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

This child page assumes you know nothing. We will name every letter, draw every picture, and only then let the parent page's equations mean something. Read top to bottom — each idea is built from the one above it.

0. What is a "vector"? (the arrow you can add)

Before any physics, the single most important object here is the velocity vector.

We write vectors with a little arrow on top: $\vec{v}$. The plain letter without the arrow, $v$, means just the length of that arrow (its size), which we call the magnitude. So $\vec{v}$ is the arrow and $v=|\vec{v}|$ is how long it is.

Figure 1 (above): on the left, one yellow arrow labelled $\vec{v}$ — notice the label of its length is the magnitude $v$. On the right, watch the blue "first" arrow and pink "second" arrow laid tip-to-tail; the thick yellow "sum" arrow runs from the very start to the very end. That yellow arrow is the whole idea of adding velocities.

Why the topic needs this: the whole slingshot is one arrow-addition, ${\vec{v}}_{sc/Sun}={\vec{V}}_p+{\vec{v}}_{\infty }$. If you cannot add two arrows tip-to-tail, no other equation will make sense. See the Reference frames & Galilean velocity addition note for the deeper rules.

1. Reference frame — "who is watching?"

Two frames run through the entire topic:

• Planet-centered frame — you ride with the planet. Here the planet sits still and the spacecraft swings past it.
• Sun-centered (heliocentric) frame — you float above the Solar System. Here the planet is racing along its orbit and the spacecraft's path is a big curve around the Sun.

Figure 2 (above): the left panel is the planet frame — the blue planet sits still and the craft traces a clean yellow hyperbola past it. The right panel is the Sun frame — now the yellow Sun sits still, and the same encounter happens on a planet (blue dot) that is itself moving along its dashed orbit (blue arrow). Look how the picture of "who is still" flips between panels; that flip is the entire reason a slingshot works.

Why the topic needs this: the "free speed boost" only appears because we switch frames. In the planet frame nothing gets faster; in the Sun frame it does. Confusing the two frames is the #1 way to think a gravity assist violates energy conservation.

2. The subscript notation ${\vec{v}}_{A/B}$ — "velocity of A relative to B"

So:

• ${\vec{v}}_{sc/Sun}$ = velocity of the spacecraft as seen from the Sun.
• ${\vec{v}}_{sc/planet}$ = velocity of the spacecraft as seen from the planet (this is the important one — it becomes ${\vec{v}}_{\infty }$).
• ${\vec{V}}_p$ = velocity of the planet as seen from the Sun (a capital $V$ just flags "this is the big fast planet").

These connect by simple arrow-adding: ${\vec{v}}_{sc/Sun}={\vec{v}}_{sc/planet}+{\vec{V}}_p$

Read it aloud: "spacecraft-from-Sun = spacecraft-from-planet + planet-from-Sun." The middle term (planet) cancels in your head like linking chains. This is Galilean velocity addition.

3. ${\vec{v}}_{\infty }$ — the "velocity at infinity"

The magnitude $v_{\infty }$ (no arrow) is a plain number — kilometres per second. The parent's central claim is: $v_{\infty }$ never changes; only the direction of ${\vec{v}}_{\infty }$ turns.

Why the topic needs this: ${\vec{v}}_{\infty }$ is the one arrow that gets rotated. Rotating it is the slingshot.

4. $r$, $r_p$, $G$, $M_p$, ${\mu }_p$ — distance and the planet's gravity strength

Why the topic needs this: $r_p$ and ${\mu }_p$ set how hard the path bends — closer approach and stronger gravity mean a sharper turn. They are the only two "knobs" (besides $v_{\infty }$) in the turn-angle formula.

5. Specific energy $\epsilon$ — energy per kilogram

The two pieces:

• $\frac{v^2}{2}$ is kinetic energy per kilogram (the standard $\frac{1}{2}m{v}^2$ with the $m$ divided out), using the planet-frame speed $v$.
• $-\frac{{\mu }_p}{r}$ is the gravitational potential energy per kilogram. It is negative and gets less negative (closer to zero) as $r$ grows, because far from the planet you have climbed out of its well.

The key move: far away $r\to \infty$, so ${\mu }_p/r\to 0$, leaving $\epsilon =v_{\infty }^2/2$. Because $\epsilon$ is conserved (gravity does no net work in the planet frame), $v_{\infty }$ is locked. That is the parent's whole proof that $v_{\infty }$ is constant. See Two-body problem & vis-viva equation.

6. Angular momentum $h$ — "how much sideways swing"

At the closest point (periapsis) the velocity is purely sideways, so $\varphi =90°$ and $\sin \varphi =1$, giving the simple special case $h=r_p{v}_p,$ where $v_p$ is the speed exactly at periapsis. We use this clean version because periapsis is the one place the formula loses its angle.

Why the topic needs this: $h$ together with $\epsilon$ fixes the shape of the path (how curved, how sharp the turn). It is the second ingredient in the eccentricity formula below. See Specific orbital energy & angular momentum.

7. Eccentricity $e$ and why the path is a hyperbola

Figure 3 (above): the same planet (white dot) with three possible paths — the blue circle ($e=0$), the pink ellipse ($0<e<1$, a closed loop), and the thick yellow hyperbola ($e>1$, an open curve that leaves forever). A slingshot is always the yellow one. Notice how the curves get progressively "more open" as $e$ grows.

Where $e=1+\frac{r_p{v}_{\infty }^2}{{\mu }_p}$ actually comes from

We are not going to memorise this formula — we are going to build it from the two conserved quantities we just met, $\epsilon$ and $h$.

