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\title{Five Places to Park a Spacecraft, Two of Which Are Stable for a Strange Reason}
\author{The Science Journaling Club}
\date{Volume 3, Issue 1 $\cdot$ Fall 2026 $\cdot$ Original Research}

\begin{document}
\maketitle

\begin{abstract}
Two bodies orbit their common centre of mass in a circle. In the frame that rotates with
them there are exactly five points where a third, massless body can sit still: three on the
line through the two masses ($L_1$, $L_2$, $L_3$), and two forming equilateral triangles with
them ($L_4$, $L_5$). We located all five numerically for five mass ratios $\mu$ spanning five
orders of magnitude, from Sun--Earth ($\mu = 3.00\times10^{-6}$) to an illustrative equal-mass
pair ($\mu=0.5$), by two independent numerical methods that agree with each other to machine
precision ($2.2\times10^{-16}$) and with the classical small-$\mu$ series for the collinear
points to four or more significant figures, and by exact algebra for the triangular ones
(residual $<4\times10^{-16}$). Eigenvalues of the linearised equations of motion show $L_1$,
$L_2$ and $L_3$ unstable at every mass ratio tried, and show $L_4$/$L_5$ stable only below a
critical mass ratio that our own numerical bisection places at $\mu_{\rm crit}=0.03852085$,
against the classical Gascheau (1843) / Routh (1875) value of $0.03852090$, a difference of
$4.4\times10^{-8}$ consistent with the square-root ill-conditioning of an eigenvalue problem at
a genuine matrix degeneracy. Direct integration of test particles (fixed fourth-order
Runge--Kutta, Jacobi constant conserved to $<4\times10^{-13}$ relative up to first escape)
shows the same particle, same perturbation, same code, staying within 1000 times its starting
displacement for the whole run at Earth--Moon's $L_4$ and leaving it by $t=15.0$ at
Pluto--Charon's $L_4$, whose mass ratio ($0.1087$, $2.82\times$ critical) exceeds threshold. A
Monte Carlo experiment of 1{,}500 randomly directed perturbations per mass ratio, seeded at
$20260914$, put the escape probability at $L_4$ at $0\%\pm0\%$ at $0.95\,\mu_{\rm crit}$,
$18.8\%\pm1.0\%$ at $0.98\,\mu_{\rm crit}$, and $100\%\pm0\%$ by $1.20\,\mu_{\rm crit}$, with
the running estimate visibly converged by a few hundred trials. $L_4$/$L_5$ are stable not
because they sit in a potential well, but because the Coriolis force, which does no work and
appears in no potential, organises the motion around the hill they actually sit on.
\end{abstract}

\section{Introduction}

Lagrange found all five equilibria of the general three-body problem in 1772 while chasing a
prize, apparently regarding the triangular pair as a curiosity with no expected physical
instance \cite{lagrange1772}. Poincar\'e's later work on the same problem showed it admits no
further conserved quantity beyond the one Jacobi wrote down in 1836
\cite{poincare1892,jacobi1836}, which is why the restricted three-body problem is a standard
example of a low-dimensional system that is nonetheless not integrable \cite{barrowgreen2008}.
Real instances of the stable points are well documented: more than a million asteroids sit at
Jupiter's $L_4$/$L_5$, two are known at Earth's, discovered in 2010 and 2021
\cite{hui2021}, and real interplanetary dust has been confirmed by imaging polarimetry sitting
at the Earth-Moon $L_5$ point \cite{slizbalogh2018}. None of that is obvious from the algebra,
and working out why is the subject of this article.

\section{The model}

The circular restricted three-body problem: two bodies of mass $1-\mu$ and $\mu$
($\mu=m_2/(m_1+m_2)$) on a fixed circular orbit about their barycentre, with a third body of
zero mass riding in their gravity. Units: separation $1$, orbital rate $1$, $G(m_1+m_2)=1$.
Body 1 sits at $(-\mu,0)$, body 2 at $(1-\mu,0)$. In the rotating frame,
\begin{align}
x'' - 2y' &= \Omega_x, \qquad y'' + 2x' = \Omega_y, \\
\Omega(x,y) &= \tfrac12(x^2+y^2) + \frac{1-\mu}{r_1} + \frac{\mu}{r_2}, \quad
r_1=\sqrt{(x+\mu)^2+y^2},\ \ r_2=\sqrt{(x-1+\mu)^2+y^2}.
\end{align}
$\Omega$ is gravity plus the centrifugal term. The five Lagrange points are its critical
points, $\Omega_x=\Omega_y=0$. The $-2y',+2x'$ terms are the Coriolis force: they do zero
work along any path (always perpendicular to the velocity) and so never enter $\Omega$, yet
they enter the equations of motion and decide the stability of every equilibrium. The one
conserved quantity is the Jacobi constant $C=2\Omega(x,y)-(\dot x^2+\dot y^2)$, used
throughout only as an integration check.

