Pork chop plots — Δv vs launch - arrival date: Visual walkthrough

3.2.27 · D2 · Physics › Orbital Mechanics & Astrodynamics › Pork chop plots — Δv vs launch - arrival date

Definition Do symbols jinpar hum depend karenge — inhe abhi define karo

Gravitational parameter $μ$ . Kisi bhi body of mass $M$ ke liye, $μ \equiv G M$ , jahan $G$ Newton's gravitational constant hai. Yeh "yeh body kitna pull karta hai" ko ek single number mein bundle karta hai (units $km^{3} / s^{2}$ ). Hum Earth ke liye $μ_{E} = G M_{Earth}$ aur target planet ke liye $μ_{T} = G M_{target}$ use karte hain. Hum $μ$ carry karte hain $G$ aur $M$ alag-alag ki jagah, kyunki har orbit formula ko sirf unka product chahiye hota hai.
Planet velocity $V$ (heliocentric frame). $V_{E}$ Earth ki apni orbital velocity hai Sun ke around, aur $V_{t a r g e t}$ target ki hai — dono Sun-centred frame mein measured (wahi frame jismein position arrows $r$ hain). Ye woh velocities hain jo planets ke paas pehle se hain, "free mein," koi bhi rocket fire hone se pehle.

Step 1 — Do dots jo rukenge nahin

KYA. Ek launch date aur ek arrival date choose karo. Un do dates par, Earth ki Sun-centred ("heliocentric") position ek arrow $r_{1}$ hai, aur target planet ki position ek arrow $r_{2}$ hai.

KYUN. Ek transfer orbit space mein ek curve hai. Koi bhi curve pin karne ke liye pehle yeh batana zaroori hai ki woh kahan se shuru hoti hai aur kahan khatam. Woh do facts hain $r_{1}$ aur $r_{2}$ . Fuel ke baare mein abhi kuch nahin — sirf geography.

PICTURE. Figure dekho. Red arrow $r_{1}$ hai: launch day par Earth ki position, Sun se measured. Black arrow $r_{2}$ arrival day par target ki position hai. Kyunki dono planets orbit karte hain, koi bhi date slide karne se us planet ke arrow ki tip apne orbit ring pe slide ho jaati hai — yahi wajah hai ki fuel cost dates par depend karegi.

Step 2 — Chord: woh seedha shortcut jo hum nahin fly karenge

KYA. $r_{1}$ ki tip se $r_{2}$ ki tip tak ek seedhi line khiincho. Iska length chord hai $c = ∣ r_{2} - r_{1} ∣.$

KYUN. Hum yeh seedhi line kabhi nahin fly karenge (gravity har cheez ko curve karti hai), lekin iska length un teen numbers mein se ek hai jo Lambert's theorem ko chahiye. Chord measure karta hai ki do dots kitne door hain, chahe dono Sun se kitni bhi door kyun na hon.

PICTURE. Red segment chord $c$ hai. Notice karo $r_{2} - r_{1}$ ka matlab literally hai "tip of $r_{1}$ par khado aur $r_{2}$ ki tip tak chalo" — vectors subtract karne se woh arrow milta hai jo do tips ke beech jaata hai. Iska length $c$ hai.

Step 3 — Lambert's question: clock ke saath kaunsa orbit fit hoga?

KYA. Ab hamare paas teen ingredients hain: $r_{1} + r_{2}$ (dono ends Sun se kitni door hain), chord $c$ (dono ek-doosre se kitni door hain), aur time of flight $Δ t = arrival - launch$ . Lambert's theorem kehta hai yeh teen orbit ka size fix karte hain: $Δ t = f (r_{1} + r_{2}, c, a) .$

KYUN yeh teen aur koi nahin? Kyunki Lambert ne prove kiya ki do points ke beech flight time sirf un teen numbers par depend karta hai — orbit ki tilt ya direction par nahin. Toh agar $r_{1} + r_{2}$ , $c$ , aur $Δ t$ pehle se pata hain, toh ek hi unknown bacha hai: $a$ , orbit ki semi-major axis (uska "size"). Hum equation ulti kar ke $a$ solve karte hain. Yahi hai Lambert's Problem.

KYUN ek naya tool yahan? Ek Hohmann Transfer "do circles ke beech sabse sasta orbit" ka jawab deta hai lekin woh aapko arrival time choose karne nahin deta — aapko wahi TOF milta hai jo Hohmann ellipse deta hai. Pork chop plot ek chosen $Δ t$ demand karta hai. Lambert woh tool hai jo arbitrary time accept karta hai aur matching orbit return karta hai. Yahi exactly woh question hai jo hamare paas hai.

PICTURE. Do orbits same do dots se hoke drawn hain. Red wali moti hai (bada $a$ ) aur zyada time leti hai; black wali patli hai (chhota $a$ ) aur jaldi hai. $Δ t$ fix karna is family mein se exactly ek ko select karta hai.

