Patched conic method — interplanetary trajectory design: Exercises

3.2.26 · D4 · Physics › Orbital Mechanics & Astrodynamics › Patched conic method — interplanetary trajectory design

Is poore page mein jo constants chahiye (wahi jo parent note mein hain parent note):

Yahan $μ$ (Greek letter "mu") $G \times (mass of the central body)$ ka shorthand hai — yeh us body ki gravitational strength hai jise hum orbit kar rahe hain. $μ_{⊙}$ Sun ke liye, $μ_{E}$ Earth ke liye, $μ_{M}$ Mars ke liye likhte hain. Neeche har formula parent note mein banaye gaye chaar tools mein se ek hai:

vis-viva $v = μ (\frac{2}{r} - \frac{1}{a})$ — semi-major axis $a$ wali orbit par kisi bhi distance $r$ par speed deta hai. Dekho Vis-viva equation.
circular speed $v_{c} = μ / r$ — vis-viva ka special case jab $a = r$ ho.
hyperbolic perigee speed $v_{p} = v_{\infty}^{2} + 2 μ / r_{p}$ — woh speed jo low position par chahiye taaki SOI edge par leftover speed $v_{\infty}$ ke saath pahuncho. Dekho Hyperbolic escape trajectories.
SOI radius $r_{S O I} = a_{p} (m_{p} / m_{⊙})^{2/5}$ . Dekho Sphere of influence.

Level 1 — Recognition

L1.1 — Earth→Mars trip ke teen patches mein se har ek mein spacecraft kis conic section par sawari karta hai, aur har ek ka central body kaun sa hai — yeh batao.

L1.2 — Equation $v = μ (\frac{2}{r} - \frac{1}{a})$ mein $v$ , $μ$ , $r$ , $a$ mein se har ek ka physical matlab kya hai? Trip ke kis patch mein $μ = μ_{⊙}$ use hota hai?

Recall Solution L1.1 & L1.2

L1.1

Patch 1 = departure hyperbola, central body = Earth.
Patch 2 = transfer ellipse (Hohmann), central body = Sun.
Patch 3 = arrival hyperbola, central body = Mars. Hyperbola ek open curve hoti hai (energy $> 0$ , escape ho jaata hai); ellipse closed hoti hai (energy $< 0$ , bound rehta hai). Dekho Hohmann transfer orbit.

L1.2

$v$ = spacecraft ki us waqt ki speed.
$μ$ = $G \times M_{central}$ , us body ki pull-strength jise tum orbit kar rahe ho.
$r$ = central body se current distance.
$a$ = orbit ka semi-major axis (sabse lambe diameter ka aadha) — yeh total energy fix karta hai. Heliocentric transfer ellipse (patch 2) $μ = μ_{⊙}$ use karta hai, kyunki wahaan Sun central hai.

Level 2 — Application

L2.1 — Mars ki circular heliocentric speed compute karo $v_{c, 2} = μ_{⊙} / r_{2}$ .

L2.2 — Transfer ellipse ki aphelion speed compute karo $v_{t, 2} = μ_{⊙} (\frac{2}{r _{2}} - \frac{1}{a _{t}})$ , jahan $a_{t} = \frac{r _{1} + r _{2}}{2}$ .

L2.3 — Upar ke dono se arrival excess speed nikalo $v_{\infty}^{a r r} = ∣ v_{c, 2} - v_{t, 2} ∣$ jise Mars arrival hyperbola ko absorb karna hoga.

Recall Solution L2.1–L2.3

L2.1 Mars ki circular speed: $v_{c, 2} = \frac{1.327 \times 1 0 ^{20}}{2.279 \times 1 0 ^{11}} = 5.823 \times 1 0^{8} \approx 2.413 \times 1 0^{4} \frac{m}{s} = 24.13 \frac{km}{s} .$ Kyun: Mars Earth se slower move karta hai (zyada door ⇒ weak pull ⇒ gentle speed), isliye yeh Earth ki 29.78 km/s se kam hai.

