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

This is the drill page for the pork chop plot topic. We take the four formulas from the parent note and run them through every kind of input — normal, extreme, zero, and degenerate — until no scenario can surprise you.

Recall The four tools we will reuse (from the parent)

Circular parking-orbit speed: $v_{c} = μ / r$ — the speed to stay on a circle of radius $r$ .
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: $\frac{1}{T _{sy n}} = \frac{1}{T _{E}} - \frac{1}{T _{T}}$ .

Every symbol is defined in the parent; if you meet one you don't recognise, chase it there first.

The scenario matrix

Before we compute anything, let's list every case class this topic can generate. Each worked example below is tagged with the cell it fills.

#	Case class	What is special about it	Example
C1	Normal cell	ordinary $v_{\infty}$ , ordinary parking orbit → one Δv	Ex 1
C2	Compare two cells	which grid point is cheaper, and by how much	Ex 2
C3	Zero input ( $v_{\infty} = 0$ )	the degenerate "already on the transfer" limit	Ex 3
C4	Large- $v_{\infty}$ limit	fast transfer, gravity well becomes negligible	Ex 4
C5	Capture side ( $μ_{T}$ , $r_{a}$ )	same formula, different body — Mars capture	Ex 5
C6	$C_{3} \leftrightarrow Δ v$ conversion	the two "cost currencies" and rounding traps	Ex 6
C7	Timing / synodic (a "when" question)	no gravity at all, pure orbital-period arithmetic	Ex 7
C8	Real-world word problem	pick the cheaper of two windows for a real mission	Ex 8
C9	Exam twist / degenerate geometry	the 180° ridge, why a cell is forbidden	Ex 9

Constants used throughout (Earth): $μ_{E} = 398600 km^{3} / s^{2}$ . Mars: $μ_{M} = 42828 km^{3} / s^{2}$ .

C1 — The normal cell

Worked example Example 1 — One departure Δv from one

$v_{\infty}$ A pork chop cell gives $v_{\infty} = 3.0$ km/s. Parking orbit $r_{p} = 6678$ km (a 300 km altitude LEO). Find $C_{3}$ and $Δ v_{d e p}$ .

Forecast: guess — will the burn be bigger or smaller than $v_{\infty} = 3.0$ ? (Most people say smaller. It's bigger — you'll see why.)

Circular speed $v_{c} = μ_{E} / r_{p} = 398600/6678 = 7.726$ km/s. Why this step? This is the speed you already have in the parking orbit — the baseline you burn away from.
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. Why this step? Energy conservation on the hyperbola: the spacecraft must be moving this fast at perigee so that, after climbing out of Earth's well, $3.0$ km/s survives.
The burn $Δ v_{d e p} = v_{p er i} - v_{c} = 11.33 - 7.726 = 3.61$ km/s. Why this step? You only pay for the change in speed at perigee, not the whole $v_{p er i}$ .
$C_{3} = v_{\infty}^{2} = 9.0 km^{2} / s^{2}$ — the datasheet number.

Verify: $Δ v = 3.61 > v_{\infty} = 3.0$ . The forecast "smaller" is wrong — the burn exceeds $v_{\infty}$ because you also fight Earth's gravity while burning deep in the well. Units: km/s throughout. ✓

C2 — Which cell is cheaper?

C3 — The zero input

C4 — The large- $v_{\infty}$ limit

Worked example Example 4 — A very fast transfer,

$v_{\infty} = 15$ km/s A rushed flyby needs $v_{\infty} = 15$ km/s. Same $r_{p}$ . Find Δv and compare it to $v_{\infty}$ itself.

Forecast: for such a big $v_{\infty}$ , does the gravity-well term still matter much?

$Δ v = 225 + 119.36 - 7.726 = 344.36 - 7.726 = 18.556 - 7.726 = 10.83$ km/s. Why this step? Same formula, large first term.
Compare to the "no-gravity" estimate $v_{\infty} - 0 = 15$ ? No — better: the burn approaches $v_{\infty}^{2} + 2 μ / r - v_{c}$ . For huge $v_{\infty}$ , $v_{\infty}^{2} + k \to v_{\infty}$ , so $Δ v \to v_{\infty} - v_{c} = 15 - 7.726 = 7.27$ km/s asymptote... but here $10.83$ , still well above it. Why this step? Shows the limiting behaviour: the $k$ term shrinks in relative importance as $v_{\infty}$ grows.

Verify: the correction $225 + 119.36 - 225 = 18.556 - 15.0 = 3.56$ km/s — smaller in fraction ( $3.56/15 = 24%$ ) than in Ex 1 where the term dominated. So gravity matters relatively less at high $v_{\infty}$ . Forecast confirmed. Units km/s. ✓

C5 — The capture side (different body)

Worked example Example 5 — Mars capture burn

Arrival excess speed $v_{\infty, a r r} = 2.5$ km/s at Mars. Capture into a circular orbit of radius $r_{a} = 3689$ km (≈ 300 km above Mars). $μ_{M} = 42828 km^{3} / s^{2}$ . Find $Δ v_{a r r}$ .

Forecast: Mars has ~1/9 of Earth's $μ$ . Will the capture burn be much smaller than an Earth burn at the same $v_{\infty}$ ?

