Three-body problem — restricted (CR3BP), characteristic equation: Foundations

This page assumes nothing. Every letter, every squiggle, every fancy word the parent topic leans on gets built here, in an order where each brick rests on the one before it.

1. A vector and its position — where things are

Picture it: stand a pencil on a table. Pick a corner as the origin. The straight arrow from that corner to the pencil tip is $r$ . The three numbers $(x, y, z)$ are just how far along, across, and up you walked to get there.

Figure — Three-body problem — restricted (CR3BP), characteristic equation

Why the topic needs it: the entire problem is "where is the little body, and where is it going?" You can't ask that without a way to name a location. That name is $r = (x, y, z)$ .

The dot on top — $\dot{r}$ — means velocity (how fast the position is changing, an arrow pointing where you're headed). Two dots, $\ddot{r}$ , means acceleration (how fast the velocity itself is changing). One dot = rate of change; two dots = rate of change of the rate of change.

Recall

What does $\overset{y}{˙}$ mean in plain words? ::: The rate at which the $y$ -coordinate is changing — the up/down component of velocity.

2. Mass and the mass parameter $μ$ — who is heavy

In this problem there are exactly three masses: $m_{1}$ (big primary, e.g. the Sun or Earth), $m_{2}$ (smaller primary, e.g. the Earth or Moon), and $m_{3}$ (the tiny third body — a spacecraft or asteroid). "Restricted" means $m_{3}$ is so light it feels the others but never tugs back on them.

Picture it: put both primaries on a seesaw. $μ$ tells you how the total weight is split. If $μ = 0.01$ , the small body is $1%$ of the pair — the big one dominates. If $μ = 0.5$ , they're twins.

Why the topic needs it: every equation later depends only on this one ratio $μ$ , not on the raw kilograms. Sun–Jupiter and Sun–Saturn behave "the same shape," differing only through their $μ$ . It compresses the whole zoo of mass pairs into a single dial.

Recall

If $m_{1} = m_{2}$ , what is $μ$ ? ::: $μ = \frac{1}{2}$ — the maximum, an equal-mass pair.

3. Nondimensional units — choosing rulers, clocks and scales so the numbers vanish

We make three independent choices, one for each basic dimension:

Length unit = the distance between the two primaries. This forces their separation $= 1$ .
Mass unit = the total mass of the two primaries. This forces $m_{1} + m_{2} = 1$ , so directly $m_{1} = 1 - μ$ and $m_{2} = μ$ (the mass parameter $μ = m_{2} / (m_{1} + m_{2})$ becomes the small mass itself).
Time unit = chosen so the gravitational constant $G = 1$ .

So the four scalings $G = 1$ , $m_{1} + m_{2} = 1$ , separation $= 1$ , and $ω = 1$ are all consistent together — the first three are our free choices, the last is their consequence.

Recall

In nondimensional units, what is $m_{2}$ equal to? ::: Exactly $μ$ — because we set $m_{1} + m_{2} = 1$ , the small mass is the mass parameter.

4. Restricting to the orbital plane

Why it matters here: although position $r = (x, y, z)$ and the gradient live happily in full 3D, for finding and picturing the equilibria we deliberately restrict the motion to $z = 0$ — the little body stays in the primaries' plane. That's why from the effective potential onward you'll see $U (x, y)$ with no $z$ : we've set $z = 0$ . (The out-of-plane $z$ direction returns only when studying vertical wobbles; the parent topic keeps to the plane for the Lagrange-point hunt.)

5. The barycentre — the balance point everything circles

Picture it: heavy adult and light child on a seesaw. The pivot sits close to the adult. Likewise the barycentre sits close to the big primary. Both primaries orbit this point, not each other's centres.

In our units the barycentre is the origin $(0, 0)$ . The big primary $m_{1} = 1 - μ$ sits at $x = - μ$ ; the small primary $m_{2} = μ$ sits at $x = 1 - μ$ . Their separation is $(1 - μ) - (- μ) = 1$ , exactly as our length unit demands.