L2.2 Pehle $a_{t} = \frac{1.496 + 2.279}{2} \times 1 0^{11} = 1.8875 \times 1 0^{11}$ m. Phir aphelion par $r = r_{2}$ : $v_{t, 2} = 1.327 \times 1 0^{20} (\frac{2}{2.279 \times 1 0 ^{11}} - \frac{1}{1.8875 \times 1 0 ^{11}}) .$ Bracket ke andar: $8.776 \times 1 0^{- 12} - 5.298 \times 1 0^{- 12} = 3.478 \times 1 0^{- 12}$ . Multiply karo: $1.327 \times 1 0^{20} \times 3.478 \times 1 0^{- 12} = 4.615 \times 1 0^{8}$ . Toh $v_{t, 2} = 4.615 \times 1 0^{8} \approx 2.148 \times 1 0^{4} \frac{m}{s} = 21.48 \frac{km}{s} .$ $v_{c, 2}$ se slower kyun? Aphelion par transfer ship apni climb ke top par hoti hai — usne kinetic energy ko potential energy mein convert kar diya hai, isliye woh dawdle kar rahi hai, wahan circle par rehne wali body se slower.

L2.3 Kyunki ship Mars se slower arrive karti hai, Mars peeche se pakad leta hai: $v_{\infty}^{a r r} = ∣24.13 - 21.48∣ = 2.65 \frac{km}{s} .$ Yeh woh speed hai jis par ship Mars ke SOI mein Mars ke relative enter karti hai, yaani arrival hyperbola ki hyperbolic excess.

Level 3 — Analysis

L3.1 — $r_{S O I} = r_{2} (μ_{M} / μ_{⊙})^{2/5}$ use karke Mars ka sphere-of-influence radius compute karo (masses ratio $μ_{M} / μ_{⊙} = m_{M} / m_{⊙}$ ke through cancel ho jaate hain). Phir isse Mars ke radius $R_{M}$ se compare karo aur comment karo ki "point-planet" approximation fine kyun hai.

L3.2 — 200 km LEO se departure $v_{\infty}$ 2.94 km/s tha jis se $Δ v \approx 3.62$ km/s mila (parent Example 2). Agar hum instead higher parking orbit se $r_{p} = 20, 000$ km par depart karein toh? Naya $Δ v$ compute karo aur Oberth effect use karke explain karo ki yeh kyon us tarah se badla.

Recall Solution L3.1 & L3.2

L3.1 $\frac{μ _{M}}{μ _{⊙}} = \frac{4.283 \times 1 0 ^{13}}{1.327 \times 1 0 ^{20}} = 3.228 \times 1 0^{- 7} .$ Raise to the $2/5$ power. Take logs: $ln (3.228 \times 1 0^{- 7}) = - 14.948$ ; times $0.4$ gives $- 5.979$ ; exponentiate: $e^{- 5.979} = 2.532 \times 1 0^{- 3}$ . Phir $r_{S O I} = 2.279 \times 1 0^{11} \times 2.532 \times 1 0^{- 3} \approx 5.77 \times 1 0^{8} m \approx 577, 000 km .$ $R_{M} = 3.39 \times 1 0^{6}$ m se compare karo: SOI Mars ke radius se lagbhag $\frac{5.77 \times 1 0 ^{8}}{3.39 \times 1 0 ^{6}} \approx 170$ guna bada hai. Comment: SOI planet se bahut bada hai, lekin $r_{2}$ ke compare mein tiny hai ( $577, 000$ km vs $228$ million km ≈ $0.25%$ ). Toh ship essentially Mars ki position par Sun frame mein SOI enter karti hai — yahi wajah hai ki hum position ko continuously "patch" kar sakte hain aur bas Mars ki orbital velocity add karni hoti hai.

