Worked examples — Patched conic method — interplanetary trajectory design

3.2.26 · D3 · Physics › Orbital Mechanics & Astrodynamics › Patched conic method — interplanetary trajectory design

Ye page ek drill ground hai. Parent note ne machinery banayi thi: SOI radii, Hohmann ellipses, vis-viva speed formula, aur hyperbolic escape jisme Oberth effect bhi shaamil hai. Yahan hum un sab tools ko har tarah ke input se guzaarte hain — inner vs outer target, zero excess speed, degenerate "same orbit" case, ek real launch-window word problem, aur ek exam twist.

Neeche sab kuch sirf wahi symbols use karta hai jo parent note ne define kiye hain:

• $\mu$ = $G$ times central body ki mass (Sun-frame ${\mu }_{\odot }$, Earth-frame ${\mu }_E$) — "pull kitna strong hai."
• $r$ = central body ke centre se doori. $a$ = semi-major axis (ellipse ki lambi width ka aadha).
• $v_c=\sqrt{\mu /r}$ = circular-orbit speed. $v_{\infty }$ = ek planet se door leftover speed us planet ke relative.

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The scenario matrix

Is topic ke har problem is grid ka ek cell hai. Baad ke worked examples har cell ko tag karte hain, toh ant tak koi cell uncovered nahi rehti.

Cell	Kya alag hai	Covered by
A. Inner → outer ($r_2>r_1$)	bahar jaane ke liye speed up; $v_{t,1}>v_{c,1}$	Ex 1
B. Outer → inner ($r_2<r_1$)	andar girne ke liye slow down; $v_{t,1}<v_{c,1}$, sign flip	Ex 2
C. Zero / degenerate ($r_2=r_1$)	limiting case, $v_{\infty }\to 0$	Ex 3
D. Escape burn from a well	$\Delta v$ with Oberth, deep vs shallow parking orbit	Ex 4
E. Limiting $v_{\infty }\to 0$	burn pure escape speed tak reduce ho jaati hai	Ex 4 (part c)
F. Arrival side (capture)	$v_{\infty }$ target SOI mein, capture burn	Ex 5
G. SOI size / sanity	kya "tiny SOI" assumption fair hai?	Ex 6
H. Real-world word problem	launch timing, phase angle	Ex 7
I. Exam twist	given $\Delta v$ budget, ulta solve karo $v_{\infty }$ ke liye	Ex 8

Constants jo puri jagah use honge (SI): ${\mu }_{\odot }=1.327\times 10^{20}$, ${\mu }_E=3.986\times 10^{14}$, ${\mu }_{Mars}=4.283\times 10^{13}$, $r_{Earth}=1.496\times 10^{11}$, $r_{Mars}=2.279\times 10^{11}$, $r_{Venus}=1.082\times 10^{11}$, $R_E=6.378\times 10^6$ (sab metres, m³/s²).

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Ex 1 — Inner → outer (Cell A)

Forecast: Mars zyada bahar hai, toh wahan pahunchne ke liye spacecraft ko departure par Earth se tez hona chahiye. Guess: $v_{\infty }$ kuch km/s, positive.

1. $a_t=\frac{r_1+r_2}{2}=\frac{1.496+2.279}{2}\times 10^{11}=1.888\times 10^{11}$ m. Ye step kyun? Sabse sasta (Hohmann) ellipse dono circular orbits ko bas chhoota hai: perihelion $r_1$ par, aphelion $r_2$ par. Iski half-width average hai.
2. $v_{c,1}=\sqrt{{\mu }_{\odot }/r_1}=\sqrt{1.327\times 10^{20}/1.496\times 10^{11}}=2.978\times 10^4$ m/s. Ye step kyun? Ye Earth ki apni Sun ke around speed hai — woh baseline jisse hum compare karte hain.
3. $v_{t,1}=\sqrt{{\mu }_{\odot }\left(\frac{2}{r_1}-\frac{1}{a_t}\right)}=3.279\times 10^4$ m/s. Ye step kyun? Vis-viva transfer-ellipse speed $v_t$ deta hai uske perihelion par, jo bilkul Earth ki orbit par baithta hai.
4. $v_{\infty }^{dep}=v_{t,1}-v_{c,1}=3.279-2.978=0.301\times 10^4\approx 3.01$ km/s (positive). Ye step kyun? Ye difference woh extra Sun-frame speed hai jo departure hyperbola inject karna chahiye. Positive ⇒ hum speed up karte hain. Figure mein red gap dekho.

