Worked examples — Halo orbits — linearized motion near Lagrange points

This is a hands-on companion to the parent topic. The parent gave you the machinery; here we drive it through every kind of input the equations can face — so that when an exam or a mission-design worksheet hands you a value of $c$, a mass ratio $\mu$, or an amplitude, you have already seen its twin.

Before we start, one plain-language refresher so nothing is used unearned.

Recall The three quantities every example uses

• $c$ ::: a single positive number that measures how strong the leftover gravity gradient is at the point; $c\equiv \frac{1-\mu }{r_1^3}+\frac{\mu }{r_2^3}$. Bigger $c$ = steeper "hill".
• $\Lambda ={\lambda }^2$ ::: the square of a growth/oscillation rate. If $\Lambda >0$ the motion runs away ($\lambda$ real); if $\Lambda <0$ the motion oscillates ($\lambda$ imaginary).
• $\kappa$ ::: how many times taller the in-plane ellipse is in $y$ than in $x$. Algebraically it is the eigenvector ratio $\kappa =\frac{{\omega }_{xy}^2+(1+2c)}{2{\omega }_{xy}}$ (derived and used in Example 3); think of it now as a single positive shape number.

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The scenario matrix

Every problem in linearized halo theory is one of these cells. The examples below are tagged with the cell they cover.

Cell	What varies	The question it forces you to answer
A. Nominal collinear	$c>1$ (typical $L_1,L_2,L_3$)	Do we always get one saddle + one oscillation?
B. Out-of-plane axis	the decoupled $\zeta$ equation	Is it always pure SHM, for any $c>0$?
C. Sign of $\Lambda$ roots	discriminant & product of roots	Which root is positive, which negative, and why never two positives?
D. Degenerate limit $c\to 1^{+}$	the boundary case	What happens to the frequencies right at $c=1$?
E. Large-$c$ limit	$c\gg 1$ (deep gravity well)	How do $\gamma$, ${\omega }_{xy}$, ${\omega }_z$ scale?
F. Amplitude ratio $\kappa$	eigenvector geometry	Shape of the planar ellipse; always $>0$?
G. Real-world word problem	Sun–Earth $L_1$ (SOHO)	Convert dimensionless rates to real days.
H. Exam twist	Lissajous closing condition	When is a loop a halo and not a Lissajous?

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Example 1 — Cell A: the nominal collinear point ($c=3.19$, Earth–Moon $L_2$)

Forecast: guess the signs — will you get two oscillations, two run-aways, or one of each?

1. Assemble the coefficients. $2-c=-1.19$, $1+2c=7.38$, $1-c=-2.19$. Why this step? The characteristic equation ${\Lambda }^2+(2-c)\Lambda +(1+2c)(1-c)=0$ (restated in the formula callout above) needs these three numbers and nothing else.
2. Discriminant. $D=(2-c{)}^2-4(1+2c)(1-c)=1.4161-4(7.38)(-2.19)=1.4161+64.6488=66.0649$, so $\sqrt{D}=8.1281$. Why this step? $D>0$ guarantees two real, distinct values of $\Lambda$ — no complex $\Lambda$, hence no spiral. Look at figure s01 (the parabola of the characteristic equation): it crosses zero twice.
3. Roots. $\Lambda =\frac{-(2-c)\pm \sqrt{D}}{2}=\frac{1.19\pm 8.1281}{2}$ → ${\Lambda }_1=+4.659$, ${\Lambda }_2=-3.469$. Why this step? One positive, one negative — that is the saddle × center signature.
4. Rates. $\gamma =\sqrt{{\Lambda }_1}=\sqrt{4.659}\approx 2.158$; ${\omega }_{xy}=\sqrt{-{\Lambda }_2}=\sqrt{3.469}\approx 1.862$. Why this step? $\Lambda >0\Rightarrow \lambda =\pm \gamma$ real (growth $e^{\gamma t}$). $\Lambda <0\Rightarrow \lambda =\pm i{\omega }_{xy}$ (oscillation).