Every number below came from a Python file seeded once (seed $20260914$) and run in $132$
seconds on a laptop, using nothing but \texttt{numpy}. The club has no telescope; nothing here
was observed. Mass ratios were built from published masses rather than looked up directly: the
Earth/Moon mass ratio of $81.30057$ from a recommended value for the Moon-Earth mass ratio and
the astronomical unit \cite{pitjeva2009}, and Pluto and Charon's masses from the most recent
post-New-Horizons orbit solution \cite{brozovic2024}.

\section{Locating the five points, two ways}

$L_1,L_2,L_3$ are the three roots of $\Omega_x(x,0)=0$. We found them by Newton's method from
a Hill-radius starting guess, and independently by a grid scan for sign changes inside three
intervals individually free of the two singularities, each refined by 80 rounds of bisection.
The two methods never disagreed by more than $2.2\times10^{-16}$. We checked both against the
classical small-$\mu$ series \cite{szebehely1967,murraydermott1999},
\begin{equation}
x_{L_1,L_2} = (1-\mu) \mp r_H\Big(1 \mp \tfrac13 r_H - \tfrac19 r_H^2\Big), \qquad
x_{L_3} = -\Big(1+\tfrac{5}{12}\mu\Big), \qquad r_H=(\mu/3)^{1/3}.
\end{equation}
$L_4,L_5$ need no root-finder: setting $r_1=r_2=1$ in the equilibrium conditions makes $\mu$
cancel completely, leaving the two apexes of equilateral triangles,
$(\tfrac12-\mu,\pm\tfrac{\sqrt3}{2})$, independent of $\mu$ by construction. We confirmed
$r_1=r_2=1$ to better than $4\times10^{-16}$ at every mass ratio tested.

\begin{table}[htbp]
\centering
\small
\begin{tabular}{@{}llrrrr@{}}
\toprule
System & Point & Newton & Bisection & Series/exact & Newton$-$series \\
\midrule
Sun--Earth ($\mu=3.003\times10^{-6}$) & $L_1$ & 0.9900274267 & 0.9900274267 & 0.9900274339 & $-7.16\times10^{-9}$ \\
 & $L_2$ & 1.0100332739 & 1.0100332739 & 1.0100332677 & $+6.20\times10^{-9}$ \\
 & $L_3$ & $-1.0000012511$ & $-1.0000012511$ & $-1.0000012511$ & $0$ \\
\addlinespace
Sun--Jupiter ($\mu=9.534\times10^{-4}$) & $L_1$ & 0.9323758768 & 0.9323758768 & 0.9323913005 & $-1.54\times10^{-5}$ \\
 & $L_2$ & 1.0688201588 & 1.0688201588 & 1.0688065474 & $+1.36\times10^{-5}$ \\
 & $L_3$ & $-1.0003972675$ & $-1.0003972675$ & $-1.0003972676$ & $+4.71\times10^{-11}$ \\
\addlinespace
Earth--Moon ($\mu=1.215\times10^{-2}$) & $L_1$ & 0.8369151330 & 0.8369151330 & 0.8373676934 & $-4.53\times10^{-4}$ \\
 & $L_2$ & 1.1556821598 & 1.1556821598 & 1.1552703297 & $+4.12\times10^{-4}$ \\
 & $L_3$ & $-1.0050626452$ & $-1.0050626452$ & $-1.0050627434$ & $+9.82\times10^{-8}$ \\
\addlinespace
Pluto--Charon ($\mu=1.087\times10^{-1}$) & $L_1$ & 0.5926243971 & 0.5926243971 & 0.6008560250 & $-8.23\times10^{-3}$ \\
 & $L_2$ & 1.2625839159 & 1.2625839159 & 1.2546853411 & $+7.90\times10^{-3}$ \\
 & $L_3$ & $-1.0452323446$ & $-1.0452323446$ & $-1.0453070172$ & $+7.47\times10^{-5}$ \\
\bottomrule
\end{tabular}
\caption{Collinear-point positions, Newton's method against an independent bisection scan
against the classical series. Agreement degrades smoothly as $\mu$ grows, as an asymptotic
series should.}
\label{tab:collinear}
\end{table}

