Patched conic method — interplanetary trajectory design: Question bank

3.2.26 · D5 · Physics › Orbital Mechanics & Astrodynamics › Patched conic method — interplanetary trajectory design

Shuru karne se pehle, is page mein baar baar aane wale symbols. Har ek yahan earn karke aata hai, trap mein aane se pehle.

Definition Wo vocabulary aur symbols jo tumhare paas hone chahiye

Patch — ek two-body conic (ek simple orbit shape) jo hum agli se jodte hain.
$μ$ ("mew"), gravitational parameter — "central body ki mass times gravity constant $G$ " ka shorthand: $μ = G m$ . Kisi body ke around orbit ke liye sirf yahi ek cheez matter karti hai, isliye hum $G$ aur $m$ ko ek number mein bundle kar dete hain. Iska subscript batata hai kaun sa body central hai: $μ_{⊙} = G m_{⊙}$ (Sun, symbol $⊙$ ), $μ_{E} = G m_{E}$ (Earth).
$r_{p}$ , perigee radius — departure hyperbola ka Earth ke centre se sabse kareeb ka distance; practically us circular parking orbit ka radius jahan se launch hote hain.
$v_{c i r c}$ , circular (parking-orbit) speed — radius $r_{p}$ par circular orbit ki speed: $v_{c i r c} = μ_{E} / r_{p}$ . Ye woh speed hai jo tumhare paas burn se pehle hi hoti hai.
$v_{esc}$ , escape speed — $r_{p}$ par wo speed jo barely infinity tak pahunch jaaye aur kuch bhi bache nahi: $v_{esc} = 2 μ_{E} / r_{p}$ . Factor $2$ note karo: escape karne mein circular speed ka square ka do guna lagtaa hai.
$v_{\infty}$ ("vee-infinity") — wo leftover speed jo spacecraft ke paas kisi planet se poori tarah escape karne ke baad bhi hoti hai, us planet ke relative mein measure ki jaati hai, uske Sphere of influence ke edge par. Doosron se energy bookkeeping se related hai: $v_{p}^{2} = v_{esc}^{2} + v_{\infty}^{2}$ .
Heliocentric — "Sun-centered"; Sun ko fixed reference maan ke measure ki gayi speed ya orbit.

Wo ek picture jo har "frame" trap ko obvious bana deti hai — patch stitch aur boundary par vector addition:

Figure — Patched conic method — interplanetary trajectory design

SOI circle dekho: hyperbola ki velocity $v_{\infty}$ Earth ke relative mein draw ki gayi hai, aur Earth ki apni orbital velocity $v_{c, 1}$ doosra arrow hai. Heliocentric transfer speed unka tip-to-tail sum hai, arithmetic addition nahi — har "Spot the error" ke liye ye figure yaad rakho.

True ya false — justify karo

Har answer mein ek reason honi chahiye, sirf verdict nahi.

Patched conic method Earth→Mars ka exact trajectory deta hai.

False. Ye idealized two-body conics ko jodata hai aur pretend karta hai ki ek waqt mein sirf ek body pull karti hai; real field many-body aur continuous hoti hai, isliye ye ek achha first guess hai jo baad mein numerically refine karte hain.

Earth ke Sphere of influence ke andar, Sun ki gravity switch off ho jaati hai.

False. Sun abhi bhi pull karta hai; hum sirf usse ignore karna choose karte hain as a small perturbation kyunki wahan Earth motion dominate karti hai. SOI ek modelling boundary hai, physical shield nahi.

Sphere of influence woh jagah hai jahan Earth aur Sun ki gravitational forces spacecraft par equal hoti hain.

False. Ye woh jagah hai jahan Earth vs Sun ko primary body treat karna equally good hota hai — primary-to-perturbing accelerations ka ratio, equal forces nahi. Equal-force point bahut zyada Earth ke kareeb hota hai.

Patch boundary ke across spacecraft ki velocity continuous hoti hai.

True, lekin sirf planet ki orbital velocity add karne ke baad. Position seedha continuous hoti hai; hyperbola ki

v_{\infty}

(planet-relative) ko heliocentric ellipse se match karne ke liye Earth ki orbital speed ke saath vector-add karna padta hai.

Hohmann transfer exactly ek ellipse ka aadha hota hai.

True. Perihelion inner orbit ko touch karta hai, aphelion outer ko, toh craft exactly

18 0^{\circ}

coast karta hai — isliye transfer time

t = π a_{t}^{3} / μ_{⊙}

(neeche box mein derive kiya gaya hai).

Outbound Earth→Mars trip mein tum Sun ke relative speed up karte ho.

