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

• x-axis = launch date (departure from Earth).
• y-axis = arrival date (at the target planet).
• The islands are regions of low cost — the innermost contour is the ==minimum Δv (or minimum $C_3$)== for that launch opportunity, the "sweet spot". There is nothing to compute here; the skill is knowing that each single cell is one Lambert problem and the whole picture is just its cost drawn over a grid. See Lambert's Problem.

$C_3\equiv v_{\infty }^2$ — the definition, no gravity involved. $C_3=(3.2{)}^2=10.24\frac{{\text{km}}^2}{{\text{s}}^2}.$ Why square it? $C_3$ is energy per unit mass at infinity ($\epsilon =v_{\infty }^2/2$, and $C_3=2\epsilon$), and launch vehicles are rated by energy delivered, not speed.

Step A — circular speed (what you already have, and what $v_c$ means). Define ==$v_c$ = the speed a spacecraft has while coasting on a circular orbit of radius $r_p$== — the speed it is already moving at before any burn. Why does $v_c=\sqrt{{\mu }_E/r_p}$? On a circle, gravity supplies exactly the centripetal force: gravitational pull $={\mu }_{E}m/r_p^2$ must equal the centripetal requirement $m{v}_c^2/r_p$. Cancel $m$ and one power of $r_p$: $\frac{{\mu }_E}{r_p^2}=\frac{v_c^2}{r_p}\Rightarrow v_c^2=\frac{{\mu }_E}{r_p}\Rightarrow v_c=\sqrt{\frac{{\mu }_E}{r_p}}.$ Numerically: $v_c=\sqrt{\frac{398600}{6678}}=7.726\text{km/s}.$ Step B — perigee speed on the departure hyperbola (what you need). Why this formula? On the hyperbola the specific energy is $\epsilon =v_{\infty }^2/2$, and energy is also $v_{peri}^2/2-{\mu }_E/r_p$. Setting them equal and solving: $v_{peri}=\sqrt{v_{\infty }^2+\frac{2{\mu }_E}{r_p}}=\sqrt{9+\frac{2(398600)}{6678}}=\sqrt{9+119.36}=11.33\text{km/s}.$ Step C — the burn (you only pay the difference between what you need and what you already have): $\Delta v_{dep}=v_{peri}-v_c=11.33-7.726=3.61\text{km/s}.$ See Hyperbolic Excess Velocity & C3 for why the $+2\mu /r$ term appears.

Same master formula, but with Mars' ${\mu }_M=42828$ and $r_a=3689$ km. The circular speed $v_c=\sqrt{{\mu }_M/r_a}$ is the speed on the target circular orbit (same balance-of-forces derivation as L2.1, now with ${\mu }_M$): $v_c=\sqrt{\frac{42828}{3689}}=3.408\text{km/s}.$ $v_{peri}=\sqrt{2.5^2+\frac{2(42828)}{3689}}=\sqrt{6.25+23.22}=\sqrt{29.47}=5.429\text{km/s}.$ $\Delta v_{arr}=5.429-3.408=2.02\text{km/s}.$ What it looks like: you arrive fast on a hyperbola passing Mars, and the burn at closest approach bends you down into a bound circle — that is the retro-burn cost.

$\frac{1}{T_{syn}}=\left|\frac{1}{365.25}-\frac{1}{686.98}\right|=|0.0027379-0.0014556|=0.0012823{\text{d}}^{-1}.$ $T_{syn}=779.9\text{d}=2.14\text{yr}.$ Why this matters: cheap geometry (planets nicely spaced for a near-Hohmann shot) recurs every $\approx 26$ months, so pork chops arrive as a repeating family. See Synodic Period.

$v_c=7.726$ km/s (same for both). $\Delta v_A=\sqrt{9+119.36}-7.726=11.33-7.726=3.607\text{km/s}.$ $\Delta v_B=\sqrt{12.25+119.36}-7.726=11.472-7.726=3.746\text{km/s}.$ A is cheaper by $\approx 0.14$ km/s. Notice $C_3$ jumped from $9.0$ to $12.25$ (a $36\%$ energy rise) but Δv rose only $\approx 4\%$ — the $+2\mu /r$ term dilutes the difference. This is the Oberth-style leverage of doing the burn deep in the well; see Oberth Effect.

The mass ratio is $R=e^{\Delta v/v_e}$. The factor between the two cases is $\frac{R_B}{R_A}=\exp \left(\frac{\Delta v_B-\Delta v_A}{v_e}\right)=\exp \left(\frac{0.139}{3.4}\right)=\exp (0.0409)=1.042.$ A $\sim 4.2\%$ heavier stack for a Δv change that looked tiny — this is why mission designers chase the innermost contour. See Tsiolkovsky Rocket Equation.

