Gravity assist (slingshot) — patched conic, v-infinity vectors: Worked examples

This page is the drill-ground for the slingshot topic. The parent note built the three master equations; here we throw every kind of situation at them — every sign, every geometry, the degenerate edges, a real mission problem, and an exam twist. By the end you should never meet a flyby scenario you have not already seen worked.

Before we start, let us re-state the toolkit in plain words so no symbol sneaks in undefined.

Definition The five symbols we will use, in words

$V_{p}$ — the planet's velocity as seen from the Sun. Length $V_{p}$ . A fat arrow the planet drags around its solar orbit.
$v_{\infty}$ — the spacecraft's velocity as seen from the planet, measured far out at the edge of the Sphere of influence (SOI) where the planet's grip fades. Its length $v_{\infty}$ ("vee-infinity") never changes during a flyby.
$v_{sc}$ — the spacecraft's velocity as seen from the Sun (heliocentric). It obeys the addition rule $v_{sc} = V_{p} + v_{\infty}$ (this is just Galilean velocity addition). We write $v_{sc}^{in}$ for its magnitude just before the flyby and $v_{sc}^{o u t}$ for its magnitude just after — the superscripts "in"/"out" always mean pre-flyby and post-flyby.
$e$ — the eccentricity of the planet-centred hyperbola: how "open" the flyby path is. For a genuine flyby $e > 1$ (the value $e = 1$ is a parabolic limit, never actually reached).
$δ$ — the turn angle: how far the planet's gravity swivels $v_{\infty}$ between coming in and going out.

Master trio: $v_{\infty} = const, e = 1 + \frac{r _{p} v _{\infty}^{2}}{μ _{p}}, sin \frac{δ}{2} = \frac{1}{e} .$ Here $r_{p}$ is the closest-approach distance and $μ_{p}$ is the planet's gravity strength ( $= G M_{p}$ ).

The scenario matrix

Every flyby problem is one (or a mix) of these cells. The worked examples below are tagged with the cell they hit, and together they touch all of them.

#	Cell class	What makes it distinct	Covered by
A	Speed-up geometry ( $v_{\infty}$ rotates toward $+ V_{p}$ )	pass behind the planet	Ex 1
B	Speed-down geometry ( $v_{\infty}$ rotates toward $- V_{p}$ )	pass in front	Ex 2
C	General oblique angle (neither aligned nor anti-aligned)	full vector addition, law of cosines	Ex 3 (figure)
D	Limiting max case $δ \to 18 0^{\circ}$	perfect reversal, gain $\to 2 v_{\infty}$	Ex 4
E	Degenerate: grazing / tiny $r_{p}$ vs huge $r_{p}$	$e \to 1$ vs $e \to \infty$	Ex 5
F	Zero input edge $v_{\infty} = 0$ (co-moving)	no boost possible	Ex 6
G	Real mission word problem	pick numbers, choose a side	Ex 7 (Voyager-style)
H	Exam twist: solve backwards	given $Δ v$ , find $r_{p}$	Ex 8

Ex 1 — Cell A: pass behind, gain speed

Forecast: guess whether the after-speed is bigger than $10$ , and by roughly how much, before reading on.

Before flyby. $v_{\infty}$ is anti-parallel to $V_{p}$ , so heliocentric speed is a straight subtraction: $v_{sc}^{in} = ∣ V_{p} - v_{\infty} ∣ = 10 - 4 = 6 km/s$ . Why this step? When two vectors point opposite ways, their sum's length is the difference of lengths.
After a $9 0^{\circ}$ turn. Now $v_{\infty}$ points perpendicular to $V_{p}$ . The two arrows form a right angle, so use Pythagoras: $v_{sc}^{o u t} = V_{p}^{2} + v_{\infty}^{2} = 100 + 16 = 116 = 10.77 km/s$ . Why this step? $v_{sc} = V_{p} + v_{\infty}$ ; for a right angle the sum's length is the hypotenuse.
Gain. $Δ v = 10.77 - 6 = 4.77 km/s$ .

