Pork chop plots — Δv vs launch - arrival date: Worked examples

3.2.27 · D3 · Physics › Orbital Mechanics & Astrodynamics › Pork chop plots — Δv vs launch - arrival date

Ye drill page hai pork chop plot topic ke liye. Hum parent note se chaar formulas lete hain aur unhe har tarah ke inputs se run karte hain — normal, extreme, zero, aur degenerate — tab tak jab tak koi bhi scenario tumhe surprise na kar sake.

Recall Chaar tools jo hum baar baar use karenge (parent se)

Circular parking-orbit speed: $v_{c} = μ / r$ — $r$ radius ke circle par rehne ki speed.
Departure/capture burn: $Δ v = v_{\infty}^{2} + \frac{2 μ}{r} - \frac{μ}{r}$ .
Characteristic energy: $C_{3} = v_{\infty}^{2}$ (units $km^{2} / s^{2}$ ).
Synodic period: $\frac{1}{T _{sy n}} = \frac{1}{T _{E}} - \frac{1}{T _{T}}$ .

Har symbol parent mein define hai; agar koi na samajh aaye to pehle wahan jaao.

The scenario matrix

Kuch bhi compute karne se pehle, har case class list kar lete hain jo ye topic generate kar sakta hai. Neeche har worked example us cell ko fill karta hai jisme wo fit hota hai.

#	Case class	Kya special hai isme	Example
C1	Normal cell	ordinary $v_{\infty}$ , ordinary parking orbit → ek Δv	Ex 1
C2	Do cells compare karo	kaun sa grid point sasta hai, aur kitna	Ex 2
C3	Zero input ( $v_{\infty} = 0$ )	degenerate "already on the transfer" limit	Ex 3
C4	Large- $v_{\infty}$ limit	fast transfer, gravity well negligible ho jaata hai	Ex 4
C5	Capture side ( $μ_{T}$ , $r_{a}$ )	same formula, alag body — Mars capture	Ex 5
C6	$C_{3} \leftrightarrow Δ v$ conversion	do "cost currencies" aur rounding traps	Ex 6
C7	Timing / synodic (ek "kab" ka sawal)	gravity bilkul nahi, sirf orbital-period arithmetic	Ex 7
C8	Real-world word problem	ek real mission ke liye do windows mein se sasta chunna	Ex 8
C9	Exam twist / degenerate geometry	180° ridge, kyun ek cell forbidden hai	Ex 9

Constants jo poore mein use honge (Earth): $μ_{E} = 398600 km^{3} / s^{2}$ . Mars: $μ_{M} = 42828 km^{3} / s^{2}$ .

C1 — The normal cell

Worked example Example 1 — Ek

$v_{\infty}$ se ek departure Δv Ek pork chop cell deta hai $v_{\infty} = 3.0$ km/s. Parking orbit $r_{p} = 6678$ km (ek 300 km altitude LEO). $C_{3}$ aur $Δ v_{d e p}$ nikalo.

Forecast: andaaza lagao — burn $v_{\infty} = 3.0$ se bada hoga ya chhota? (Zyaadatar log kehte hain chhota. Bada hoga — abhi dekhoge kyun.)

Circular speed $v_{c} = μ_{E} / r_{p} = 398600/6678 = 7.726$ km/s. Ye step kyun? Ye wo speed hai jo parking orbit mein tumhare paas already hai — wo baseline jisse tum burn karke door jaate ho.
Perigee speed on the escape hyperbola $v_{p er i} = v_{\infty}^{2} + 2 μ_{E} / r_{p} = 9 + 119.36 = 11.33$ km/s. Ye step kyun? Hyperbola par energy conservation: spacecraft ko perigee par itni tezi se move karna hoga taaki Earth ke well se nikalne ke baad $3.0$ km/s bach sake.
The burn $Δ v_{d e p} = v_{p er i} - v_{c} = 11.33 - 7.726 = 3.61$ km/s. Ye step kyun? Tum sirf perigee par speed ke change ka cost dete ho, poore $v_{p er i}$ ka nahi.
$C_{3} = v_{\infty}^{2} = 9.0 km^{2} / s^{2}$ — datasheet ka number.

Verify: $Δ v = 3.61 > v_{\infty} = 3.0$ . Forecast "chhota" galat tha — burn $v_{\infty}$ se zyaada hai kyunki tum well ke andar deeply burn karte waqt Earth ki gravity se bhi ladte ho. Units: km/s throughout. ✓

C2 — Kaun sa cell sasta hai?

C3 — Zero input

Worked example Example 3 — Agar

$v_{\infty} = 0$ ho toh? Ek magical cell require karta hai $v_{\infty} = 0$ km/s — tum Earth ke escape boundary par exactly zero leftover speed ke saath pahunchte ho. $Δ v_{d e p}$ nikalo.

Forecast: zero excess speed — kya burn bhi zero hai?

