Exercises — Lagrange points L1–L5 — derivation, stability

• L1: on the line between $M_1$ and $M_2$, between them (nearer the lighter body $M_2$).
• L2: on the same line, beyond $M_2$ (outside the smaller body).
• L3: on the same line, beyond $M_1$, roughly opposite $M_2$.
• L4: off the line, forming an equilateral triangle with the two bodies, leading $M_2$ by $60^{\circ }$.
• L5: the mirror of L4, trailing $M_2$ by $60^{\circ }$.

Look at the map below — the three collinear points lie on the black dashed axis, the two triangular ones cap the equilateral triangles.

• Unstable: L1, L2, L3 (they are saddle points of $\Omega$ — a nudge grows).
• Can be stable: L4, L5, even though they are maxima of $\Omega$.
• The rescuer is the Coriolis force — it acts only on a moving particle, curving a drifting test mass into a small closed loop instead of letting it escape. See Coriolis and Centrifugal Forces (Rotating Frames).

False. A particle sitting at rest in the rotating frame has velocity $v=0$, and Coriolis $=-2\omega \times v=0$. So locating the points needs only gravity + centrifugal force. Coriolis appears only in the stability (dynamics) analysis, where the particle actually moves.

WHAT: plug the numbers into the Hill-radius-scale formula. $\frac{\mu }{3}=\frac{3.0\times 10^{-6}}{3}=1.0\times 10^{-6},(1.0\times 10^{-6}{)}^{1/3}=1.0\times 10^{-2}.$ $r\approx 1.50\times 10^8\times 10^{-2}=1.5\times 10^6\text{km}.$ Sanity: SOHO sits at L1 and JWST at L2, both ~1.5 million km from Earth. ✓ See Halo Orbits & Station-Keeping (JWST, SOHO).

$\frac{25+\sqrt{621}}{2}=\frac{25+24.92}{2}\approx \mathrm{24.96.}$ Since $1047\gg 24.96$, the criterion is satisfied with enormous margin ⇒ Jupiter's Trojans are stable. This is exactly why thousands of Trojan Asteroids cluster at Jupiter's L4 and L5. ✓

${\omega }^2=\frac{G(M_1+M_2)}{D^3}=\frac{1}{1^3}=1⟹\omega =1.$ Why it helps: with $\omega =1$ the centrifugal potential $-\frac{1}{2}{\omega }^2(x^2+y^2)$ loses its clutter and becomes simply $-\frac{1}{2}(x^2+y^2)$, so $\Omega$ is clean to differentiate. This is Kepler's Third Law applied to the two big bodies.

WHAT: the barycenter is at $x_{\mathrm{cm}}=\frac{M_1{x}_1+M_2{x}_2}{M_1+M_2}$. In our units $M_1=1-\mu$ and $M_2=\mu$. $M_1{x}_1+M_2{x}_2=(1-\mu )(-\mu )+\mu (1-\mu )=-\mu (1-\mu )+\mu (1-\mu )=0.$ So $x_{\mathrm{cm}}=0$ ✓ — the two contributions cancel exactly, which is the defining property of the barycenter.

WHAT: compute the two distances. $r_1^2=(x+\mu {)}^2+y^2={\left(\frac{1}{2}-\mu +\mu \right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{4}+\frac{3}{4}=1.$ $r_2^2=(x-(1-\mu ){)}^2+y^2={\left(\frac{1}{2}-\mu -1+\mu \right)}^2+\frac{3}{4}={\left(-\frac{1}{2}\right)}^2+\frac{3}{4}=1.$ Both give $r_1=r_2=1=D$. WHAT IT LOOKS LIKE: an equilateral triangle — all three sides equal length $1$ (figure s01). This is why L4/L5 are called the triangular points.

