Patched conic method — interplanetary trajectory design

3.2.26 · Physics › Orbital Mechanics & Astrodynamics

KYA patch ho raha hai?

Hum poori trip ko teen conic patches mein divide karte hain:

Departure hyperbola — Earth ke SOI ke andar, Earth central body hai. Spacecraft ek hyperbola pe nikal jaati hai.
Heliocentric transfer ellipse — SOIs ke beech mein, Sun central body hai. Zyaadatar ek Hohmann-type ellipse hoti hai.
Arrival hyperbola — target ke SOI ke andar, target planet central body hai.

Figure — Patched conic method — interplanetary trajectory design

SOI radius mein woh $2/5$ power KYU hai? (Derivation)

Formula

$r_{S O I}$ nikalna Spacecraft ko planet ke paas consider karo, usase $r$ distance pe, aur Sun se $\approx a_{p}$ distance pe. Do frames mein Sun vs planet ki gravity ko primary vs perturbing ke roop mein compare karo.

Frame A (planet-centered): planet primary hai → uska acceleration $\sim \frac{G m _{p}}{r ^{2}}$ ; Sun ek perturbation hai → tidal-type disturbing acceleration $\sim \frac{G m _{⊙} r}{a _{p}^{3}}$ . Perturbation ratio: $A_{A} = \frac{G m _{⊙} r / a _{p}^{3}}{G m _{p} / r ^{2}} = \frac{m _{⊙}}{m _{p}} \frac{r ^{3}}{a _{p}^{3}}$ .

Frame B (Sun-centered): Sun primary hai → $\sim \frac{G m _{⊙}}{a _{p}^{2}}$ ; planet perturbation hai → $\sim \frac{G m _{p}}{r ^{2}}$ . Perturbation ratio: $A_{B} = \frac{G m _{p} / r ^{2}}{G m _{⊙} / a _{p}^{2}} = \frac{m _{p}}{m _{⊙}} \frac{a _{p}^{2}}{r ^{2}}$ .

SOI boundary woh jagah hai jahan planet ko primary maanna utna hi achha hai jitna Sun ko primary maanna — dono disturbance ratios ko equal rakho: $A_{A} = A_{B}$ . $\frac{m _{⊙}}{m _{p}} \frac{r ^{3}}{a _{p}^{3}} = \frac{m _{p}}{m _{⊙}} \frac{a _{p}^{2}}{r ^{2}}$ $r^{5} = a_{p}^{5} (\frac{m _{p}}{m _{⊙}})^{2} \Rightarrow r = a_{p} (\frac{m _{p}}{m _{⊙}})^{2/5} ■$

YEH kyun sahi lagta hai: planet ke paas ( $r$ chota) $A_{A} \to 0$ (planet clearly jeet ta hai); door jaane pe $A_{B} \to 0$ (Sun clearly jeetta hai). Yahi crossover SOI hai.

Transfer design KAISE karte hain: heliocentric leg

Departure/arrival hyperbola KAISE design karte hain

Formula Parking orbit se escape

Earth ke SOI ke andar ( $μ_{E} = G m_{E}$ ), craft ek hyperbola pe hai jiska excess speed $v_{\infty}$ hai. Hyperbola pe energy hai $ε = \frac{v ^{2}}{2} - \frac{μ _{E}}{r} = \frac{v _{\infty}^{2}}{2} (> 0) .$ Perigee radius $r_{p}$ (parking orbit) pe required speed hai $v_{p} = v_{\infty}^{2} + \frac{2 μ _{E}}{r _{p}} .$ Speed $v_{c i r c} = μ_{E} / r_{p}$ ki circular parking orbit se injection burn (departure $Δ v$ ) hai $Δ v = v_{p} - v_{c i r c} = v_{\infty}^{2} + \frac{2 μ _{E}}{r _{p}} - \frac{μ _{E}}{r _{p}}$

$2 μ_{E} / r_{p}$ KYU hai? $r_{p}$ se infinity tak escape karne ke liye $v_{esc} = 2 μ_{E} / r_{p}$ chahiye ( $v_{\infty} = 0$ ke liye). Infinity pe leftover speed $v_{\infty}$ bhi rakhne ke liye, energies add hoti hain: $v_{p}^{2} = v_{esc}^{2} + v_{\infty}^{2}$ . Yeh sirf energy bookkeeping hai — "Oberth effect" baked in hai: deep mein ( $r_{p}$ chota) burn karne se $v_{\infty}$ sasta pad ta hai.

Worked Example 1 — Earth→Mars Hohmann heliocentric speeds

Use $μ_{⊙} = 1.327 \times 1 0^{20} m^{3} / s^{2}$ , $r_{1} = 1.496 \times 1 0^{11}$ m, $r_{2} = 2.279 \times 1 0^{11}$ m.

$a_{t} = \frac{r _{1} + r _{2}}{2} = 1.888 \times 1 0^{11}$ m. Kyun? Ellipse dono orbits ko touch karti hai.
$v_{c, 1} = μ_{⊙} / r_{1} = 1.327 \times 1 0^{20} /1.496 \times 1 0^{11} \approx 2.978 \times 1 0^{4}$ m/s. Kyun? Earth ki circular speed baseline set karti hai.
$v_{t, 1} = μ_{⊙} (2/ r_{1} - 1/ a_{t}) \approx 3.279 \times 1 0^{4}$ m/s. Kyun? Transfer ke perihelion pe vis-viva.
$v_{\infty}^{d e p} = v_{t, 1} - v_{c, 1} \approx 2.94 \times 1 0^{3}$ m/s ≈ 2.94 km/s. Kyun? Woh speed jo departure hyperbola ko deliver karni hai.

