Halo orbits — linearized motion near Lagrange points: Exercises

Before we start, one shared reminder of the toolkit, so nothing below is used un-defined:

Figure — Halo orbits — linearized motion near Lagrange points

Level 1 — Recognition

Exercise 1.1

The linearized out-of-plane equation is $\ddot{ζ} = - c ζ$ . Without solving anything, name the type of motion and write its frequency.

Recall Solution

The equation has the form $\ddot{ζ} = - (positive number) \cdot ζ$ . That is the exact signature of Simple Harmonic Motion (see Simple Harmonic Motion): acceleration proportional to displacement and pointing back toward zero. Matching $\ddot{ζ} = - ω^{2} ζ$ gives $ω^{2} = c$ , so $ω_{z} = c .$ Why the minus sign matters: the negative sign is the "restoring" arrow. If it were $+ c$ , the force would push you away and you'd get exponential runaway, not oscillation.

Exercise 1.2

In the eigenvalue analysis you find one $Λ_{1} > 0$ and one $Λ_{2} < 0$ . For each, state whether $λ = Λ$ is real or imaginary, and what motion it produces.

Recall Solution

$Λ_{1} > 0$ : $λ = \pm Λ_{1} = \pm γ$ is real. Real exponents give $e^{+ γ t}$ (grows) and $e^{- γ t}$ (decays) → a saddle, i.e. unstable.
$Λ_{2} < 0$ : $λ = \pm Λ_{2} = \pm i ω_{x y}$ is imaginary. Imaginary exponents give $e^{\pm i ω_{x y} t} = cos \pm i sin$ → pure oscillation. Picture: on figure s01, the red arrows point away (saddle), the violet loop is the oscillation.

Level 2 — Application

Exercise 2.1

For Sun–Earth $L_{1}$ a representative value is $c = 4.06$ . Compute $ω_{z}$ , and the in-plane roots $Λ_{1}, Λ_{2}$ .

Recall Solution

Out-of-plane. $ω_{z} = 4.06 \approx 2.015$ . In-plane. Build the three ingredients: $2 - c = - 2.06$ , $1 + 2 c = 9.12$ , $1 - c = - 3.06$ . Discriminant $D = (2 - c)^{2} - 4 (1 + 2 c) (1 - c) = (- 2.06)^{2} - 4 (9.12) (- 3.06) = 4.2436 + 111.63 = 115.87$ . $D \approx 10.765$ . $Λ = \frac{- ( 2 - c ) \pm D}{2} = \frac{2.06 \pm 10.765}{2} .$ $Λ_{1} = \frac{2.06 + 10.765}{2} \approx 6.41$ , $Λ_{2} = \frac{2.06 - 10.765}{2} \approx - 4.35$ . Why: $Λ_{1} > 0$ (saddle) and $Λ_{2} < 0$ (oscillation) — the same structure as every collinear point.

Exercise 2.2

Using the $L_{1}$ numbers above, compute the instability rate $γ$ and the in-plane frequency $ω_{x y}$ . Then find the in-plane oscillation period $T = 2 π / ω_{x y}$ (dimensionless time).

Recall Solution

$γ = Λ_{1} = 6.41 \approx 2.532.$ $ω_{x y} = - Λ_{2} = 4.35 \approx 2.086.$ $T = \frac{2 π}{ω _{x y}} = \frac{6.2832}{2.086} \approx 3.012.$ Interpretation: a displacement along the unstable eigenvector multiplies by $e^{γ} \approx e^{2.53} \approx 12.6$ per dimensionless time unit — the reason $L_{1}$ probes need station-keeping (Station-keeping and Orbit Maintenance).

Level 3 — Analysis

Exercise 3.1

Show algebraically that for every collinear point with $c > 1$ , the product $Λ_{1} Λ_{2} < 0$ (one saddle, one oscillation). Use the fact that for $A Λ^{2} + B Λ + C = 0$ the product of roots is $C / A$ .

Recall Solution

The quartic in $Λ$ is $Λ^{2} + (2 - c) Λ + (1 + 2 c) (1 - c) = 0$ , so $A = 1$ , $C = (1 + 2 c) (1 - c)$ . Product of roots $= \frac{C}{A} = (1 + 2 c) (1 - c)$ . For $c > 1$ : $(1 + 2 c) > 0$ always, and $(1 - c) < 0$ . So the product is $(+) (-) < 0$ . A negative product means the two roots have opposite signs → exactly one positive $Λ_{1}$ and one negative $Λ_{2}$ . This is the saddle × center structure, proven without numbers.

Exercise 3.2

Compute the amplitude ratio $κ$ for Sun–Earth $L_{1}$ ( $ω_{x y} = 2.086$ , $c = 4.06$ ) and state which axis the in-plane ellipse is stretched along.

Recall Solution

$κ = \frac{ω _{x y}^{2} + ( 1 + 2 c )}{2 ω _{x y}} = \frac{( 2.086 ) ^{2} + 9.12}{2 ( 2.086 )} = \frac{4.351 + 9.12}{4.172} = \frac{13.471}{4.172} \approx 3.229.$ Since $κ > 1$ , the $η$ (i.e. $y$ ) amplitude is $\approx 3.23 \times$ the $ξ$ (i.e. $x$ ) amplitude. The planar ellipse is stretched along $y$ — elongated across the Sun–Earth line, squashed along it. Look at figure s02: the violet ellipse is tall, not round.

