Pork chop plots — Δv vs launch - arrival date

WHY do we even need this plot?

So each cell of the grid is one Lambert problem, and its solution gives one Δv value. The plot is just Δv drawn over the whole grid.

WHAT exactly is on the axes and contours?

HOW is each grid cell computed? (Derivation from scratch)

We build the machinery from first principles.

Step 1 — Where are the planets?

Given a date, ephemerides give position vectors $\vec r_1$ (Earth at launch) and $\vec r_2$ (target at arrival), both heliocentric.

Why this step? The transfer orbit must start at $\vec r_1$ and end at $\vec r_2$; these are the boundary conditions.

Step 2 — Lambert's problem

We seek the heliocentric conic joining $\vec r_1 \to \vec r_2$ in the chosen time of flight $\Delta t$. Lambert's theorem says the TOF depends only on: $\Delta t = f\big(r_1 + r_2,\; c,\; a\big)$ where $c = |\vec r_2 - \vec r_1|$ is the chord and $a$ the semi-major axis.

Why this step? Two positions + a time fully constrain the orbit (up to short-way / long-way choice). Solving Lambert's equation gives $a$, hence the whole conic.

From the conic we get the required velocity vectors:

• $\vec v_1$ = spacecraft velocity at departure (on transfer orbit),
• $\vec v_2$ = spacecraft velocity at arrival.

Step 3 — Convert to Δv

The spacecraft is already moving with Earth's orbital velocity $\vec V_{\!E}$. So the velocity it must gain relative to Earth is: $\vec v_\infty^{\,dep} = \vec v_1 - \vec V_{\!E}, \qquad C_3 = \big|\vec v_\infty^{\,dep}\big|^2$

Why this step? You don't pay for the speed Earth gives you for free; you only pay for the difference. That difference, once out of Earth's gravity well, is $v_\infty$.

At arrival the excess speed relative to the target planet is: $\vec v_\infty^{\,arr} = \vec v_2 - \vec V_{\!target}$

Step 4 — Departure & arrival Δv from parking orbits

Leaving a circular parking orbit of radius $r_p$ around Earth, using $v_{esc}^2 = 2\mu_E/r_p$ and energy conservation on the departure hyperbola: $v_{peri} = \sqrt{v_\infty^2 + \frac{2\mu_E}{r_p}}$

Why? On the hyperbola, specific energy $\varepsilon = v_\infty^2/2 = v_{peri}^2/2 - \mu_E/r_p$. Rearranging gives the perigee speed.

Then the burn from the circular parking-orbit speed $v_c=\sqrt{\mu_E/r_p}$: $\boxed{\Delta v_{dep} = \sqrt{v_\infty^2 + \frac{2\mu_E}{r_p}} - \sqrt{\frac{\mu_E}{r_p}}}$

Similarly for capture at the target (radius $r_a$, gravity $\mu_T$): $\Delta v_{arr} = \sqrt{v_{\infty,arr}^2 + \frac{2\mu_T}{r_a}} - \sqrt{\frac{\mu_T}{r_a}}$

Total cost for that cell: $\Delta v_{tot} = \Delta v_{dep} + \Delta v_{arr}$

Why this is the whole plot: run Steps 1–4 for every (launch, arrival) pair → a Δv value per cell → contour it. Done.

Reading the plot

Worked examples

Common mistakes (Steel-manned)

Forecast-then-Verify

Active-recall flashcards

Connections

• Lambert's Problem — the solver behind every grid cell.
• Hohmann Transfer — the minimum-energy limit sitting near the chop's center.
• Hyperbolic Excess Velocity & C3 — the departure-cost quantity.
• Oberth Effect — why the $2\mu/r_p$ term makes deep-gravity burns efficient.
• Synodic Period — why windows repeat every ~26 months.
• Patched Conic Approximation — the framework that lets us treat departure/cruise/arrival separately.
• Tsiolkovsky Rocket Equation — why small Δv savings translate to big payload gains.

Concept Map

Hinglish (regional understanding)