Visual walkthrough — Gravity assist (slingshot) — patched conic, v-infinity vectors

Step 1 — An arrow is a speed and a direction

WHAT. Before any physics, we agree on one tool: a velocity vector. Draw it as an arrow. Its length means "how fast" (say, in km/s), and where it points means "which way it's going."

WHY this tool and not just a number? A plain number (a "speed") can only get bigger or smaller. But a slingshot's whole trick is to change direction while keeping length the same — and a bare number cannot describe a direction. So we need arrows. We will spend the entire page manipulating two arrows.

PICTURE. Look at the figure. The blue arrow is the spacecraft's velocity. Its foot (tail) is where the craft is; its head (tip) points where it heads next; its physical length is the speed. Two arrows are "equal" only if they match in both length and direction.

Step 2 — Two frames = two people watching

WHAT. We introduce reference frames: a frame is simply "who is watching, and are they moving?" We use two watchers. One rides along with the planet (the planet frame). One stands still relative to the Sun (the Sun frame, or heliocentric frame).

WHY two frames? Because the same flyby looks totally different to the two watchers, and the entire paradox ("free speed") lives in the gap between their two views. To the planet-rider, nothing surprising happens. To the Sun-watcher, the craft speeds up. Both are correct.

PICTURE. The planet moves through space with its own velocity — call it ${\vec{V}}_p$ (green arrow, the Planet's velocity in the Sun frame). Separately, the planet-rider watches the craft and measures its own arrow for the craft; call that arrow $\vec{u}$ for now (blue). To convert what the planet-rider sees into what the Sun-watcher sees, we add the green arrow back.

The subscript $sc/\text{Sun}$ reads "spacecraft, relative to the Sun." Read every "$a/b$" as "$a$ as seen by $b$." In Step 3 we will give this planet-frame arrow $\vec{u}$ its proper name.

Step 3 — In the planet's frame, only the direction can change

WHAT. Meet the star of the show — the planet-frame arrow $\vec{u}$ from Step 2, now measured "far away." At the edge of the region where the planet's gravity still matters, the sphere of influence, this arrow gets its official name: ${\vec{v}}_{\infty }$, the craft's velocity as the planet-rider sees it at infinity. Its length $v_{\infty }$ is the hyperbolic excess speed. (So $\vec{u}$ and ${\vec{v}}_{\infty }$ are the same arrow; from here on we only use ${\vec{v}}_{\infty }$.)

WHY does the length stay fixed? Gravity is conservative: swing toward the planet and back out and you leave with exactly the speed you came in with (relative to the planet). Let's prove it with the one energy statement of the two-body problem:

$\epsilon =\underset{\begin{array}{c}\text{kinetic energy}\\ \text{per unit mass}\end{array}}{\underbrace{\frac{v^2}{2}}}-\underset{\begin{array}{c}\text{gravity’s pull, }{\mu }_p=G{M}_p,\\ r=\text{distance to planet}\end{array}}{\underbrace{\frac{{\mu }_p}{r}}}$

Here $\epsilon$ is the specific orbital energy (energy per kilogram) and it is constant the whole flyby. Far away, $r\to \infty$, so the second term dies (${\mu }_p/\infty =0$), leaving

$\epsilon =\frac{v_{\infty }^2}{2}⟹v_{\infty }=\sqrt{2\epsilon }.$

Since $\epsilon$ never changes, the incoming and outgoing lengths are equal: $v_{\infty }^{\text{in}}=v_{\infty }^{\text{out}}$.

PICTURE. In the planet frame the incoming blue arrow and outgoing blue arrow have the same length — they both just touch a dashed circle of radius $v_{\infty }$. All the flyby can do is swing the arrow around that circle.

Step 4 — The path bends: how much? (the turn angle)

WHAT. The craft doesn't pass in a straight line — the planet's gravity bends it. The angle between the incoming blue arrow and the outgoing blue arrow is the turn angle $\delta$ (Greek "delta," here a swing angle).