What to notice in the figure: the big black dot (mass $1 - μ$ ) sits only a whisker left of the origin at $x = - μ$ , while the small dot ( $μ$ ) sits far to the right at $x = 1 - μ$ . The red barycentre dot is at $0$ — hugging the heavy body, exactly like the seesaw pivot near the adult. Confirm with your eye that the two dashed distances from the red dot are inversely weighted: the heavy body is close, the light body is far.

Why the topic needs it: anchoring the origin at the barycentre makes the rotating-frame equations clean and symmetric — the centrifugal "bowl" (next section) is centred right here.

Related vault reading: Two-body problem & Kepler orbits builds why two masses orbit a common barycentre in the first place.

6. The rotating reference frame — spinning the camera

Picture it: you're on a merry-go-round. To someone on the ground (inertial view) the horses whirl past. To you (rotating view) the horse beside you is frozen still, but the whole outside world seems to wheel around.

The CR3BP trick: ride a frame that turns at exactly $ω$ , the primaries' orbital rate. In that frame the two primaries are nailed in place. See Rotating reference frames — Coriolis & centrifugal forces for the full machinery.

The price — fictitious forces. Spinning your viewpoint makes objects appear to be pushed by forces that aren't "real" pulls:

Centrifugal — the outward fling you feel on the merry-go-round; it grows with distance from the spin axis.
Coriolis — the sideways deflection of anything moving in the rotating frame; it always shoves perpendicular to your velocity, so it bends your path without ever speeding you up or slowing you down.

What to notice in the figure: the body (black dot) is moving with the black velocity arrow, but the red Coriolis arrow points at a perfect right angle to it. Trace both arrows with your finger — they meet at $9 0^{\circ}$ . That right angle is the entire reason Coriolis can steer a path without ever adding or removing speed.

Where the centrifugal term comes from. In the rotating frame the outward centrifugal force per mass at distance $ρ = x^{2} + y^{2}$ from the spin axis is $ω^{2} ρ$ , pointing outward. A force that is the (positive) gradient of a potential must come from $\frac{1}{2} ω^{2} ρ^{2} = \frac{1}{2} (x^{2} + y^{2})$ (using $ω = 1$ ), since differentiating $\frac{1}{2} (x^{2} + y^{2})$ with respect to $x$ gives back $x$ , i.e. the outward push. That is exactly why the term $\frac{1}{2} (x^{2} + y^{2})$ appears in the effective potential below — it is the centrifugal force written as a landscape.

Recall

Why can the Coriolis force never change a body's speed? ::: It always points perpendicular to the velocity, so it does zero work — it only turns the path.

7. Gradient $\nabla$ and partial derivatives — the slope of a landscape

Picture it: stand on a hillside. Face due east and ask how steep the ground is that way — that's $\partial U / \partial x$ . Turn to face north — that's $\partial U / \partial y$ . Each partial derivative is the slope in one chosen direction.

Why the topic needs it: with this convention, force per mass is $\nabla U$ — objects are pushed along the landscape's slope toward higher $U$ . Where the ground is flat, $\nabla U = 0$ : that's an equilibrium, a possible resting spot. Those flat spots are the Lagrange points.

The double subscripts — $U_{xx} = \partial^{2} U / \partial x^{2}$ , $U_{x y} = \partial^{2} U / \partial x \partial y$ — are second derivatives: the curvature of the landscape. $U_{xx} > 0$ = curves up along $x$ ; $U_{xx} < 0$ = curves down. Curvature tells us whether a flat spot is a bowl or a saddle.

Recall

What does $\nabla U = 0$ mean physically? ::: The landscape is locally flat — no net force — an equilibrium point.