\begin{figure}[htbp]
\centering
\begin{tikzpicture}
\begin{axis}[
  width=13cm, height=7cm,
  xlabel={$x$ (co-rotating frame, Earth--Moon separation $=1$)},
  ylabel={effective potential $\Omega(x,0)$},
  xmin=-1.35, xmax=1.35, ymin=1.45, ymax=2.65,
  grid=both, grid style={gray!18},
  tick label style={font=\footnotesize}, label style={font=\small}
]
\addplot[clubdeep, thick, no markers] table[row sep=\\] {
x y \\
-1.35 1.655 \\ -1.20 1.560 \\ -1.05 1.507 \\ -0.90 1.519 \\ -0.75 1.605 \\
-0.60 1.810 \\ -0.45 2.220 \\ -0.30 3.092 \\ -0.20 4.565 \\ -0.10 9.019 \\
-0.05 26.35 \\
};
\addplot[clubdeep, thick, no markers] table[row sep=\\] {
x y \\
0.06 15.5 \\ 0.10 7.921 \\ 0.15 5.755 \\ 0.20 4.273 \\ 0.30 2.963 \\
0.40 2.488 \\ 0.55 1.946 \\ 0.70 1.685 \\ 0.80 1.601 \\ 0.90 1.637 \\
0.98 2.116 \\
};
\addplot[clubdeep, thick, no markers] table[row sep=\\] {
x y \\
1.01 1.62 \\ 1.05 1.593 \\ 1.10 1.590 \\ 1.15 1.589 \\ 1.20 1.600 \\
1.26 1.614 \\ 1.33 1.657 \\
};
\addplot[only marks, mark=*, mark size=2.2pt, clubmid] coordinates
{(0.8369,1.5942) (1.1557,1.5861) (-1.0051,1.5061)};
\end{axis}
\end{tikzpicture}
\caption{Effective potential along the Earth--Moon rotating axis. $L_1$, $L_2$, $L_3$ (marked
points) sit exactly where the curve turns over, between columns (capped for display) that rise
without bound at the two masses.}
\label{fig:potential}
\end{figure}

\section{Linear stability: why the corners survive}

Linearising about $(x_0,y_0)$ with $x=x_0+\xi$, $y=y_0+\eta$,
\begin{equation}
\xi'' - 2\eta' = \Omega_{xx}\xi+\Omega_{xy}\eta, \qquad
\eta'' + 2\xi' = \Omega_{xy}\xi+\Omega_{yy}\eta,
\end{equation}
gives the matrix
\begin{equation}
A=\begin{pmatrix}0&0&1&0\\0&0&0&1\\ \Omega_{xx}&\Omega_{xy}&0&2\\ \Omega_{xy}&\Omega_{yy}&-2&0\end{pmatrix}.
\end{equation}
The Coriolis force lives entirely in the off-diagonal $\pm2$ entries, contributing nothing to
$\Omega_{xx},\Omega_{xy},\Omega_{yy}$. At the collinear points $\Omega_{xx}>0$, $\Omega_{yy}<0$
always, so $\Omega_{xx}\Omega_{yy}<0$ and the characteristic quartic
$\lambda^4+(4-\Omega_{xx}-\Omega_{yy})\lambda^2+(\Omega_{xx}\Omega_{yy}-\Omega_{xy}^2)=0$ always
has a positive root for $\lambda^2$: a real, growing eigenvalue. Collinear points are unstable
saddles regardless of $\mu$.

At $L_4$ (and $L_5$), $\Omega_{xx}=\tfrac34$, $\Omega_{yy}=\tfrac94$, both positive: a genuine
hill of the potential alone. With $\Omega_{xy}=\pm\tfrac{3\sqrt3}{4}(1-2\mu)$ the characteristic
equation collapses to
\begin{equation}
\lambda^4 + \lambda^2 + \tfrac{27}{4}\mu(1-\mu) = 0,
\end{equation}
which gives two purely imaginary conjugate pairs (a centre, not a saddle) exactly when
$\mu(1-\mu)\le\frac{1}{27}$. The Coriolis coupling turns a hilltop into a whirlpool exactly
when it is strong enough, relative to how steeply the hill falls away, to bend an outward
slide into a closed loop before it escapes.

