Patched conic method — interplanetary trajectory design: Visual walkthrough

3.2.26 · D2 · Physics › Orbital Mechanics & Astrodynamics › Patched conic method — interplanetary trajectory design

Intuition Yeh page kya karta hai

Parent note ek boxed formula ke saath khatam hoti hai interplanetary departure burn ke liye: $Δ v = v_{\infty}^{2} + \frac{2 μ _{E}}{r _{p}} - \frac{μ _{E}}{r _{p}}$ Is page par hum us formula ko draw karte hain — ek picture at a time — bilkul "speed kya hoti hai?" se shuru karke. Neeche kuch bhi assumed nahi hai. Har symbol ko use karne se pehle earn kiya gaya hai. Last picture tak aap dekh paoge ki deep burns saste kyun hote hain, planet-relative leftover speed $v_{\infty}$ Earth ki apni motion ke saath kaise combine hoti hai, aur number real Earth→Mars launch ke liye approximately $3.6 km/s$ kyun aata hai.

Yeh parent note ka visual companion hai. Yeh Vis-viva equation, Hyperbolic escape trajectories, Oberth effect, Sphere of influence, aur Hohmann transfer orbit par lean karta hai.

Definition Subscript legend (poore page ke liye fixed)

Symbols ko wobble hone se rokne ke liye hum ek rule use karte hain: subscript reference ka naam batata hai — ya toh woh circle jiska woh hissa hai, ya woh jagah jahan measure kiya gaya hai.

$v_{c, 1}$ — Earth ki apni circular speed Sun ke around (planet 1 = Earth). Sun ke relative measure ki gayi.
$v_{c i r c}$ — spacecraft ki circular speed apne parking orbit mein Earth ke around. Earth ke relative.
$v_{p}$ — spacecraft ki speed perigee par (escape path ka low point). Earth ke relative.
$v_{esc}$ — radius $r_{p}$ se escape speed. Earth ke relative.
$v_{\infty}$ — infinity par leftover speed, yaani Earth ko poori tarah escape karne ke baad. Earth ke relative.
$Δ v$ — engine jo speed change deta hai perigee par. Earth ke relative.

$v_{c, 1}$ ko chhodkar baaki sab Earth ke relative measure hote hain; $v_{c, 1}$ ek hi Sun-relative speed hai, sirf final vector-addition step mein use hoti hai.

Step 1 — "Speed the trip needs" ka matlab kya hai

KYA. Kisi bhi formula se pehle, hum characters ka cast pictures of motion ke roop mein set up karte hain. Earth Sun ke around near-circle par speed $v_{c, 1}$ se move karti hai. Earth ke atmosphere ke thoda upar parked ek spacecraft bhi move karta hai — Earth ke around — speed $v_{c i r c}$ se. Do alag "central bodies", do alag speeds jo straight rakhni hain.

KYUN. Is pure subject mein confusion ka sabse bada source "speed relative to the Sun" aur "speed relative to Earth" ko mix karna hai. Isliye hum unhe naam dete hain aur har ek ko pehle ek picture se anchor karte hain, phir unhe kabhi blur nahi hone dete. Hum $v_{c, 1}$ se milte hain, ise ek taraf rakh dete hain, aur yeh Step 8 mein crucial vector-addition ke liye wapas aata hai.

PICTURE. Neeche ke figure mein, bada straight arrow Earth ki motion Sun ke around hai ( $v_{c, 1}$ ). Chhota loop spacecraft ka parking orbit hai Earth ke around ( $v_{c i r c}$ ). Do speeds, do centres.

Do speeds, do pictures, phir kabhi confuse nahi hongey.

Step 2 — $v_{\infty}$ : woh speed jo climbing out ke baad bachi rehti hai

KYA. Hum departure problem ki sabse important quantity introduce karte hain: $v_{\infty}$ , padho "vee-infinity." Yeh woh speed hai jo craft ke paas tab bhi hoti hai, Earth ke relative, jab woh Earth se infinitely door hoti hai — yaani Earth ki pakad se poori tarah escape karne ke baad.

