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

Before anything, let us re-anchor four symbols so nothing appears unearned.

The scenario matrix

Every worked example below is tagged with the cell it fills.

Cell	Case class	Distinguishing feature	Example
A	Collinear points $L_{1}, L_{2}, L_{3}$ , generic $μ$	$U_{x y} = 0$ , $c < 0$ → saddle (ALL three)	Ex 1 (+ note on $L_{2}, L_{3}$ ), Ex 8 ( $L_{2}$ )
B	Triangular point, small $μ$	$c > 0, b^{2} - 4 c > 0$ → stable	Ex 2
C	Triangular point, large $μ$	$b^{2} - 4 c < 0$ → unstable	Ex 3
D	Degenerate boundary $μ = μ_{crit}$	$b^{2} - 4 c = 0$ (repeated root)	Ex 4
E	Limit $μ \to 0$	one primary vanishes	Ex 5
F	Limit $μ \to \frac{1}{2}$	equal masses, most unstable triangular	Ex 6
G	Real-world word problem	pick a real system, decide fate	Ex 7 (Sun–Jupiter)
H	Exam twist / trap	recover the growth rate, not just sign	Ex 8 ( $L_{2}$ )

The triangular curvatures are fixed by geometry (parent note): at $L_{4}, L_{5}$ $U_{xx} = \frac{3}{4}, U_{y y} = \frac{9}{4}, U_{x y} = \pm \frac{3 3}{4} (1 - 2 μ) .$ So for ALL triangular examples: $b = 4 - \frac{3}{4} - \frac{9}{4} = 1$ , and $c = \frac{3}{4} \cdot \frac{9}{4} - \frac{27}{16} (1 - 2 μ)^{2} = \frac{27}{16} [1 - (1 - 2 μ)^{2}] = \frac{27}{16} \cdot 4 μ (1 - μ) = \frac{27}{4} μ (1 - μ) .$ And the stability discriminant is $b^{2} - 4 c = 1 - 27 μ (1 - μ)$ , which is the parent's condition. Keep those two facts in view; the triangular examples are all one formula with different $μ$ .

The figure below is our map for the whole page: its horizontal axis is the mass parameter $μ$ (running from $0$ to $\frac{1}{2}$ , dimensionless), and its vertical axis is the stability discriminant $1 - 27 μ (1 - μ)$ (also dimensionless — a pure number). Where the yellow curve sits above zero (blue shading) triangular points are stable; below zero (pink shading) they are unstable. The pink dotted line marks the critical mass $μ_{crit}$ where the curve crosses zero, and each coloured dot marks the $μ$ of one worked example so you can see where every case lives before we compute it.

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

Figure s01 — Triangular-point stability across all $μ$ : the discriminant curve, the stable/unstable shading, $μ_{crit}$ , and the marked example cases.

Worked examples

Worked example Ex 1 — Cell A · Collinear point

$L_{1}$ (Earth–Moon), a saddle Statement. For Earth–Moon $L_{1}$ the parent note quotes $U_{xx} \approx + 11.3$ , $U_{y y} \approx - 4.2$ , and $U_{x y} = 0$ (collinear points sit on the symmetry axis). Classify the equilibrium and find one growth rate $λ$ .

Forecast: $U_{y y}$ is negative — the landscape is a ridge in the $y$ direction. Guess: unstable. But how unstable — real $λ$ of what size?

