Exercises — Three-body problem — restricted (CR3BP), characteristic equation

WHAT we do: plug the two masses into the definition of $\mu$. $\mu =\frac{1.898\times 10^{27}}{1.989\times 10^{30}+1.898\times 10^{27}}=\frac{1.898\times 10^{27}}{1.9909\times 10^{30}}\approx 9.534\times 10^{-4}.$ WHY: $\mu$ is just the fraction of the total mass carried by the smaller body — nothing more. Big primary (Sun): $x=-\mu \approx -9.53\times 10^{-4}$. Small primary (Jupiter): $x=1-\mu \approx 0.99905$. The Sun sits a hair off the origin; the barycentre (origin) is almost inside the Sun.

There are five Lagrange points.

• $L_1,L_2,L_3$ lie on the $x$-axis (the $y=0$ line joining the primaries) → collinear.
• $L_4,L_5$ lie off-axis, each forming an equilateral triangle with the two primaries (because equilibrium off the axis forces $r_1=r_2=1$, and the primaries are also separated by $1$).

Step 1 — distances to each primary (parent formulas, $z=0$): $r_1=\sqrt{(x+\mu {)}^2+y^2}=\sqrt{(0.5+0.0123{)}^2+0.3^2}=\sqrt{0.5123^2+0.09}=\sqrt{0.3524}=0.5936,$ $r_2=\sqrt{(x-1+\mu {)}^2+y^2}=\sqrt{(0.5-0.9877{)}^2+0.09}=\sqrt{(-0.4877{)}^2+0.09}=\sqrt{0.3288}=\mathrm{0.5734.}$ Step 2 — effective potential: $U=\frac{1}{2}(x^2+y^2)+\frac{1-\mu }{r_1}+\frac{\mu }{r_2}=\frac{1}{2}(0.25+0.09)+\frac{0.9877}{0.5936}+\frac{0.0123}{0.5734}.$ $U=0.17+1.6640+0.02145=\mathrm{1.8555.}$ Step 3 — speed-squared and assemble: $v^2={\dot{x}}^2+{\dot{y}}^2=0.01+0.04=0.05,C_J=2(1.8555)-0.05=\boxed{3.661}.$ WHY it matters: the value $C_J$ is fixed for the whole trajectory (Coriolis does no work). Wherever the craft later goes, $2U\ge C_J$ must hold — that carves out the forbidden regions.

$27\mu (1-\mu )=27(9.534\times 10^{-4})(0.99905)=\mathrm{0.02572.}$ Since $1-0.02572=0.9743>0$, the discriminant is positive ⇒ both $\Lambda$ roots real and negative ⇒ $\lambda =\pm i\omega$ (pure oscillation) ⇒ linearly stable. This is exactly why the Trojan asteroids sit contentedly at Jupiter's $L_4$ and $L_5$.

The characteristic equation (parent) with $U_{xy}=0$: ${\Lambda }^2+(4-U_{xx}-U_{yy})\Lambda +U_{xx}{U}_{yy}=0.$ Coefficients: $4-11.3-(-4.2)=4-11.3+4.2=-3.1$; constant $=11.3\times (-4.2)=-47.46$. ${\Lambda }^2-3.1\Lambda -47.46=0\Rightarrow \Lambda =\frac{3.1\pm \sqrt{3.1^2+4(47.46)}}{2}=\frac{3.1\pm \sqrt{9.61+189.84}}{2}=\frac{3.1\pm 14.123}{2}.$ ${\Lambda }_1=8.611,{\Lambda }_2=-\mathrm{5.511.}$ Read off $\lambda =\pm \sqrt{\Lambda }$:

• ${\Lambda }_1=+8.611>0\Rightarrow \lambda =\pm 2.935$ (real, one positive → exponential blow-up).
• ${\Lambda }_2=-5.511<0\Rightarrow \lambda =\pm i2.348$ (pure oscillation). WHY unstable: the mere existence of a real positive $\lambda$ means a nudge grows like $e^{+2.935t}$. $L_1$ is a saddle — stable in one direction, runaway in another. This is why halo orbits around $L_1$ need constant correction.

For ${\Lambda }^2+b\Lambda +c=0$, the product of roots is $c=U_{xx}{U}_{yy}-U_{xy}^2=U_{xx}{U}_{yy}$ (since $U_{xy}=0$). With $U_{xx}>0$ and $U_{yy}<0$, the product $c<0$. Two numbers whose product is negative must have opposite signs → one $\Lambda >0$, one $\Lambda <0$. The positive $\Lambda$ gives a real positive $\lambda$ ⇒ always unstable. No dependence on $\mu$ — collinear points are unstable for every mass ratio. See Eigenvalues & linear stability analysis.