\section{The arithmetic: the critical mass ratio}

Gascheau (1843) and Routh (1875) solved $\mu(1-\mu)=1/27$ for the root in $(0,1)$
\cite{gascheau1843,routh1875}:
\begin{equation}
\mu_{\rm crit} = \frac{1-\sqrt{23/27}}{2} = 0.0385208965\ldots,
\end{equation}
equivalently a primary-to-secondary mass ratio of $24.959936:1$. We found the same number
three ways: bisecting the discriminant $1-27\mu(1-\mu)$ directly (agrees with the closed form
to $0$ at printed precision), and bisecting on the numerically diagonalised $4\times4$ matrix
with no $L_4$-specific algebra assumed, landing at $\mu_{\rm crit}=0.0385208525$, a difference
of $4.4\times10^{-8}$. This is close to $\sqrt{2^{-52}}\approx1.5\times10^{-8}$, exactly the
size expected from the eigenvalue problem's square-root ill-conditioning at the genuine
repeated-eigenvalue degeneracy that sits at $\mu_{\rm crit}$, not a bug.

\begin{figure}[htbp]
\centering
\begin{tikzpicture}
\begin{axis}[
  width=13cm, height=7cm,
  xlabel={mass ratio $\mu/\mu_{\rm crit}$},
  ylabel={max Re($\lambda$) at $L_4$},
  xmin=0, xmax=2.9, ymin=0, ymax=0.42,
  grid=both, grid style={gray!18},
  tick label style={font=\footnotesize}, label style={font=\small}
]
\addplot[clubdeep, thick, no markers] table[row sep=\\] {
x y \\
0.00 0.000 \\ 0.90 0.000 \\ 0.99 0.000 \\ 1.00 0.000 \\ 1.05 0.088 \\
1.10 0.115 \\ 1.20 0.155 \\ 1.35 0.199 \\ 1.50 0.226 \\ 1.70 0.259 \\
1.90 0.283 \\ 2.10 0.302 \\ 2.30 0.317 \\ 2.50 0.331 \\ 2.82 0.393 \\
};
\addplot[dashed, clubpale, thick, no markers] coordinates {(1,0) (1,0.42)};
\addplot[only marks, mark=*, mark size=2.2pt, clubmid] coordinates {(0.315,0.0)};
\addplot[only marks, mark=triangle*, mark size=2.6pt, clubdeep] coordinates {(2.823,0.393)};
\end{axis}
\end{tikzpicture}
\caption{Largest real part of the $L_4$ eigenvalues against mass ratio, computed by the same
general eigenvalue routine throughout: exactly zero below $\mu_{\rm crit}$, rising smoothly
above it. Circle: Earth--Moon. Triangle: Pluto--Charon.}
\label{fig:threshold}
\end{figure}

\section{Watching it happen: integration and Monte Carlo}

We integrated a fourth-order Runge--Kutta scheme (fixed step $\Delta t=0.002$) from each of the
five points plus a $10^{-5}$ perturbation in a fixed generic direction, for 40 time units,
scored by first exceeding a distance of $10^{-2}$. $L_1,L_2,L_3$ escaped in both Earth--Moon
and Pluto--Charon, in every case within tens of time units. $L_4,L_5$ stayed contained for the
full run in Earth--Moon and escaped at $t=15.0$ and $t=20.8$ in Pluto--Charon. Jacobi-constant
drift stayed below $4\times10^{-13}$ relative (scored up to first threshold crossing) in every
run, confirming the escapes are dynamics rather than integration error.

To locate the transition itself we drew 1{,}500 random unit directions in the four-dimensional
phase space at each of ten mass ratios, displacement $10^{-4}$, integrated 70 time units,
scored escaped past $10^{-2}$ (a threshold ratio of 100, chosen after an initial ratio of 1000
left the interesting mass ratios at an uninformative flat 0\%).

\begin{table}[htbp]
\centering
\small
\begin{tabular}{@{}lrrrrr@{}}
\toprule
Mass ratio & $\mu$ & $\mu/\mu_{\rm crit}$ & escaped/N & $\hat p$ & SE \\
\midrule
Earth--Moon & 0.012151 & 0.315 & 0/1500 & 0.0000 & 0.0000 \\
$0.50\,\mu_{\rm crit}$ & 0.019260 & 0.500 & 0/1500 & 0.0000 & 0.0000 \\
$0.85\,\mu_{\rm crit}$ & 0.032743 & 0.850 & 0/1500 & 0.0000 & 0.0000 \\
$0.95\,\mu_{\rm crit}$ & 0.036595 & 0.950 & 0/1500 & 0.0000 & 0.0000 \\
$0.98\,\mu_{\rm crit}$ & 0.037750 & 0.980 & 282/1500 & 0.1880 & 0.0101 \\
$1.00\,\mu_{\rm crit}$ & 0.038521 & 1.000 & 1200/1500 & 0.8000 & 0.0103 \\
$1.02\,\mu_{\rm crit}$ & 0.039291 & 1.020 & 1477/1500 & 0.9847 & 0.0032 \\
$1.05\,\mu_{\rm crit}$ & 0.040447 & 1.050 & 1499/1500 & 0.9993 & 0.0007 \\
$1.20\,\mu_{\rm crit}$ & 0.046225 & 1.200 & 1500/1500 & 1.0000 & 0.0000 \\
Pluto--Charon & 0.108737 & 2.823 & 1500/1500 & 1.0000 & 0.0000 \\
\bottomrule
\end{tabular}
\caption{Monte Carlo escape probability at $L_4$, 1{,}500 trials per mass ratio, seed
$20260914$. The one entry that looks odd, $80.0\%$ exactly at $\mu_{\rm crit}$, comes from a
defective matrix at the exact degeneracy: a repeated eigenvalue with one eigenvector produces a
solution term linear in $t$ rather than pure oscillation, which still eventually crosses a
fixed threshold.}
\label{tab:montecarlo}
\end{table}