L3.2 $r_{p} = 2.0 \times 1 0^{7}$ m aur $v_{\infty} = 2.94 \times 1 0^{3}$ m/s ke saath: $v_{c i r c} = \frac{3.986 \times 1 0 ^{14}}{2.0 \times 1 0 ^{7}} = 1.993 \times 1 0^{7} \approx 4.464 \times 1 0^{3} \frac{m}{s} .$ $v_{p} = v_{\infty}^{2} + \frac{2 μ _{E}}{r _{p}} = (2940)^{2} + \frac{2 ( 3.986 \times 1 0 ^{14} )}{2.0 \times 1 0 ^{7}} = 8.644 \times 1 0^{6} + 3.986 \times 1 0^{7} .$ $= 4.850 \times 1 0^{7} \approx 6.964 \times 1 0^{3} \frac{m}{s} .$ $Δ v = 6.964 - 4.464 = 2.50 \frac{km}{s} .$ Ruko — yeh 3.62 km/s se kam hai? Dobara dekho: burn sasta ho gaya km/s mein? Nahin — yeh ulta lesson hai. Oberth effect kehta hai ki deeper (smaller $r_{p}$ ) burn karna $v_{\infty}$ ko sasta banata hai. Yahan hum higher gaye ( $r_{p}$ bada). High orbit par $v_{c i r c}$ aur $v_{p}$ dono shrink karte hain, lekin gap $Δ v$ bhi numerically yahan shrink karta hai kyunki well shallow hai aur $v_{\infty}$ kam dominate karta hai... dhyan raho: har m/s burn mein deliver hoti useful energy compare karo. Deep burns, burn- $Δ v$ ko bahut zyada $v_{\infty}$ mein convert karte hain. $20, 000$ km par raw $Δ v$ number ka chhota hona Oberth ko contradict nahin karta — Oberth efficiency ke baare mein hai (energy per burn), aur fairly compare karne ke liye tumhe parking orbit raise karne ka $Δ v$ cost bhi account karna hoga. Woh raising cost add karne par, low orbit overall win karta hai.

Level 4 — Synthesis

L4.1 — Full Mars arrival capture budget. Ship Mars ke SOI mein $v_{\infty}^{a r r} = 2.65$ km/s ke saath enter karti hai (L2.3 se). Woh $r_{p} = R_{M} + 500 km = 3.890 \times 1 0^{6}$ m par circular capture orbit aim karta hai. Compute karo: (a) hyperbolic periapsis speed $v_{p}$ , (b) wahan chahiye final circular speed $v_{c i r c}$ , (c) capture burn $Δ v_{c a p} = v_{p} - v_{c i r c}$ (ek retro-burn, toh yeh speed remove karta hai).

L4.2 — L4.1 ka capture $Δ v$ ko parent ke departure $Δ v = 3.62$ km/s mein add karo taaki impulsive Hohmann Earth→Mars trip ka rough total mission $Δ v$ mile (plane changes aur mid-course corrections ignore karte hue).

Recall Solution L4.1 & L4.2

L4.1(a) Hyperbolic periapsis speed Mars ke $μ_{M}$ use karta hai: $v_{p} = (v_{\infty}^{a r r})^{2} + \frac{2 μ _{M}}{r _{p}} = (2650)^{2} + \frac{2 ( 4.283 \times 1 0 ^{13} )}{3.890 \times 1 0 ^{6}} .$ $= 7.0225 \times 1 0^{6} + 2.2021 \times 1 0^{7} = 2.9044 \times 1 0^{7} \approx 5.389 \times 1 0^{3} \frac{m}{s} = 5.39 \frac{km}{s} .$ Kyun: ek slow SOI entry bhi (2.65 km/s) Mars ke gravity well mein girte girte enormously speed up ho jaati hai — $2 μ_{M} / r_{p}$ term dominate karta hai.

L4.1(b) Wahan circular speed: $v_{c i r c} = \frac{μ _{M}}{r _{p}} = \frac{4.283 \times 1 0 ^{13}}{3.890 \times 1 0 ^{6}} = 1.1011 \times 1 0^{7} \approx 3.318 \times 1 0^{3} \frac{m}{s} = 3.32 \frac{km}{s} .$

L4.1(c) Capture burn (retro): $Δ v_{c a p} = v_{p} - v_{c i r c} = 5.389 - 3.318 = 2.07 \frac{km}{s} .$ Yeh burn periapsis par 2.07 km/s remove karta hai taaki fly-by hyperbola se bound circle mein aa sako.

