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

KYA karte hain: dono masses ko $\mu$ ki definition mein plug karo. $\mu =\frac{1.898\times 10^{27}}{1.989\times 10^{30}+1.898\times 10^{27}}=\frac{1.898\times 10^{27}}{1.9909\times 10^{30}}\approx 9.534\times 10^{-4}.$ KYUN: $\mu$ bas chhote body ka total mass mein fraction hai — aur kuch nahi. Bada primary (Sun): $x=-\mu \approx -9.53\times 10^{-4}$. Chhota primary (Jupiter): $x=1-\mu \approx 0.99905$. Sun thoda sa origin se door hai; barycentre (origin) lagbhag Sun ke andar hi hai.

Paanch Lagrange points hote hain.

• $L_1,L_2,L_3$ $x$-axis par hote hain (woh $y=0$ line jo primaries ko join karti hai) → collinear.
• $L_4,L_5$ axis se door hote hain, har ek dono primaries ke saath ek equilateral triangle banata hai (kyunki axis ke bahar equilibrium force karta hai $r_1=r_2=1$, aur primaries bhi $1$ se separate hain).

Step 1 — har primary se distances (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 aur assemble karo: $v^2={\dot{x}}^2+{\dot{y}}^2=0.01+0.04=0.05,C_J=2(1.8555)-0.05=\boxed{3.661}.$ KYUN matter karta hai: $C_J$ ki value poori trajectory ke liye fixed rehti hai (Coriolis koi kaam nahi karta). Craft baad mein jahan bhi jaaye, $2U\ge C_J$ hold karna chahiye — yahi forbidden regions ko carve karta hai.

$27\mu (1-\mu )=27(9.534\times 10^{-4})(0.99905)=\mathrm{0.02572.}$ Kyunki $1-0.02572=0.9743>0$, discriminant positive hai ⇒ dono $\Lambda$ roots real aur negative ⇒ $\lambda =\pm i\omega$ (pure oscillation) ⇒ linearly stable. Yahi reason hai ki Trojan asteroids Jupiter ke $L_4$ aur $L_5$ par aaraam se baithe hain.

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

• ${\Lambda }_1=+8.611>0\Rightarrow \lambda =\pm 2.935$ (real, ek positive → exponential blow-up).
• ${\Lambda }_2=-5.511<0\Rightarrow \lambda =\pm i2.348$ (pure oscillation). KYUN unstable: ek real positive $\lambda$ ka hona hi kaafi hai — ek chhota sa nudge $e^{+2.935t}$ ki tarah grow karta hai. $L_1$ ek saddle hai — ek direction mein stable, doosre mein runaway. Yahi reason hai ki halo orbits ko $L_1$ ke around constant correction chahiye.

${\Lambda }^2+b\Lambda +c=0$ ke liye, roots ka product $c=U_{xx}{U}_{yy}-U_{xy}^2=U_{xx}{U}_{yy}$ hai (kyunki $U_{xy}=0$). Jab $U_{xx}>0$ aur $U_{yy}<0$ ho, toh product $c<0$ hoga. Do numbers jinka product negative ho, unke opposite signs honge → ek $\Lambda >0$, ek $\Lambda <0$. Positive $\Lambda$ se ek real positive $\lambda$ milta hai ⇒ hamesha unstable. $\mu$ par koi dependence nahi — collinear points har mass ratio ke liye unstable hain. Dekho Eigenvalues & linear stability analysis.

Step 1 — coefficients. $U_{xx}+U_{yy}=\frac{3}{4}+\frac{9}{4}=3$, toh $4-U_{xx}-U_{yy}=4-3=1$. Constant term: $U_{xx}{U}_{yy}-U_{xy}^2=\frac{3}{4}\cdot \frac{9}{4}-{\left(\frac{3\sqrt{3}}{4}\right)}^2(1-2\mu {)}^2=\frac{27}{16}-\frac{27}{16}(1-2\mu {)}^2=\frac{27}{16}[1-(1-2\mu {)}^2].$ Ab $1-(1-2\mu {)}^2=1-(1-4\mu +4{\mu }^2)=4\mu -4{\mu }^2=4\mu (1-\mu )$, toh constant $=\frac{27}{16}\cdot 4\mu (1-\mu )=\frac{27}{4}\mu (1-\mu )$. Step 2 — $\Lambda$-quadratic. ${\Lambda }^2+\Lambda +\frac{27}{4}\mu (1-\mu )=0.$ Step 3 — real, negative roots demand karo. Roots ka sum $=-1<0$ ✓ (dono negative agar real hain), product $=\frac{27}{4}\mu (1-\mu )>0$ ✓ (same sign). Jo fail ho sakta hai woh reality hai: hum discriminant $\ge 0$ chahte hain: $1^2-4\cdot \frac{27}{4}\mu (1-\mu )=1-27\mu (1-\mu )\ge 0.$ Yahi criterion hai. KYUN discriminant: agar woh negative hota, toh $\Lambda$ complex hota, jisse $\lambda$ mein positive real part aata → growth. Step 4 — $27\mu (1-\mu )=1$ solve karo, yaani $27{\mu }^2-27\mu +1=0$: $\mu =\frac{27\pm \sqrt{27^2-4\cdot 27}}{2\cdot 27}=\frac{27\pm \sqrt{729-108}}{54}=\frac{27\pm \sqrt{621}}{54}.$ Chhota root lete hain (physical wala, $\mu \le \frac{1}{2}$): ${\mu }_{\text{crit}}=\frac{27-\sqrt{621}}{54}=\frac{1}{2}\left(1-\sqrt{\frac{23}{27}}\right)\approx \mathrm{0.03852.}$ Koi bhi system jisme $\mu <0.03852$ ho, uske triangular points stable hain.