Set up each region's signs (the test mass at position $x$):

• At L1 ($-\mu <x<1-\mu$): $x+\mu >0$ (mass is to the right of $M_1$, so $M_1$ pulls it left / negative-$x$), and $x-(1-\mu )<0$ (mass is to the left of $M_2$, so $M_2$ pulls it right / positive-$x$). The two gravities point in opposite directions along the axis → they partly cancel, and the leftover plus centrifugal $+x$ balances to zero.
• At L2 ($x>1-\mu$): now $x+\mu >0$ and $x-(1-\mu )>0$. Both gravities pull the mass back toward the bodies (negative-$x$) → they add. Centrifugal $+x$ (outward) alone must balance both. That's why L2 is farther out than a naive single-body estimate.

What it looks like: on figure s02 the two gravity arrows at L1 point toward each other; at L2 they point the same way (inward), and the centrifugal arrow (outward) balances them.

Setup. Put the test mass at distance $r$ from $M_2$, on the line, with $r\ll D$. Work in real units, $D$ general. Three forces along the line, taking toward $M_1$ (inward) as positive:

1. Gravity of $M_2$ (magnitude $G{M}_2/r^2$) pulls the mass toward $M_2$ — i.e. outward for L2, so it enters as $-G{M}_2/r^2$.
2. Tidal gravity of $M_1$. At $M_2$'s location the mass would feel $G{M}_1/D^2$; a step $r$ farther out changes it by the gradient $\frac{d}{dR}\left(\frac{G{M}_1}{R^2}\right)=-\frac{2G{M}_1}{R^3}$. Over distance $r$: an extra inward change of magnitude $\frac{2G{M}_{1}r}{D^3}$... but relative to circular motion at $M_2$ we keep only the difference.
3. Centrifugal at radius $\approx D+r$: ${\omega }^2(D+r)$ with ${\omega }^2=G{M}_1/D^3$ (dominant body). Relative to $M_2$'s orbit this contributes $+{\omega }^{2}r=\frac{G{M}_{1}r}{D^3}$ outward.

Combine the two $M_1$ terms (tidal $2$ + centrifugal $1$): the net restoring pull from the large body's environment is $(2+1)\frac{G{M}_{1}r}{D^3}=\frac{3G{M}_{1}r}{D^3}.$ That is the source of the "3" — tidal gradient contributes $2$, centrifugal contributes $1$.

Balance against $M_2$'s local gravity: $\frac{G{M}_2}{r^2}=\frac{3G{M}_{1}r}{D^3}⟹r^3=\frac{M_2}{3M_1}{D}^3⟹r=D{\left(\frac{M_2}{3M_1}\right)}^{1/3}.$ For small $\mu$, $M_2/M_1\approx \mu$, so $\boxed{r\approx D{\left(\frac{\mu }{3}\right)}^{1/3}.}$ This is the same length scale as the Hill Sphere radius — no coincidence, both come from tidal + centrifugal balance.

WHAT: substitute $s={\lambda }^2$ to turn the quartic into a quadratic: $s^2+s+\frac{27}{4}\mu (1-\mu )=0.$ WHY: motion stays bounded only if every $\lambda$ is purely imaginary, i.e. every $s={\lambda }^2$ is real and negative (so $\lambda =\pm i\sqrt{|s|}$, pure oscillation). For $s$ to be real, the discriminant of the quadratic must be non-negative: $\Delta =1-4\cdot \frac{27}{4}\mu (1-\mu )=1-27\mu (1-\mu )\ge 0.$ That's the required condition: $27\mu (1-\mu )\le 1$. Solve the boundary $27\mu (1-\mu )=1$, i.e. $27{\mu }^2-27\mu +1=0$: $\mu =\frac{27\pm \sqrt{27^2-4\cdot 27}}{2\cdot 27}=\frac{27\pm \sqrt{621}}{54}=\frac{1}{2}\left(1\pm \sqrt{\frac{621}{729}}\right)=\frac{1}{2}\left(1\pm \sqrt{\frac{23}{27}}\right).$ The physically relevant (small) root is $\boxed{{\mu }_c=\frac{1}{2}\left(1-\sqrt{\frac{23}{27}}\right)\approx \mathrm{0.0385.}}$ Stability holds for $\mu <{\mu }_c$. (When $s$ is real-negative we also need both roots negative — sum $=-1<0$ and product $=\frac{27}{4}\mu (1-\mu )>0$ guarantee it. ✓)

