Patched conic method — interplanetary trajectory design: Visual walkthrough

Definition Subscript legend (fixed for the whole page)

To stop symbols wobbling, we use one rule: the subscript names the reference — either the circle it belongs to or the place it is measured.

$v_{c, 1}$ — Earth's own circular speed around the Sun (planet 1 = Earth). Measured relative to the Sun.
$v_{c i r c}$ — the spacecraft's circular speed in its parking orbit around Earth. Relative to Earth.
$v_{p}$ — spacecraft speed at perigee (the low point of the escape path). Relative to Earth.
$v_{esc}$ — the escape speed from radius $r_{p}$ . Relative to Earth.
$v_{\infty}$ — the leftover speed at infinity, i.e. after fully escaping Earth. Relative to Earth.
$Δ v$ — the speed change the engine supplies at perigee. Relative to Earth.

Everything except $v_{c, 1}$ is measured relative to Earth; $v_{c, 1}$ is the one Sun-relative speed, used only in the final vector-addition step.

Step 1 — What "speed the trip needs" even means

WHAT. Before any formula, we set up the cast of characters as pictures of motion. Earth moves around the Sun on a near-circle at speed $v_{c, 1}$ . A spacecraft parked just above Earth's atmosphere also moves — around Earth — at speed $v_{c i r c}$ . Two different "central bodies", two different speeds to keep straight.

WHY. The single biggest source of confusion in this whole subject is mixing up "speed relative to the Sun" with "speed relative to Earth." So we name them and anchor each to a picture first, then never let them blur. We meet $v_{c, 1}$ now, set it aside, and it returns for the crucial vector-addition in Step 8.

PICTURE. In the figure below, the big straight arrow is Earth's motion around the Sun ( $v_{c, 1}$ ). The small loop is the spacecraft's parking orbit around Earth ( $v_{c i r c}$ ). Two speeds, two centres.

Two speeds, two pictures, never confused again.

Step 2 — $v_{\infty}$ : the speed you have left over after climbing out

WHAT. We introduce the single most important quantity of the departure problem: $v_{\infty}$ , read "vee-infinity." It is the speed the craft still has, relative to Earth, once it is infinitely far from Earth — i.e. after it has completely escaped Earth's grip.

WHY this symbol and not just "the escape speed"? Escaping means reaching infinity with zero speed left. But we don't want to arrive at the edge of Earth's Sphere of influence exhausted — we need to be still moving, because the Sun-centred transfer needs that motion. So we invent a name for the leftover: $v_{\infty}$ . The subscript $\infty$ literally means "measured out at infinity, where Earth no longer pulls."

PICTURE. Roll a ball up a hill. If it just barely crests, it arrives at the top with zero speed. If you push harder, it crests and keeps rolling — that leftover roll is $v_{\infty}$ . The hill is Earth's gravity well.

Before we can write the energy equation, we need one bookkeeping quantity. As a ball climbs the hill it trades kinetic energy (energy of motion) for potential energy (energy of height). Their sum — call it the specific orbital energy $ε$ (Greek letter epsilon; "specific" = per kilogram, so the craft's mass cancels) — stays constant along any engine-off coast, because gravity does no net accounting mischief: whatever kinetic energy you lose climbing, you gain back as height, and vice versa. We give this conserved sum the single letter $ε$ so we can write it once and reuse it at two different points. Let $v$ be the craft's speed in the Earth-centred frame at radius $r$ (distance from Earth's centre). Then:

Formula The energy of an escape path

$ε = kinetic, per kg, Earth-frame speed v \frac{v ^{2}}{2} - potential well, depth at radius r \frac{μ _{E}}{r} = what’s left at infinity \frac{v _{\infty}^{2}}{2}$

$ε$ — the specific orbital energy (energy per kilogram), constant along any coast. This constancy is the whole reason we can compare two points.
$v$ — the craft's speed measured in the Earth-centred frame, at radius $r$ .
$r$ — distance from Earth's centre to the craft.
$\frac{v ^{2}}{2}$ — kinetic energy per kg at speed $v$ .
$μ_{E} = G m_{E}$ — Earth's gravitational parameter. $G$ is Newton's constant, $m_{E}$ is Earth's mass; bundling them saves ink and is known very precisely.
$\frac{μ _{E}}{r}$ — the depth of the well at distance $r$ . Bigger when $r$ is small (deep in the well).
As $r \to \infty$ , $\frac{μ _{E}}{r} \to 0$ , so all that survives is $\frac{v _{\infty}^{2}}{2}$ . That is the definition of $v_{\infty}$ made mathematical.

