Pork chop plots — Δv vs launch - arrival date: Question bank

True or false — justify

Each of these is a statement people believe. Decide true/false and say why.

The x-axis of a pork chop plot is the time of flight.

False. The x-axis is the launch date and the y-axis the arrival date; time of flight is the difference (arrival − launch), shown as diagonal lines running across the plot.

A single cell of the grid can have several different Δv values depending on which orbit you pick.

Essentially false for a fixed transfer type. Once launch and arrival dates are set, the two planet positions and the flight time are fixed, so Lambert's problem returns a unique connecting conic (per short-way / long-way choice) — one definite Δv.

Lower time of flight always means lower Δv.

False. A short flight forces a steeply-curved, high-energy transfer with large

v_{\infty}

. The cheapest transfers sit near a slow Hohmann geometry, so minimum Δv is usually not minimum time.

C_{3}

and departure Δv carry the same units and are interchangeable.

False.

C_{3} = v_{\infty}^{2}

has units

km^{2} / s^{2}

(an energy per unit mass), while Δv has units

km/s

and folds in the

2 μ_{E} / r_{p}

gravity-well term. They are linked by

Δ v = C_{3} + 2 μ_{E} / r_{p} - μ_{E} / r_{p}

, not equal.

If you double

v_{\infty}

, the departure Δv also doubles.

False. The

+ 2 μ_{E} / r_{p}

term inside the square root makes Δv grow sublinearly with

v_{\infty}

. Doubling

v_{\infty}

raises

C_{3}

fourfold but Δv rises much more gently.

Pork chops repeat because Earth returns to the same point each year.

False. They repeat every synodic period — the time for the two planets to return to the same relative geometry (~26 months for Earth–Mars), not every Earth year.

Every point on a constant-TOF diagonal costs the same Δv.

False. Constant TOF only fixes the flight duration; as you slide along the diagonal the actual planet positions change, so the geometry and Δv vary.

The two lobes of a pork chop are just a plotting artifact of drawing two contour sets.

False. They are physically distinct: Lambert Type I sweeps < 180° of heliocentric angle, Type II sweeps > 180°. The ridge between them is a real Δv spike, not a glitch.

The launch-vehicle datasheet number you compare against the plot is the total Δv.

False. Rockets are rated by the

C_{3}

they can deliver for a given payload mass, so the launch-side comparison is done in

C_{3}

, not Δv.

Spot the error

Each statement contains one flawed step. Name it and fix it.

"To leave Earth I need the transfer speed

v_{1}

, so Δv =

v_{1}

Wrong: Earth already supplies its orbital velocity

V_{E}

for free. You only pay for

v_{\infty} = v_{1} - V_{E}

, and then add the gravity-well cost via the hyperbola energy relation.

v_{\infty}

is the spacecraft's speed relative to the Sun after departure."

Wrong frame:

v_{\infty}

is the excess speed relative to Earth once far enough that Earth's gravity is negligible. See Hyperbolic Excess Velocity & C3.

v_{p er i} = v_{\infty} + 2 μ_{E} / r_{p}

, just add the speeds."

Speeds don't add linearly here; energies do. Specific energy gives

v_{p er i} = v_{\infty}^{2} + 2 μ_{E} / r_{p}

— a Pythagorean-style combination under the root, not a sum.

"The synodic period is just

T_{E} + T_{T}

Wrong: it comes from angular rates,

1/ T_{sy n} = ∣1/ T_{E} - 1/ T_{T} ∣

. It's about how fast one planet laps the other, giving ~780 days for Earth–Mars, not the sum of periods.

"I'll pick the exact 180° transfer since it's the pure Hohmann geometry and cheapest."

Near-180° is cheap, but exactly 180° makes the transfer plane undefined, so any orbital-plane mismatch forces a huge plane-change spike. Pick a point slightly inside a lobe, not on the ridge.

"Capture Δv uses Earth's

μ_{E}

At arrival you're in the target planet's gravity, so capture uses

μ_{T}

and the target parking radius

r_{a}

Δ v_{a r r} = v_{\infty, a r r}^{2} + 2 μ_{T} / r_{a} - μ_{T} / r_{a}

Why questions

Answer the "why" in one or two sentences of real reasoning.

