Halo orbits — linearized motion near Lagrange points: Foundations

This is the ground-floor page for the parent topic. Every arrow, letter, and dot that the parent uses is defined here from absolute zero, in the order that each one needs the previous one.

1. Position, and what a dot over a letter means

Before anything moves, we must say where it is.

Now, motion means position changing in time. Physicists mark "rate of change" with a dot on top of the letter — this is Newton's shorthand.

Why the topic needs this: the equations of motion are rules about acceleration — they say "given where you are, here is your $\overset{x}{¨}$ ." Everything downstream is an acceleration bookkeeping exercise.

Figure s01. Three curves against time for a body that speeds up steadily: the blue curve is position $x$ (rising ever faster), the yellow line is its slope $\overset{x}{˙}$ (velocity, rising straight), and the flat red line is $\overset{x}{¨}$ (acceleration, constant). The white vertical arrows show that each dot answers a "how fast is the one below it changing?" question.

2. The mass ratio $μ$

Two big bodies (the primaries) pull on the spacecraft. Before we can fix units or place them, we need one number describing how their masses compare.

We define $μ$ first because the very next step (choosing units) will make the two masses come out as $1 - μ$ and $μ$ .

3. The spin rate $ω$

The two primaries orbit their common balance point at a steady rate. That rate has a name.

We define $ω$ before using it in the normalization below, because the whole point of the normalization is to set $ω = 1$ .

4. Normalizing the units so that $ω = 1$

Before we write any equation we must fix our rulers and clocks. The whole CR3BP is written in dimensionless (normalized) units — a bookkeeping trick that hides messy constants.

Definition The normalization convention

We choose our units so that four things equal $1$ :

Unit length = the constant distance between the two primaries (so they sit at fixed spots).
Unit mass = the total mass of the pair, $m_{1} + m_{2}$ . With this choice the bigger body's mass becomes $\frac{m _{1}}{m _{1} + m _{2}} = 1 - μ$ and the smaller body's becomes $\frac{m _{2}}{m _{1} + m _{2}} = μ$ (using $μ$ from §2), and they add to $1$ .
Gravitational constant $G = 1$ (folded into the units).
Unit time = chosen so the pair completes one radian of its orbit per time unit, which forces the spin rate $ω = 1$ (using $ω$ from §3). Picture rescaling every ruler and clock in the problem until the two bodies are exactly one metre apart, weigh one kilogram between them, and turn at one radian per second. Every formula on this page lives in these units.

Why the topic needs this: with $G$ , the masses, the separation, and $ω$ all set to $1$ , the equations lose their clutter and the balance points sit at clean, fixed coordinates. Answers come out as pure numbers (like "period $\approx 3.4$ ") with no kilograms or seconds attached.

5. Where things sit: origin, axes, and the positions of the primaries

Units alone are not enough — we must say where the zero is and which way $x$ points.

Figure s05. The rotating-frame layout seen from above. The white cross at the origin is the barycentre; the yellow dot (big body, mass $1 - μ$ ) sits at $x = - μ$ and the red dot (small body, mass $μ$ ) sits at $x = 1 - μ$ , one unit apart. The blue arrow shows the positive $x$ -axis pointing from big toward small. The blue spacecraft dot has distances $r_{1}$ (to the big body) and $r_{2}$ (to the small body) drawn as dashed lines.

Now the distances the parent uses have explicit formulas.

Why the topic needs this: without these explicit formulas we could not differentiate the potential $Ω$ (§7) — the slopes $\partial Ω/ \partial x$ would be undefined.

6. The rotating frame and its two extra forces

The Sun and a planet (or Earth and Moon) go round each other. Instead of watching from outside, we ride along, spinning with them at rate $ω = 1$ , so the two big bodies sit still in our view (exactly the fixed positions of §5).

Why the topic needs this: in a still (non-rotating) frame the two big bodies keep moving, and there is no fixed balance point to hover near. Spinning with them freezes them in place, which is the only way "sit next to $L_{1}$ forever" even makes sense.

But riding a merry-go-round has a price: two extra apparent (fictitious) forces appear.

Definition Coriolis effect and where the factor of

$2$ comes from A sideways deflection that only acts on things that are moving in the rotating frame. The general fictitious-force law for a frame spinning at $ω$ is $a_{Coriolis} = - 2 ω \times v,$ where $v = (\overset{x}{˙}, \overset{y}{˙}, \overset{z}{˙})$ is the velocity and $\times$ is the cross product (a rule that produces a vector perpendicular to both inputs). The factor $2$ is not a fudge — it drops out of the calculus of transforming acceleration into a rotating frame (differentiating a rotating position vector twice produces the rotation rate twice). With $ω = (0, 0, 1)$ (spin about $z$ at unit rate), the cross product gives $- 2 (0, 0, 1) \times (\overset{x}{˙}, \overset{y}{˙}, \overset{z}{˙}) = - 2 (- \overset{y}{˙}, \overset{x}{˙}, 0) = (2 \overset{y}{˙}, - 2 \overset{x}{˙}, 0) .$ Moving these to the left-hand side of "acceleration = force" flips their signs, producing exactly the parent's $- 2 \overset{y}{˙}$ in the $x$ -line and $+ 2 \overset{x}{˙}$ in the $y$ -line. The $z$ -line gets nothing because the spin axis is along $z$ : rotating about $z$ cannot deflect vertical motion.

Figure s02. A spinning disk seen from above. The blue dot is the spacecraft, its blue arrow is how it is moving. The red arrow (centrifugal) always points straight outward from the yellow spin axis and needs only where the body is. The green arrow (Coriolis) points sideways to the motion and appears only because the body is moving. This is the visual root of the two force families.