Step 1 — WHAT the general shape formula is. Solving the two-body problem (done fully in Two-body problem & vis-viva equation) gives eccentricity in terms of the two conserved specific quantities: $e=\sqrt{1+\frac{2\epsilon h^2}{{\mu }_p^2}}.$ Why this shape? $\epsilon$ carries "how much energy" and $h$ carries "how much sideways swing" — together they pin down exactly one conic shape. More energy or more swing → more open curve → bigger $e$.

Step 2 — WHY plug in our two special values. We already found $\epsilon =\frac{v_{\infty }^2}{2}$ (far-away energy) and, at periapsis, $h=r_p{v}_p$. Substituting: $e=\sqrt{1+\frac{2\left(\frac{v_{\infty }^2}{2}\right)(r_p{v}_p{)}^2}{{\mu }_p^2}}=\sqrt{1+\frac{v_{\infty }^2{r}_p^2{v}_p^2}{{\mu }_p^2}}.$

Step 3 — WHY evaluate the speed at periapsis. At periapsis, energy conservation ties $v_p$ to $v_{\infty }$: $\frac{v_p^2}{2}-\frac{{\mu }_p}{r_p}=\frac{v_{\infty }^2}{2}\Rightarrow v_p^2=v_{\infty }^2+\frac{2{\mu }_p}{r_p}.$ This is just "kinetic + potential is the same far out and at periapsis." Putting $v_p^2$ back in and simplifying the algebra collapses the square root exactly into the tidy result: $\boxed{e=1+\frac{r_p{v}_{\infty }^2}{{\mu }_p}}$

WHAT it says in words: a bigger leftover speed $v_{\infty }$, or a farther/faster pass (bigger $r_p{v}_{\infty }^2$), gives a more open (larger $e$) hyperbola that barely bends. A slow, close pass keeps $e$ near 1 — the sharpest turn.

8. Asymptotes and the turn angle $\delta$

Figure 4 (above): the yellow hyperbola sweeps past the blue planet. The pink arrow is ${\vec{v}}_{\infty }$ coming in along the lower asymptote; the blue arrow is ${\vec{v}}_{\infty }$ going out along the upper asymptote. The white arc between them is the turn angle $\delta$ — the rotation the encounter delivered. The dashed line marks the closest approach $r_p$. Everything in the slingshot is "how big is that white arc?"

Where $\sin \left(\frac{\delta }{2}\right)=\frac{1}{e}$ comes from

This is the payoff, so let us earn it geometrically.

Step 1 — WHAT an asymptote angle is. A hyperbola has two straight asymptotes. Each one makes a fixed angle ${\theta }_{\infty }$ with the axis of the hyperbola (the symmetry line through the planet and periapsis). Standard conic geometry (see Hyperbolic orbits & orbital eccentricity) gives the direction of that asymptote by $\cos {\theta }_{\infty }=-\frac{1}{e}.$ Why $-1/e$? On a hyperbola the spacecraft can only reach angles up to ${\theta }_{\infty }$ before it is infinitely far away; setting the orbit's radius to infinity in the conic equation forces exactly this cosine.

Step 2 — WHY the turn angle is the supplement. The incoming and outgoing ${\vec{v}}_{\infty }$ lie along the two asymptotes. Because the two asymptotes are mirror images about the axis, the angle the velocity actually rotates through, $\delta$, is related to ${\theta }_{\infty }$ by $\delta =\pi -2\left(\pi -{\theta }_{\infty }\right)=2{\theta }_{\infty }-\pi .$ In words: the velocity started aimed inward along one asymptote and ends aimed outward along the other; the leftover rotation is $\delta$.

Step 3 — WHY it collapses to a clean sine. Take the boxed relation from Step 1, $\cos {\theta }_{\infty }=-1/e$, and rewrite the half-turn. From $\delta =2{\theta }_{\infty }-\pi$ we get ${\theta }_{\infty }=\frac{\delta +\pi }{2}$, so $\cos {\theta }_{\infty }=\cos \left(\frac{\delta }{2}+\frac{\pi }{2}\right)=-\sin \left(\frac{\delta }{2}\right).$ Setting the two expressions for $\cos {\theta }_{\infty }$ equal, $-\sin (\delta /2)=-1/e$, the minus signs cancel and we land on $\boxed{\sin \left(\frac{\delta }{2}\right)=\frac{1}{e}}$

WHAT it says in words: a hyperbola close to $e=1$ (a slow, close pass) has $\sin (\delta /2)$ near 1, so $\delta$ near $180°$ — a huge turn. A very open hyperbola (large $e$) has $\sin (\delta /2)$ near 0, so $\delta$ near $0°$ — barely deflected. This is exactly the parent's "closer and slower = bigger turn."

9. The infinity symbol, the Greek letters, and the "delta" of change

A quick glossary so no squiggle trips you up:

• $\infty$ — "infinity", meaning "arbitrarily far away."
• $\mu$ — Greek "mu", used for the gravity parameter.
• $\epsilon$ — Greek "epsilon", used for specific energy.
• $\delta$ — Greek "delta", the turn angle.
• $\Delta$ (capital delta) — "the change in." So $\Delta \vec{v}={\vec{v}}_{out}-{\vec{v}}_{in}$ is "final arrow minus initial arrow," the net kick.

How it all feeds the topic

Equipment checklist

Test yourself — cover the right side and answer aloud.