Pork chop plots — Δv vs launch - arrival date: Visual walkthrough

Step 1 — Two dots that will not hold still

WHAT. Pick a launch date and an arrival date. On those two dates, the Sun-centred ("heliocentric") position of Earth is an arrow $r_{1}$ , and the position of the target planet is an arrow $r_{2}$ .

WHY. A transfer orbit is a curve through space. To pin any curve down we must first say where it starts and where it ends. Those two facts are $r_{1}$ and $r_{2}$ . Nothing about fuel yet — just geography.

PICTURE. Look at the figure. The red arrow is $r_{1}$ : Earth's position on launch day, measured from the Sun. The black arrow $r_{2}$ is the target's position on arrival day. Because both planets orbit, sliding either date slides the tip of its arrow along its orbit ring — that is the whole reason the fuel cost will depend on dates.

Step 2 — The chord: the straight shortcut we will not fly

WHAT. Draw the straight line from the tip of $r_{1}$ to the tip of $r_{2}$ . Its length is the chord $c = ∣ r_{2} - r_{1} ∣.$

WHY. We will never fly this straight line (gravity curves everything), but its length is one of the three numbers that Lambert's theorem needs. The chord measures how far apart the two dots are, regardless of how far each is from the Sun.

PICTURE. The red segment is the chord $c$ . Notice $r_{2} - r_{1}$ literally means "stand at the tip of $r_{1}$ and walk to the tip of $r_{2}$ " — subtracting vectors gives the arrow between the two tips. Its length is $c$ .

Step 3 — Lambert's question: which orbit fits the clock?

WHAT. We now have three ingredients: $r_{1} + r_{2}$ (how far the two ends sit from the Sun), the chord $c$ (how far apart they are), and the time of flight $Δ t = arrival - launch$ . Lambert's theorem says these three fix the orbit's size: $Δ t = f (r_{1} + r_{2}, c, a) .$

WHY these three and not others? Because Lambert proved the flight time between two points depends only on those three numbers — not on the orbit's tilt or where it points. So if we already know $r_{1} + r_{2}$ , $c$ , and $Δ t$ , the only unknown left is $a$ , the orbit's semi-major axis (its "size"). We turn the equation around and solve for $a$ . This is Lambert's Problem.

WHY a new tool here? A Hohmann Transfer answers "cheapest orbit between two circles" but it does not let you choose the arrival time — you get whatever TOF the Hohmann ellipse gives. A pork chop plot demands a chosen $Δ t$ . Lambert is the tool that accepts an arbitrary time and returns the matching orbit. That is exactly the question we have.

PICTURE. Two orbits are drawn through the same two dots. The red one is fatter (bigger $a$ ) and takes longer; the black one is leaner (smaller $a$ ) and is quicker. Fixing $Δ t$ picks exactly one of this family.

Step 4 — From the orbit, read off the two velocities

WHAT. Once $a$ is known the whole conic is known, and geometry hands us two velocity arrows:

$v_{1}$ = how fast (and which way) the spacecraft moves at the start of the transfer,
$v_{2}$ = its velocity at the end (arrival).

WHY. Fuel pays for changes in velocity, not position. So the moment we have the transfer orbit, the first thing worth extracting is the velocity it demands at each end. Everything downstream is comparing these to the velocities the planets already have.

PICTURE. At the start-dot, the red arrow $v_{1}$ is tangent to the transfer curve (velocity is always along the direction of travel). At the end-dot, $v_{2}$ points along the curve there. These are required velocities — what the spacecraft must have to ride this orbit.

Step 5 — Subtract what Earth gives for free ( $v_{\infty}$ )

WHAT. The spacecraft is already flying along with Earth at Earth's heliocentric orbital velocity $V_{E}$ (defined in the opening block). The velocity we must supply is the leftover: $v_{\infty}^{d e p} = v_{1} - V_{E}, C_{3} \equiv v_{\infty}^{d e p}^{2} .$

WHY subtract? You are standing on a moving train (Earth). To match a passing car you only need the speed difference, not the car's full speed. Earth donates $V_{E}$ at no cost, so the honest cost is the difference $v_{1} - V_{E}$ . Once the craft is far enough that Earth's gravity has faded, this leftover speed is called the hyperbolic excess speed $v_{\infty}$ . Its square is $C_{3}$ — see Hyperbolic Excess Velocity & C3.

WHY a vector subtraction (why not just subtract speeds)? Because two arrows can have the same length but point differently. If $v_{1}$ and $V_{E}$ point apart by even a small angle, the leftover arrow can be large. Direction matters, so we subtract as arrows.

PICTURE. Tail-to-tail: the black arrow is $V_{E}$ (Earth's free velocity), the other black arrow is the required $v_{1}$ , and the red arrow closing the triangle is $v_{\infty}^{d e p} = v_{1} - V_{E}$ — the only piece you pay for.