Compute $c = U_{xx} U_{y y} - U_{x y}^{2} = (11.3) (- 4.2) - 0 = - 47.46$ . Why this step? $c$ is the product of the two $Λ$ -roots; its sign alone decides saddle vs centre.
Since $c < 0$ , the roots $Λ_{\pm}$ have opposite signs. Compute $b = 4 - 11.3 - (- 4.2) = - 3.1$ . Why this step? We need $b$ to solve the quadratic and pull out the actual positive $Λ$ .
Solve $Λ^{2} - 3.1 Λ - 47.46 = 0$ : $Λ = \frac{3.1 \pm 3. 1 ^{2} + 4 ( 47.46 )}{2} = \frac{3.1 \pm 199.45}{2}$ . $199.45 = 14.123$ , so $Λ_{+} = 8.611$ , $Λ_{-} = - 5.511$ . Why this step? The positive root gives the real exponential; the negative gives an oscillation.
Growth rate $λ = Λ_{+} = 8.611 = 2.935$ (on the $ω = 1$ clock defined above — so $λ$ is measured per radian of orbital angle). Why this step? The small offset obeys $ξ \propto e^{λ t}$ (recall $ξ$ is the tiny displacement off $L_{1}$ ). With $λ = 2.935$ , after one time unit — one radian of the Earth–Moon orbit — the nudge grows by $e^{2.935} \approx 18.8 \times$ . That relentless amplification is why $L_{1}$ needs active station-keeping.

Verify: Product of roots $Λ_{+} Λ_{-} = 8.611 \times (- 5.511) = - 47.46 = c$ . ✓ Sum $= 8.611 - 5.511 = 3.1 = - b$ . ✓ Opposite signs confirm the saddle.

Worked example Ex 2 — Cell B · Triangular

$L_{4}$ (Earth–Moon), stable Statement. Earth–Moon $μ = 0.0123$ . Classify $L_{4}$ and give the two oscillation frequencies.

Forecast: $L_{4}$ is a maximum of $U$ (a hilltop). Naively "balls roll off hilltops" ⇒ unstable. But Coriolis rescues small- $μ$ triangular points. With $μ = 0.0123$ well under $0.0385$ , guess: stable, two distinct real oscillation frequencies.

$b = 1$ (always, for triangular). Compute $c = \frac{27}{4} μ (1 - μ) = \frac{27}{4} (0.0123) (0.9877) = 0.08202$ . Why this step? $c > 0$ is the first stability gate; here it passes.
Discriminant $b^{2} - 4 c = 1 - 4 (0.08202) = 1 - 0.32808 = 0.67192 > 0$ . Why this step? Equivalent to $1 - 27 μ (1 - μ)$ ; positive ⇒ both $Λ$ real. Combined with $c > 0, b > 0$ ⇒ both negative ⇒ stable.
Roots: $Λ_{\pm} = \frac{- 1 \pm 0.67192}{2} = \frac{- 1 \pm 0.81971}{2}$ , giving $Λ_{+} = - 0.09015$ , $Λ_{-} = - 0.90985$ . Why this step? Both negative confirms pure imaginary $λ$ .
Frequencies $ω_{1, 2} = - Λ_{\pm}$ : $ω_{1} = 0.09015 = 0.3003$ (long-period libration), $ω_{2} = 0.90985 = 0.9539$ (short-period). These are measured on the same $ω = 1$ clock, so $ω_{2} \approx 0.95$ means the fast wobble is almost as quick as one orbital radian. Why this step? These are the two libration frequencies real Trojan-like bodies wobble at.

Verify: $Λ_{+} Λ_{-} = (- 0.09015) (- 0.90985) = 0.08202 = c$ . ✓ Sum $= - 1 = - b$ . ✓ Both negative ⇒ stable, matching Ex 2 of the parent ( $27 μ (1 - μ) = 0.328 < 1$ ).

Worked example Ex 3 — Cell C · Triangular point, mass just over critical → unstable

Statement. A hypothetical binary has $μ = 0.05$ (above $μ_{crit} \approx 0.03852$ ). Classify $L_{4}$ .

Forecast: $μ$ exceeds the critical value, so guess: the discriminant flips negative → complex $Λ$ → unstable (spiralling growth).