Figure s01 — the characteristic-equation parabola $f(\Lambda )$ for $c=3.19$: its two zero-crossings (amber dots) are the roots ${\Lambda }_1>0$ (saddle) and ${\Lambda }_2<0$ (oscillation).

Verify: product ${\Lambda }_1{\Lambda }_2=4.659\times (-3.469)\approx -16.16$, and $(1+2c)(1-c)=7.38\times (-2.19)=-16.16$. ✓ Sum ${\Lambda }_1+{\Lambda }_2=1.190=-(2-c)$. ✓

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Example 2 — Cell B: the out-of-plane axis is pure SHM (any $c>0$)

Forecast: does the out-of-plane bob ever run away like the in-plane saddle?

1. Compare to the SHM template. SHM is $\ddot{s}=-{\omega }^{2}s$. Here $\ddot{\zeta }=-c\zeta$, so ${\omega }_z^2=c$. Why this step? Matching to the template lets us read off the frequency without re-solving — the restoring force is proportional to displacement and points back, since $c>0$.
2. Frequency. ${\omega }_z=\sqrt{3.19}\approx 1.786$.
3. Period. $T_z=2\pi /{\omega }_z=6.2832/1.786\approx 3.518$ dimensionless time units. Why this step? The bob repeats every $T_z$; halos need this to line up with the in-plane period.

Verify: ${\omega }_z^2=1.786^2=3.190=c$. ✓ $T_z{\omega }_z=3.518\times 1.786=6.283=2\pi$. ✓

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Example 3 — Cell F: the in-plane amplitude ratio $\kappa$ (ellipse shape)

Forecast: is the loop wider along $x$ or along $y$?

1. Where the formula comes from. The oscillatory root gives an eigenvector $(A,B)$; forcing $\eta =\kappa \xi$ in amplitude and using the first linearized equation yields $\kappa =\frac{{\omega }_{xy}^2+(1+2c)}{2{\omega }_{xy}}$. Why this step? $\kappa$ is not free — it is fixed by the eigenvector of the oscillatory root. For a unit-$\xi$ amplitude the $\eta$ amplitude is exactly $\kappa$.
2. Plug in. $\kappa =\frac{{\omega }_{xy}^2+(1+2c)}{2{\omega }_{xy}}=\frac{3.469+7.38}{3.724}=\frac{10.849}{3.724}\approx 2.913$. Why this step? We already have ${\omega }_{xy}^2=3.469$ (it is $-{\Lambda }_2$ from Example 1) and $1+2c=7.38$; substituting the known numbers is the only arithmetic left to turn the eigenvector formula into a concrete shape factor.
3. Interpret geometry. Since $\kappa \approx 2.91>1$, the ellipse is stretched along $y$ by nearly a factor of three. See figure s02. Why this step? The parametric pair $\xi =-A_x\cos ,\eta =\kappa A_x\sin$ traces an ellipse with semi-axes $A_x$ (in $x$) and $\kappa A_x$ (in $y$).

Figure s02 — the parametric in-plane ellipse $(\xi ,\eta )$: the $y$ semi-axis is $\kappa \approx 2.91$ times the $x$ semi-axis, so the loop is elongated vertically around the Lagrange point.

Verify: $\kappa \times 2{\omega }_{xy}=2.913\times 3.724=10.849={\omega }_{xy}^2+1+2c$. ✓

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Example 4 — Cell D: the degenerate limit $c\to 1^{+}$

Forecast: does the instability strengthen or vanish as we approach $c=1$?