KYUN yeh symbol aur sirf "escape speed" nahi? Escaping ka matlab hai infinity par zero speed ke saath pahunchna. Lekin hum Earth ke Sphere of influence ki edge par thak-haare nahi pahunchna chahte — humein tab bhi move karna hai, kyunki Sun-centred transfer ko woh motion chahiye. Isliye hum leftover ke liye ek naam invent karte hain: $v_{\infty}$ . Subscript $\infty$ literally matlab hai "infinity par measure kiya gaya, jahan Earth ab nahi kheenchti."

PICTURE. Ek ball ko hill par roll karo. Agar woh barely crest kare, toh top par zero speed se pahunchi. Agar tum zyada push karo, toh woh crest karti hai aur chalti rehti hai — woh leftover roll $v_{\infty}$ hai. Hill Earth ka gravity well hai.

Energy equation likhne se pehle, humein ek bookkeeping quantity chahiye. Jab ball hill par chadhti hai toh woh kinetic energy (motion ki energy) ko potential energy (height ki energy) mein trade karti hai. Unka sum — ise specific orbital energy $ε$ (Greek letter epsilon; "specific" = per kilogram, toh craft ka mass cancel ho jata hai) kehte hain — kisi bhi engine-off coast mein constant rehta hai, kyunki gravity koi net accounting mischief nahi karta: chadhte waqt jo bhi kinetic energy tum khote ho, woh height ke roop mein wapas milti hai, aur vice versa. Hum is conserved sum ko single letter $ε$ dete hain taaki hum ise ek baar likh sakein aur do alag points par reuse kar sakein. $v$ ko craft ki Earth-centred frame mein radius $r$ par speed (Earth ke centre se distance) maano. Toh:

Formula Ek escape path ki energy

$ε = kinetic, per kg, Earth-frame speed v \frac{v ^{2}}{2} - potential well, depth at radius r \frac{μ _{E}}{r} = infinity par jo bacha \frac{v _{\infty}^{2}}{2}$

$ε$ — specific orbital energy (energy per kilogram), kisi bhi coast mein constant. Yahi constancy hai jis wajah se hum do points compare kar sakte hain.
$v$ — craft ki speed Earth-centred frame mein radius $r$ par measure ki gayi.
$r$ — Earth ke centre se craft tak ki distance.
$\frac{v ^{2}}{2}$ — speed $v$ par kinetic energy per kg.
$μ_{E} = G m_{E}$ — Earth ka gravitational parameter. $G$ Newton ka constant hai, $m_{E}$ Earth ka mass hai; inhe bundle karna ink bachata hai aur yeh bahut precisely jaana jaata hai.
$\frac{μ _{E}}{r}$ — distance $r$ par well ki depth. Jab $r$ chhota hota hai (well mein deep) toh bada hota hai.
Jab $r \to \infty$ , $\frac{μ _{E}}{r} \to 0$ , toh sirf $\frac{v _{\infty}^{2}}{2}$ bachta hai. Yahi $v_{\infty}$ ki mathematical definition hai.

Kyunki $ε$ ek coasting path ke har point par same hai, hum ise perigee aur infinity par evaluate kar sakte hain aur equal set kar sakte hain. Yahi trick poora game hai.

Step 3 — "Leftover speed" ko "speed at perigee" mein convert karna

KYA. Ab hum $v_{p}$ nikalte hain — woh speed jo craft ko perigee par honi chahiye — escape path ka low point, radius $r_{p}$ par (parking-orbit radius) — taaki uske paas leftover speed $v_{\infty}$ bache.

KYUN. Hum engine ko "infinity par" fire nahi kar sakte. Hum ise neeche, radius $r_{p}$ par fire karte hain, jahan craft physically hai. Toh humein goal ( $v_{\infty}$ wahan bahar) ko requirement ( $v_{p}$ yahan neeche) mein translate karna hoga. Conservation of $ε$ exactly woh tool hai jo same coasting path ke do points ko connect karta hai — yahi iske liye hai.

PICTURE. Step 2 jaisi hi hill, lekin ab hum do points mark karte hain: bottom (perigee $r_{p}$ , fast) aur infinitely far right (flat, $v_{\infty}$ par move karta hua). Climb ki gayi height kinetic energy ko potential energy mein convert karti hai.