Verify: After a behind pass the craft should be faster than the naive $6$ — it is ( $10.77$ ), so the geometry (rotating toward $+ V_{p}$ ) matches Cell A. Units: all km/s throughout. ✓

Ex 2 — Cell B: pass in front, lose speed

Forecast: will the craft end faster or slower than it started?

Before. $v_{\infty}$ parallel to $V_{p}$ : straight addition, $v_{sc}^{in} = V_{p} + v_{\infty} = 10 + 4 = 14 km/s$ . Why this step? Same-direction vectors add in length.
After $9 0^{\circ}$ . $v_{\infty}$ is now perpendicular: $v_{sc}^{o u t} = V_{p}^{2} + v_{\infty}^{2} = 116 = 10.77 km/s$ . Why this step? The magnitude of the vector sum with a right angle is the hypotenuse — same formula as Ex 1, but here it represents a drop from 14.
Change. $Δ v = 10.77 - 14 = - 3.23 km/s$ — a loss.

Verify: Front-side pass must slow the craft (used by MESSENGER to reach Mercury). We got a negative $Δ v$ . ✓ Note: same $v_{\infty}$ , same $δ$ as Ex 1, but opposite sign of result because the starting orientation and side differ — this is exactly why "which side" (Cell A vs B) is the whole game.

Ex 3 — Cell C: general oblique angle (figure)

Forecast: with such an oblique geometry, is the gain bigger or smaller than the tidy $9 0^{\circ}$ cases above?

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

The figure shows both addition triangles sharing the same black $V_{p}$ base. The red dashed arrow is the outgoing heliocentric velocity $v_{sc}^{o u t}$ (the key object); the black dashed arrow is the inbound $v_{sc}^{in}$ . Notice the red arrow points more along $V_{p}$ than the black one — that is the visual meaning of "gained speed."

We use the law of cosines on the triangle $v_{sc} = V_{p} + v_{\infty}$ . If the angle between the two arrows is $ϕ$ , then $v_{sc}^{2} = V_{p}^{2} + v_{\infty}^{2} + 2 V_{p} v_{\infty} cos ϕ .$ Why the plus sign? Because we are adding the vectors tip-to-tail; the interior triangle angle is $18 0^{\circ} - ϕ$ , and $cos (18 0^{\circ} - ϕ) = - cos ϕ$ , which flips the usual minus into a plus.

Inbound, $ϕ_{in} = 15 0^{\circ}$ , $cos 15 0^{\circ} = - 0.8660$ : $v_{sc}^{in} = 169 + 36 + 2 (13) (6) (- 0.8660) = 205 - 135.1 = 69.9 = 8.36 km/s .$ Why this step? The obtuse angle means $v_{\infty}$ largely opposes $V_{p}$ , so we expect a low heliocentric speed.
Outbound, $ϕ_{o u t} = 8 0^{\circ}$ , $cos 8 0^{\circ} = 0.1736$ : $v_{sc}^{o u t} = 169 + 36 + 2 (13) (6) (0.1736) = 205 + 27.1 = 232.1 = 15.23 km/s .$ Why this step? The near-right angle means $v_{\infty}$ now sides with $V_{p}$ , so heliocentric speed jumps.
Gain. $Δ v = 15.23 - 8.36 = 6.87 km/s$ .

Verify: The gain must respect the cap $2 v_{\infty} = 12 km/s$ (see Ex 4) — and $6.87 < 12$ . ✓ The red outgoing vector in the figure leans further toward $V_{p}$ than the black incoming one, matching the speed increase.

Ex 4 — Cell D: the limiting maximum, $δ \to 18 0^{\circ}$

Forecast: can a flyby ever give more than $2 v_{\infty}$ ? Commit to yes or no.