8. Distances $r_{1}, r_{2}$ and the effective potential $U$ — the frozen landscape itself

Before writing $U$ we need the two distances it depends on. From the little body at $(x, y)$ (we're in the plane, $z = 0$ ) to each primary:

Reading the gravity term consistently. With the parent's flipped sign convention (see §7), gravity raises $U$ as you approach a primary: as $r_{1} \to 0$ , $(1 - μ) / r_{1} \to + \infty$ . So in this landscape each primary is an infinitely tall spike, not a well — objects are pushed up toward the primary because that's the direction of the real gravitational pull. To keep the metaphor clean we will consistently call them gravity spikes (tall peaks), never "wells." Everything in this topic points uphill = toward stronger attraction.

Why "effective": the Coriolis force can't be folded into a landscape (it depends on velocity, not position), so $U$ captures only gravity + centrifugal. Coriolis stays as a separate term in the motion equations. This landscape is exactly what has Lagrange points as its flat spots — and see Jacobi constant & zero-velocity curves for how the level sets of $2 U$ become walls the spacecraft cannot cross.

Downstream this feeds directly into Lagrange points and Eigenvalues & linear stability analysis.

9. Exponentials $e^{λ t}$ and eigenvalues $λ$ — grow, oscillate, or decay?

Picture it: a marble in a bowl oscillates around the bottom (imaginary $λ$ ). A marble balanced on a dome rolls off ever faster (real positive $λ$ ).

The parent's characteristic equation $λ^{4} + (4 - U_{xx} - U_{y y}) λ^{2} + (U_{xx} U_{y y} - U_{x y}^{2}) = 0$ is nothing but "which growth-rates $λ$ are consistent with the curvature of $U$ plus the Coriolis coupling (the lone $4$ )?"

Recall

A Lagrange point yields a real positive $λ$ . Stable or unstable? ::: Unstable — a nudge grows like $e^{λ t}$ and runs away.

The prerequisite map

Read it top-down: naming where and how fast comes first; picking units and restricting to the plane sets the stage; spinning the camera freezes the scene; the frozen landscape $U$ has flat spots (Lagrange points); nudging them and testing with $e^{λ t}$ gives the characteristic equation.

Equipment checklist

I can describe what the arrow $r$ and its dots $\dot{r}, \ddot{r}$ mean ::: $r$ = position arrow from origin; one dot = velocity; two dots = acceleration.
I can compute $μ$ from two masses ::: $μ = m_{2} / (m_{1} + m_{2})$ , the smaller body's fraction of the pair.
I can list the three unit choices and the quantity they force ::: Length = separation (→1), mass = total (→1), time so that $G = 1$ ; together these force $ω = 1$ via Kepler's third law.
I can place both primaries in nondimensional units ::: $m_{1} = 1 - μ$ at $x = - μ$ , $m_{2} = μ$ at $x = 1 - μ$ , separation $1$ , barycentre at origin.
I know why $U$ has no $z$ in the parent's analysis ::: We restrict the motion to the primaries' orbital plane, $z = 0$ .
I can write $r_{1}$ and $r_{2}$ from $(x, y)$ ::: $r_{1} = (x + μ)^{2} + y^{2}$ , $r_{2} = (x - (1 - μ))^{2} + y^{2}$ .
I can say where the centrifugal term $\frac{1}{2} (x^{2} + y^{2})$ comes from ::: It is the potential whose gradient gives the outward centrifugal force $ω^{2} ρ$ with $ω = 1$ .
I know the sign convention for $U$ here ::: This topic uses acceleration $= + \nabla U$ (push toward higher $U$ ), the opposite of the usual $F = - \nabla V$ ; so primaries are spikes, not wells.
I can name the two fictitious forces and one key fact about each ::: Centrifugal flings outward (grows with distance); Coriolis deflects moving bodies perpendicular to velocity (does no work).
I can read $\partial U / \partial x$ and $U_{xx}$ ::: First partial = slope along $x$ ; second $U_{xx}$ = curvature.
I know what $\nabla U = 0$ signals ::: A flat spot — no net force — an equilibrium (Lagrange point).
I can interpret the sign/type of $λ$ in $e^{λ t}$ ::: Real positive → runs away (unstable); pure imaginary → oscillates (stable).