Level 4 — Synthesis

Exercise 4.1

Explain, using the numbers from Sun–Earth $L_{1}$ ( $ω_{x y} \approx 2.086$ , $ω_{z} \approx 2.015$ ), why linear theory produces a Lissajous, not a halo, and quantify how mismatched the two frequencies are as a percentage.

Recall Solution

The in-plane wobble repeats every $T_{x y} = 2 π / ω_{x y}$ and the out-of-plane bob every $T_{z} = 2 π / ω_{z}$ . A path closes into a single 3-D loop only if these two periods are commensurate — for a simple halo we need the 1:1 resonance $ω_{x y} = ω_{z}$ . Mismatch: $\frac{ω _{x y} - ω _{z}}{ω _{z}} = \frac{2.086 - 2.015}{2.015} \approx 0.0352 = 3.52%$ . Because $ω_{x y} \neq = ω_{z}$ (they differ by $\approx 3.5%$ ), the in-plane and out-of-plane motions drift out of phase and the trajectory fills a torus-like band → a Lissajous orbit, quasi-periodic, never closing. See figure s03: the two frequencies weave a non-closing ribbon. What fixes it: nonlinear frequency-amplitude corrections (Richardson's third-order expansion). At a specific amplitude the corrected frequencies satisfy $ω_{x y} = ω_{z}$ , and the loop snaps shut into a halo. Linear theory gives the ingredients ( $ω_{x y}, ω_{z}, κ$ , ellipse shape); the closing condition is nonlinear.

Exercise 4.2

A designer wants the out-of-plane bob to be as slow as possible relative to the in-plane wobble, to make a very flat Lissajous. Between Sun–Earth $L_{1}$ ( $c = 4.06$ ) and Earth–Moon $L_{2}$ ( $c = 3.19$ ), which gives the smaller ratio $ω_{z} / ω_{x y}$ ? Compute both.

Recall Solution

Sun–Earth $L_{1}$ : $ω_{z} = 2.015$ , $ω_{x y} = 2.086$ (Ex 2.2/2.1). Ratio $= 2.015/2.086 \approx 0.9660$ . Earth–Moon $L_{2}$ : from the parent worked example, $ω_{z} = 3.19 \approx 1.786$ , $ω_{x y} = 3.47 \approx 1.863$ . Ratio $= 1.786/1.863 \approx 0.9587$ . Earth–Moon $L_{2}$ has the smaller ratio ( $0.959 < 0.966$ ), so its out-of-plane bob is proportionally the slowest — the flatter Lissajous of the two.

Level 5 — Mastery

Exercise 5.1

For a collinear point define the stability index loosely as the exponential growth factor per in-plane period, $Σ \equiv e^{γ T_{x y}}$ with $T_{x y} = 2 π / ω_{x y}$ . Compute $Σ$ for Earth–Moon $L_{2}$ ( $γ \approx 2.16$ , $ω_{x y} \approx 1.863$ ) and interpret it for station-keeping cadence.

Recall Solution

$T_{x y} = \frac{2 π}{ω _{x y}} = \frac{6.2832}{1.863} \approx 3.372.$ $Σ = e^{γ T_{x y}} = e^{2.16 \times 3.372} = e^{7.284} \approx 1454.$ Meaning: any error along the unstable eigenvector grows by a factor of $\approx 1450$ over a single in-plane period. A 1 km error becomes $\sim$ 1450 km within one cycle. This is why maintenance burns must be scheduled many times per orbit, not once — the manifold structure (Invariant Manifolds and Low-Energy Transfers) that carries you away is also the leverage station-keeping uses to correct cheaply.

Exercise 5.2

Show that the eigenvector amplitude ratio $κ$ can be written using $Λ_{2}$ instead of $ω_{x y}$ , and verify numerically for Earth–Moon $L_{2}$ that both forms agree. Recall $ω_{x y}^{2} = - Λ_{2}$ .

Recall Solution

Start from $κ = \frac{ω _{x y}^{2} + ( 1 + 2 c )}{2 ω _{x y}}$ . Substitute $ω_{x y}^{2} = - Λ_{2}$ and $ω_{x y} = - Λ_{2}$ : $κ = \frac{- Λ _{2} + ( 1 + 2 c )}{2 - Λ _{2}} .$ Verify (Earth–Moon $L_{2}$ , $c = 3.19$ , $Λ_{2} = - 3.47$ ): Numerator $= 3.47 + 7.38 = 10.85$ . Denominator $= 2 3.47 = 2 (1.863) = 3.726$ . $κ = 10.85/3.726 \approx 2.912$ — matching the parent note's $κ \approx 2.92$ . ✓ Why this matters: it lets you get $κ$ straight from the eigenvalue $Λ_{2}$ without separately taking a square root for $ω_{x y}$ — fewer rounding steps in a design pipeline.