Exercises — Pork chop plots — Δv vs launch - arrival date

• x-axis = launch date (Earth se departure).
• y-axis = arrival date (target planet par).
• Islands low-cost ke regions hain — sabse andar wala contour us launch opportunity ka ==minimum Δv (ya minimum $C_3$)== hai, yaani "sweet spot". Yahan compute karne ko kuch nahi hai; skill ye jaanna hai ki har ek cell ek Lambert problem hai aur poori picture bas uski cost ko ek grid par draw karti hai. Dekho Lambert's Problem.

$C_3\equiv v_{\infty }^2$ — ye definition hai, koi gravity involve nahi. $C_3=(3.2{)}^2=10.24\frac{{\text{km}}^2}{{\text{s}}^2}.$ Square kyun karte hain? $C_3$ infinity par unit mass ke relative energy hai ($\epsilon =v_{\infty }^2/2$, aur $C_3=2\epsilon$), aur launch vehicles ko energy delivered ke basis par rate kiya jaata hai, speed ke basis par nahi.

Step A — circular speed (jo tumhare paas already hai, aur $v_c$ ka matlab kya hai). Define karo ==$v_c$ = woh speed jo ek spacecraft ki hoti hai jab woh radius $r_p$ ki circular orbit par coast kar rahi hoti hai== — woh speed jis par woh kisi bhi burn se pehle already move kar rahi hoti hai. $v_c=\sqrt{{\mu }_E/r_p}$ kyun? Ek circle par, gravity exactly centripetal force supply karti hai: gravitational pull $={\mu }_{E}m/r_p^2$ centripetal requirement $m{v}_c^2/r_p$ ke equal honi chahiye. $m$ aur $r_p$ ki ek power cancel karo: $\frac{{\mu }_E}{r_p^2}=\frac{v_c^2}{r_p}\Rightarrow v_c^2=\frac{{\mu }_E}{r_p}\Rightarrow v_c=\sqrt{\frac{{\mu }_E}{r_p}}.$ Numerically: $v_c=\sqrt{\frac{398600}{6678}}=7.726\text{km/s}.$ Step B — departure hyperbola par perigee speed (jo tumhe chahiye). Ye formula kyun? Hyperbola par specific energy $\epsilon =v_{\infty }^2/2$ hai, aur energy $v_{peri}^2/2-{\mu }_E/r_p$ bhi hai. Inhe equal set karke solve karo: $v_{peri}=\sqrt{v_{\infty }^2+\frac{2{\mu }_E}{r_p}}=\sqrt{9+\frac{2(398600)}{6678}}=\sqrt{9+119.36}=11.33\text{km/s}.$ Step C — burn (tum sirf woh difference pay karte ho jo tumhe chahiye aur jo tumhare paas already hai uske beech): $\Delta v_{dep}=v_{peri}-v_c=11.33-7.726=3.61\text{km/s}.$ $+2\mu /r$ term kyun appear hota hai ye jaanne ke liye dekho Hyperbolic Excess Velocity & C3.

Wahi master formula, lekin ab Mars ke ${\mu }_M=42828$ aur $r_a=3689$ km ke saath. Circular speed $v_c=\sqrt{{\mu }_M/r_a}$ target circular orbit par speed hai (wahi balance-of-forces derivation jaise L2.1 mein, ab ${\mu }_M$ ke saath): $v_c=\sqrt{\frac{42828}{3689}}=3.408\text{km/s}.$ $v_{peri}=\sqrt{2.5^2+\frac{2(42828)}{3689}}=\sqrt{6.25+23.22}=\sqrt{29.47}=5.429\text{km/s}.$ $\Delta v_{arr}=5.429-3.408=2.02\text{km/s}.$ Ye kaisa dikhta hai: tum Mars se guzarte hue ek hyperbola par fast arrive karte ho, aur closest approach par burn tumhe neeche ek bound circle mein le aata hai — yahi retro-burn cost hai.

$\frac{1}{T_{syn}}=\left|\frac{1}{365.25}-\frac{1}{686.98}\right|=|0.0027379-0.0014556|=0.0012823{\text{d}}^{-1}.$ $T_{syn}=779.9\text{d}=2.14\text{yr}.$ Ye kyun matter karta hai: sasti geometry (planets nicely spaced near-Hohmann shot ke liye) har $\approx 26$ months mein recur hoti hai, isliye pork chops ek repeating family ke roop mein aate hain. Dekho Synodic Period.