True. Mars bahar hai, isliye transfer ellipse ko Earth ki circular speed se zyada perihelion speed chahiye; burn heliocentric speed add karta hai (

v_{t, 1} > v_{c, 1}

Departure burn

Δ v

equals

v_{\infty}

hota hai.

False.

v_{\infty}

infinity par measure hota hai jahan craft slow hota hai; tum perigee par burn karte ho jahan wo fast hota hai, aur coast par energy (speed nahi) conserve hoti hai.

Δ v = v_{\infty}^{2} + 2 μ_{E} / r_{p} - μ_{E} / r_{p}

, jo usually

v_{\infty}

se chhota hota hai.

Departure conic hamesha hyperbola hoti hai, kabhi parabola ya ellipse nahi.

True, interplanetary trip ke liye. Doosre planet ke SOI tak pahunchne ke liye

v_{\infty} > 0

chahiye, aur positive excess energy matlab hyperbola;

v_{\infty} = 0

parabola hogi (barely escapes), aur negative matlab bound ellipse.

Vis-viva equation hyperbolic departure leg ke liye bhi kaam karti hai.

True. Vis-viva

v = μ (2/ r - 1/ a)

kisi bhi conic ke liye hold karti hai; hyperbola ke liye semi-major axis

a

negative hoti hai, jo

v^{2}

ko infinity par bhi positive rakhti hai.

Formula Sidebar 1 —

$2/5$ power actually kahan se aata hai $r_{S O I} = a_{p} (m_{p} / m_{⊙})^{2/5}$ ek honest sketch deserve karta hai, kyunki " $\propto r^{3}$ vs $\propto 1/ r^{2}$ " thoda jaldi keh diya tha. Yahan clean version hai, $μ = G m$ use karte hue.

Planet se $r$ distance par aur Sun se $\approx a_{p}$ distance par, hum compare karte hain ki doosra body pehle ki clean two-body motion ko kitna disturb karta hai. Disturbance ratio hai (perturbing acceleration) / (primary acceleration).

Planet as primary: primary pull $= \frac{μ _{p}}{r ^{2}}$ . Sun ka disturbing effect uska poora pull nahi balki chhoti si distance $r$ par tidal difference hai, jo scale karta hai $\frac{μ _{⊙} r}{a _{p}^{3}}$ ki tarah (ek pull jo $\sim \frac{d}{d a _{p}} (\frac{μ _{⊙}}{a _{p}^{2}}) \cdot r$ se change hoti hai). Toh $A_{A} = \frac{μ _{⊙} r / a _{p}^{3}}{μ _{p} / r ^{2}} = \frac{m _{⊙}}{m _{p}} \frac{r ^{3}}{a _{p}^{3}} (\propto r^{3}) .$

Sun as primary: primary pull $= \frac{μ _{⊙}}{a _{p}^{2}}$ . Ab planet ka poora pull $\frac{μ _{p}}{r ^{2}}$ disturbance hai (craft uske kareeb hai), toh $A_{B} = \frac{μ _{p} / r ^{2}}{μ _{⊙} / a _{p}^{2}} = \frac{m _{p}}{m _{⊙}} \frac{a _{p}^{2}}{r ^{2}} (\propto 1/ r^{2}) .$

Dono descriptions equally good hain jahan $A_{A} = A_{B}$ : $\frac{m _{⊙}}{m _{p}} \frac{r ^{3}}{a _{p}^{3}} = \frac{m _{p}}{m _{⊙}} \frac{a _{p}^{2}}{r ^{2}} \Rightarrow r^{5} = a_{p}^{5} (\frac{m _{p}}{m _{⊙}})^{2} \Rightarrow r = a_{p} (\frac{m _{p}}{m _{⊙}})^{2/5} .$ Squared mass ratio isliye aata hai kyunki ek side mein $m_{⊙} / m_{p}$ hai aur doosri mein $m_{p} / m_{⊙}$ ; $r^{5}$ isolate karne ke liye cross multiply karne par ratio do baar stack hota hai. Us square ka fifth root hi $2/5$ hai. Isliye $2/5$ hai aur $1/2$ nahi: primary/perturbing accelerations $r^{3}$ aur $1/ r^{2}$ ki tarah scale karti hain, aur $3 - (- 2) = 5$ hi $r$ par power hai.

Spot the error

Har line mein ek galat claim hai; reveal batata hai kyun wo fail karta hai.

"Kyunki

v_{\infty} = 2.94

km/s hai aur parking speed

7.78

km/s hai, bas add karo:

Δ v = 2.94

km/s."