7. The effective potential $Ω$ and its gradient $\nablaΩ$

We now package "centrifugal + both gravities" into one landscape.

To talk about slopes we need the gradient.

Why "partial" and not ordinary derivative? Because $Ω$ depends on three variables at once. To isolate how it changes with one of them, we freeze the others. The curly $\partial$ is the flag that says "others held fixed."

Now we can finally write the parent's central equations — the rules for acceleration in the rotating frame.

Figure s03. Contour lines of $Ω$ in the $x$ – $y$ plane. The yellow dot is the big body, the red dot the small body — each sits in a tight nest of rings (a tall spike of $Ω$ ). Contours crowd where the landscape is steep. The green $\times$ marks a Lagrange point: the one spot where the rings open out and the ground is momentarily flat ( $\nablaΩ = 0$ ) — the balance we hover near.

8. Displacement $(ξ, η, ζ)$ — the "zoom-in" coordinates

We do not care about the spacecraft's absolute position; we care how far it strays from the balance point.

Why the topic needs this: small quantities are what let us linearize (next section). If we tried to keep the full nonlinear $Ω$ , we could not solve it.

9. Taylor expansion and "linear"

This is the engine that turns an unsolvable problem into a solvable one.

Because the balance point is flat ( $\nablaΩ = 0$ ), the constant term of the expansion is zero, and the first surviving term is the linear one — proportional to $(ξ, η, ζ)$ .

10. From $Ω$ to $U$ : second derivatives and the constant $c$

The stiffness of our "spring" is set by how the slope itself changes — that is a second derivative.

Formula Full derivation of

$U_{xx} = 1 + 2 c, U_{y y} = 1 - c, U_{z z} = - c$ We differentiate $Ω = \frac{1}{2} (x^{2} + y^{2}) + \frac{1 - μ}{r _{1}} + \frac{μ}{r _{2}}$ twice and then evaluate on the $x$ -axis, where $y = z = 0$ so both bodies line up with the point along $x$ .

Piece A — the centrifugal bowl $\frac{1}{2} (x^{2} + y^{2})$ .

\frac{\partial^2}{\partial y^2}[\cdots] = 1,\quad \frac{\partial^2}{\partial z^2}[\cdots] = 0$$ (there is no $z$ in it). Contribution: $(+1, +1, 0)$. **Piece B — one gravity well $\frac{m}{r}$** with $r=\sqrt{(x-x_0)^2+y^2+z^2}$. A standard result for the inverse-distance potential is that its second derivatives, evaluated on the axis through the source (so the offset is purely along $x$), are $$\frac{\partial^2}{\partial x^2}\!\left(\frac{m}{r}\right) = +\frac{2m}{r^3}\ \text{(along the line to the body)},\qquad \frac{\partial^2}{\partial y^2}\!\left(\frac{m}{r}\right) = \frac{\partial^2}{\partial z^2}\!\left(\frac{m}{r}\right) = -\frac{m}{r^3}\ \text{(perpendicular)}.$$ The $+2$ along the line and $-1$ perpendicular are the classic **tidal** pattern (stretch along the line to the mass, squeeze across it), and they must sum to zero — $2 - 1 - 1 = 0$ — because the $1/r$ potential satisfies Laplace's equation away from the mass. **Sum over both bodies.** With masses $1-\mu$ (at $r_1$) and $\mu$ (at $r_2$), the gravity contribution to each curvature is: - along $x$: $+\frac{2(1-\mu)}{r_1^3} + \frac{2\mu}{r_2^3} = +2c$, - along $y$: $-\frac{1-\mu}{r_1^3} - \frac{\mu}{r_2^3} = -c$, - along $z$: $-\frac{1-\mu}{r_1^3} - \frac{\mu}{r_2^3} = -c$, using the definition $c = \frac{1-\mu}{r_1^3} + \frac{\mu}{r_2^3}$. **Add Piece A + Piece B:** $$U_{xx} = 1 + 2c,\qquad U_{yy} = 1 - c,\qquad U_{zz} = 0 + (-c) = -c.$$ Since $c>1$, only $U_{xx}$ is positive (a valley/spring in $x$); $U_{yy}$ and $U_{zz}$ are negative (hills/anti-springs). That mixed sign is exactly what makes the point a **saddle**.

11. Exponentials, sinusoids, $λ$ , $γ$ , $ω_{x y}$ , $ω_{z}$

When we guess a solution to a spring-like equation, we try $e^{λ t}$ .

Figure s04. Two displacement-versus-time curves showing what $λ$ decides. The red curve is $e^{γ t}$ for a real positive $λ$ — it runs away without bound (instability). The green curve is the wobble for an imaginary $λ$ — a bounded sine that oscillates forever at frequency $ω$ . Collinear Lagrange points carry both at once, which is the saddle signature.

12. Amplitudes, $κ$ , Lissajous vs halo

The prerequisite map

graph TD
  A["Position x y z and dots"] --> V["Vector r and velocity v"]
  V --> MU["Mass ratio mu"]
  MU --> W["Spin rate omega"]
  W --> N["Normalized units omega equals 1"]
  N --> O["Origin at barycentre, x axis big to small"]
  O --> R["Distances r1 and r2 with formulas"]
  R --> B["Rotating frame"]
  B --> C["Centrifugal and Coriolis minus 2 omega cross v"]
  C --> E["Effective potential Omega"]
  E --> EQ["Equations of motion"]
  EQ --> F["Gradient equals zero = Lagrange point"]
  F --> G["Displacement xi eta zeta"]
  G --> H["Taylor expand to linear"]
  H --> I["