Geometry. Only the direction of $v_{\infty}$ can change; its tip lies on a circle of radius $v_{\infty}$ . The biggest difference between two points on that circle is its diameter, $2 v_{\infty}$ . Why this step? $Δ v = v_{\infty}^{o u t} - v_{\infty}^{in}$ ; the longest chord of a circle is its diameter.
Best case realized as $δ \to 18 0^{\circ}$ : $v_{\infty}$ flips from anti-parallel to parallel with $V_{p}$ . Before $= V_{p} - v_{\infty} = 13 - 6 = 7$ ; after $= V_{p} + v_{\infty} = 13 + 6 = 19 km/s$ . Why this step? Full reversal maximizes the along- $V_{p}$ component swing.
Change. $Δ v = 19 - 7 = 12 = 2 v_{\infty}$ . Exactly the cap.

Verify: $2 v_{\infty} = 2 (6) = 12 = Δ v$ . ✓ A full $18 0^{\circ}$ turn needs $sin (δ /2) = sin 9 0^{\circ} = 1 = 1/ e \Rightarrow e = 1$ , the parabolic limit — not an achievable hyperbola. From $e = 1 + r_{p} v_{\infty}^{2} / μ_{p}$ , reaching $e \to 1$ means $r_{p} \to 0$ ; but physically $r_{p}$ can never fall below the planet's surface/atmosphere ( $r_{p} \geq R_{m i n}$ ). So the true best turn is set by that smallest safe $r_{p}$ , and $2 v_{\infty}$ is an unattainable upper bound, approached but never met. ✓

Ex 5 — Cell E: degenerate radii (grazing vs distant)

Forecast: which pass bends the trajectory more — close or far?

(a) grazing. $e = 1 + \frac{( 71 , 500 ) ( 36 )}{1.267 \times 1 0 ^{8}} = 1 + 0.02032 = 1.02032$ . Then $sin (δ /2) = 1/1.02032 = 0.9801 \Rightarrow δ /2 = 78.5 5^{\circ} \Rightarrow δ = 157. 1^{\circ} .$ Why this step? Small $r_{p}$ → $e$ near $1$ → $sin (δ /2)$ near $1$ → huge turn.
(b) distant. $e = 1 + \frac{( 5 \times 1 0 ^{6} ) ( 36 )}{1.267 \times 1 0 ^{8}} = 1 + 1.4207 = 2.4207$ . Then $sin (δ /2) = 1/2.4207 = 0.4131 \Rightarrow δ /2 = 24.4 0^{\circ} \Rightarrow δ = 48. 8^{\circ} .$ Why this step? Large $r_{p}$ → $e$ big → small turn.

Verify: Closer pass turns far more ( $15 7^{\circ}$ vs $4 9^{\circ}$ ), matching the rule "fly close to bend a lot." As $r_{p} \to \infty$ , $e \to \infty$ and $δ \to 0$ (no bend, straight line) — the correct degenerate limit. ✓

Ex 6 — Cell F: zero input, $v_{\infty} = 0$

Forecast: does a very close, very massive planet still help here?

Eccentricity and turn. $e = 1 + \frac{r _{p} \cdot 0}{μ _{p}} = 1$ , so $sin (δ /2) = 1/ e = 1 \Rightarrow δ /2 = 9 0^{\circ} \Rightarrow δ = 18 0^{\circ}$ . Why this step? Formally the turn is a full reversal — but it is a reversal of a zero-length vector, so it moves nothing. The equations flag the degenerate parabolic limit rather than a real hyperbola.
Boost cap. Maximum change $= 2 v_{\infty} = 2 (0) = 0 km/s$ . Why this step? The tip of $v_{\infty}$ sits at the centre of a zero-radius circle — no chord exists, so $δ = 18 0^{\circ}$ turning nothing yields nothing.

Verify: A craft co-moving with the planet has nothing to rotate, so no matter how massive the planet, $Δ v = 0$ . ✓ This is why you cannot slingshot off a planet you are drifting alongside — you need genuine relative speed. Independent of $μ_{p}$ and $r_{p}$ , correctly.