Step 4 — Orbit se do velocities nikalo

KYA. Jab $a$ pata ho jaata hai to poora conic pata ho jaata hai, aur geometry hamare haath mein do velocity arrows deta hai:

$v_{1}$ = spacecraft transfer ke start par kitna fast (aur kis direction mein) move karta hai,
$v_{2}$ = end (arrival) par uski velocity.

KYUN. Fuel velocity mein changes ke liye pay karta hai, position ke liye nahin. Toh jis moment hume transfer orbit pata chale, sabse pehle kaam yeh hai ki har end par jo velocity demand hai woh nikalo. Aage ki sab cheez in velocities ki planets ki already-existing velocities se comparison hai.

PICTURE. Start-dot par, red arrow $v_{1}$ transfer curve ke tangent hai (velocity hamesha travel ki direction mein hoti hai). End-dot par, $v_{2}$ wahan curve ke saath point karta hai. Yeh required velocities hain — spacecraft ko is orbit par ride karne ke liye yahi honi chahiye.

Step 5 — Earth jo free mein deta hai woh subtract karo ( $v_{\infty}$ )

KYA. Spacecraft pehle se Earth ke heliocentric orbital velocity $V_{E}$ ke saath fly kar raha hai (opening block mein define kiya). Woh velocity jo humein supply karni hai woh bacha hua hissa hai: $v_{\infty}^{d e p} = v_{1} - V_{E}, C_{3} \equiv v_{\infty}^{d e p}^{2} .$

KYUN subtract? Tum ek chalti train (Earth) par khade ho. Ek passing car match karne ke liye tumhe sirf speed difference chahiye, car ki poori speed nahin. Earth $V_{E}$ bina cost ke donate karta hai, toh actual cost difference $v_{1} - V_{E}$ hai. Jab craft itna door chali jaati hai ki Earth ki gravity fade ho jaaye, toh yeh leftover speed hyperbolic excess speed $v_{\infty}$ kehlati hai. Iska square $C_{3}$ hai — Hyperbolic Excess Velocity & C3 dekho.

KYUN vector subtraction (sirf speeds kyun nahin subtract karte)? Kyunki do arrows ki same length ho sakti hai lekin direction alag. Agar $v_{1}$ aur $V_{E}$ thodi si bhi angle se alag point karein, toh leftover arrow bada ho sakta hai. Direction matter karta hai, isliye hum arrows ki tarah subtract karte hain.

PICTURE. Tail-to-tail: black arrow $V_{E}$ hai (Earth ki free velocity), doosra black arrow required $v_{1}$ hai, aur red arrow jo triangle close karta hai woh $v_{\infty}^{d e p} = v_{1} - V_{E}$ hai — woh akela hissa jo tum pay karte ho.

Step 6 — Gravity well se bahar niklo (departure burn)

KYA. Reality mein tum "door se" shuru nahin karte — tum Earth ke around radius $r_{p}$ ki ek circular parking orbit mein shuru karte ho, $v_{c} = μ_{E} / r_{p}$ par chalte ho (yaad karo $μ_{E} = G M_{Earth}$ ). Escape (hyperbolic) orbit par energy conservation se woh speed milti hai jo tumhe perigee par chahiye, aur burn $v_{c}$ se us tak ka jump hai: $Δ v_{d e p} = v_{\infty}^{2} + \frac{2 μ _{E}}{r _{p}} - \frac{μ _{E}}{r _{p}} .$

KYUN $2 μ_{E} / r_{p}$ term? Earth ki gravity well se bahar jaane mein speed lagti hai — jaise ek ball ko bowl se upar roll karne ke liye extra effort chahiye. Hum isko specific orbital energy $ε$ se track karte hain — total energy (kinetic + gravitational) spacecraft ke per kilogram, units $km^{2} / s^{2}$ . Escape orbit ke liye $ε = v_{\infty}^{2} /2$ (door se saari energy leftover kinetic $v_{\infty}^{2} /2$ hai, kyunki gravity fade ho chuki hai). Wahi $ε$ , radius $r_{p}$ par evaluate kiya, $v_{p er i}^{2} /2 - μ_{E} / r_{p}$ hai. Kyunki energy conserved hai, yeh do expressions equal hain: $\frac{v _{\infty}^{2}}{2} = \frac{v _{p er i}^{2}}{2} - \frac{μ _{E}}{r _{p}} \Rightarrow v_{p er i} = v_{\infty}^{2} + \frac{2 μ _{E}}{r _{p}} .$ Yahi wajah hai ki $C_{3}$ akela burn under-state karta hai: gravity square root lene se pehle $2 μ_{E} / r_{p}$ add karta hai. Perigee par fire karna (well mein deep, sabse fast chalte hue) sabse sasta hai — Oberth Effect.

PICTURE. Ek circle (parking orbit, speed $v_{c}$ ) aur ek red hyperbola us se peeling away. Shared perigee par do speed markers stacked hain; unke beech ka chhota red bracket hi $Δ v_{d e p}$ hai.