Galat kyunki

v_{\infty}

infinity par rehta hai, perigee par nahi; pehle perigee speed

v_{p} = v_{\infty}^{2} + v_{esc}^{2}

nikaloni hogi jahan

v_{esc} = 2 μ_{E} / r_{p}

hai, phir

Δ v = v_{p} - v_{c i r c}

. Escape term hi banaata hai ki

v_{\infty}

ko gravity well mein deep "sasta" kharida ja sake (Oberth effect).

v_{\infty}

spacecraft ki Sun ke around speed hai Earth chodne ke baad."

Galat reference frame —

v_{\infty}

Earth ke relative measure hota hai. Heliocentric transfer speed

v_{c, 1} \pm v_{\infty}

hoti hai, jo Earth ki orbital velocity ko vector-add karke milti hai (upar stitch figure dekho).

"SOI radius

r_{S O I} = a_{p} (m_{p} / m_{⊙})^{2/5}

m_{p}

ke saath scale karta hai, isliye Jupiter ka SOI mainly isliye enormous hai kyunki wo massive hai."

Dono factors matter karte hain, lekin

2/5

power mass dependence ko weak karti hai, jabki linear

a_{p}

(Jupiter Sun se door hai) strongly contribute karta hai; distance bahut kaam kar raha hai.

"Earth se Mars jaane ke liye hume Mars ki current position tak burn karna chahiye."

Galat — ~259-day coast ke dauran Mars move karta hai. Tum aim karte ho jahan Mars arrival par hoga, isliye launch windows exist karti hain.

"Kyunki hyperbola ek escape orbit hai, uski total energy zero hoti hai."

Zero energy parabola hoti hai. Hyperbola ki strictly positive specific energy hoti hai

ε = v_{\infty}^{2} /2 > 0

; wo leftover positive energy exactly infinity par excess speed hai.

"Transfer ellipse par craft ki speed constant hoti hai."

Kepler's second law isse forbid karta hai — craft equal time mein equal areas sweep karta hai, isliye wo perihelion par (Earth ke paas) fastest aur aphelion par (Mars ke paas) slowest hota hai.

"Parking orbit radius

r_{p}

departure

Δ v

ko affect nahi karta."

Karta hai: chhota

r_{p}

matlab zyada

v_{esc} = 2 μ_{E} / r_{p}

, toh burn zyada deep aur cheap hota hai — Oberth effect low perigees ko reward karta hai.

Formula Sidebar 2 — transfer time

$t = π a_{t}^{3} / μ_{⊙}$ kyun hota hai Ye formula parent page par har jagah assert kiya gaya hai; yahan kyun $π$ aur $3/2$ power aate hain wo batate hain.

Step 1 — pehle full period. Kepler's third law kisi bhi orbit ke liye semi-major axis $a$ aur parameter $μ$ ke saath kehta hai ki full period hai $T = 2 π \frac{a ^{3}}{μ} .$ $3/2$ kahan se aata hai: ek orbit gravity ( $\propto 1/ a^{2}$ ) aur uspar bane rehne ke liye zaruri speed ka balance hai; geometry work through karne par, ghoomne ka time $a^{3/2}$ ki tarah scale karta hai — bade orbits slower aur longer hote hain, aur $3/2$ exactly woh hai jaise dono effects compound hote hain. $2 π$ "ek full lap" factor hai, bilkul jaise circle ki circumference mein $2 π$ hota hai.

Step 2 — Hohmann transfer half lap hota hai. Transfer ellipse Earth ko perihelion par aur Mars ko aphelion par touch karti hai, isliye craft exactly $18 0^{\circ}$ coast karta hai — aadha ellipse. Aadha period: $t = \frac{1}{2} T = \frac{1}{2} \cdot 2 π \frac{a _{t}^{3}}{μ _{⊙}} = π \frac{a _{t}^{3}}{μ _{⊙}} .$ $2 π$ half hoke akela $π$ ban jaata hai — exactly isliye transfer-time formula mein single $π$ hota hai (na ki $2 π$ ). Neeche ki figure half-ellipse aur $18 0^{\circ}$ sweep dikhati hai:

Why questions

Mechanism answer karo, sirf outcome nahi.

SOI derivation mein

2/5

power kyun aata hai, na ki koi aur jaise

1/2

Kyunki planet-as-primary disturbance ratio

r^{3}

ki tarah scale karta hai aur Sun-as-primary ratio

1/ r^{2}

ki tarah; unhe equal set karne par

r^{5} \propto a_{p}^{5} (m_{p} / m_{⊙})^{2}

milta hai, aur squared mass ratio ka fifth root

2/5

hai (Sidebar 1 mein poori sketch).