From $\mu$ to $M_1/M_2$. Since $\mu =\frac{M_2}{M_1+M_2}$, write $q=M_1/M_2$. Then $\mu =\frac{1}{q+1}.$ The boundary $27\mu (1-\mu )=1$ becomes, with $\mu =1/(q+1)$ and $1-\mu =q/(q+1)$: $27\cdot \frac{1}{q+1}\cdot \frac{q}{q+1}=1⟹27q=(q+1{)}^2⟹q^2-25q+1=0.$ $q=\frac{25\pm \sqrt{625-4}}{2}=\frac{25\pm \sqrt{621}}{2}.$ The stable regime is the larger root: $q>\frac{25+\sqrt{621}}{2}\approx \mathrm{24.96.}$

Earth–Moon: $q=81.3>24.96$ ✓ ⇒ the Moon's L4/L5 are (weakly) stable. This is the theoretical home of the faint Kordylewski dust clouds.

Evaluate $f(\mu )=27\mu (1-\mu )$ and compare to $1$ (stable when $f\le 1$).

• (a) $\mu \to 0$: $f\to 0\le 1$ ✓ — stable, with huge margin. Makes sense: a vanishingly light second body barely perturbs the equilateral point, and Coriolis easily traps the grain.
• (b) $\mu =\frac{1}{2}$: $f=27\cdot \frac{1}{2}\cdot \frac{1}{2}=6.75>1$ ✗ — unstable. With two equal masses, the perturbing pulls on the triangular point are too strong for Coriolis to tame; the point cannot hold a body.
• Boundary check: stability is lost somewhere in between, precisely at ${\mu }_c\approx 0.0385$ (from Problem 4.2). So the stable window is $0\le \mu <0.0385$ — only lopsided pairs (one body much heavier) get stable Trojans. That matches reality: Sun–Jupiter, Sun–Neptune, Earth–Moon all have tiny $\mu$.

Both scales are set by the competition between a body's own gravity and the tidal + centrifugal stretch from a larger neighbour.

• At the Hill/L1 scale $r\sim D(\mu /3{)}^{1/3}$, the question is about a third small body's orbit: inside this radius $M_2$ can hold a satellite in a bound orbit; at $r$ (the L1 gate) the satellite can leak out toward $M_1$. It marks the edge of gravitational ownership.
• At the Roche limit, the question is about a fluid or rubble-pile satellite's own cohesion: when the tidal stretch across the satellite exceeds its self-gravity, the satellite is torn apart (rings form). The Roche limit scales as $R_{\mathrm{Roche}}\sim R_{M_2}(2{\rho }_1/{\rho }_2{)}^{1/3}$ — depending on densities, not the mass ratio directly. Contrast: Hill/L1 asks "can $M_2$ keep an orbiting companion?"; Roche asks "will a companion survive intact?". Same tidal machinery, different thing being weighed — orbit stability versus material strength.

Which two Lagrange points can be stable, and why despite being potential maxima? ::: L4 and L5; the velocity-dependent Coriolis force curves a drifting mass into a bounded loop, but only if $M_1/M_2>24.96$ (equivalently $\mu <0.0385$). Where does the "3" in $r=D(\mu /3{)}^{1/3}$ come from? ::: Tidal gradient contributes 2, centrifugal contributes 1, summing to 3. Is L4 stable for equal masses ($\mu =\frac{1}{2}$)? ::: No — $27\mu (1-\mu )=6.75>1$, so motion is unbounded.