Step 6 — Climb out of the gravity well (the departure burn)

WHAT. In reality you don't start "far away" — you start in a circular parking orbit of radius $r_{p}$ around Earth, moving at $v_{c} = μ_{E} / r_{p}$ (recall $μ_{E} = G M_{Earth}$ ). Energy conservation on the escape (hyperbolic) orbit gives the speed you need at perigee, and the burn is the jump from $v_{c}$ to that: $Δ v_{d e p} = v_{\infty}^{2} + \frac{2 μ _{E}}{r _{p}} - \frac{μ _{E}}{r _{p}} .$

WHY the $2 μ_{E} / r_{p}$ term? Leaving Earth's gravity well costs speed — like needing extra effort to roll a ball up out of a bowl. We track this with the specific orbital energy $ε$ — the total energy (kinetic + gravitational) per kilogram of spacecraft, units $km^{2} / s^{2}$ . For the escape orbit $ε = v_{\infty}^{2} /2$ (far away all energy is the leftover kinetic $v_{\infty}^{2} /2$ , since gravity has faded to zero). That same $ε$ , evaluated down at radius $r_{p}$ , is $v_{p er i}^{2} /2 - μ_{E} / r_{p}$ . Because energy is conserved, these two expressions are equal: $\frac{v _{\infty}^{2}}{2} = \frac{v _{p er i}^{2}}{2} - \frac{μ _{E}}{r _{p}} \Rightarrow v_{p er i} = v_{\infty}^{2} + \frac{2 μ _{E}}{r _{p}} .$ This is why $C_{3}$ alone under-states the burn: gravity adds the $2 μ_{E} / r_{p}$ before you take the square root. Firing at perigee (deep in the well, moving fastest) is cheapest — the Oberth Effect.

PICTURE. A circle (parking orbit, speed $v_{c}$ ) and a red hyperbola peeling away from it. At the shared perigee two speed markers sit stacked; the short red bracket between them is $Δ v_{d e p}$ .

Step 7 — Do the same at arrival, then add up

WHAT. Repeat Steps 5–6 at the far end. The excess speed relative to the target is $v_{\infty}^{a r r} = v_{2} - V_{t a r g e t}$ (with $V_{t a r g e t}$ the target's heliocentric velocity from the opening block), and the capture burn into an orbit of radius $r_{a}$ around a planet of gravity parameter $μ_{T} = G M_{t a r g e t}$ is: $Δ v_{a r r} = v_{\infty, a r r}^{2} + \frac{2 μ _{T}}{r _{a}} - \frac{μ _{T}}{r _{a}} .$ Total cost of this one grid cell: $Δ v_{t o t} = Δ v_{d e p} + Δ v_{a r r}$

WHY identical in form? Arrival is departure run backwards: instead of subtracting a burn to escape one well, you subtract a burn to drop into another. Same energy bookkeeping (same $ε$ argument, now with $μ_{T}$ and $r_{a}$ ), so the same formula with the target's numbers.

WHY add? The two burns happen at opposite ends of the trip and can't help each other — total fuel is their sum. This single number is what gets contoured over the (launch, arrival) grid to make the pork chop. Feeding $Δ v_{t o t}$ into the Tsiolkovsky Rocket Equation then tells you deliverable payload, which is why saving even $0.1$ km/s matters.

PICTURE. Two mirror-image "burn brackets": red at Earth (escape) on the left, black at the target (capture) on the right, joined by a plus sign into the total.

Step 8 — The degenerate case: the 180° ridge

WHAT. What if $r_{1}$ and $r_{2}$ are exactly opposite the Sun (a 180° transfer)? Then the two dots and the Sun are collinear, and infinitely many planes contain that line.

WHY it breaks. The orbit plane is no longer defined — you could tilt the transfer any way and still pass through both dots. Real planets are slightly out of a shared plane (recall the planar assumption we made up front is only an approximation), so threading a 180° transfer forces a large plane change, which is expensive. This is the sharp ridge splitting the pork chop's two lobes: not a rendering glitch, a genuine wall.

PICTURE. The Sun with two dots dead opposite; a red fan of tilted orbits all passing through both — showing the plane is undetermined. The reader should see why Δv spikes here.

The one-picture summary

This single figure chains all eight steps: two dots → chord → Lambert's orbit → required velocities → subtract planet velocity → climb the well → sum. Follow the red thread left to right.

Recall Feynman retelling — say it in plain words

I mark two spots in space: where Earth is when I leave, where my target is when I arrive. I draw the straight gap between them — the chord — and note my chosen flight time. Lambert's rule says those three facts fix exactly one orbit size, so I solve for it. That orbit tells me how fast I must be going at each end. But I'm already coasting with Earth, so I only pay for the difference in velocity — that leftover, once gravity has faded, is $v_{\infty}$ , and its square is the $C_{3}$ on a rocket's spec sheet. To actually get there I must climb out of Earth's gravity bowl from a parking orbit, which adds a $2 μ / r$ toll under the square root; I do the mirror-image thing to drop into the target. Add the two burns and I have the fuel bill for this one launch/arrival pair. Do that for every pair of dates and contour the numbers — and the cheap valleys, split by the forbidden 180° ridge, come out shaped like pork chops.

Recall Quick self-test

What three numbers does Lambert's theorem need? ::: $r_{1} + r_{2}$ , the chord $c$ , and the time of flight $Δ t$ . Why subtract $V_{E}$ from $v_{1}$ ? ::: Earth gives its heliocentric orbital velocity for free; you only pay for the difference $v_{\infty}$ . What does the $2 μ_{E} / r_{p}$ term represent? ::: The extra speed needed to climb out of Earth's gravity well from radius $r_{p}$ ; here $μ_{E} = G M_{Earth}$ . What distinguishes Type I from Type II transfers? ::: Type I sweeps less than 180° of heliocentric angle (short-way), Type II sweeps more than 180° (long-way). Why is the ridge between lobes expensive? ::: At a 180° transfer the orbit plane is undefined, forcing a costly plane change.