$v_c=7.726$ km/s (dono ke liye same). $\Delta v_A=\sqrt{9+119.36}-7.726=11.33-7.726=3.607\text{km/s}.$ $\Delta v_B=\sqrt{12.25+119.36}-7.726=11.472-7.726=3.746\text{km/s}.$ A sasta hai $\approx 0.14$ km/s se. Notice karo $C_3$ $9.0$ se $12.25$ tak jump kiya (ek $36\%$ energy rise) lekin Δv sirf $\approx 4\%$ badha — $+2\mu /r$ term difference ko dilute karta hai. Ye burn ko well mein deep karne ka Oberth-style leverage hai; dekho Oberth Effect.

Mass ratio hai $R=e^{\Delta v/v_e}$. Dono cases ke beech factor hai $\frac{R_B}{R_A}=\exp \left(\frac{\Delta v_B-\Delta v_A}{v_e}\right)=\exp \left(\frac{0.139}{3.4}\right)=\exp (0.0409)=1.042.$ Ek Δv change ke liye jo chhoti lagti thi — $\sim 4.2\%$ bhaari stack — isliye mission designers innermost contour ke peeche bhagte hain. Dekho Tsiolkovsky Rocket Equation.

Pehle, transfer angle define karo. ==$\Delta \theta$ = woh heliocentric angle jo do radius vectors ke beech sweep hota hai== ko define karo, yaani Sun par ${\vec{r}}_1$ (launch par Earth) aur ${\vec{r}}_2$ (arrival par target) ke beech ka angle. Short-way ("Type I") transfer mein $\Delta \theta <180°$ hota hai; long-way ("Type II") transfer mein $\Delta \theta >180°$ hota hai. Ridge exactly $\Delta \theta =180°$ par hai.

Figure dekho. Sun center par hai; teal dot ${\vec{r}}_1$ (launch par Earth) hai aur orange dots do possible arrival points ${\vec{r}}_2$ hain. Plum arc 180° se kam sweep karta hai ($\Delta \theta <180°$, Type I); dotted arc 180° se zyada sweep karta hai ($\Delta \theta >180°$, Type II). Thick faded orange line woh special case hai jahan ${\vec{r}}_1$ aur ${\vec{r}}_2$ anti-parallel hain — ek seedha $\Delta \theta =180°$ sweep.

Δv wahan kyun spike karta hai — sahi reason. Transfer angle $\Delta \theta$ Lambert's geometry mein chord $c=|{\vec{r}}_2-{\vec{r}}_1|$ ke through enter karta hai aur, crucially, us angular momentum $\vec{h}=\vec{r}\times \vec{v}$ ke through jitna transfer conic ko carry karna hoga. Jab $\Delta \theta \to 180°$ hota hai, do radius vectors line up ho jaate hain, aur connecting ellipses ka family degenerate ho jaata hai: minimum-energy ellipse ka required semi-major axis $a$ us point ki taraf push karta hai jahan orbit almost ek seedha, radial "fall" ban jaati hai do points ke beech. Physically transverse (sideways) velocity jo available time mein corner round karne ke liye chahiye woh collapse ho jaati hai, toh solver ek extreme conic par force ho jaata hai jiske departure aur arrival speeds — aur isliye $v_{\infty }$ — rapidly badhte hain. Co-planar idealisation mein ye ek sharp energy spike ke roop mein dikhta hai; real 3-D problem mein do planet orbits thodi si tilted hain, isliye exactly $180°$ par transfer plane sirf weakly constrained hai aur ek modest tilt energy spike ke upar ek bade required plane change mein turn ho jaata hai. Kisi bhi tarah root cause degenerate transfer geometry hai jab $\Delta \theta \to 180°$ (angular momentum → uski singular value, $a$ apne extreme par jaata hai), na ki akela "kai planes through a line."

• Ridge ke neeche (shorter TOF): sweep $\Delta \theta <180°$ = Type I transfer.
• Ridge ke upar (longer TOF): sweep $\Delta \theta >180°$ = Type II transfer.