L4.2 Rough total: $Δ v_{t o t a l} \approx 3.62 + 2.07 = 5.69 \frac{km}{s} .$ Yahi classic "≈ 5.7 km/s in-space budget" hai impulsive Hohmann Mars orbiter ke liye (atmospheric-drag aerobraking se pehle, jo capture cost ko dramatically slash kar sakta hai).

Level 5 — Mastery

L5.1 — Venus flyby departure leg scratch se design karo. Venus $r_{V} = 1.082 \times 1 0^{11}$ m par orbit karta hai. Tum ek Hohmann transfer Earth→Venus chahte ho (ek inward transfer). Order mein nikalo: (a) Earth→Venus ellipse ke liye $a_{t}$ , (b) Earth par transfer speed (is ellipse ka aphelion, kyunki Venus Earth ki orbit ke andar hai) $v_{t, 1} = μ_{⊙} (\frac{2}{r _{1}} - \frac{1}{a _{t}})$ , (c) Earth ki circular speed $v_{c, 1} = μ_{⊙} / r_{1}$ , (d) departure excess $v_{\infty}^{d e p} = ∣ v_{t, 1} - v_{c, 1} ∣$ , aur Earth ki motion ke relative iska direction batao.

L5.2 — Earth→Venus transfer time $t = π a_{t}^{3} / μ_{⊙}$ days mein compute karo.

Geometry seedha rakhne ke liye neeche ke figures use karo.

Figure — Patched conic method — interplanetary trajectory design

Recall Solution L5.1 & L5.2

L5.1(a) Ek inward Hohmann ke liye ellipse Earth ki orbit ko touch karta hai (iska aphelion, door wala point) aur Venus ki orbit ko (iska perihelion, paas wala point): $a_{t} = \frac{r _{1} + r _{V}}{2} = \frac{1.496 \times 1 0 ^{11} + 1.082 \times 1 0 ^{11}}{2} = 1.289 \times 1 0^{11} m .$

L5.1(b) Earth par hum aphelion par hain ( $r = r_{1}$ ): $v_{t, 1} = 1.327 \times 1 0^{20} (\frac{2}{1.496 \times 1 0 ^{11}} - \frac{1}{1.289 \times 1 0 ^{11}}) .$ Bracket: $1.3369 \times 1 0^{- 11} - 7.7580 \times 1 0^{- 12} = 5.611 \times 1 0^{- 12}$ . Product: $1.327 \times 1 0^{20} \times 5.611 \times 1 0^{- 12} = 7.446 \times 1 0^{8}$ . $v_{t, 1} = 7.446 \times 1 0^{8} \approx 2.729 \times 1 0^{4} \frac{m}{s} = 27.29 \frac{km}{s} .$

L5.1(c) Earth ki circular speed (parent jaisi hi): $v_{c, 1} = \frac{1.327 \times 1 0 ^{20}}{1.496 \times 1 0 ^{11}} \approx 2.978 \times 1 0^{4} \frac{m}{s} = 29.78 \frac{km}{s} .$

L5.1(d) Departure excess: $v_{\infty}^{d e p} = ∣27.29 - 29.78∣ = 2.49 \frac{km}{s} .$ Direction: transfer ke liye ship ko Sun frame mein Earth se slower hona chahiye ( $v_{t, 1} < v_{c, 1}$ ), isliye $v_{\infty}^{d e p}$ Earth ki orbital velocity ke opposite point karta hai (ek "retrograde" departure). Tum Earth ki motion ke against brake karte ho taaki Sun ki taraf Venus ki direction mein giro — yeh outward Mars case ka mirror image hai jahan tum speed up karte ho.

L5.2 Transfer time: $t = π \frac{( 1.289 \times 1 0 ^{11} ) ^{3}}{1.327 \times 1 0 ^{20}} = π \frac{2.142 \times 1 0 ^{33}}{1.327 \times 1 0 ^{20}} = π 1.614 \times 1 0^{13} .$ $= π \times 4.018 \times 1 0^{6} \approx 1.262 \times 1 0^{7} s \approx 146 days .$ 259-day Mars trip se chhota, kyunki inward ellipse chhoti hai. Aise window kab khulte hain, yeh dekhne ke liye Launch windows & synodic period dekho.