Because $ε$ is the same at every point of a coasting path, we can evaluate it at perigee and at infinity and set them equal. That trick is the whole game.

Step 3 — Turning "leftover speed" into "speed at perigee"

WHAT. We now find $v_{p}$ , the speed the craft must have at perigee — the low point of its escape path, at radius $r_{p}$ (the parking-orbit radius) — in order to end up with leftover speed $v_{\infty}$ .

WHY. We can't fire the engine "at infinity." We fire it low, at radius $r_{p}$ , where the craft physically is. So we must translate the goal ( $v_{\infty}$ out there) into a requirement ( $v_{p}$ down here). Conservation of $ε$ is exactly the tool that connects two points on the same coasting path — that is what it is for.

PICTURE. Same hill as Step 2, but now we mark two points: the bottom (perigee $r_{p}$ , fast) and infinitely far right (flat, moving at $v_{\infty}$ ). The height climbed converts kinetic energy into potential energy.

Here $r_{p}$ means the perigee radius = Earth's radius $+$ parking-orbit altitude (the closest the escape path comes to Earth's centre). With that defined, set the energy at the two points equal:

Step 4 — The two ingredients side by side

WHAT. We already have $v_{c i r c}$ from Step 1 (the speed the craft has before the burn) and now $v_{p}$ from Step 3 (the speed it needs after the burn). The burn is the difference.

WHY. An engine changes speed. The change we must pay for is exactly "final minus initial" at the burn point — both measured in the same frame (relative to Earth), both at the same place (perigee $r_{p}$ ), and (by our co-planar assumption) both pointing the same direction. Because they share frame, place, and direction, subtracting magnitudes is legal.

PICTURE. Two arrows starting from the same point: the shorter blue one is $v_{c i r c}$ , the longer orange one is $v_{p}$ . The orange sliver of difference is $Δ v$ , the burn.

Step 5 — Why deep burns are cheap (the Oberth heart of it)

WHAT. We look at what the formula does as $r_{p}$ shrinks — i.e. as we burn deeper in Earth's gravity well, closer to the surface.

WHY. This is the non-obvious payoff. Naively you'd think it costs the same $Δ v$ to reach a given $v_{\infty}$ no matter where you burn. It does not. The square-root structure means burning low gives you more $v_{\infty}$ per unit $Δ v$ . This is the Oberth effect, and it falls straight out of "energies add."

PICTURE. Plot $Δ v$ needed versus perigee altitude for a fixed target $v_{\infty}$ . The curve slopes downward as you go lower — deeper is cheaper.

Step 6 — Edge cases: don't get caught out

WHAT. We check the corners of the formula so no scenario surprises the reader.

WHY. A formula you trust only in the "typical" case is a trap. We test four degenerate inputs: $v_{\infty} = 0$ , $r_{p} \to \infty$ , the naive "just add speeds," and the sign of $v_{\infty}$ (departure vs capture).

PICTURE. Mini-panels, one per edge case, each showing the escape path degenerating.

Step 7 — Plug in real numbers (Earth → Mars, 200 km LEO)

WHAT. We run the parent note's worked examples through the pictures we just built, so the abstract steps land as concrete kilometres per second.

WHY. A derivation you can't turn into a number is decoration. Here we confirm $v_{\infty} \approx 2.94 km/s$ , $v_{p} \approx 11.40 km/s$ , $Δ v \approx 3.62 km/s$ .