Har $\mu$ ko ${\mu }_{\text{crit}}=0.03852$ se compare karo:

• Sun–Jupiter: $9.5\times 10^{-4}<0.03852$ → stable (real Trojans exist karte hain).
• Earth–Moon: $0.0123<0.03852$ → stable (marginally; dust "Kordylewski clouds" debated hain).
• Pluto–Charon: $0.104>0.03852$ → $L_4/L_5$ par unstable; Charon Pluto ke relative bahut massive hai. KYUN cutoff exist karta hai: Coriolis term (woh "$4$" → cancellation ke baad yahan "$1$") ek potential maximum ko tabhi stabilise kar sakta hai jab dono masses kaafi lopsided hon; near-equal pair yeh steering advantage kho deta hai.

Constant term: $\frac{27}{4}\mu (1-\mu )=\frac{27}{4}(9.534\times 10^{-4})(0.99905)=6.750\times 9.534\times 10^{-4}\times 0.99905=6.4304\times 10^{-3}.$ ${\Lambda }^2+\Lambda +6.4304\times 10^{-3}=0\Rightarrow \Lambda =\frac{-1\pm \sqrt{1-4(6.4304\times 10^{-3})}}{2}=\frac{-1\pm \sqrt{0.974278}}{2}.$ $\sqrt{0.974278}=0.987055$, toh ${\Lambda }_1=\frac{-1+0.987055}{2}=-6.4726\times 10^{-3},{\Lambda }_2=\frac{-1-0.987055}{2}=-\mathrm{0.993528.}$ Dono negative → stable, jaise expected tha. Frequencies ${\omega }_i=\sqrt{-{\Lambda }_i}$ hain (kyunki $\lambda =\pm i\omega$): ${\omega }_{\text{long}}=\sqrt{6.4726\times 10^{-3}}=0.08045,{\omega }_{\text{short}}=\sqrt{0.993528}=\mathrm{0.99676.}$ Interpretation: ${\omega }_{\text{short}}\approx 1$ matlab ek libration mode ka roughly Jupiter ka orbital period hai (yaad karo hamare units mein $\omega =1$). Doosra, ${\omega }_{\text{long}}\approx 0.080$, ek slow drift hai jiska period $2\pi /0.08045\approx 78$ Jupiter-orbits hai — Trojan asteroids ki famous long-period tadpole libration.

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

• Sum: ${\Lambda }_1+{\Lambda }_2=-1$. Kyunki ${\omega }_i^2=-{\Lambda }_i$, hum paate hain ${\omega }_{\text{short}}^2+{\omega }_{\text{long}}^2=-({\Lambda }_1+{\Lambda }_2)=1$. ✓
• Product: ${\Lambda }_1{\Lambda }_2=\frac{27}{4}\mu (1-\mu )$, toh ${\omega }_{\text{short}}^2{\omega }_{\text{long}}^2={\Lambda }_1{\Lambda }_2=\frac{27}{4}\mu (1-\mu )$. ✓ Small-$\mu$ limit: short mode $\to$ orbital frequency, ${\omega }_{\text{short}}^2\to 1$. Phir product se, ${\omega }_{\text{long}}^2=\frac{\frac{27}{4}\mu (1-\mu )}{{\omega }_{\text{short}}^2}\to \frac{27}{4}\mu \Rightarrow {\omega }_{\text{long}}\approx \frac{3\sqrt{3}}{2}\sqrt{\mu }.$ 5.1 se check karo: $\frac{3\sqrt{3}}{2}\sqrt{9.534\times 10^{-4}}=2.598\times 0.03088=0.08023$ — exact $0.08045$ se teen digits tak match karta hai. Jaise $\mu$ chhota hota hai, tadpole libration aur bhi slow hoti jaati hai.

Ek Lagrange point ke linearly stable hone ke liye dono $\Lambda$ roots mein kya hona chahiye? ::: Dono real aur negative hone chahiye, taaki $\lambda =\pm i\omega$ ho (pure oscillation, koi growth nahi). $L_1,L_2,L_3$ kyun har mass ratio ke liye unstable hain? ::: Unka constant term $U_{xx}{U}_{yy}-U_{xy}^2<0$ hota hai, jisse ek $\Lambda >0$ force hota hai, hence ek real positive $\lambda$. Triangular-point stability ke liye critical mass parameter kya hai? ::: ${\mu }_{\text{crit}}=\frac{1}{2}\left(1-\sqrt{23/27}\right)\approx 0.03852$. $L_4/L_5$ par $U$ ke maximum ko kaun si fictitious force stabilise karti hai? ::: Coriolis force, jo runaway ko ek closed libration mein bend kar deta hai.