Step 7 — Arrival par bhi yahi karo, phir add karo

KYA. Steps 5–6 ko dusre end par repeat karo. Target ke relative excess speed $v_{\infty}^{a r r} = v_{2} - V_{t a r g e t}$ hai (jahan $V_{t a r g e t}$ opening block se target ki heliocentric velocity hai), aur target planet ki gravity parameter $μ_{T} = G M_{t a r g e t}$ wale planet ke around radius $r_{a}$ ki orbit mein capture burn hai: $Δ v_{a r r} = v_{\infty, a r r}^{2} + \frac{2 μ _{T}}{r _{a}} - \frac{μ _{T}}{r _{a}} .$ Is ek grid cell ki total cost: $Δ v_{t o t} = Δ v_{d e p} + Δ v_{a r r}$

KYUN same form? Arrival departure ka reverse hai: ek well se escape karne ke liye burn subtract karne ki jagah, tum drop into karne ke liye burn subtract karte ho. Same energy bookkeeping (same $ε$ argument, ab $μ_{T}$ aur $r_{a}$ ke saath), toh target ke numbers ke saath same formula.

KYUN add karo? Dono burns trip ke opposite ends par hote hain aur ek-doosre ki help nahin kar sakte — total fuel unka sum hai. Yeh single number pork chop banane ke liye (launch, arrival) grid par contour kiya jaata hai. $Δ v_{t o t}$ ko Tsiolkovsky Rocket Equation mein feed karne se deliverable payload pata chalta hai, yahi wajah hai ki $0.1$ km/s bachana bhi matter karta hai.

PICTURE. Do mirror-image "burn brackets": left mein Earth par red (escape), right mein target par black (capture), plus sign se total mein join.

Step 8 — Degenerate case: 180° ridge

KYA. Kya ho agar $r_{1}$ aur $r_{2}$ Sun ke exact opposite hon (180° transfer)? Phir do dots aur Sun collinear hain, aur us line mein infinitely many planes hain.

KYUN breaks hota hai. Orbit plane ab defined nahin — tum transfer kisi bhi taraf tilt kar sakte ho aur phir bhi dono dots se guzar sakte ho. Real planets ek shared plane se thode bahar hote hain (yaad karo planar assumption jo humne shuru mein ki thi woh sirf approximation hai), toh 180° transfer thread karne ke liye ek badi plane change force hoti hai, jo expensive hai. Yahi woh sharp ridge hai jo pork chop ke do lobes ko split karta hai: yeh rendering glitch nahin, ek genuine wall hai.

PICTURE. Sun jisme do dots dead opposite hain; ek red fan of tilted orbits jo dono se hoke jaati hain — dikhata hai ki plane undetermined kyun hai. Reader ko dikhna chahiye ki Δv yahan spike kyun karta hai.

Ek-picture summary

Yeh single figure saare aath steps chain karta hai: do dots → chord → Lambert's orbit → required velocities → planet velocity subtract karo → well climb karo → sum karo. Red thread ko left se right follow karo.

Recall Feynman retelling — plain words mein bolo

Main space mein do spots mark karta hoon: jab main leave karun tab Earth kahan hai, jab main pahunchuun tab mera target kahan hai. Main unke beech seedha gap kheenchta hoon — chord — aur apna chosen flight time note karta hoon. Lambert's rule kehta hai woh teen facts exactly ek orbit size fix karte hain, toh main use solve karta hoon. Woh orbit mujhe bataati hai ki mujhe har end par kitna fast hona chahiye. Lekin main pehle se Earth ke saath coast kar raha hoon, toh main sirf velocity mein difference pay karta hoon — woh leftover, jab gravity fade ho jaaye, $v_{\infty}$ hai, aur iska square woh $C_{3}$ hai jo rocket ki spec sheet par hota hai. Wahan actually pahunchne ke liye mujhe parking orbit se Earth ki gravity bowl se bahar nikalna hoga, jo square root ke andar $2 μ / r$ toll add karta hai; main mirror-image kaam target mein drop hone ke liye karta hoon. Dono burns add karo aur mujhe is ek launch/arrival pair ka fuel bill mil jaata hai. Yeh har date pair ke liye karo aur numbers contour karo — aur saste valleys, forbidden 180° ridge se split, pork chops ke shaped aate hain.

Recall Quick self-test

Lambert's theorem ko kaunse teen numbers chahiye? ::: $r_{1} + r_{2}$ , chord $c$ , aur time of flight $Δ t$ . $v_{1}$ se $V_{E}$ kyun subtract karte hain? ::: Earth apni heliocentric orbital velocity free mein deta hai; tum sirf difference $v_{\infty}$ pay karte ho. $2 μ_{E} / r_{p}$ term kya represent karta hai? ::: Radius $r_{p}$ se Earth ki gravity well se bahar nikalne ke liye extra speed chahiye; yahan $μ_{E} = G M_{Earth}$ . Type I aur Type II transfers mein kya fark hai? ::: Type I 180° se kam heliocentric angle sweep karta hai (short-way), Type II 180° se zyada sweep karta hai (long-way). Lobes ke beech ridge expensive kyun hai? ::: 180° transfer par orbit plane undefined hai, jo ek costly plane change force karta hai.