PICTURE. A labelled number line: $v_{c i r c} = 7.78$ , then the escape rung $v_{esc} = 11.01$ , then the perigee target $v_{p} = 11.40$ , with the burn $Δ v$ marked as the jump from $v_{c i r c}$ to $v_{p}$ .

Step 8 — Where $v_{c, 1}$ finally returns: vector-adding to Earth's motion

WHAT. We promised $v_{\infty}$ is measured relative to Earth. But the transfer ellipse lives in the Sun's frame. So at the edge of the Sphere of influence we must vector-add the craft's Earth-relative leftover $v_{\infty}$ to Earth's own Sun-relative velocity $v_{c, 1}$ to get the craft's heliocentric speed.

WHY. This is the "patch" that stitches the departure hyperbola to the Hohmann transfer orbit. It answers the question the whole page's headline promised: why do we add speeds the way we do? Because velocities are arrows (they have direction), not just numbers — you add them tip-to-tail, not by piling up magnitudes.

PICTURE. Earth's velocity $v_{c, 1}$ (long teal arrow) with the craft's $v_{\infty}$ (orange) laid tip-to-tail. For a Hohmann departure we aim $v_{\infty}$ along $v_{c, 1}$ , so the magnitudes happen to add here: $v_{h e l i o} = v_{c, 1} + v_{\infty}$ .

The one-picture summary

This single figure stacks the whole story: the gravity-well hill, the perigee point where we burn, the two Earth-frame speeds $v_{c i r c}$ (before) and $v_{p}$ (after), the leftover $v_{\infty}$ way out at the flat top, and — off to the side — the tip-to-tail vector sum with Earth's $v_{c, 1}$ that hands the craft to the Sun-frame transfer, all tied together by the boxed formula.

Recall Feynman retelling — say it to a friend

You're on a fast-spinning merry-go-round (your parking orbit) at the bottom of a big smooth hill (Earth's gravity), and the whole playground is itself sailing around a lamppost (the Sun) at speed $v_{c, 1}$ . You want to leap off and still be jogging when you reach the flat plain at the top of the hill — that leftover jog is $v_{\infty}$ , measured against the playground. Question: how much extra push ( $Δ v$ ) does your leap need? Answer: energy is conserved as you climb, so the push you need at the bottom isn't "leftover speed plus climbing speed" piled up — it's added up as energies, under a square root. Because you're already moving fast at the bottom, a small extra push buys a big chunk of energy, so leaping from deep down (low orbit) is cheapest. Finally, to see how fast you're going relative to the lamppost, you add your jog $v_{\infty}$ as an arrow onto the playground's own motion $v_{c, 1}$ — point it the same way and the speeds add, which is exactly the road to Mars. Crunch the numbers and the leap costs about 3.6 km/s from a 200 km orbit.

Recall Active recall — cover the answers

What does $v_{\infty}$ mean physically, and in which frame? ::: The speed still left relative to Earth once infinitely far from Earth (out of its well).
Why do we set energy-at-perigee equal to energy-at-infinity? ::: The specific energy $ε$ is conserved along the engine-off coast, so it's equal at every point of the escape path.
Why isn't $Δ v = v_{\infty} + (v_{esc} - v_{c i r c})$ ? ::: That adds speeds; the truth adds energies, giving $v_{\infty}^{2} + v_{esc}^{2} - v_{c i r c}$ , which is smaller (Oberth).
Why are deep burns cheap? ::: You're fastest at perigee, and adding $Δ v$ to a big speed adds the most kinetic energy, which becomes $v_{\infty}^{2}$ .
What happens to $Δ v$ as $r_{p} \to \infty$ ? ::: It approaches $v_{\infty}$ exactly — no Oberth discount at all.
How does $v_{\infty}$ become a Sun-frame speed? ::: Vector-add it to Earth's orbital velocity $v_{c, 1}$ tip-to-tail; aimed prograde, magnitudes add to $v_{t, 1}$ .
What does $v_{\infty}^{2} < 0$ mean? ::: The craft can't escape — it's bound to Earth (capture), not departure.