$c = \frac{27}{4} (0.05) (0.95) = 0.320625$ ; $b = 1$ . Why this step? Set up the quadratic in $Λ$ .
Discriminant $b^{2} - 4 c = 1 - 1.2825 = - 0.2825 < 0$ . Why this step? Negative discriminant ⇒ $Λ$ are complex conjugates, not real. Then $λ = Λ$ has a nonzero real part ⇒ growth.
Complex roots $Λ = \frac{- 1 \pm i 0.2825}{2} = - 0.5 \pm 0.26575 i$ . Why is there always a growing direction? Write a complex $Λ$ in polar form as $Λ = R e^{i θ}$ with $R > 0$ and angle $- π < θ < π$ . Its two square roots are $λ = \pm R e^{i θ /2}$ . Their real parts are $\pm R cos (θ /2)$ , and since $- \frac{π}{2} < θ /2 < \frac{π}{2}$ we have $cos (θ /2) > 0$ — so one root has a strictly positive real part whenever $θ \neq = 0$ (i.e. whenever $Λ$ is not real). That positive real part is the exponential growth. This is oscillatory instability: the body spirals outward, unlike the pure exponential saddle of the collinear points.

Verify: discriminant sign check: $27 μ (1 - μ) = 27 (0.05) (0.95) = 1.2825 > 1$ , so $1 - 27 μ (1 - μ) = - 0.2825 < 0$ . ✓ Unstable confirmed.

Worked example Ex 4 — Cell D · The degenerate boundary

$μ = μ_{crit}$ Statement. At exactly $μ_{crit} = \frac{1}{2} (1 - 23/27)$ the discriminant is zero. Find $μ_{crit}$ numerically and the (repeated) root $Λ$ .

Forecast: Boundary between stable and unstable — expect a repeated real root, marginal case. Guess $μ_{crit} \approx 0.0385$ .

Solve $1 - 27 μ (1 - μ) = 0$ , i.e. $27 μ^{2} - 27 μ + 1 = 0$ . Then $μ = \frac{27 - 2 7 ^{2} - 4 \cdot 27}{2 \cdot 27} = \frac{27 - 621}{54}$ . Why this step? The stability discriminant, set to zero, is a plain quadratic in $μ$ .
$621 = 24.9199$ , so $μ_{crit} = \frac{27 - 24.9199}{54} = \frac{2.0801}{54} = 0.038520$ . Why this step? Confirms the parent's $0.03852$ and shows the algebraic form $\frac{1}{2} (1 - 23/27)$ is identical (since $621 = 2 7^{2} - 108$ and $23/27 = 621/729$ ).
At the boundary $b^{2} - 4 c = 0$ , so $Λ = - b /2 = - 1/2$ (double root). Then $λ = \pm i / 2$ (repeated). Why this step? A repeated imaginary pair means the two libration frequencies merge; beyond this $μ$ they split into a complex pair and destabilise. This coalescence is the mathematical signature of the stability boundary.

Verify: $\frac{1}{2} (1 - 23/27) = \frac{1}{2} (1 - 0.922958) = 0.038521$ . ✓ matches step 2. And $27 μ_{crit} (1 - μ_{crit}) = 1$ . ✓

Worked example Ex 5 — Cell E · Limit

$μ \to 0$ (second primary vanishes) Statement. Let $μ \to 0^{+}$ . What happens to $c$ and to the triangular-point stability? Physically interpret.

Forecast: With $m_{2} \to 0$ we are back to (nearly) a two-body / one-body situation. Guess: $L_{4}$ becomes marginally, then trivially stable; the libration frequencies approach $0$ and $1$ .

$c = \frac{27}{4} μ (1 - μ) \to 0$ as $μ \to 0$ . Why this step? $c$ is the product of roots; if it $\to 0$ , one root $Λ \to 0$ .
With $b = 1$ and $c \to 0$ : roots $Λ_{+} \to 0$ and $Λ_{-} \to - 1$ . Frequencies $ω_{1} = - Λ_{+} \to 0$ , $ω_{2} = - Λ_{-} \to 1$ . Why this step? $ω_{2} \to 1$ is the orbital frequency itself (on the $ω = 1$ clock) — the tiny body just orbits the lone primary at the mean motion. $ω_{1} \to 0$ means the long-period libration becomes infinitely slow (no second mass to hold it).
Discriminant $1 - 27 μ (1 - μ) \to 1 > 0$ : deeply inside the stable region. Why this step? Confirms every real planetary system with a tiny secondary keeps stable Trojans.