Worked Example 2 — 200 km LEO se Departure $Δ v$

$μ_{E} = 3.986 \times 1 0^{14}$ m³/s², $r_{p} = R_{E} + 200 km = 6.578 \times 1 0^{6}$ m, $v_{\infty} = 2.94$ km/s.

$v_{c i r c} = μ_{E} / r_{p} = 3.986 \times 1 0^{14} /6.578 \times 1 0^{6} \approx 7.784$ km/s. Kyun? Parking-orbit speed.
$v_{p} = v_{\infty}^{2} + 2 μ_{E} / r_{p} = 294 0^{2} + 2 (3.986 \times 1 0^{14}) /6.578 \times 1 0^{6}$ $= 8.64 \times 1 0^{6} + 1.212 \times 1 0^{8} \approx 1.140 \times 1 0^{4}$ m/s ≈ 11.40 km/s. Kyun? Hyperbolic perigee speed.
$Δ v = 11.40 - 7.784 \approx 3.62$ km/s. Kyun? Injection burn.

Worked Example 3 — Transfer time

Ellipse ki period ka aadha: $t = π a_{t}^{3} / μ_{⊙}$ . Kyun? Hohmann aadha orbit hota hai. $t = π (1.888 \times 1 0^{11})^{3} /1.327 \times 1 0^{20} \approx 2.24 \times 1 0^{7}$ s ≈ 259 days.

Common mistake Classic errors ko steel-man karna

Error 1: " $Δ v$ nikalne ke liye bas $v_{\infty}$ ko $v_{c i r c}$ mein add kar do." Kyun sahi lagta hai: ek hi direction mein velocities add honi chahiye. Fix: $v_{\infty}$ infinity pe measure hoti hai, lekin tum perigee pe burn karte ho jahan craft fast hoti hai. Energy (velocity nahi) coast ke along conserved hai: $v_{p} = v_{\infty}^{2} + v_{esc}^{2}$ , isliye $Δ v = v_{p} - v_{c i r c}$ , jo $v_{\infty} + (v_{esc} - v_{c i r c})$ se kam hai. Deep burns saste hote hain (Oberth).

Error 2: " $v_{\infty}$ = spacecraft ki heliocentric speed." Kyun sahi lagta hai: dono "nikalte waqt ki speed" lagte hain. Fix: $v_{\infty}$ planet ke relative hoti hai. Heliocentric transfer speed $v_{c, 1} \pm v_{\infty}$ hai — tumhe planet ki orbital velocity vector-add karni padegi.

Error 3: SOI ko ek real hard wall maanna. Kyun sahi lagta hai: math wahan central bodies switch karta hai. Fix: yeh ek approximation hai; asli field continuous hai. Patched conics ek zabardast first guess deta hai, phir tum numerically refine karte ho.

Recall Feynman: 12-saal ke bachhe ko explain karo

Socho ek fast-ghoomte merry-go-round (Earth) se ek aur merry-go-round (Mars) pe ball phenko jo ek lamppost (Sun) ke around ghoom raha hai. Lamppost, apna merry-go-round, aur doosra — teeno ek saath sochna bahut mushkil hai. Isliye hum trip ko chapters mein kaatte hain: Chapter 1 — sirf tumhara merry-go-round matter karta hai, tum dodte ho aur choddte ho (ek curved escape path). Chapter 2 — ab sirf lamppost matter karta hai, ball Mars ki taraf ek lambe looping arc pe swing karti hai. Chapter 3 — pakad te waqt sirf Mars ka merry-go-round matter karta hai. Har chapter ek easy shape hai jo hum pehle se draw karna jaante hain. Phir hum chapters ko aise jod te hain ki ball kabhi jump na kare. Yahi jodhna "patch" hai.

Connections

Vis-viva equation — yahan har speed ka engine.
Hohmann transfer orbit — sabse sasta heliocentric ellipse.
Hyperbolic escape trajectories — departure/arrival legs.
Oberth effect — deep burns efficient kyun hote hain.
Sphere of influence — woh boundary jo har patch define karti hai.
Kepler's laws — guarantee karti hain ki har patch ek conic hai.
Launch windows & synodic period — jab Mars sahi jagah ho.

Patched conic method ki core assumption

Kisi bhi instant pe sirf ek body ki gravity dominate karti hai, isliye motion ek two-body conic hai.

Sphere-of-influence radius ka formula

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

2/5 power kyun hai?

Planet-frame aur Sun-frame perturbation ratios ko equal set karne pe

r^{5} = a_{p}^{5} (m_{p} / m_{⊙})^{2}

milta hai.

Vis-viva equation

v = μ (2/ r - 1/ a)

, energy

\frac{1}{2} v^{2} - μ / r = - μ /2 a

se.

Hohmann semi-major axis

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

Hyperbolic excess speed

v_{\infty}

ka matlab

SOI boundary pe planet ke relative speed (planet frame mein

r \to \infty

jaane pe).

Departure hyperbola pe perigee speed

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

Departure injection

Δ v

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

Δ v < v_{\infty} + (v_{esc} - v_{c i r c})

kyun hai?

Oberth effect: energy conservation ka matlab hai ki well ke deep mein burn karne se

v_{\infty}

sasta pad ta hai.

Earth→Mars ke liye heliocentric departure excess

v_{\infty} = ∣ v_{t, 1} - v_{c, 1} ∣ \approx 2.94

km/s.

Hohmann transfer time

t = π a_{t}^{3} / μ_{⊙}

(ellipse period ka aadha), Earth→Mars ke liye ~259 days.

Kya SOI ek real physical wall hai?

Nahi — yeh ek approximation hai; asli gravity field continuous hai.