Worked examples — Lagrange points L1–L5 — derivation, stability

Everything here uses the same normalized units from the parent: total mass $=1$, body separation $D=1$, $G(M_1+M_2)=1$, so the co-rotating angular speed is $\omega =1$. The two heavy bodies sit at $x=-\mu$ (the big one, $M_1$) and $x=+(1-\mu )$ (the small one, $M_2$), where $\mu =\frac{M_2}{M_1+M_2}.$

Prereqs you may want open: Restricted Three-Body Problem, Effective Potential & Jacobi Constant, Coriolis and Centrifugal Forces (Rotating Frames), Hill Sphere, Kepler's Third Law.

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The scenario matrix

Every problem this topic throws is one of these cells. The examples that follow are tagged with the cell they cover.

Cell	Case class	What is special / degenerate	Covered by
A	Collinear L1, small $\mu$	between the bodies; use Hill-radius approx	Ex 1
B	Collinear L2, small $\mu$	beyond the small body; sign of $r$ flips	Ex 2
C	Collinear L3, small $\mu$	far side, near $-1$; needs its own expansion	Ex 3
D	Triangular L4/L5 location	off-axis, $y\ne 0$, equilateral triangle	Ex 4
E	Stability decision, small ratio	is $M_1/M_2>24.96$?	Ex 5
F	Degenerate equal masses $\mu =\frac{1}{2}$	symmetry, L4/L5 unstable, L3 at center-ish	Ex 6
G	Real word problem (JWST at L2)	plug real km, convert units	Ex 7
H	Exam twist: Jacobi/energy sign	zero-velocity gate at L1	Ex 8

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Example 1 — Cell A: Sun–Earth L1

Step 1 — Pick the right tool: the Hill-radius approximation. Because $\mu \ll 1$, L1 sits very near the small body. The full collinear equation is a quintic, but near $M_2$ it collapses to the leading balance $r\approx D{\left(\frac{\mu }{3}\right)}^{1/3}.$ Why this step? We don't need the exact quintic root; the small-$\mu$ expansion keeps only the dominant tidal + centrifugal terms (the "3" = tidal 2 + centrifugal 1). See Hill Sphere.

Step 2 — Plug in. $r\approx 1.496\times 10^8{\left(\frac{3.0\times 10^{-6}}{3}\right)}^{1/3}=1.496\times 10^8\times (10^{-6}{)}^{1/3}.$ Why this step? $\mu /3=10^{-6}$ exactly here — the cube root is clean.

Step 3 — Evaluate the cube root. $(10^{-6}{)}^{1/3}=10^{-2},r\approx 1.496\times 10^8\times 10^{-2}=1.496\times 10^6\text{ km}.$

Verify: $\approx 1.5$ million km — this is where SOHO lives. That is $\sim 1\%$ of $D$, not halfway. Forecast "hugging Earth" was right. Units: km $\times$ (dimensionless) $=$ km ✓.

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Example 2 — Cell B: Sun–Earth L2 and why the sign flips

Step 1 — Same leading balance, opposite side. For L2 the small body's gravity now points toward the Sun (inward) while centrifugal points outward. Expanding the collinear equation on the far side gives the same leading form $r\approx D{\left(\frac{\mu }{3}\right)}^{1/3}.$ Why this step? To leading order in $\mu$, L1 and L2 are symmetric about $M_2$ — both at $\pm r$. The difference only appears at the next order.

Step 2 — Numbers are identical to leading order. $r\approx 1.496\times 10^6\text{ km, but now on the anti-Sun side.}$ Why this step? Same $\mu$, same cube root; the geometry (which side) is what flips, not the magnitude.

Verify: JWST and Gaia orbit Sun–Earth L2 at $\approx 1.5$ million km anti-Sunward. Consistent with the real mission (Halo Orbits & Station-Keeping (JWST, SOHO)). ✓

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Example 3 — Cell C: Sun–Earth L3 (far side)

Step 1 — Different expansion. L3 is near $x=-1$, far from the tiny body. The small-$\mu$ result here is $x_{L3}\approx -1-\frac{5}{12}\mu .$ Why this step? Near the Sun the small body is a distant perturber, so we expand the quintic about $x=-1$ instead of about $M_2$. The coefficient $5/12$ is standard.

Step 2 — Plug $\mu =3.0\times 10^{-6}$. $\delta =\frac{5}{12}\mu =\frac{5}{12}(3.0\times 10^{-6})=1.25\times 10^{-6}.$ So $x_{L3}\approx -1.00000125$. Why this step? The offset $\delta$ is the shift beyond the antipode in normalized units.

Verify: $\delta \times D=1.25\times 10^{-6}\times 1.496\times 10^8\approx 187$ km. So L3 is $\sim 190$ km past the exact opposite point — a truly tiny nudge, as forecast. ✓

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Example 4 — Cell D: L4/L5 geometry

Step 1 — Use the exact triangular solution. The parent showed $r_1=r_2=1$ solves both ${\partial }_x\Omega =0$ and ${\partial }_y\Omega =0$, giving $x=\frac{1}{2}-\mu ,y=+\frac{\sqrt{3}}{2}.$ Why this step? No approximation is needed — L4/L5 are exact regardless of $\mu$. The equilateral triangle is a geometric identity, not a small-$\mu$ trick.