$Δ v = 0 + 2 μ_{E} / r_{p} - μ_{E} / r_{p} = 2 v_{c} - v_{c} = (2 - 1) v_{c}$ . Ye step kyun? $v_{\infty} = 0$ set karo aur $v_{c} = μ_{E} / r_{p}$ factor out karo.
Numerically: $(2 - 1) (7.726) = 0.4142 \times 7.726 = 3.20$ km/s. Ye step kyun? Ye exactly escape burn hai — $2 v_{c}$ escape speed hai, toh burn escape speed minus circular speed hai.

Verify: forecast "zero" galat tha. $v_{\infty} = 0$ ke baad bhi tumhe Earth se escape karna padega, jiska cost hai $(2 - 1) v_{c} \approx 3.20$ km/s. Ye floor hai: is parking orbit ke liye koi bhi interplanetary Δv kabhi isse neeche nahi hogi. Units km/s. ✓

C4 — Large- $v_{\infty}$ limit

Worked example Example 4 — Bahut fast transfer,

$v_{\infty} = 15$ km/s Ek rushed flyby ko $v_{\infty} = 15$ km/s chahiye. Same $r_{p}$ . Δv nikalo aur $v_{\infty}$ se compare karo.

Forecast: itne bade $v_{\infty}$ ke liye, kya gravity-well term ab bhi zyaada matter karta hai?

$Δ v = 225 + 119.36 - 7.726 = 344.36 - 7.726 = 18.556 - 7.726 = 10.83$ km/s. Ye step kyun? Same formula, pehla term bada hai.
"No-gravity" estimate $v_{\infty} - 0 = 15$ se compare karo? Nahi — better: burn approach karta hai $v_{\infty}^{2} + 2 μ / r - v_{c}$ . Bade $v_{\infty}$ ke liye, $v_{\infty}^{2} + k \to v_{\infty}$ , toh $Δ v \to v_{\infty} - v_{c} = 15 - 7.726 = 7.27$ km/s asymptote... lekin yahan $10.83$ , ab bhi us se kaafi upar. Ye step kyun? Limiting behaviour dikhata hai: $k$ term ki relative importance $v_{\infty}$ ke badhne par ghaatti hai.

Verify: correction $225 + 119.36 - 225 = 18.556 - 15.0 = 3.56$ km/s — fraction mein chhota ( $3.56/15 = 24%$ ) Ex 1 se jahan term dominate karta tha. Toh high $v_{\infty}$ par gravity relatively kam matter karta hai. Forecast confirm. Units km/s. ✓

C5 — Capture side (alag body)

Worked example Example 5 — Mars capture burn

Mars par arrival excess speed $v_{\infty, a r r} = 2.5$ km/s. $r_{a} = 3689$ km radius ki circular orbit mein capture (Mars se ≈ 300 km upar). $μ_{M} = 42828 km^{3} / s^{2}$ . $Δ v_{a r r}$ nikalo.

Forecast: Mars ka $μ$ Earth ka ~1/9 hai. Kya same $v_{\infty}$ par capture burn Earth burn se bahut chhota hoga?

Mars circular speed $v_{c} = 42828/3689 = 11.61 = 3.407$ km/s. Ye step kyun? Final Mars parking orbit ki baseline speed.
Periapsis speed $v_{p er i} = 2. 5^{2} + 2 (42828) /3689 = 6.25 + 23.22 = 29.47 = 5.428$ km/s. Ye step kyun? Same hyperbola energy relation, ab Mars ke $μ$ aur $r$ ke saath.
$Δ v_{a r r} = v_{p er i} - v_{c} = 5.428 - 3.407 = 2.02$ km/s. Ye step kyun? Capture burn = hyperbola periapsis speed se circular speed tak drop karke excess shed karo.

Verify: $v_{\infty} = 2.5$ near Earth pe burn hota $6.25 + 119.36 - 7.726 = 11.208 - 7.726 = 3.48$ km/s. Mars deta hai $2.02$ km/s — chhota, kyunki uska shallower well ( $2 μ / r$ term $23$ vs $119$ hai). Forecast confirm. Units km/s. ✓

C6 — Do currencies ke beech convert karo

C7 — Pure timing (synodic) question

C8 — Real-world word problem

Worked example Example 8 — Mission planner ka choice

Mars orbiter ke liye do candidate launch cells: Window P deta hai $v_{\infty}^{d e p} = 3.2$ km/s aur $v_{\infty}^{a r r} = 2.4$ km/s; Window Q deta hai $v_{\infty}^{d e p} = 2.9$ km/s lekin $v_{\infty}^{a r r} = 3.1$ km/s. Earth $r_{p} = 6678$ km ( $μ_{E} = 398600$ ); Mars $r_{a} = 3689$ km ( $μ_{M} = 42828$ ). Kaun sa window kam total Δv deta hai?

Forecast: Q launch sasta karta hai lekin arrive mehenga. Kya total flip hoga? P ya Q guess karo.