Ex 7 — Cell G: real mission word problem (Voyager-style)

Forecast: will one Jupiter pass be enough to reach Solar-System escape (heliocentric $≳ 18.5$ km/s at Jupiter's distance)?

Eccentricity. $e = 1 + \frac{( 2 \times 1 0 ^{5} ) ( 100 )}{1.267 \times 1 0 ^{8}} = 1 + 0.15785 = 1.15785.$ Why this step? Feed the mission's $r_{p}$ and $v_{\infty}$ into the eccentricity relation.
Turn angle. $sin (δ /2) = 1/1.15785 = 0.8636 \Rightarrow δ /2 = 59.7 2^{\circ} \Rightarrow δ = 119. 4^{\circ} .$ Why this step? This is the real (not idealized $18 0^{\circ}$ ) rotation the geometry allows.
Heliocentric speed. Inbound $v_{\infty}$ is anti-parallel to $V_{p}$ (angle $18 0^{\circ}$ ); after turning by $δ = 119. 4^{\circ}$ toward $V_{p}$ , the outbound angle is $ϕ = 18 0^{\circ} - 119. 4^{\circ} = 60. 6^{\circ}$ . Law of cosines: $v_{sc}^{o u t} = 1 3^{2} + 1 0^{2} + 2 (13) (10) cos 60. 6^{\circ} = 169 + 100 + 260 (0.4909) = 396.6 = 19.91 km/s .$ Why this step? Convert the rotated $v_{\infty}$ back to the Sun frame via the addition triangle.

Verify: Inbound heliocentric was $13 - 10 = 3 km/s$ ; outbound $19.91 km/s$ — a gain of $16.9$ , below the cap $2 v_{\infty} = 20$ . ✓ And $19.91 > 18.5$ , so yes, a single Jupiter assist can fling a probe to Solar-System escape — historically true for the Voyagers. ✓

Ex 8 — Cell H: exam twist, solve backwards

Forecast: will the required $r_{p}$ be tight (near the planet) or loose?

From $δ$ to $e$ . $sin (δ /2) = sin 3 0^{\circ} = 0.5 = 1/ e \Rightarrow e = 2.$ Why this step? Invert $sin (δ /2) = 1/ e$ ; we work the master trio in reverse.
From $e$ to $r_{p}$ . Rearrange $e = 1 + \frac{r _{p} v _{\infty}^{2}}{μ _{p}}$ : $r_{p} = \frac{( e - 1 ) μ _{p}}{v _{\infty}^{2}} = \frac{( 1 ) ( 3.79 \times 1 0 ^{7} )}{5. 5 ^{2}} = \frac{3.79 \times 1 0 ^{7}}{30.25} = 1.253 \times 1 0^{6} km .$ Why this step? Solve algebraically for the one unknown.

Verify: Plug back: $e = 1 + \frac{( 1.253 \times 1 0 ^{6} ) ( 30.25 )}{3.79 \times 1 0 ^{7}} = 1 + 1.000 = 2.00$ , giving $sin (δ /2) = 0.5$ , $δ = 6 0^{\circ}$ . ✓ Round-trip consistent. That $r_{p} \approx 1.253$ million km is about $21$ Saturn radii — a loose pass, correct for a modest $6 0^{\circ}$ bend. ✓

Recall Which cell decides

sign of the heliocentric change, and which decides its size? Sign: the side you fly (behind → gain, Cell A; front → loss, Cell B) plus your inbound orientation. Size: the turn angle $δ$ (set by $r_{p}$ , $v_{\infty}$ , $μ_{p}$ ), capped at $2 v_{\infty}$ (Cell D).

Recall Why can't a

$v_{\infty} = 0$ flyby ever help, no matter how massive the planet? The heliocentric change is bounded by $2 v_{\infty} = 0$ ; with no relative speed there is no vector to rotate. Mass and $r_{p}$ are irrelevant.