Patch boundary par velocities "add" kyun hoti hain lekin simply nahi?

Position shared hoti hai, lekin hyperbola ki velocity planet-relative hai jabki ellipse ki Sun-relative; frames ke beech convert karne ke liye planet ki orbital velocity ko vector-add karte hain, isliye magnitudes arithmetically tab tak add nahi hote jab tak vectors aligned na hon.

Gravity well mein deep burn karna (chhota

r_{p}

) zyada efficient kyun hai?

Chhote

r_{p}

par craft fast move karta hai, aur kinetic energy

\propto v^{2}

hoti hai; ek fixed

Δ v

jo large

v

mein add hoti hai,

v^{2}

(aur thus

v_{\infty}^{2}

) ko zyada raise karti hai, wahi

Δ v

slow craft mein add hone se zyada — yahi Oberth effect hai.

Hohmann transfer orbit exactly do tangential burns kyun use karta hai?

Tangential burns speed change karte hain bina direction changes mein energy waste kiye, aur dono circular orbits ko touch karna (perihelion

r_{1}

par, aphelion

r_{2}

par) coplanar circles ke beech minimum-energy two-impulse connection hai.

Hum craft ko "Earth ki position par SOI chodna" kyun treat kar sakte hain?

Kyunki

r_{S O I}

Earth ke Sun distance

a_{p}

se kaafi chhota hai (~1000× chhota), isliye heliocentric scale par poora SOI ek point mein collapse ho jaata hai Earth ki location par.

Transfer time

r_{1}

aur

r_{2}

choose karte hi fixed kyun ho jaata hai?

Time sirf

a_{t} = (r_{1} + r_{2}) /2

par depend karta hai

t = π a_{t}^{3} / μ_{⊙}

ke through (Kepler's third law); geometry akela coast duration determine karta hai.

Edge cases

Woh cases jo topic invite karta hai lekin rarely dikhata hai.

Agar departure par

v_{\infty} = 0

ho toh?

Departure conic parabola ban jaati hai (marginal escape); craft barely Earth ka SOI zero relative speed se chodta hai, isliye sirf Earth ki orbit share karne wali bodies tak drift kar sakta hai — koi real transfer nahi.

Agar target planet ki orbit Earth ke andar ho (jaise Venus)?

Transfer ellipse ka aphelion Earth par aur perihelion Venus par hoga, isliye tumhe Sun ke relative slow down karna hoga (

v_{t, 1} < v_{c, 1}

); departure burn retrograde hoga, aur

v_{\infty} = ∣ v_{t, 1} - v_{c, 1} ∣

abhi bhi magnitude use karta hai.

Agar dono planets ki orbits coplanar na hon toh?

Basic Hohmann patch fail ho jaata hai — ek plane-change component chahiye, jisse

Δ v

aur badhta hai; pure patched conic idealization ke roop mein coplanar circular orbits assume karta hai.

m_{p} \to 0

(ek tiny asteroid) hone par

r_{S O I}

ka kya hoga?

r_{S O I} = a_{p} (m_{p} / m_{⊙})^{2/5} \to 0

, matlab body ka practically koi region nahi jahan wo dominate kare — Sun har jagah motion control karta hai.

a_{p} \to 0

(Sun ke bahut kareeb ek body) hone par

r_{S O I}

ka kya hoga?

SOI linearly zero tak shrink ho jaata hai kyunki Sun ka overwhelming nearby pull koi aisa region nahi chhodta jahan chhoti body dominate kar sake.

Agar

r_{1} = r_{2}

ho (same orbit) toh?

Tab

a_{t} = r_{1}

, "transfer ellipse" degenerate hokar circular orbit ban jaati hai,

v_{t, 1} = v_{c, 1}

, aur

v_{\infty} = 0

— koi burn aur koi transfer ki zaroorat nahi.

Mars arrival par, kya

v_{\infty}

wahi value hai jo Earth departure par thi?

Nahi — arrival

v_{\infty}

Mars par planet-relative speed hai, jo aphelion par

∣ v_{t, 2} - v_{c, 2} ∣

se compute hoti hai; ye capture karta hai ki ellipse Mars se kitna fast/slow move kar rahi hai, aur arrival hyperbola ko drive karta hai.

Recall Ek-line self-test

Wo single fact batao jo parent page par Errors 1–3 mein se har ek ko fix karta hai. ::: $v_{\infty}$ planet ke relative, infinity par measure hota hai — ye yaad rakho aur tum "sirf speeds add karo," "heliocentric = $v_{\infty}$ ," aur frame-mixing mistakes ek saath fix kar loge.