Window P departure: $3. 2^{2} + 119.36 - 7.726 = 129.60 - 7.726 = 11.384 - 7.726 = 3.658$ km/s. Ye step kyun? Earth master formula.
Window P arrival: $2. 4^{2} + 23.22 - 3.407 = 28.98 - 3.407 = 5.384 - 3.407 = 1.977$ km/s. Ye step kyun? Mars capture formula ( $v_{c} = 3.407$ Ex 5 se).
P total: $3.658 + 1.977 = 5.635$ km/s.
Window Q departure: $2. 9^{2} + 119.36 - 7.726 = 127.77 - 7.726 = 11.303 - 7.726 = 3.577$ km/s.
Window Q arrival: $3. 1^{2} + 23.22 - 3.407 = 32.83 - 3.407 = 5.730 - 3.407 = 2.323$ km/s.
Q total: $3.577 + 2.323 = 5.900$ km/s.

Verify: P $= 5.635$ km/s $<$ Q $= 5.900$ km/s. Window P chunno. Q ka sasta departure ( $- 0.081$ ) uske mehenga arrival ( $+ 0.346$ ) se overwhelm ho jaata hai. Total Δv hi rocket equation ke zariye cost deta hai, na koi ek half akela. ✓

C9 — Exam twist: degenerate geometry

Worked example Example 9 — 180° ridge cell "forbidden" kyun hai

Ek exam cell Earth aur Mars ko tumhare chosen launch/arrival dates par heliocentric angle mein exactly 180° apart rakhta hai, lekin unke orbital planes $i = 1.85°$ (Mars) se inclined hain. Extra plane-change Δv estimate karo, heliocentric transfer speed $v_{t} \approx 24$ km/s par, aur explain karo ye cell kyun unusable hai.

Forecast: do positions aur Sun almost ek seedhi line banate hain. Kya transfer easy hoga (Hohmann jaisa) ya impossible?

Geometry: $r_{1}$ aur $r_{2}$ almost antiparallel through the Sun hone par, $r_{1}, r_{2}$ aur Sun ko contain karne wala plane undefined hai — ek line se infinitely many planes guzarti hain. Dekho Lambert's Problem: short-way/long-way split yahan collapse ho jaata hai. Ye step kyun? Pork chop ke do lobes (Type I, Type II) exactly is 180° ridge par milte hain; transfer plane pin nahi ho sakta.
Plane-change cost: angle $Δ i$ ka plane change, heliocentric speed $v_{t}$ par, cost karta hai $Δ v_{pl an e} = 2 v_{t} sin (Δ i /2)$ . Ye step kyun? Length $v_{t}$ ke velocity vector ko $Δ i$ rotate karne ke liye us rotation ke chord ke barabar burn chahiye — $2 v_{t} sin (Δ i /2)$ .
$Δ i = 1.85°$ ke saath: $Δ v_{pl an e} = 2 (24) sin (0.925°) = 48 \times 0.016144 = 0.775$ km/s — aur ye blow up ho jaata hai jab exact ridge ke paas geometry $Δ i \to 90°$ force karta hai, jahan $2 v_{t} sin 45° \approx 34$ km/s. Ye step kyun? Ek modest inclination bhi ~0.78 km/s cost karti hai; singular ridge par required plane change diverge ho jaata hai.

Verify: $2 (24) sin (0.925°) = 0.775$ km/s (chhota lekin real). Twist ka point: jaise jaise transfer angle → 180°, plane-change Δv ek wall ki taraf spike karta hai — ye ridge hai jo do pork-chop lobes ko alag karti hai, ek physical forbidden zone, plot glitch nahi. Forecast "impossible" exact ridge par sahi tha. ✓

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

Recall check

Reveal karne se pehle answer karo.

Δ v_{d e p}

ek low LEO parking orbit ke liye

v_{\infty}

se bada kyun hota hai?

Kyunki tum

(2 - 1) v_{c} \approx 3.2

km/s Earth ke gravity well se nikalne ke liye bhi dete ho; root ke andar

2 μ / r

term

v_{\infty}

mein add ho jaata hai.

v_{\infty} = 0

ke liye, departure burn kya hai?

Pure escape burn

(2 - 1) μ / r

— kabhi zero nahi.

High

v_{\infty}

par

C_{3}

double karne par Δv barely kyun change hota hai?

Kyunki

v_{\infty}^{2} + k \to v_{\infty}

; gravity term

k = 2 μ / r

relatively negligible ho jaata hai (square-root compression).

Window choose karne ka sahi tarika total-Δv comparison kyun hai, launch-

C_{3}

comparison nahi?

Capture burn dominate kar sakta hai; rocket equation total Δv par respond karta hai.

180° cell forbidden kyun hai?

Transfer plane undefined ho jaata hai, ek huge plane change force hoti hai; Δv Type I aur Type II lobes ke beech ridge par spike karta hai.

Recall Prerequisites jo is drill ko feed karte hain

Lambert's Problem (har cell), Hyperbolic Excess Velocity & C3 ( $v_{\infty}$ , $C_{3}$ ), Hohmann Transfer (cheap near-180° reference), Oberth Effect (well ke andar burn karna kyun help karta hai), Patched Conic Approximation (Earth-frame vs Sun-frame handoff), Synodic Period (Ex 7), Tsiolkovsky Rocket Equation (kyun Δv mass mein exponential hai).