Step 1 — coefficients. $U_{xx}+U_{yy}=\frac{3}{4}+\frac{9}{4}=3$, so $4-U_{xx}-U_{yy}=4-3=1$. Constant term: $U_{xx}{U}_{yy}-U_{xy}^2=\frac{3}{4}\cdot \frac{9}{4}-{\left(\frac{3\sqrt{3}}{4}\right)}^2(1-2\mu {)}^2=\frac{27}{16}-\frac{27}{16}(1-2\mu {)}^2=\frac{27}{16}[1-(1-2\mu {)}^2].$ Now $1-(1-2\mu {)}^2=1-(1-4\mu +4{\mu }^2)=4\mu -4{\mu }^2=4\mu (1-\mu )$, so the constant $=\frac{27}{16}\cdot 4\mu (1-\mu )=\frac{27}{4}\mu (1-\mu )$. Step 2 — the $\Lambda$-quadratic. ${\Lambda }^2+\Lambda +\frac{27}{4}\mu (1-\mu )=0.$ Step 3 — demand real, negative roots. Sum of roots $=-1<0$ ✓ (both negative if real), product $=\frac{27}{4}\mu (1-\mu )>0$ ✓ (same sign). What can fail is reality: we need the discriminant $\ge 0$: $1^2-4\cdot \frac{27}{4}\mu (1-\mu )=1-27\mu (1-\mu )\ge 0.$ That is the criterion. WHY discriminant: if it were negative, $\Lambda$ would be complex, giving $\lambda$ with a positive real part → growth. Step 4 — solve $27\mu (1-\mu )=1$, i.e. $27{\mu }^2-27\mu +1=0$: $\mu =\frac{27\pm \sqrt{27^2-4\cdot 27}}{2\cdot 27}=\frac{27\pm \sqrt{729-108}}{54}=\frac{27\pm \sqrt{621}}{54}.$ Taking the smaller root (the physical one, $\mu \le \frac{1}{2}$): ${\mu }_{\text{crit}}=\frac{27-\sqrt{621}}{54}=\frac{1}{2}\left(1-\sqrt{\frac{23}{27}}\right)\approx \mathrm{0.03852.}$ Any system with $\mu <0.03852$ has stable triangular points.

Compare each $\mu$ to ${\mu }_{\text{crit}}=0.03852$:

• Sun–Jupiter: $9.5\times 10^{-4}<0.03852$ → stable (real Trojans exist).
• Earth–Moon: $0.0123<0.03852$ → stable (marginally; dust "Kordylewski clouds" debated).
• Pluto–Charon: $0.104>0.03852$ → unstable at $L_4/L_5$; Charon is too massive relative to Pluto. WHY the cutoff exists: the Coriolis term (the "$4$" → here "$1$" after cancellation) can stabilise a potential maximum only while the two masses are lopsided enough; a near-equal pair loses that steering advantage.

Constant term: $\frac{27}{4}\mu (1-\mu )=\frac{27}{4}(9.534\times 10^{-4})(0.99905)=6.750\times 9.534\times 10^{-4}\times 0.99905=6.4304\times 10^{-3}.$ ${\Lambda }^2+\Lambda +6.4304\times 10^{-3}=0\Rightarrow \Lambda =\frac{-1\pm \sqrt{1-4(6.4304\times 10^{-3})}}{2}=\frac{-1\pm \sqrt{0.974278}}{2}.$ $\sqrt{0.974278}=0.987055$, so ${\Lambda }_1=\frac{-1+0.987055}{2}=-6.4726\times 10^{-3},{\Lambda }_2=\frac{-1-0.987055}{2}=-\mathrm{0.993528.}$ Both negative → stable, as expected. The frequencies are ${\omega }_i=\sqrt{-{\Lambda }_i}$ (since $\lambda =\pm i\omega$): ${\omega }_{\text{long}}=\sqrt{6.4726\times 10^{-3}}=0.08045,{\omega }_{\text{short}}=\sqrt{0.993528}=\mathrm{0.99676.}$ Interpretation: ${\omega }_{\text{short}}\approx 1$ means one libration mode has roughly the orbital period of Jupiter (recall $\omega =1$ in our units). The other, ${\omega }_{\text{long}}\approx 0.080$, is a slow drift with period $2\pi /0.08045\approx 78$ Jupiter-orbits — the famous long-period tadpole libration of Trojan asteroids.

From ${\Lambda }^2+\Lambda +\frac{27}{4}\mu (1-\mu )=0$ with roots ${\Lambda }_1,{\Lambda }_2$:

• Sum: ${\Lambda }_1+{\Lambda }_2=-1$. Since ${\omega }_i^2=-{\Lambda }_i$, we get ${\omega }_{\text{short}}^2+{\omega }_{\text{long}}^2=-({\Lambda }_1+{\Lambda }_2)=1$. ✓
• Product: ${\Lambda }_1{\Lambda }_2=\frac{27}{4}\mu (1-\mu )$, so ${\omega }_{\text{short}}^2{\omega }_{\text{long}}^2={\Lambda }_1{\Lambda }_2=\frac{27}{4}\mu (1-\mu )$. ✓ Small-$\mu$ limit: the short mode $\to$ the orbital frequency, ${\omega }_{\text{short}}^2\to 1$. Then from the product, ${\omega }_{\text{long}}^2=\frac{\frac{27}{4}\mu (1-\mu )}{{\omega }_{\text{short}}^2}\to \frac{27}{4}\mu \Rightarrow {\omega }_{\text{long}}\approx \frac{3\sqrt{3}}{2}\sqrt{\mu }.$ Check against 5.1: $\frac{3\sqrt{3}}{2}\sqrt{9.534\times 10^{-4}}=2.598\times 0.03088=0.08023$ — matches the exact $0.08045$ to three digits. As $\mu$ shrinks, the tadpole libration grows ever slower.

What must be true of both $\Lambda$ roots for a Lagrange point to be linearly stable? ::: Both real and negative, so that $\lambda =\pm i\omega$ (pure oscillation, no growth). Why are $L_1,L_2,L_3$ unstable for every mass ratio? ::: Their constant term $U_{xx}{U}_{yy}-U_{xy}^2<0$, forcing one $\Lambda >0$ hence a real positive $\lambda$. What is the critical mass parameter for triangular-point stability? ::: ${\mu }_{\text{crit}}=\frac{1}{2}\left(1-\sqrt{23/27}\right)\approx 0.03852$. Which fictitious force stabilises the maximum of $U$ at $L_4/L_5$? ::: The Coriolis force, which bends the runaway into a closed libration.