Why does each grid cell correspond to exactly one Lambert problem?

Because fixing the launch and arrival dates fixes both endpoint positions

r_{1}, r_{2}

and the flight time

Δ t

— precisely the boundary data Lambert's problem needs to return the connecting conic.

Why do we subtract the planet's velocity before computing cost?

The spacecraft rides along with the planet's orbital velocity for free; only the extra velocity

v_{\infty} = v_{1} - V_{E}

must be paid for by the rocket.

Why is Δv exponentially important even for tiny savings?

The rocket equation relates mass ratio to Δv exponentially, so a small Δv cut can translate into a large gain in deliverable payload mass.

Why are launch windows spaced ~26 months for Mars rather than every year?

Because favourable near-Hohmann geometry recurs once per synodic period (~780 days), which is when Earth catches back up to the right relative angle with Mars. See Synodic Period.

Why is

C_{3}

, and not Δv, printed on a launch vehicle's performance chart?

The rocket delivers the interplanetary energy (the escape hyperbola's

v_{\infty}^{2}

); the burn from a parking orbit into Δv depends on the chosen

r_{p}

, which is mission-specific, so the vehicle-independent number is

C_{3}

Why does the minimum-Δv point sit at the center of a pork-chop island rather than at its edge?

Contours are nested closed curves of increasing Δv; the innermost point is the single lowest-cost (launch, arrival) pair for that opportunity — the local optimum.

Why does firing near perigee (deep in the gravity well) make departure more efficient?

The Oberth effect: a burn at high speed adds more kinetic energy for the same Δv, so departing from low perigee squeezes more

v_{\infty}

per unit fuel.

Edge cases

Boundary and degenerate scenarios the plot quietly contains.

What happens to Δv along the ridge where the heliocentric transfer angle is exactly 180°?

The transfer plane becomes undefined, so any inclination difference between the planets demands an enormous plane change — Δv spikes to a wall separating Type I and Type II.

What does a cell where arrival date ≤ launch date represent?

A non-physical region (you'd arrive before or when you leave). The valid plot lives strictly above the launch = arrival diagonal, where TOF > 0.

If both planets had identical orbital periods, what happens to the synodic period?

1/ T_{sy n} = ∣1/ T_{E} - 1/ T_{T} ∣ \to 0

, so

T_{sy n} \to \infty

: the geometry never repeats favourably, and there'd be no periodic family of pork chops.

What happens to

Δ v_{d e p}

v_{\infty} \to 0

It reduces to the pure escape burn:

Δ v_{d e p} \to 2 μ_{E} / r_{p} - μ_{E} / r_{p} = (2 - 1) μ_{E} / r_{p}

, the cost just to barely escape Earth with zero leftover speed.

What happens to

Δ v_{d e p}

as the parking radius

r_{p} \to \infty

(starting far from Earth)?

The gravity term

2 μ_{E} / r_{p} \to 0

, so

Δ v_{d e p} \to v_{\infty}

: with no well to climb out of, you simply pay the excess speed directly — and you lose the Oberth benefit.

If a cell demands a very short TOF, what does that imply about

C_{3}

and payload?

Short TOF forces a high-energy transfer with large

v_{\infty}

, hence large

C_{3}

; the rocket must throw a smaller mass to that higher energy, cutting deliverable payload.

Along a constant-TOF diagonal that passes through a pork-chop lobe, why isn't every point equally good?

Fixing TOF doesn't fix the endpoint geometry; only the sub-segment inside the lobe's low contours is cheap, and one point on it is the true minimum for that flight time.

What distinguishes a Type I from a Type II transfer at the same launch and arrival dates?

They are the short-way (< 180°) and long-way (> 180°) solutions of the same Lambert problem — same endpoints and time, different swept angle and thus different Δv.

Recall One-line self-test

Name the axes, the contour quantity, why islands recur, and why the central point wins. ::: x = launch date, y = arrival date; contours = $C_{3}$ or total Δv; islands recur each synodic period; the center is the local minimum-Δv (nested closed contours).