First, define the transfer angle. Let ==$\Delta \theta$ = the heliocentric angle swept between the two radius vectors==, i.e. the angle at the Sun between ${\vec{r}}_1$ (Earth at launch) and ${\vec{r}}_2$ (target at arrival). A short-way ("Type I") transfer has $\Delta \theta <180°$; a long-way ("Type II") transfer has $\Delta \theta >180°$. The ridge is exactly $\Delta \theta =180°$.

Look at the figure. The Sun is at the centre; the teal dot is ${\vec{r}}_1$ (Earth at launch) and the orange dots are two possible arrival points ${\vec{r}}_2$. The plum arc sweeps less than 180° ($\Delta \theta <180°$, Type I); the dotted arc sweeps more than 180° ($\Delta \theta >180°$, Type II). The thick faded orange line is the special case where ${\vec{r}}_1$ and ${\vec{r}}_2$ are anti-parallel — a straight $\Delta \theta =180°$ sweep.

Why Δv spikes there — the correct reason. The transfer angle $\Delta \theta$ enters Lambert's geometry through the chord $c=|{\vec{r}}_2-{\vec{r}}_1|$ and, crucially, through the amount of angular momentum $\vec{h}=\vec{r}\times \vec{v}$ the transfer conic must carry. As $\Delta \theta \to 180°$, the two radius vectors line up, and the family of connecting ellipses degenerates: the minimum-energy ellipse's required semi-major axis $a$ pushes toward the point where the orbit becomes a nearly straight, radial "fall" between the two points. Physically the transverse (sideways) velocity needed to round the corner in the available time collapses, so the solver is forced onto an extreme conic whose departure and arrival speeds — and hence $v_{\infty }$ — grow rapidly. In the co-planar idealisation this shows up as a sharp energy spike; in the real 3-D problem the two planet orbits are slightly tilted, so at exactly $180°$ the transfer plane is only weakly constrained and a modest tilt turns into a large required plane change on top of the energy spike. Either way the root cause is the degenerate transfer geometry as $\Delta \theta \to 180°$ (angular momentum → its singular value, $a$ runs to its extreme), not "many planes through a line" alone.

• Below the ridge (shorter TOF): sweep $\Delta \theta <180°$ = Type I transfer.
• Above the ridge (longer TOF): sweep $\Delta \theta >180°$ = Type II transfer.

Edge case — multi-revolution ($N>0$) solutions. Everything above is the zero-revolution ($N=0$) family, where the spacecraft flies straight from ${\vec{r}}_1$ to ${\vec{r}}_2$ without completing a full loop of the Sun. But for long times of flight Lambert's equation admits extra solutions in which the spacecraft loops the Sun $N=1,2,\dots$ full turns before arriving. Each $N$ opens its own pair of Type I / Type II branches, so a full pork chop plot over a wide TOF range can show additional islands stacked at longer flight times, each with its own ridge. Practical Earth–Mars plots usually stay in the $N=0$ family (looping the Sun wastes propellant and time), but a complete treatment must acknowledge these multi-rev branches so you are never surprised by an extra contour cluster high on the plot. See Lambert's Problem.

You pick a lobe and stay off the ridge. See Lambert's Problem and Hohmann Transfer (the ideal $180°$ Hohmann case is cheap only because it is coplanar by construction, dodging the tilt penalty).

Why we can add the two burns at all — the patched-conic idea. The trip is really governed by three bodies (Earth, Sun, Mars), which has no clean solution. The Patched Conic Approximation cuts the journey into three separate two-body problems: (1) a hyperbola escaping Earth's gravity, (2) an ellipse around the Sun, (3) a hyperbola captured by Mars. They are "patched" at the edges of each planet's sphere of influence, where the leftover speed is exactly $v_{\infty }$. Because each piece is a distinct burn in a distinct gravity well, their Δv's do add — but only after each is computed in its own well. That is why we can treat departure and arrival independently below.

Departure (Earth, ${\mu }_E=398600$) — the master burn formula, because you are climbing out of Earth's well from a circular parking orbit: $\Delta v_{dep}=\sqrt{3.2^2+\frac{2(398600)}{6678}}-\sqrt{\frac{398600}{6678}}=\sqrt{10.24+119.36}-7.726=11.384-7.726=3.658\text{km/s}.$ Arrival (Mars, ${\mu }_M=42828$) — same formula, now dropping into Mars' well to a circular capture orbit: $\Delta v_{arr}=\sqrt{2.8^2+\frac{2(42828)}{3689}}-\sqrt{\frac{42828}{3689}}=\sqrt{7.84+23.22}-3.408=5.573-3.408=2.165\text{km/s}.$ Total (patched-conic sum): $\Delta v_{tot}=3.658+2.165=5.82\text{km/s}.$ This single number is one contour value — the plot is this computation repeated across every (launch, arrival) cell.