Yahan $r_{p}$ ka matlab hai perigee radius = Earth ka radius $+$ parking-orbit altitude (escape path Earth ke centre ke kitna karib aata hai). Iske define hone ke baad, do points par energy equal set karo:

Step 4 — Do ingredients side by side

KYA. Hmare paas pehle se $v_{c i r c}$ hai Step 1 se (woh speed jo craft ke paas burn se pehle hai) aur ab $v_{p}$ hai Step 3 se (woh speed jo burn ke baad chahiye). Burn difference hai.

KYUN. Engine speed change karta hai. Jis change ke liye hum pay karte hain woh exactly "final minus initial" hai burn point par — dono same frame mein measured (Earth ke relative), dono same place par (perigee $r_{p}$ ), aur (hamare co-planar assumption se) dono same direction mein point karte hue. Kyunki woh frame, jagah, aur direction share karte hain, magnitudes subtract karna legal hai.

PICTURE. Do arrows ek hi point se shuru hote hain: chhota blue arrow $v_{c i r c}$ hai, lamba orange arrow $v_{p}$ hai. Orange ka woh chhota sa sliver $Δ v$ hai, the burn.

Step 5 — Deep burns saste kyun hote hain (Oberth ka dil)

KYA. Hum dekhte hain ki formula kya karta hai jab $r_{p}$ chhota hota hai — yaani jab hum Earth ke gravity well mein deeper burn karte hain, surface ke karib.

KYUN. Yeh non-obvious payoff hai. Naively tum sochoge ki ek diye gaye $v_{\infty}$ tak pahunchne ke liye same $Δ v$ lagta hai chahe tum kahan burn karo. Aisa nahi hota. Square-root structure ka matlab hai ki neeche burn karna tumhe zyada $v_{\infty}$ per unit $Δ v$ deta hai. Yeh Oberth effect hai, aur yeh seedha "energies add" se nikalta hai.

PICTURE. $Δ v$ needed vs. perigee altitude plot karo ek fixed target $v_{\infty}$ ke liye. Curve neeche dhalta hai jab tum lower jaate ho — deeper sasta hai.

Step 6 — Edge cases: pakde mat jaana

KYA. Hum formula ke corners check karte hain taaki koi scenario reader ko surprise na kare.

KYUN. Ek formula jis par tum sirf "typical" case mein trust karte ho woh ek trap hai. Hum char degenerate inputs test karte hain: $v_{\infty} = 0$ , $r_{p} \to \infty$ , naive "just add speeds," aur $v_{\infty}$ ka sign (departure vs capture).

PICTURE. Mini-panels, ek per edge case, har ek escape path ko degenerate hota dikhata hai.

Step 7 — Real numbers daalo (Earth → Mars, 200 km LEO)

KYA. Hum parent note ke worked examples ko unhi pictures mein run karte hain jo hum ne abhi banai hain, taaki abstract steps concrete kilometres per second mein land ho sakein.

KYUN. Ek derivation jise tum number mein turn nahi kar sakte woh decoration hai. Yahan hum confirm karte hain $v_{\infty} \approx 2.94 km/s$ , $v_{p} \approx 11.40 km/s$ , $Δ v \approx 3.62 km/s$ .

PICTURE. Ek labelled number line: $v_{c i r c} = 7.78$ , phir escape rung $v_{esc} = 11.01$ , phir perigee target $v_{p} = 11.40$ , with burn $Δ v$ $v_{c i r c}$ se $v_{p}$ tak ke jump ke roop mein marked.

Step 8 — $v_{c, 1}$ finally wapas aata hai: Earth ki motion ke saath vector-add karna

KYA. Humne promise kiya tha ki $v_{\infty}$ Earth ke relative measure hoti hai. Lekin transfer ellipse Sun ke frame mein rehti hai. Toh Sphere of influence ki edge par humein craft ki Earth-relative leftover $v_{\infty}$ ko Earth ki apni Sun-relative velocity $v_{c, 1}$ mein vector-add karna hoga taaki craft ki heliocentric speed mile.