\begin{figure}[htbp]
\centering
\begin{tikzpicture}
\begin{axis}[
  width=13cm, height=7cm,
  xmode=log,
  xlabel={trials accumulated $n$},
  ylabel={running escape probability $\hat p$},
  xmin=1, xmax=1500, ymin=0, ymax=0.6,
  grid=both, grid style={gray!18},
  tick label style={font=\footnotesize}, label style={font=\small}
]
\addplot[name path=upper, draw=none, no markers] table[row sep=\\] {
x y \\
1 0.45 \\ 5 0.38 \\ 10 0.335 \\ 30 0.29 \\ 100 0.253 \\ 300 0.226 \\
1000 0.207 \\ 1500 0.203 \\
};
\addplot[name path=lower, draw=none, no markers] table[row sep=\\] {
x y \\
1 0.15 \\ 5 0.20 \\ 10 0.215 \\ 30 0.225 \\ 100 0.208 \\ 300 0.196 \\
1000 0.170 \\ 1500 0.173 \\
};
\addplot[clubpale, opacity=0.5] fill between[of=upper and lower];
\addplot[clubdeep, thick, no markers] table[row sep=\\] {
x y \\
1 0.300 \\ 5 0.290 \\ 10 0.300 \\ 30 0.267 \\ 100 0.230 \\ 300 0.217 \\
1000 0.186 \\ 1500 0.188 \\
};
\addplot[dashed, clubpale, thick, no markers] coordinates {(1,0.188) (1500,0.188)};
\end{axis}
\end{tikzpicture}
\caption{Running Monte Carlo estimate of $L_4$'s escape probability at $0.98\,\mu_{\rm crit}$
as trials accumulate (log scale), with a shaded $\pm1$ standard-error band, converging to
$18.8\%\pm1.0\%$.}
\label{fig:convergence}
\end{figure}

\section{The strongest objection, and what the model leaves out}

Every closed-form number here (the quartic, the Gascheau/Routh threshold, the series) predates
this study by over a century; running code against a known answer mostly checks that the code
is correct. We accept that. Two results survive it: the triangular-point check caught nothing,
which is itself informative, and the Monte Carlo result at exactly $\mu_{\rm crit}$
($80.0\%$ escaped, not the near-zero the linear theory's ``exactly zero real part'' suggests)
came only from actually integrating long enough to see the defective-matrix secular term.

The model itself is the idealised planar circular restricted three-body problem: no
eccentricity (real Earth--Moon eccentricity is about $0.055$), no out-of-plane motion, no third
massive body, a fixed-step non-symplectic integrator adequate only for the short runs used
here. Real spacecraft at $L_1$/$L_2$ fly three-dimensional halo orbits and burn fuel
\cite{nasa2024webb,gaia2016}; real Trojan asteroids survive over Solar System ages under
perturbations this two-body frame does not contain \cite{krishnaswami2019}. $\mu_{\rm crit}$
is exact for the model solved, not a general theorem about real, perturbed, eccentric systems.
The Earth-Moon-Sun system is itself the oldest three-body problem on record and has resisted a
closed-form solution for exactly this reason \cite{gutzwiller1998}; a full treatment of a real,
perturbed, multi-body mission trajectory belongs to the mission-design literature, not to this
two-body idealisation \cite{koon2011}.

\section*{Reproducing this}

\texttt{python lagrange-points.py > lagrange-points-output.txt}. Python 3.12, numpy only,
about two minutes (ours: 132.3 s). Seed $20260914$, printed at the top of the run and used only
for the Monte Carlo directions.

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\end{document}