Mars circular speed $v_{c} = 42828/3689 = 11.61 = 3.407$ km/s. Why this step? Baseline speed of the final Mars parking orbit.
Periapsis speed $v_{p er i} = 2. 5^{2} + 2 (42828) /3689 = 6.25 + 23.22 = 29.47 = 5.428$ km/s. Why this step? Same hyperbola energy relation, now with Mars's $μ$ and $r$ .
$Δ v_{a r r} = v_{p er i} - v_{c} = 5.428 - 3.407 = 2.02$ km/s. Why this step? Capture burn = shed the excess by dropping from hyperbola periapsis speed to circular speed.

Verify: with $v_{\infty} = 2.5$ near Earth the burn would be $6.25 + 119.36 - 7.726 = 11.208 - 7.726 = 3.48$ km/s. Mars gives $2.02$ km/s — smaller, because its shallower well ( $2 μ / r$ term is $23$ vs $119$ ). Forecast confirmed. Units km/s. ✓

C6 — Converting between the two currencies

C7 — A pure timing (synodic) question

C8 — Real-world word problem

Worked example Example 8 — Mission planner's choice

Two candidate launch cells for a Mars orbiter: Window P gives $v_{\infty}^{d e p} = 3.2$ km/s and $v_{\infty}^{a r r} = 2.4$ km/s; Window Q gives $v_{\infty}^{d e p} = 2.9$ km/s but $v_{\infty}^{a r r} = 3.1$ km/s. Earth $r_{p} = 6678$ km ( $μ_{E} = 398600$ ); Mars $r_{a} = 3689$ km ( $μ_{M} = 42828$ ). Which window has the lower total Δv?

Forecast: Q launches cheaper but arrives dearer. Does the total flip? Guess P or Q.

Window P departure: $3. 2^{2} + 119.36 - 7.726 = 129.60 - 7.726 = 11.384 - 7.726 = 3.658$ km/s. Why this step? 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. Why this step? Mars capture formula ( $v_{c} = 3.407$ from Ex 5).
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. Choose Window P. The cheaper departure of Q ( $- 0.081$ ) is overwhelmed by its dearer arrival ( $+ 0.346$ ). Total Δv is what pays via the rocket equation, not either half alone. ✓

C9 — The exam twist: a degenerate geometry

Worked example Example 9 — Why the 180° ridge cell is "forbidden"

An exam cell places Earth and Mars exactly 180° apart in heliocentric angle at your chosen launch/arrival dates, but their orbital planes are inclined by $i = 1.85°$ (Mars). Estimate the extra plane-change Δv the transfer demands, at a heliocentric transfer speed of $v_{t} \approx 24$ km/s, and explain why this cell is unusable.

Forecast: the two positions and the Sun nearly form a straight line. Does that make the transfer easy (like Hohmann) or impossible?

Geometry: with $r_{1}$ and $r_{2}$ nearly antiparallel through the Sun, the plane containing $r_{1}, r_{2}$ and the Sun is undefined — infinitely many planes pass through a line. See Lambert's Problem: the short-way/long-way split collapses here. Why this step? A pork chop's two lobes (Type I, Type II) meet exactly at this 180° ridge; the transfer plane can't be pinned down.
Plane-change cost: a plane change of angle $Δ i$ at heliocentric speed $v_{t}$ costs $Δ v_{pl an e} = 2 v_{t} sin (Δ i /2)$ . Why this step? Rotating a velocity vector of length $v_{t}$ by $Δ i$ needs a burn equal to the chord of that rotation — $2 v_{t} sin (Δ i /2)$ .
With $Δ i = 1.85°$ : $Δ v_{pl an e} = 2 (24) sin (0.925°) = 48 \times 0.016144 = 0.775$ km/s — and this blows up as the geometry forces $Δ i \to 90°$ near the exact ridge, where $2 v_{t} sin 45° \approx 34$ km/s. Why this step? Even a modest inclination costs ~0.78 km/s; at the singular ridge the required plane change diverges.

Verify: $2 (24) sin (0.925°) = 0.775$ km/s (small but real). The point of the twist: as the transfer angle → 180°, the plane-change Δv spikes toward a wall — this is the ridge separating the two pork-chop lobes, a physical forbidden zone, not a plot glitch. Forecast "impossible" is right at the exact ridge. ✓

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

Recall check

Answer before revealing.

Why is

Δ v_{d e p}

larger than

v_{\infty}

for a low LEO parking orbit?

Because you also pay

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

km/s to climb out of Earth's gravity well; the

2 μ / r

term inside the root adds to

v_{\infty}

For

v_{\infty} = 0

, what is the departure burn?

The pure escape burn

(2 - 1) μ / r

— never zero.

Why does doubling

C_{3}

barely change Δv at high

v_{\infty}

Because

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

; the gravity term

k = 2 μ / r

becomes relatively negligible (square-root compression).

Why is a total-Δv comparison, not a launch-

C_{3}

comparison, the right way to pick a window?

The capture burn can dominate; the rocket equation responds to total Δv.

What makes the 180° cell forbidden?

The transfer plane is undefined, forcing a huge plane change; Δv spikes at the ridge between Type I and Type II lobes.

Recall Prerequisites feeding this drill

Lambert's Problem (each cell), Hyperbolic Excess Velocity & C3 ( $v_{\infty}$ , $C_{3}$ ), Hohmann Transfer (cheap near-180° reference), Oberth Effect (why burning deep helps), Patched Conic Approximation (Earth-frame vs Sun-frame handoff), Synodic Period (Ex 7), Tsiolkovsky Rocket Equation (why Δv is exponential in mass).