WHY a hyperbola, and why does that set $\delta$? Because $\epsilon >0$ (the craft has leftover speed at infinity), the orbit is unbound — it is a hyperbola, an open curve with two straight-line asymptotes. The incoming and outgoing ${\vec{v}}_{\infty }$ point exactly along those two asymptotes. So "how sharply does the velocity turn" is "how wide is the hyperbola" — captured by the eccentricity $e$ (how open the curve is; $e=1$ is a parabola, $e\to \infty$ is a straight line).

The shape follows from energy and angular momentum; evaluated at closest approach it collapses to

$e=1+\underset{\begin{array}{c}{r}_p=\text{closest-approach distance}\\ v_{\infty }=\text{excess speed},{\mu }_p=G{M}_p\end{array}}{\underbrace{\frac{r_p{v}_{\infty }^2}{{\mu }_p}}}$

and the geometry of a hyperbola's asymptotes gives the clean link

$\sin \left(\frac{\delta }{2}\right)=\frac{1}{e}.$

WHY the sine, and why half of $\delta$? Drop a line from the planet to the corner where the two asymptotes cross. That line splits the turn into two equal halves of $\delta /2$. In the right triangle formed there, the side "opposite" the half-angle over the "hypotenuse" is $1/e$ — that ratio is the sine of the half-angle. So $\sin (\delta /2)=1/e$ answers "which half-angle has this openness?"

PICTURE. The hyperbola with its two asymptotes; the blue in-arrow and out-arrow lie along them; the turn angle $\delta$ is marked between them, and $r_p$ is the closest the curve gets to the planet.

Step 5 — Add the planet's arrow back: the speed changes

WHAT. Now we cross back into the Sun frame using Step 2's rule. We add the green planet arrow ${\vec{V}}_p$ to the blue ${\vec{v}}_{\infty }$ — before and after the turn.

WHY does the length now change even though $v_{\infty }$ didn't? Because we rotated the blue arrow (Step 4) and then added a fixed green arrow. Rotating one side of a triangle changes the length of the third side. The third side is the Sun-frame velocity — that's the one we care about.

PICTURE. Two triangles share the same green base ${\vec{V}}_p$.

• Incoming: blue ${\vec{v}}_{\infty }^{\text{in}}$ points against the green ${\vec{V}}_p$, so the tip-to-tail sum (the red Sun-frame arrow) is short — the craft is slow.
• Outgoing: the same-length blue arrow, now rotated by $\delta$, points more along green ${\vec{V}}_p$, so the red sum is long — the craft is fast.

Same blue length, different red length. That difference is the free boost.

Step 6 — The hard ceiling: you can never beat $2v_{\infty }$

WHAT. How big can the boost get? The best case is a full flip: the blue arrow reverses direction ($\delta =180^{\circ }$).

WHY exactly $2v_{\infty }$? The blue arrow's tip is stuck on a circle of radius $v_{\infty }$ (Step 3). The largest possible difference between two points on that circle is its diameter — from one side clean across to the other — which is $2v_{\infty }$. No flyby, no planet, no matter how massive, can move the tip farther than across the circle.

PICTURE. The $v_{\infty }$-circle. Incoming blue points left, outgoing blue points right after a $180^{\circ }$ flip; the change $\Delta \vec{v}$ (red) spans the full diameter $=2v_{\infty }$.

Step 7 — Which side you pass decides gain vs. loss

WHAT. The turn is fixed in size by $e$, but you choose its sense by choosing which side of the planet you fly.

WHY does side matter? Passing behind the planet (through its wake) rotates the blue arrow toward $+{\vec{V}}_p$ — the outgoing red arrow lengthens → speed gain. Passing in front rotates it toward $-{\vec{V}}_p$ → speed loss (deliberately used to slow down, e.g. MESSENGER braking into Mercury).

PICTURE. Same planet, two trajectories. Behind-pass: red Sun-frame arrow grows. Front-pass: red arrow shrinks. Identical physics, opposite outcome — your choice of geometry.

Worked numbers, re-checked visually

The one-picture summary

Everything on this page is one image: a fixed-length blue arrow whose tip rides a circle, plus a fixed green planet arrow added at both ends. The turn angle $\delta$ (set by how close and slow you fly) swings the blue arrow; adding green converts blue's rotation into a red length change; the red change can never exceed the circle's diameter, $2v_{\infty }$.