1. Constant term at $c=1$. $(1+2c)(1-c)=3\times 0=0$. Why this step? The product of the roots is this term; if it is zero, one root is exactly zero.
2. Solve at $c=1$. ${\Lambda }^2+(2-1)\Lambda +0=\Lambda (\Lambda +1)=0$ → ${\Lambda }_1=0,{\Lambda }_2=-1$. Why this step? Reading the factored form directly gives the boundary values.
3. Interpret. $\gamma =\sqrt{{\Lambda }_1}=0$: the exponential instability rate collapses to zero at $c=1$. Meanwhile ${\omega }_{xy}=\sqrt{1}=1$ and ${\omega }_z=\sqrt{c}=1$. Why this step? $c=1$ is precisely the borderline between saddle (collinear, $c>1$) and pure-center (triangular-like, $c<1$) behaviour. See figure s03: the amber parabola touches zero at the origin.

Figure s03 — the product-of-roots curve $(1+2c)(1-c)$ versus $c$: it is negative for $c>1$ (saddle), zero exactly at $c=1$ (marginal, $\gamma \to 0$), and positive for $c<1$ (pure oscillation).

Verify: At $c=1$, roots $0$ and $-1$; sum $=-1=-(2-c)$; product $=0=(1+2c)(1-c)$. ✓ And ${\omega }_z=\sqrt{1}=1$. ✓

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Example 5 — Cell E: the deep-well limit $c\gg 1$

Forecast: which grows faster with $c$ — the instability rate or the in-plane frequency?

1. Approximate the roots for large $c$. With $2-c\approx -c$ and constant term $\approx (2c)(-c)=-2c^2$: $\Lambda \approx \frac{c\pm \sqrt{c^2+8c^2}}{2}=\frac{c\pm 3c}{2}.$ Why this step? Keeping only the top powers of $c$ isolates the dominant behaviour; $\sqrt{c^2+8c^2}=\sqrt{9c^2}=3c$.
2. Two roots. ${\Lambda }_1\approx \frac{c+3c}{2}=2c$, ${\Lambda }_2\approx \frac{c-3c}{2}=-c$. Why this step? Now $\gamma \approx \sqrt{2c}$ and ${\omega }_{xy}\approx \sqrt{c}$, while ${\omega }_z=\sqrt{c}$ exactly.
3. Test at $c=100$ (exact). $D=(2-100{)}^2-4(201)(-99)=9604+79596=89200$, $\sqrt{D}=298.66$. $\Lambda =\frac{98\pm 298.66}{2}$ → ${\Lambda }_1=198.33$, ${\Lambda }_2=-100.33$. So $\gamma =14.083$, ${\omega }_{xy}=10.017$, ${\omega }_z=10$. Why this step? Compare to the approximations: $\sqrt{2\cdot 100}=14.14$ vs $14.08$ ✓; $\sqrt{100}=10$ vs $10.017$ ✓.

Verify: at $c=100$: ${\Lambda }_1+{\Lambda }_2=198.33-100.33=98=-(2-c)$ ✓; product $=198.33\times (-100.33)=-19898\approx (201)(-99)=-19899$ ✓.

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Example 6 — Cell G: real-world word problem (Sun–Earth $L_1$, SOHO)

Forecast: will an uncontrolled SOHO drift off in weeks, months, or years?

1. Dimensionless in-plane rates. $2-c=-2.061$, $1+2c=9.122$, $1-c=-3.061$. $D=(-2.061{)}^2-4(9.122)(-3.061)=4.2477+111.69=115.94$, $\sqrt{D}=10.768$. ${\Lambda }_1=\frac{2.061+10.768}{2}=6.4145$, ${\Lambda }_2=-4.3535$. So $\gamma =2.5327$, ${\omega }_{xy}=2.0865$. Why this step? Same machine as Example 1, just a new $c$.
2. The time unit. One full frame revolution is $2\pi$ normalized time units and takes $365.25$ real days, so $1$ normalized time unit $=\frac{365.25}{2\pi }=58.13$ days. Why this step? Every rate we computed is per normalized time unit; to speak in days we multiply time-units by 58.13 days.
3. $e$-folding time of the instability. The unstable mode grows as $e^{\gamma t}$; it grows by a factor $e$ in $t=1/\gamma =1/2.5327=0.3948$ time units $=0.3948\times 58.13\approx 22.9$ days. Why this step? $1/\gamma$ is the natural "how fast do I run away" clock. ~23 days → SOHO needs station-keeping every few weeks.
4. In-plane period in days. $T_{xy}=2\pi /{\omega }_{xy}=6.2832/2.0865=3.0114$ time units $=3.0114\times 58.13\approx 175.1$ days. Why this step? Converts the wobble period to a calendar figure — roughly half a year per in-plane loop.

