$ compute karo, phir infinite bi-elliptic bound, aur batao kaun si strategy aur investigate karni chahiye.

Recall Solution

$v_{c 1} = 398600/6600 = 7.771$ km/s aur $R = 90000/6600 = 13.636$ use karo. Hohmann: $a_{t} = (6600 + 90000) /2 = 48300$ km. $v_{p} = 398600 (2/6600 - 1/48300) = 398600 (3.0303 \times 1 0^{- 4} - 2.0704 \times 1 0^{- 5}) = 112.98 = 10.629 km/s$ $Δ v_{1} = 10.629 - 7.771 = 2.858$ $v_{a} = 398600 (2/90000 - 1/48300) = 398600 (2.2222 \times 1 0^{- 5} - 2.0704 \times 1 0^{- 5}) = 0.6053 = 0.7780 km/s$ $v_{c 2} = 398600/90000 = 2.104 km/s, Δ v_{2} = 2.104 - 0.778 = 1.326$ $Δ v_{H} = 2.858 + 1.326 = 4.184 km/s$ Infinite bi-elliptic bound: $Δ v_{B E}^{\infty} = (2 - 1) v_{c 1} (1 + \frac{1}{R}) = 0.41421 \times 7.771 \times (1 + 1/ 13.636)$ $= 3.219 \times (1 + 0.27080) = 3.219 \times 1.27080 = 4.091 km/s$ Verdict: Kyunki $R = 13.6 > 11.94$ , bound already Hohmann se neeche hai ( $4.09 < 4.18$ ). Ek finite- $r_{b}$ bi-elliptic aur bhi behtar karega — isliye yeh raise ek genuine candidate hai Bi-elliptic transfer ke liye. Lekin transfer time ke against weigh karo, jo bi-elliptic ke liye kai zyada lambi ho sakti hai.

Recall

Recall One-line self-test

$r$ par circular speed? ::: $μ / r$ .
Transfer ke periapsis par speed? ::: vis-viva at $r_{1}$ with $a = a_{t}$ .
Door ऊपर same-height raise sasta kyun hai? ::: speeds $1/ r$ ki tarah fall karti hain, isliye same $Δ r$ tiny speed change cover karta hai.
Kya lowering utna hi cost karti hai jitna raising? ::: haan, same magnitudes — tum sirf boost ki jagah brake karte ho.
Bi-elliptic threshold? ::: $r_{2} / r_{1} \approx 11.94$ .
$Δ v_{1}$ ka limit jab $r_{2} \to \infty$ ? ::: $(2 - 1) v_{c 1}$ , escape boost.

Related: 3.2.19 Hohmann transfer — derivation, minimum energy transfer (Hinglish) · Vis-viva equation · Kepler's Third Law · Oberth effect · Two-body problem.