Recall Verify

Units: $\sqrt{({\text{m}}^3/{\text{s}}^2)/\text{m}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=$ m/s ✓. Sanity: $v_{t,1}>v_{c,1}$ kyunki perihelion par ek ellipse sabse tez chalti hai, aur ye ellipse Mars tak jaati hai (Kepler's 2nd law). Positive $v_{\infty }$ confirm karta hai "outer targets ke liye speed up karo."

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Ex 2 — Outer → inner (Cell B: the sign flip)

Forecast: Venus Sun ke zyada kareeb hai. Andar girne ke liye spacecraft ko Earth ke relative slow down karna hoga. Guess: $v_{\infty }$ Earth ki motion ke opposite point karta hai — lekin iski magnitude hi woh hai jo burn supply karta hai.

1. $a_t=\frac{1.496+1.082}{2}\times 10^{11}=1.289\times 10^{11}$ m. Kyun? Ab Earth ki orbit transfer ellipse ka aphelion hai (door wala point), Venus perihelion.
2. $v_{c,1}=2.978\times 10^4$ m/s (same Earth baseline jaise Ex 1). Kyun? Departure planet abhi bhi Earth hai.
3. $v_{t,1}=\sqrt{{\mu }_{\odot }\left(\frac{2}{r_1}-\frac{1}{a_t}\right)}=2.734\times 10^4$ m/s. Kyun? Vis-viva transfer-ellipse speed $v_t$ deta hai Earth ki orbit par, jo ab ellipse ka sabse slow point (aphelion) hai. Toh $v_{t,1}<v_{c,1}$.
4. $v_{\infty }^{dep}=v_{t,1}-v_{c,1}=2.734-2.978=-0.244\times 10^4=-2.44$ km/s. Kyun? Negative sign = transfer speed Earth ki speed se kam hai. Physically excess velocity vector Earth ki motion ke peeche point karta hai. Rocket ko jo burn deliver karna hai uski magnitude $|v_{\infty }|=2.44$ km/s hai — Earth ki orbital velocity ke relative retrograde.

Recall Verify

Units m/s ✓. Sanity: outer target (Ex 1) ne $+3.01$ diya, inner target ne $-2.44$ diya — opposite signs, exactly expected slow-down ✓. Dono sirf kuch km/s hain $\sim$30 km/s orbital speed ke against, toh Hohmann ek badi orbit par ek choti si nudge hai ✓.

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Ex 3 — Degenerate case $r_2=r_1$ (Cell C)

Forecast: Same orbit ⇒ same speed ⇒ kuch change nahi karna. Guess $v_{\infty }=0$.

1. $a_t=\frac{r_1+r_1}{2}=r_1$. Kyun? Jab dono radii equal hain toh "ellipse" degenerate hokar circle ban jaata hai — semi-major axis radius ke equal hoti hai.
2. $v_{t,1}=\sqrt{{\mu }_{\odot }\left(\frac{2}{r_1}-\frac{1}{r_1}\right)}=\sqrt{{\mu }_{\odot }/r_1}=v_{c,1}$. Kyun? $a_t=r_1$ ke saath vis-viva circular-speed formula mein collapse ho jaata hai. Transfer speed $v_t$ hi circular speed hai.
3. $v_{\infty }^{dep}=v_{t,1}-v_{c,1}=0$. Kyun? Koi radius change nahi ⇒ koi energy change nahi ⇒ zero excess. Limiting case check out hota hai aur confirm karta hai formulas blow up nahi karti.

Recall Verify

$2/r_1-1/r_1=1/r_1$ exactly, toh $v_{t,1}=v_{c,1}$ identically ⇒ $v_{\infty }=0$. Koi units issue nahi (ye zero hai) ✓.

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Ex 4 — Escape burn with Oberth, deep vs shallow (Cells D & E)

Forecast: Well mein deep burn karna (chota $r_p$, fast local speed) zyada sasta hona chahiye per unit $v_{\infty }$ — Oberth effect. Guess: (a) ko (b) se kam $\Delta v$ chahiye, even though dono same $v_{\infty }$ add karte hain.