Verify: $1/\gamma =0.3948$ tu; $\times 58.13=22.95$ days. $T_{xy}=3.0114$ tu $\times 58.13=175.05$ days. $58.13\times 2\pi =365.25$ ✓.

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Example 7 — Cell H: Lissajous vs Halo (the exam twist)

Forecast: are the two frequencies equal? If not, does the path close?

1. Compare the frequencies. ${\omega }_{xy}=1.862$ (in-plane wobble) and ${\omega }_z=1.786$ (out-of-plane bob). Ratio ${\omega }_{xy}/{\omega }_z=1.862/1.786=1.0426\ne 1$. Why this step? A 3-D loop retraces itself only if the two frequencies are commensurate; the cleanest closure is a 1:1 ratio. Since $1.0426$ is not $1$ (nor any simple ratio), the periods do not line up.
2. Compute both periods to see the mismatch. $T_{xy}=2\pi /{\omega }_{xy}=3.375$ tu, $T_z=2\pi /{\omega }_z=3.518$ tu. They differ by $0.143$ tu each loop. Why this step? Explicit periods make the drift concrete: after one in-plane loop the out-of-plane bob is not yet back to its start, so the 3-D curve lands at a new point every turn.
3. Name the resulting motion. Because the curve never returns exactly to its starting state, it slowly precesses and densely fills a band on a torus — a quasi-periodic Lissajous orbit, not a closed halo. See figure s04. Why this step? This is the direct geometric consequence of unequal, incommensurate frequencies.
4. State the closing condition. A genuine halo requires $\boxed{{\omega }_{xy}={\omega }_z}$ (a 1:1 resonance). Linear theory fixes these two frequencies at unequal values, so it cannot meet the condition. Only the nonlinear, amplitude-dependent frequency corrections (Richardson's third-order expansion) shift ${\omega }_{xy}$ and ${\omega }_z$ until, at one special amplitude $A_z$, they coincide and the loop snaps shut. Why this step? This is the whole moral of the topic — linear theory supplies the ingredients (frequencies, $\kappa$, ellipse shape); the halo's existence is a nonlinear amplitude constraint.

Figure s04 — the $(\eta ,\zeta )$ projection traced over many periods: because ${\omega }_{xy}\ne {\omega }_z$ the curve never retraces itself and densely fills a band — the signature of a Lissajous orbit, not a closed halo.

Verify: ${\omega }_{xy}/{\omega }_z=1.862/1.786=1.0426\ne 1$; $T_{xy}=3.375$ tu, $T_z=3.518$ tu, difference $0.143$ tu — periods genuinely unequal, so no closure. ✓

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Recall check

Recall

Product of the two in-plane roots equals which quantity, and what sign for a collinear point? ::: $(1+2c)(1-c)$; negative since $c>1$, forcing one saddle + one center. Out-of-plane motion is stable for which values of $c$? ::: All $c>0$, because $U_{zz}=-c<0$ gives pure SHM with ${\omega }_z=\sqrt{c}$. What does linear theory FAIL to produce, and what fixes it? ::: A closed halo; the fix is nonlinear amplitude tuning forcing ${\omega }_{xy}={\omega }_z$.

See also: Circular Restricted Three-Body Problem · Lagrange Points L1–L5 · Jacobi Integral and Zero-Velocity Curves · Invariant Manifolds and Low-Energy Transfers · Station-keeping and Orbit Maintenance · Simple Harmonic Motion