Verify: at $μ = 0$ , $c = 0$ , roots $0$ and $- 1$ ; $ω_{2} = 1 = ω$ (orbital rate). ✓

Worked example Ex 6 — Cell F · Limit

$μ \to \frac{1}{2}$ (equal masses) Statement. Equal-mass binary, $μ = 0.5$ . Classify $L_{4}$ and find the complex $Λ$ .

Forecast: Two equal stars — the most symmetric, and (since $0.5 > μ_{crit}$ ) the most unstable triangular case. Guess: strongly complex $Λ$ .

$c = \frac{27}{4} (0.5) (0.5) = \frac{27}{16} = 1.6875$ ; $b = 1$ . Why this step? $c$ is largest possible here, since $μ (1 - μ)$ peaks at $μ = \frac{1}{2}$ .
Discriminant $1 - 4 (1.6875) = 1 - 6.75 = - 5.75 < 0$ ⇒ complex conjugate $Λ$ . Why this step? Confirms strong oscillatory instability; note $U_{x y} = \frac{3 3}{4} (1 - 2 μ) = 0$ here, so the twist vanishes yet instability remains — instability is not about the twist term.
$Λ = \frac{- 1 \pm i 5.75}{2} = - 0.5 \pm 1.19896 i$ . Why this step? Nonzero real part in $λ = Λ$ ⇒ runaway (by the polar-form argument of Ex 3). Equal-mass binaries have no stable Trojan points.

Verify: $∣ Λ ∣ = 0. 5^{2} + 1.1989 6^{2} = 0.25 + 1.4375 = 1.6875 = c$ . ✓ (product of a conjugate pair $= ∣Λ ∣^{2} = c = 1.6875$ .)

Worked example Ex 7 — Cell G · Real-world word problem: Sun–Jupiter Trojans

Statement. Jupiter is $9.55 \times 1 0^{- 4}$ of the Sun's mass. Real asteroids (the Trojans) sit at $L_{4}, L_{5}$ . Show the theory permits them and find the two libration periods (in units of Jupiter's orbital period).

Forecast: $μ \approx 0.00095 ≪ 0.0385$ , so stable — matches observation. Expect one very long-period and one nearly-orbital-period libration.

$μ = \frac{9.55 \times 1 0 ^{- 4}}{1 + 9.55 \times 1 0 ^{- 4}} = 9.541 \times 1 0^{- 4}$ . Why this step? Convert mass ratio to the mass parameter $μ = m_{2} / (m_{1} + m_{2})$ .
$c = \frac{27}{4} μ (1 - μ) = \frac{27}{4} (9.541 \times 1 0^{- 4}) (0.99905) = 6.4327 \times 1 0^{- 3}$ ; $b = 1$ . Why this step? Tiny $c$ ⇒ one root near $0$ , one near $- 1$ .
Roots $Λ_{\pm} = \frac{- 1 \pm 1 - 4 c}{2} = \frac{- 1 \pm 0.974269}{2}$ . $0.974269 = 0.987051$ . So $Λ_{+} = - 0.0064746$ , $Λ_{-} = - 0.993525$ . Why this step? Both negative ⇒ stable; feed to frequencies.
Frequencies $ω_{1} = - Λ_{+} = 0.0064746 = 0.080465$ , $ω_{2} = - Λ_{-} = 0.993525 = 0.996757$ (on the $ω = 1$ clock). Why this step? These are the two libration rates; now convert to periods.
On the $ω = 1$ clock the primaries' orbital period is $T_{orb} = 2 π / ω = 2 π$ . A libration period is $T = 2 π / ω_{i}$ , so its ratio to the orbital period is $T / T_{orb} = ω / ω_{i} = 1/ ω_{i}$ . Hence the long-period libration lasts $1/ ω_{1} = 1/0.080465 = 12.43$ Jupiter orbits, and the short one $1/ ω_{2} = 1/0.996757 = 1.003$ Jupiter orbits. Why this step? The ~12.4-orbit long libration matches the observed slow "tadpole" wobble of the real Trojans.