Part (a) — deep (LEO):

1. $v_{circ}=\sqrt{{\mu }_E/r_p}=\sqrt{3.986\times 10^{14}/6.578\times 10^6}=7.784$ km/s. Kyun? Wo speed jo tumhare paas circle karte waqt already hai.
2. $v_p=\sqrt{v_{\infty }^2+2{\mu }_E/r_p}=\sqrt{3010^2+2(3.986\times 10^{14})/6.578\times 10^6}=1.145\times 10^4$ m/s $=11.45$ km/s. Kyun? Hyperbola par energy add hoti hai: escape energy ($2{\mu }_E/r_p$) plus leftover $v_{\infty }^2$. Yahan $v_p$ upar define ki gayi perigee speed hai.
3. $\Delta v_a=v_p-v_{circ}=11.45-7.784=3.67$ km/s. Kyun? Burn circular se hyperbolic-perigee speed tak ka jump hai.

Part (b) — shallow (high orbit): 4. $v_{circ}=\sqrt{{\mu }_E/r_p}=\sqrt{3.986\times 10^{14}/4.0\times 10^7}=3.157$ km/s. Kyun? Same circular-speed formula $v_{circ}=\sqrt{{\mu }_E/r_p}$ — bas bade radius par evaluate ki, toh kam nikli (cheezein upar slower orbit karti hain). 5. $v_p=\sqrt{v_{\infty }^2+2{\mu }_E/r_p}=\sqrt{3010^2+2(3.986\times 10^{14})/4.0\times 10^7}=5.38\times 10^3$ m/s $=5.38$ km/s. Kyun? Same hyperbolic-perigee energy formula $v_p=\sqrt{v_{\infty }^2+2{\mu }_E/r_p}$ — escape energy plus leftover $v_{\infty }^2$, ab zyada $r_p$ par jahan escape term $2{\mu }_E/r_p$ choti hai. 6. $\Delta v_b=5.38-3.157=2.23$ km/s. Kyun compare karein? Departure burn khud yahan chota hai ($2.23<3.67$) — lekin ye is high orbit tak pahunchne ki cost ignore karta hai, jo aage compute karenge.

Net-cost argument poora karna: parking orbit ko LEO ($r_{LEO}=6.578\times 10^6$ m) se $r_{high}=4.0\times 10^7$ m tak raise karne ka apna $\Delta v$ cost hai. Un dono circles ke beech Hohmann raise ka $a_{raise}=\frac{6.578\times 10^6+4.0\times 10^7}{2}=2.329\times 10^7\text{ m}.$

• Burn 1 (LEO chhodna): transfer perigee speed $v_{peri}=\sqrt{{\mu }_E(2/r_{LEO}-1/a_{raise})}=\sqrt{3.986\times 10^{14}(2/6.578\times 10^6-1/2.329\times 10^7)}=9.10\times 10^3=9.10$ km/s, toh $\Delta v_1=v_{peri}-v_{circ,LEO}=9.10-7.784=1.32$ km/s. Ye step kyun? Low circle se raising ellipse par jaana khud LEO ke perigee ($r_{LEO}$) par ek vis-viva speed jump hai; burn LEO circular speed ke upar woh jump hai.
• Burn 2 (high par circularise): transfer apogee speed $v_{apo}=\sqrt{{\mu }_E(2/r_{high}-1/a_{raise})}=\sqrt{3.986\times 10^{14}(2/4.0\times 10^7-1/2.329\times 10^7)}=1.496\times 10^3=1.50$ km/s, toh $\Delta v_2=v_{circ,high}-v_{apo}=3.157-1.50=1.66$ km/s. Ye step kyun? High circle par pahunchte waqt ellipse wahan circle se slow hai (apogee ellipse ka slowest point hai), toh ek prograde burn speed ko circular tak top up karta hai.
• Raising cost $\approx 1.32+1.66=2.98$ km/s.
• Total shallow path: $2.98+2.23=5.21$ km/s, direct deep burn $3.67$ km/s se bahut zyada.

Ye kyun important hai: jab tum wapas add karte ho upar jaane ki cost, shallow route bahut bura hai ($5.21$ vs $3.67$ km/s). Asli Oberth saving deep burn karne se aati hai jab tum pehle se wahan ho, pehle ek higher orbit par jaane se nahi.

Oberth efficiency comparison — har burn per km/s kharch karne par kitna $v_{\infty }$ milta hai:

Part (c) — limiting $v_{\infty }\to 0$: 7. (a) mein $v_{\infty }=0$ rakho: $v_p\to \sqrt{2{\mu }_E/r_p}=\sqrt{2}{v}_{circ}=11.01$ km/s, escape speed. Kyun? Zero leftover speed matlab "bas escape ho jao." Hyperbola parabola mein degenerate ho jaata hai, aur $\Delta v=\sqrt{2}{v}_{circ}-v_{circ}=0.414v_{circ}=3.22$ km/s. Ye kyun matter karta hai: general formula textbook escape burn mein reduce ho jaata hai — limit behave karta hai.