KYUN. Yahi "patch" hai jo departure hyperbola ko Hohmann transfer orbit se stitch karta hai. Yeh us sawaal ka jawab deta hai jo poore page ki headline ne promise kiya tha: hum speeds ko jis tarah add karte hain, woh kyun? Kyunki velocities arrows hain (unki direction hoti hai), sirf numbers nahi — tum unhe tip-to-tail add karte ho, magnitudes pile up karke nahi.

PICTURE. Earth ki velocity $v_{c, 1}$ (lamba teal arrow) jiske saath craft ka $v_{\infty}$ (orange) tip-to-tail rakh diya. Hohmann departure ke liye hum $v_{\infty}$ ko along $v_{c, 1}$ aim karte hain, toh magnitudes yahan add hote hain: $v_{h e l i o} = v_{c, 1} + v_{\infty}$ .

Ek-picture summary

Yeh single figure poori kahani stack karta hai: gravity-well hill, perigee point jahan hum burn karte hain, do Earth-frame speeds $v_{c i r c}$ (before) aur $v_{p}$ (after), leftover $v_{\infty}$ flat top par bahut door, aur — side mein — tip-to-tail vector sum Earth ke $v_{c, 1}$ ke saath jo craft ko Sun-frame transfer ko hand off karta hai, sab kuch boxed formula se tied together.

Recall Feynman retelling — ek dost ko batao

Tum ek fast-spinning merry-go-round par ho (tumhara parking orbit) ek bade smooth hill ke bottom par (Earth ki gravity), aur poora playground khud ek lamppost ke around speed $v_{c, 1}$ se sail kar raha hai (Sun). Tum leap off karna chahte ho aur hill ke top par flat plain par pahunchne ke baad tab bhi jogging karna chahte ho — woh leftover jog $v_{\infty}$ hai, playground ke against measure ki gayi. Sawaal: tumhare leap ko kitna extra push ( $Δ v$ ) chahiye? Jawab: jab tum chadhte ho toh energy conserve hoti hai, toh tumhe bottom par jo push chahiye woh "leftover speed plus climbing speed" pile up nahi hai — yeh energies ke roop mein add hoti hai, ek square root ke neeche. Kyunki tum bottom par already fast move kar rahe ho, ek chhoti si extra push bahut badi chunk of energy khareedti hai, toh deep down se leap karna (low orbit) sabse sasta hai. Finally, yeh dekhne ke liye ki tum lamppost ke relative kitni tezi se ja rahe ho, tum apna jog $v_{\infty}$ playground ki apni motion $v_{c, 1}$ par ek arrow ke roop mein add karo — ise same direction mein point karo aur speeds add hoti hain, jo exactly Mars ka raasta hai. Numbers crunch karo aur leap 200 km orbit se roughly 3.6 km/s lagti hai.

Recall Active recall — jawab chhupao

$v_{\infty}$ physically kya mean karta hai, aur kis frame mein? ::: Earth se infinitely door hone ke baad (well se bahar) Earth ke relative jo speed bachi rehti hai.
Hum energy-at-perigee ko energy-at-infinity ke equal kyun set karte hain? ::: Specific energy $ε$ engine-off coast mein conserve hoti hai, toh escape path ke har point par equal hai.
$Δ v = v_{\infty} + (v_{esc} - v_{c i r c})$ kyun nahi hai? ::: Woh speeds add karta hai; sach energies add karta hai, deta hai $v_{\infty}^{2} + v_{esc}^{2} - v_{c i r c}$ , jo chhota hai (Oberth).
Deep burns saste kyun hote hain? ::: Tum perigee par sabse fast ho, aur $Δ v$ ko badi speed mein add karna sabse zyada kinetic energy add karta hai, jo $v_{\infty}^{2}$ ban jaati hai.
$r_{p} \to \infty$ par $Δ v$ ka kya hota hai? ::: Yeh exactly $v_{\infty}$ approach karta hai — bilkul Oberth discount nahi.
$v_{\infty}$ Sun-frame speed kaise banta hai? ::: Ise Earth ki orbital velocity $v_{c, 1}$ mein tip-to-tail vector-add karo; prograde aim karo, magnitudes add hokar $v_{t, 1}$ dete hain.
$v_{\infty}^{2} < 0$ ka kya matlab hai? ::: Craft escape nahi kar sakta — woh Earth se bound hai (capture), departure nahi.