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

WHY does a "free" energy boost even exist?

WHAT is the puzzle? Gravity is conservative. If you fall toward a planet and swing back out, you leave with the same speed you came in with (relative to the planet). So how can a slingshot speed you up?

The resolution — frames matter. Energy and speed depend on the reference frame.

• In the planet-centered frame: the flyby is a hyperbola. Speed in = speed out $=v_\infty$. Only the direction changes.
• In the Sun-centered (heliocentric) frame: the planet is moving at $\vec V_p$. When we transform the rotated velocity back to the Sun frame, the magnitude changes.

The Patched-Conic Method

WHY it works: Inside the SOI the planet's pull dominates the Sun's; outside, vice versa. Because the SOI is tiny compared to the heliocentric orbit, we idealize the whole flyby as happening at a single point — so the spacecraft's heliocentric position doesn't change, only its velocity rotates.

v-infinity: the key vector

The link between the two frames is a simple vector addition at the patch point:

$\vec v_{sc/Sun} = \vec V_p + \vec v_\infty$

where $\vec V_p$ is the planet's heliocentric velocity.

Deriving conservation of $v_\infty$ from first principles

Energy in the planet frame (specific orbital energy): $\varepsilon = \frac{v^2}{2} - \frac{\mu_p}{r}$ This is constant (gravity is conservative). As $r\to\infty$ the potential term $\to 0$, so $\varepsilon = \frac{v_\infty^2}{2} \quad\Rightarrow\quad v_\infty = \sqrt{2\varepsilon}.$ Since $\varepsilon$ is fixed by the incoming conditions and doesn't change, ==$v_\infty^{in}=v_\infty^{out}$==. Only the direction of $\vec v_\infty$ rotates.

Deriving the turn (deflection) angle

Inside the SOI the path is a hyperbola. WHY a hyperbola? Because $\varepsilon>0$ (excess speed), an unbound orbit → eccentricity $e>1$.

The asymptotes of a hyperbola make an angle with each other. The incoming and outgoing $\vec v_\infty$ point along these asymptotes, so the velocity rotates by the turn angle $\delta$:

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

Derivation sketch: For a hyperbola the asymptote makes angle $\theta_\infty$ with the axis where $\cos\theta_\infty = -1/e$. The total exterior turn between asymptotes gives $\delta = \pi - 2(\pi-\theta_\infty)$... more usefully, using the orbit shape:

$e = 1 + \frac{r_p v_\infty^2}{\mu_p}$

where $r_p$ = closest approach (periapsis) radius. WHY this form? From the vis-viva/energy relation $e = \sqrt{1 + \frac{2\varepsilon h^2}{\mu_p^2}}$, with $\varepsilon = v_\infty^2/2$ and angular momentum $h = r_p v_p$; algebra at periapsis reduces it to the boxed expression. So:

$\sin\!\left(\frac{\delta}{2}\right)=\frac{1}{1+\dfrac{r_p v_\infty^2}{\mu_p}}$

WHY closer & slower = bigger turn: Small $r_p$ or small $v_\infty$ → smaller $e$ → larger $\delta$. Fly close and slow to bend a lot; scream past fast and far to barely bend.

Worked Example 1 — How much can you gain?

Planet moves at $V_p = 13\ \text{km/s}$. Spacecraft arrives with $v_\infty = 5\ \text{km/s}$. Best-case flyby fully reverses $\vec v_\infty$ (turn by $\delta=180°$, only possible if $v_\infty$ small enough / $r_p$ tiny) so it flips from anti-parallel to parallel with $\vec V_p$.

• Before: $v_{helio} = V_p - v_\infty = 13-5 = 8\ \text{km/s}$ Why this step? Choose incoming $\vec v_\infty$ opposite to $\vec V_p$ (worst inbound speed).
• After: $v_{helio} = V_p + v_\infty = 13+5 = 18\ \text{km/s}$ Why this step? Full reversal makes $\vec v_\infty$ align with $\vec V_p$.
• Gain: $\Delta v = 10\ \text{km/s} = 2v_\infty$. Why this step? The maximum heliocentric change is $2v_\infty$ — you can never gain more than twice your excess speed from one flyby.

Worked Example 2 — Compute the turn angle

Jupiter: $\mu_p = 1.267\times10^{8}\ \text{km}^3/\text{s}^2$. Flyby at $r_p = 100{,}000$ km with $v_\infty = 6$ km/s.

• $e = 1 + \dfrac{r_p v_\infty^2}{\mu_p} = 1 + \dfrac{(10^5)(36)}{1.267\times10^8} = 1 + 0.0284 = 1.0284$ Why this step? Plug periapsis and excess speed into the eccentricity relation.
• $\sin(\delta/2) = 1/1.0284 = 0.9724 \Rightarrow \delta/2 = 76.5° \Rightarrow \delta = 153°$. Why this step? Huge turn because Jupiter is massive and $v_\infty$ is modest. This is why Jupiter is the king of gravity assists (Voyager, Cassini, Juno, Pioneer).

Worked Example 3 — Which side to fly?

You want to speed up heliocentrically.

• Pass behind the planet (planet moving away from you as you cross its wake) → $\vec v_\infty$ rotates toward $+\vec V_p$ → speed gain.
• Pass in front → rotates toward $-\vec V_p$ → speed loss (used to slow down, e.g. MESSENGER to Mercury).

Why this step? The geometry of which asymptote you exit on decides whether $\vec v_\infty^{out}$ has a bigger component along $\vec V_p$.

Active Recall

Flashcards

Connections

• Two-body problem & vis-viva equation
• Hyperbolic orbits & orbital eccentricity
• Sphere of influence (SOI)
• Specific orbital energy & angular momentum
• Interplanetary transfer & Hohmann transfer
• Reference frames & Galilean velocity addition
• Oberth effect (powered flybys)
• Voyager & Cassini mission trajectories

Concept Map

Hinglish (regional understanding)