Recall Verify

Units m/s ✓. Deep-burn Oberth check: LEO par, $3.67$ km/s se $v_{\infty }=3.01$ milta hai; poora climb-and-go shallow route total $\approx 5.21$ km/s — bahut bura. Limit: $v_{\infty }=0$ ke saath, $v_p=\sqrt{2}{v}_{circ}$ exactly ✓.

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Ex 5 — Arrival side / capture (Cell F)

Forecast: Departure ka ulta lekin peeche ki taraf: hum speed shed karte hain hyperbola se bound circle mein girne ke liye. Guess $\Delta v$ order 1–2 km/s.

1. $v_p=\sqrt{v_{\infty }^2+2{\mu }_{Mars}/r_p}=\sqrt{2650^2+2(4.283\times 10^{13})/3.6\times 10^6}=5.55\times 10^3$ m/s $=5.55$ km/s. Kyun? Same hyperbolic-perigee energy formula — $v_p$ incoming hyperbola ka fastest (closest) point hai.
2. $v_{circ}=\sqrt{{\mu }_{Mars}/r_p}=\sqrt{4.283\times 10^{13}/3.6\times 10^6}=3.449$ km/s. Kyun? Wo circular orbit ki speed jis par hum end up karna chahte hain.
3. $\Delta v_{capture}=v_p-v_{circ}=5.55-3.449=2.10$ km/s (ek braking burn, retrograde). Kyun? Hum exactly utni kinetic energy remove karte hain ki open hyperbola ko perigee par ek closed circle mein convert kar sakein.

Recall Verify

Units m/s ✓. $v_p>v_{circ}$ (hyperbola hamesha same radius par circle se tez hota hai) toh ek positive braking magnitude — physical ✓. Order 2 km/s typical Mars-orbit-insertion budgets se match karta hai.

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Ex 6 — SOI size sanity check (Cell G)

Forecast: Patch method assume karta hai ki SOI orbit ke comparison mein negligibly chota hai, toh hum pretend kar sakte hain ki craft "Earth ki exact position se" nikla. Guess: ratio 1% se kaafi kam.

1. $r_{SOI}=a_p{\left(\frac{m_E}{m_{\odot }}\right)}^{2/5}=1.496\times 10^{11}\times (3.003\times 10^{-6}{)}^{0.4}$. Kyun? Ye woh crossover radius hai jo parent ne dono perturbation ratios barabar set karke derive ki thi.
2. $(3.003\times 10^{-6}{)}^{0.4}=6.17\times 10^{-3}$, toh $r_{SOI}=1.496\times 10^{11}\times 6.17\times 10^{-3}=9.24\times 10^8$ m. Kyun? Ek chote number ko $2/5$ power tak uthane se woh soft ho jaata hai — SOI $m_E/m_{\odot }$ akele se zyada badi hai lekin phir bhi choti hai.
3. Ratio $r_{SOI}/a_p=6.17\times 10^{-3}\approx 0.6\%$. Ye kyun matter karta hai: 1% se kaafi kam. Toh craft ko exactly Earth ki orbital position se nikla treat karne se sub-percent error aati hai — "tiny SOI" assumption justified hai.

Recall Verify

Units m ✓. Known Earth SOI $\approx 9.2\times 10^5$ km $=9.2\times 10^8$ m — step 2 se two figures tak match karta hai ✓. Ratio dimensionless, $0.6\%\ll 1$ ✓.

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Ex 7 — Real-world word problem: Mars ko pakadna (Cell H)

Forecast: Mars transfer ke arrival point se slower hai, toh launch ke waqt Mars arrival spot se aage hona chahiye, lekin poore 180° nahi. Guess: Mars Earth se roughly 40–50° aage.

1. Mars ki period Kepler's 3rd se: $T_{Mars}=2\pi \sqrt{r_2^3/{\mu }_{\odot }}=2\pi \sqrt{(2.279\times 10^{11}{)}^3/1.327\times 10^{20}}=5.94\times 10^7$ s $=687$ days. Kyun? Humein Mars ki angular speed chahiye taaki pata chale ki wo flight ke dauran kitna drift karta hai.
2. Mars 259 days mein kitna angle sweep karta hai: ${\theta }_{Mars}=360°\times \frac{259}{687}=135.7°$. Kyun? Angular speed $\times$ time = angle covered.
3. Probe Earth se $180°$ around ek point par pahunchta hai (Hohmann = half an ellipse). Kyun? Transfer ka aphelion launch point ke diametrically opposite baithta hai.
4. Required lead angle = $180°-{\theta }_{Mars}=180°-135.7°=44.3°$. Kyun? Mars ko bas bacha hua arc cover karna hai meeting point tak theek tab pahunchne ke liye jab probe pahunche. Toh launch ke waqt Mars Earth se 44.3° aage hona chahiye apni orbit mein — ye ek launch window condition hai.