Verify: $Λ_{+} Λ_{-} = (- 0.0064746) (- 0.993525) = 6.4327 \times 1 0^{- 3} = c$ . ✓ Both roots negative ⇒ stable, $μ ≪ μ_{crit}$ . ✓ Long-period ratio $1/ ω_{1} \approx 12.43$ . ✓

Worked example Ex 8 — Cell H (and cell A, point

$L_{2}$ ) · Exam twist: get the number of e-foldings at $L_{2}$ Statement. Sun–Earth $L_{2}$ (home of space telescopes) has effective curvatures $U_{xx} \approx + 7.38$ , $U_{y y} \approx - 2.19$ , $U_{x y} = 0$ . A spacecraft is nudged and left uncontrolled. By what factor has the displacement grown after one dimensionless time unit ( $t = 1$ , i.e. one orbital radian)?

Forecast: $L_{2}$ is collinear ⇒ (by cell A) unstable ⇒ real positive $λ$ . The trap: many stop at "unstable." The exam wants the growth factor $e^{λ}$ .

$c = U_{xx} U_{y y} - U_{x y}^{2} = (7.38) (- 2.19) - 0 = - 16.1622 < 0$ ⇒ saddle, exactly like the other collinear points. Why this step? Confirms a real positive $λ$ exists (cell A behaviour for $L_{2}$ ).
$b = 4 - 7.38 - (- 2.19) = - 1.19$ . Solve $Λ^{2} - 1.19Λ - 16.1622 = 0$ : $Λ_{+} = \frac{1.19 + 1.4161 + 64.6488}{2} = \frac{1.19 + 66.0649}{2} = \frac{1.19 + 8.1281}{2} = 4.6591$ . Why this step? Positive root gives the unstable direction's rate.
$λ = 4.6591 = 2.1585$ . Growth factor of the offset $ξ$ after $t = 1$ : $e^{λ \cdot 1} = e^{2.1585} = 8.658$ . Why this step? In one time unit (one radian of the Sun–Earth orbit, ~58 s-scaled days) an unmanaged deviation grows ~8.7×. That is the quantitative reason station-keeping burns happen roughly every few weeks.

Verify: $Λ_{+} = 4.6591$ ; check $Λ_{+}^{2} - 1.19 Λ_{+} + (- 16.1622) \approx 0$ : $21.707 - 5.544 - 16.162 = 0.001 \approx 0$ . ✓ Growth factor $e^{2.1585} = 8.66$ . ✓

Recall Self-test (reveal after guessing)

Collinear points $L_{1}, L_{2}, L_{3}$ are always ::: unstable saddles, because $U_{x y} = 0$ with $U_{xx} > 0, U_{y y} < 0$ gives $c < 0$ at each of them. Triangular points are stable iff ::: $μ \leq μ_{crit} \approx 0.03852$ , equivalently $27 μ (1 - μ) \leq 1$ . What single quantity decides saddle vs centre for a collinear point? ::: the sign of $c = U_{xx} U_{y y}$ (product of the $Λ$ -roots). Why don't equal-mass binaries have stable Trojans? ::: $μ = 0.5 > μ_{crit}$ , so the discriminant $1 - 27 μ (1 - μ) = - 5.75 < 0$ , giving complex $Λ$ and oscillatory runaway. The "4" in the characteristic equation comes from ::: the Coriolis coupling terms $2 \overset{η}{˙}, 2 \dot{ξ}$ squared into the determinant. "One time unit" means ::: one radian of the primaries' orbital angle (the clock where $ω = 1$ ).

See also: Lagrange points, Jacobi constant & zero-velocity curves, Rotating reference frames — Coriolis & centrifugal forces.