Edge case — multi-revolution ($N>0$) solutions. Upar sab kuch zero-revolution ($N=0$) family ke liye hai, jahan spacecraft ${\vec{r}}_1$ se ${\vec{r}}_2$ tak seedha jaata hai bina Sun ka poora loop complete kiye. Lekin lambe times of flight ke liye Lambert's equation extra solutions admit karta hai jisme spacecraft Sun ke $N=1,2,\dots$ full turns loop karta hai arrive karne se pehle. Har $N$ apna khud ka Type I / Type II branches ka pair kholta hai, toh ek wide TOF range par poore pork chop plot mein additional islands stacked at longer flight times dikh sakte hain, har ek ka apna ridge hota hai. Practical Earth–Mars plots usually $N=0$ family mein rehte hain (Sun ko loop karna propellant aur time dono waste karta hai), lekin ek complete treatment in multi-rev branches ko acknowledge karna chahiye taaki tum kabhi ek extra contour cluster se surprise na ho jo plot par upar dikh raha ho. Dekho Lambert's Problem.

Tum ek lobe choose karte ho aur ridge se door rehte ho. Dekho Lambert's Problem aur Hohmann Transfer (ideal $180°$ Hohmann case sirf isliye sasta hai kyunki woh coplanar by construction hai, tilt penalty se bachta hua).

Do burns add kyun kar sakte hain — patched-conic idea. Trip actually teen bodies (Earth, Sun, Mars) se govern hoti hai, jiska koi clean solution nahi hai. Patched Conic Approximation journey ko teen alag two-body problems mein cut karta hai: (1) Earth's gravity se escape hoti ek hyperbola, (2) Sun ke around ek ellipse, (3) Mars se captured hoti ek hyperbola. Inhe har planet ke sphere of influence ke edges par "patch" kiya jaata hai, jahan leftover speed exactly $v_{\infty }$ hoti hai. Kyunki har piece ek distinct burn hai ek distinct gravity well mein, unke Δv's add hote hain — lekin sirf tab, jab har ek apne well mein compute ho chuki hoti hai. Isliye hum departure aur arrival ko neeche independently treat kar sakte hain.

Departure (Earth, ${\mu }_E=398600$) — master burn formula, kyunki tum Earth's well se ek circular parking orbit se bahar climb kar rahe ho: $\Delta v_{dep}=\sqrt{3.2^2+\frac{2(398600)}{6678}}-\sqrt{\frac{398600}{6678}}=\sqrt{10.24+119.36}-7.726=11.384-7.726=3.658\text{km/s}.$ Arrival (Mars, ${\mu }_M=42828$) — wahi formula, ab Mars' well mein ek circular capture orbit mein drop karte hue: $\Delta v_{arr}=\sqrt{2.8^2+\frac{2(42828)}{3689}}-\sqrt{\frac{42828}{3689}}=\sqrt{7.84+23.22}-3.408=5.573-3.408=2.165\text{km/s}.$ Total (patched-conic sum): $\Delta v_{tot}=3.658+2.165=5.82\text{km/s}.$ Ye single number ek contour value hai — plot ye computation har (launch, arrival) cell par repeat karke bana hai.

Vis-viva kyun? Humein Sun-centred transfer ellipse par do points par spacecraft ki speed chahiye. Kisi bhi orbit par energy conservation kehta hai specific energy $\epsilon =\frac{v^2}{2}-\frac{\mu }{r}$ constant hai aur $-\frac{\mu }{2a}$ ke equal hai. $v$ ke liye solve karne par vis-viva equation milti hai $v=\sqrt{\mu \left(\frac{2}{r}-\frac{1}{a}\right)}$ — ye jawab deti hai "radius $r$ par size $a$ ki orbit par mein kitni fast move kar raha hoon?" bina poori trajectory ke. Exactly yahi sawaal ek Hohmann burn poochhti hai.

Do planets ki circular speeds (har ek apne circle par, isliye $a=r$, aur vis-viva reduce ho kar $\sqrt{\mu /r}$ ban jaati hai): $v_1=\sqrt{\frac{{\mu }_{\odot }}{r_1}}=\sqrt{\frac{1.327\times 10^{11}}{1.496\times 10^8}}=29.78\text{km/s},v_2=\sqrt{\frac{{\mu }_{\odot }}{r_2}}=24.13\text{km/s}.$ Transfer ellipse Earth's circle ko perihelion par aur Mars' circle ko aphelion par touch karti hai, isliye uska semi-major axis average radius hai: $a=\frac{r_1+r_2}{2}=1.8875\times 10^8\text{km}.$ Har end par transfer speeds, vis-viva se: $v_p=\sqrt{1.327\times 10^{11}\left(\frac{2}{1.496\times 10^8}-\frac{1}{1.8875\times 10^8}\right)}=32.73\text{km/s}(\text{at Earth’s radius}),$ $v_a=\sqrt{1.327\times 10^{11}\left(\frac{2}{2.279\times 10^8}-\frac{1}{1.8875\times 10^8}\right)}=21.48\text{km/s}(\text{at Mars’ radius}).$ Burns (har ek planet ke circle aur transfer ellipse ke beech speed gap hai us radius par): $\Delta v_1=32.73-29.78=2.95\text{km/s},$ $\Delta v_2=24.13-21.48=2.65\text{km/s}.$ Sum — total heliocentric Hohmann Δv: $\Delta v_{tot}^{Hohmann}=\Delta v_1+\Delta v_2=2.95+2.65=5.60\text{km/s}.$ Ye theoretical cheapest heliocentric transfer hai — pork chop ka deepest island iske paas hota hai. Dekho Hohmann Transfer.