Recall Verify

$360\times 259/687=135.7°$ ✓. Lead $=180-135.7=44.3°$, forecast 40–50° band mein ✓. Textbook Earth→Mars phase angle $\approx 44°$ hai ✓.

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Ex 8 — Exam twist: ulta solve karo (Cell I)

Forecast: Hum normally $v_{\infty }\to \Delta v$ jaate hain. Yahan hum invert karte hain: given $\Delta v$, $v_{\infty }$ nikalo, phir transfer ellipse nikalo. Expect karo $v_{\infty }$ thoda 3 km/s se kam (Ex 4a se, $3.67$ km/s ne $3.01$ diya tha).

1. $v_p=v_{circ}+\Delta v=7.784+3.60=11.384$ km/s. Kyun? $\Delta v=v_p-v_{circ}$ ka reverse: circular speed par burn wapas add karo perigee speed $v_p$ recover karne ke liye.
2. $v_{\infty }=\sqrt{v_p^2-2{\mu }_E/r_p}=\sqrt{11384^2-2(3.986\times 10^{14})/6.578\times 10^6}=2.86\times 10^3$ m/s $=2.86$ km/s. Kyun? Hyperbolic energy relation $v_p^2=v_{\infty }^2+2{\mu }_E/r_p$ ko invert karo escape term strip karne ke liye.
3. Heliocentric transfer speed at Earth: $v_{t,1}=v_{c,1}+v_{\infty }=29.78+2.86=32.64$ km/s. Kyun? Earth ki orbital velocity add karo (patch condition) — Sun-frame transfer-ellipse speed $v_t$ perihelion par.
4. Vis-viva se $a_t$ back out karo: $v_{t,1}^2={\mu }_{\odot }\left(\frac{2}{r_1}-\frac{1}{a_t}\right)\Rightarrow \frac{1}{a_t}=\frac{2}{r_1}-\frac{v_{t,1}^2}{{\mu }_{\odot }}$. Numerically $\frac{2}{r_1}=1.3369\times 10^{-11}$, $\frac{v_{t,1}^2}{{\mu }_{\odot }}=\frac{(32640{)}^2}{1.327\times 10^{20}}=8.029\times 10^{-12}$, toh $\frac{1}{a_t}=5.340\times 10^{-12}$, $a_t=1.873\times 10^{11}$ m. Kyun? Ye Ex 1 step 3 ka algebraic inverse hai — same equation, alag unknown.
5. Reachable aphelion: $r_{aph}=2a_t-r_1=2(1.873\times 10^{11})-1.496\times 10^{11}=2.250\times 10^{11}$ m. Kyun? Ek ellipse ka aphelion $2a_t$ minus perihelion $r_1$ hota hai. Ye Mars se thoda short hai ($2.279\times 10^{11}$), bata raha hai ki $3.60$ km/s LEO se Mars ki orbit tak lagbhag — lekin poori tarah nahi — pahunchta hai.

Recall Verify

Consistency: $2.86$ km/s ($3.60$ budget se $v_{\infty }$) $3.01$ km/s se thoda kam hai jiske liye $3.67$ chahiye tha — monotonic ✓. Aphelion $2.250\times 10^{11}$ m Mars ke $2.279\times 10^{11}$ m ke andar baitha hai, "3.60 km/s ek sach mein Mars Hohmann se thoda short hai" se match karta hai ✓.

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Recall Feynman: poori page ek 12-year-old ko explain karo

• $r_1$ aur $r_2$ ka kya matlab tha? ::: Sun se doori jahan se nikle aur jahan jaana hai.
• Outer target ko speed up kyun chahiye lekin inner target ko slow down? ::: Bahar chadhne ke liye zyada speed chahiye; andar girne ke liye speed shed karte hain.
• Gravity well mein deep burn efficient kyun hai? ::: Wahan tum pehle se tez chal rahe ho, toh same burn zyada usable energy add karta hai (Oberth).