Why vis-viva? We need the speed of the spacecraft at two points on the Sun-centred transfer ellipse. Conservation of energy on any orbit says the specific energy $\epsilon =\frac{v^2}{2}-\frac{\mu }{r}$ is constant and equals $-\frac{\mu }{2a}$. Solving for $v$ gives the vis-viva equation $v=\sqrt{\mu \left(\frac{2}{r}-\frac{1}{a}\right)}$ — it answers "how fast am I moving at radius $r$ on an orbit of size $a$?" without needing the full trajectory. That is exactly the question a Hohmann burn asks.

Circular speeds of the two planets (each on its own circle, so $a=r$, and vis-viva reduces to $\sqrt{\mu /r}$): $v_1=\sqrt{\frac{{\mu }_{\odot }}{r_1}}=\sqrt{\frac{1.327\times 10^{11}}{1.496\times 10^8}}=29.78\text{km/s},v_2=\sqrt{\frac{{\mu }_{\odot }}{r_2}}=24.13\text{km/s}.$ The transfer ellipse touches Earth's circle at perihelion and Mars' circle at aphelion, so its semi-major axis is the average radius: $a=\frac{r_1+r_2}{2}=1.8875\times 10^8\text{km}.$ Transfer speeds at each end, from vis-viva: $v_p=\sqrt{1.327\times 10^{11}\left(\frac{2}{1.496\times 10^8}-\frac{1}{1.8875\times 10^8}\right)}=32.73\text{km/s}(\text{at Earth’s radius}),$ $v_a=\sqrt{1.327\times 10^{11}\left(\frac{2}{2.279\times 10^8}-\frac{1}{1.8875\times 10^8}\right)}=21.48\text{km/s}(\text{at Mars’ radius}).$ Burns (each is the speed gap between the planet's circle and the transfer ellipse at that radius): $\Delta v_1=32.73-29.78=2.95\text{km/s},$ $\Delta v_2=24.13-21.48=2.65\text{km/s}.$ Sum — the total heliocentric Hohmann Δv: $\Delta v_{tot}^{Hohmann}=\Delta v_1+\Delta v_2=2.95+2.65=5.60\text{km/s}.$ This is the theoretical cheapest heliocentric transfer — the pork chop's deepest island sits near it. See Hohmann Transfer.

Opportunities recur every synodic period. Number in 10 years ($3652.5$ d): $N=\lfloor \frac{3652.5}{779.9}\rfloor +1=\lfloor 4.68\rfloor +1=4+1=5\text{opportunities}.$ (The $+1$ counts the window open at $t=0$.) Each opens a fresh pork chop; the $\pm 20$-day window is a vertical strip you slide inside each island to find the cheapest cell. Design insight: with only $\sim 5$ shots per decade, missing a window costs $\sim 26$ months — this is why Δv margin and schedule are negotiated together.

$\Delta v_A=\sqrt{9+119.36}-7.726=11.330-7.726=3.604\text{km/s}.$ $\Delta v_B=\sqrt{4.4^2+119.36}-7.726=\sqrt{19.36+119.36}-7.726=11.778-7.726=4.052\text{km/s}.$ Extra Δv for B: $\Delta v_B-\Delta v_A=0.448$ km/s. Days saved: $270-180=90$ d. $\text{penalty}=\frac{0.448\text{km/s}}{90\text{d}}=0.0050\frac{\text{km/s}}{\text{day}}\approx 5.0\frac{\text{m/s}}{\text{day}}.$ Reading it: you pay $\sim 5$ m/s of Δv for each day of flight-time you shave off. For a crewed mission (radiation dose scales with TOF) that may be worth it; for a cheap probe it is not. This is precisely the fast-vs-cheap trade the pork chop plot exists to visualize.

Look at the figure below, which plots departure Δv against time of flight along a single launch column (a vertical slice through one lobe).

Left end of the curve (short TOF). To reach ${\vec{r}}_2$ in very little time the spacecraft must sweep the heliocentric arc fast, which demands a highly eccentric, high-energy conic. High energy means large $v_{\infty }$, and through $\Delta v=\sqrt{v_{\infty }^2+2\mu /r}-\sqrt{\mu /r}$ that means large Δv. So the curve is high on the left (plum note).

Middle of the curve (near-Hohmann). As TOF grows toward roughly $259$ days for Earth–Mars — half the period of the minimum-energy transfer ellipse — the required conic relaxes toward that minimum-energy ellipse. The transfer angle is near the ideal $\sim 180°$ Hohmann geometry (planets well-spaced), $v_{\infty }$ bottoms out, and Δv reaches its minimum: the orange dot, the "eye" of the pork chop.

Right end of the curve (long TOF). Push TOF longer still and the geometry becomes awkward again: the transfer must loiter, the arc overshoots toward the $>180°$ Type-II regime and eventually toward the $180°$ ridge singularity of L3.3, so $v_{\infty }$ and Δv climb back up.

The three regimes together make a bowl, whose bottom sits at intermediate TOF — never on the shortest-TOF diagonal (that is the expensive left edge). On the 2-D pork chop plot this bowl is what closes the contours into an island: the innermost ring is the bowl's floor, off both TOF extremes.