Opportunities har synodic period mein recur hoti hain. 10 years mein count ($3652.5$ d): $N=\lfloor \frac{3652.5}{779.9}\rfloor +1=\lfloor 4.68\rfloor +1=4+1=5\text{opportunities}.$ ($+1$ $t=0$ par khule window ko count karta hai.) Har ek fresh pork chop kholti hai; $\pm 20$-day window ek vertical strip hai jo tum har island ke andar slide karte ho sabse sasta cell dhundhne ke liye. Design insight: sirf $\sim 5$ shots per decade ke saath, ek window miss karna $\sim 26$ months ka cost hota hai — isliye Δv margin aur schedule ek saath negotiate hote hain.

$\Delta v_A=\sqrt{9+119.36}-7.726=11.330-7.726=3.604\text{km/s}.$ $\Delta v_B=\sqrt{4.4^2+119.36}-7.726=\sqrt{19.36+119.36}-7.726=11.778-7.726=4.052\text{km/s}.$ B ke liye extra Δv: $\Delta v_B-\Delta v_A=0.448$ km/s. Days saved: $270-180=90$ d. $\text{penalty}=\frac{0.448\text{km/s}}{90\text{d}}=0.0050\frac{\text{km/s}}{\text{day}}\approx 5.0\frac{\text{m/s}}{\text{day}}.$ Ise padhna: tum flight-time ka har ek din jo tum shave karte ho uske liye $\sim 5$ m/s Δv pay karte ho. Ek crewed mission ke liye (radiation dose TOF ke saath scale hoti hai) ye worth it ho sakta hai; ek cheap probe ke liye nahi. Exactly yahi fast-vs-cheap trade hai jo pork chop plot visualize karne ke liye exist karta hai.

Neeche di figure dekho, jo ek single launch column (ek lobe ke through ek vertical slice) par departure Δv ko time of flight ke against plot karta hai.

Curve ka left end (short TOF). ${\vec{r}}_2$ tak bahut kam time mein pahunchne ke liye spacecraft ko heliocentric arc fast sweep karna hoga, jo ek highly eccentric, high-energy conic ki demand karta hai. High energy ka matlab large $v_{\infty }$ hai, aur $\Delta v=\sqrt{v_{\infty }^2+2\mu /r}-\sqrt{\mu /r}$ ke through woh large Δv ka matlab hai. Isliye curve left par high hai (plum note).

Curve ka middle (near-Hohmann). Jab TOF Earth–Mars ke liye roughly $259$ days ki taraf badhta hai — minimum-energy transfer ellipse ke period ka aadha — required conic us minimum-energy ellipse ki taraf relax hota hai. Transfer angle ideal $\sim 180°$ Hohmann geometry ke paas hota hai (planets well-spaced), $v_{\infty }$ bottom out karta hai, aur Δv apna minimum reach karta hai: orange dot, pork chop ka "eye."

Curve ka right end (long TOF). TOF ko aur zyada push karo aur geometry dobara awkward ho jaati hai: transfer ko loiter karna padta hai, arc $>180°$ Type-II regime ki taraf overshoot hota hai aur eventually L3.3 ke $180°$ ridge singularity ki taraf, isliye $v_{\infty }$ aur Δv waapis climb karte hain.

Teeno regimes milke ek bowl banaate hain, jiska bottom intermediate TOF par baitta hai — kabhi bhi shortest-TOF diagonal par nahi (woh expensive left edge hai). 2-D pork chop plot par ye bowl hi hai jo contours ko ek island mein close karta hai: innermost ring bowl ka floor hai, dono TOF extremes se door.