Step 2 — Plug $\mu =9.54\times 10^{-4}$. $x=0.5-0.000954=0.499046,y=\mathrm{0.866025.}$

Step 3 — Check both distances equal 1. Distance to $M_1$ at $(-\mu ,0)$: $r_1=\sqrt{(x+\mu {)}^2+y^2}=\sqrt{0.5^2+(\sqrt{3}/2{)}^2}=\sqrt{0.25+0.75}=1$. Distance to $M_2$ at $(1-\mu ,0)$: $r_2=\sqrt{(x-(1-\mu ){)}^2+y^2}=\sqrt{(-0.5{)}^2+0.75}=1$. Why this step? Confirms the equilateral claim: both legs $=1=$ the base $D$.

Verify: $r_1=r_2=D=1$ ✓ — equilateral triangle confirmed. This is where the Trojan Asteroids cluster.

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Example 5 — Cell E: Stability decision

Step 1 — State the criterion. L4/L5 are stable iff $\frac{M_1}{M_2}>\frac{25+\sqrt{621}}{2}\approx \mathrm{24.9599.}$ Why this step? This comes from the characteristic equation ${\lambda }^4+{\lambda }^2+\frac{27}{4}\mu (1-\mu )=0$ needing real-negative roots in $s={\lambda }^2$ — the Coriolis force (see Coriolis and Centrifugal Forces (Rotating Frames)) must be strong enough to whirl a drifting mass back.

Step 2 — Compute the threshold. $\sqrt{621}=24.9199,\frac{25+24.9199}{2}=\mathrm{24.9599.}$

Step 3 — Compare. Sun–Jupiter: $M_1/M_2\approx 1047.6>24.96$ → stable. Earth–Moon: $M_1/M_2\approx 81.3>24.96$ → stable (weakly; Kordylewski dust clouds). Why this step? Both ratios clear the bar with room to spare.

Verify: Threshold $=24.9599$; both $1047.6$ and $81.3$ exceed it. Forecast "yes/yes" holds. ✓

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Example 6 — Cell F: Degenerate equal masses $\mu =\frac{1}{2}$

Step 1 — L4/L5 location. $x=\frac{1}{2}-\mu =\frac{1}{2}-\frac{1}{2}=0$, $y=\pm \frac{\sqrt{3}}{2}$. Why this step? With equal masses the triangle apex sits exactly on the $y$-axis (above the barycenter). The equilateral geometry still holds — that never depends on $\mu$.

Step 2 — Stability check. Need $\mu <{\mu }_c=\frac{1}{2}(1-\sqrt{23/27})\approx 0.0385$. Here $\mu =0.5\gg 0.0385$. Why this step? The Coriolis-strength condition fails badly; a nudge grows without bound.

Verify: ${\mu }_c\approx 0.0385$ and $0.5>0.0385$ → unstable. Equal-mass binaries cannot trap Trojans at L4/L5. The symmetry that creates the points does not stabilize them. ✓

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Example 7 — Cell G: Real word problem — sizing JWST's station-keeping region

Step 1 — Hill radius. $R_H=D{\left(\frac{\mu }{3}\right)}^{1/3}=1.496\times 10^8\times (10^{-6}{)}^{1/3}=1.496\times 10^8\times 10^{-2}.$ Why this step? The Hill radius (see Hill Sphere) sets the scale of both L1 and L2 for small $\mu$ — they sit essentially at $\pm R_H$ from Earth.

Step 2 — Evaluate. $R_H\approx 1.496\times 10^6\text{ km}.$

Verify: $R_H\approx 1.5\times 10^6$ km $\approx$ L2's actual distance. JWST does one loop inside roughly this scale. Units: km ✓. Matches Ex 1/Ex 2 as it must (L1, L2, and $R_H$ share the leading order). ✓

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Example 8 — Cell H: Exam twist — the Jacobi/zero-velocity gate

Step 1 — Recall the potential. $\Omega (x,0)=-\frac{1-\mu }{|x+\mu |}-\frac{\mu }{|x-(1-\mu )|}-\frac{1}{2}{x}^2.$ Why this step? On the axis $y=0$, so $r_1=|x+\mu |$, $r_2=|x-(1-\mu )|$. This is the Effective Potential & Jacobi Constant restricted to the line.

Step 2 — Locate L1 and L2 for $\mu =0.1$. Solving the collinear quintic numerically gives approximately $x_{L1}\approx 0.6090,x_{L2}\approx \mathrm{1.2668.}$ Why this step? We need the roots first to know where to evaluate $\Omega$.

Step 3 — Evaluate $\Omega$. $\Omega (x_{L1})\approx -1.8871,\Omega (x_{L2})\approx -\mathrm{1.8541.}$ Why this step? Higher $\Omega$ = a lower "barrier" in the zero-velocity picture ($C_J=-2\Omega$ is smaller). The inner gate (L1) is more negative → opens last; the outer neck at L2 opens before L1 here.

Verify: Numerically $\Omega (L2)>\Omega (L1)$ (both near $-1.85$ to $-1.89$). The first neck to open as energy rises is the one with the largest $\Omega$; among these two it is L2. This is exactly why a spacecraft can leak out past L2 before it can cross inward at L1. ✓

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Recall Which cell does each example cover?

Ex 1 collinear L1 small $\mu$ ::: Cell A Ex 2 collinear L2 small $\mu$ (sign flip) ::: Cell B Ex 3 collinear L3 far side ::: Cell C Ex 4 triangular L4/L5 geometry ::: Cell D Ex 5 stability decision ::: Cell E Ex 6 degenerate equal masses ::: Cell F Ex 7 real word problem (JWST/Hill) ::: Cell G Ex 8 exam twist (Jacobi gate) ::: Cell H