Worked examples — Halo orbits — linearized motion near Lagrange points

3.2.31 · D3 · Physics › Orbital Mechanics & Astrodynamics › Halo orbits — linearized motion near Lagrange points

Yeh parent topic ka ek hands-on companion hai. Parent ne tumhe machinery di thi; yahan hum use har tarah ke input ke through drive karte hain jo equations face kar sakti hain — taaki jab koi exam ya mission-design worksheet tumhe $c$ ki value, mass ratio $\mu$, ya koi amplitude de, tum pehle se uska twin dekh chuke ho.

Shuru karne se pehle, ek plain-language refresher taaki kuch bhi bina samjhe use na ho.

Recall Har example mein use hone waali teen quantities

• $c$ ::: ek single positive number jo measure karta hai ki us point par leftover gravity gradient kitna strong hai; $c\equiv \frac{1-\mu }{r_1^3}+\frac{\mu }{r_2^3}$. Bada $c$ = zyada steep "hill".
• $\Lambda ={\lambda }^2$ ::: growth/oscillation rate ka square. Agar $\Lambda >0$ toh motion run away karta hai ($\lambda$ real); agar $\Lambda <0$ toh motion oscillate karta hai ($\lambda$ imaginary).
• $\kappa$ ::: in-plane ellipse $y$ mein $x$ se kitni baar zyada tall hai. Algebraically yeh eigenvector ratio hai $\kappa =\frac{{\omega }_{xy}^2+(1+2c)}{2{\omega }_{xy}}$ (Example 3 mein derive aur use hua); abhi isko ek single positive shape number ki tarah socho.

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Scenario matrix

Linearized halo theory mein har problem in cells mein se kisi ek mein aata hai. Neeche ke examples us cell ke saath tagged hain jo woh cover karte hain.

Cell	Kya vary karta hai	Woh question jo yeh force karta hai
A. Nominal collinear	$c>1$ (typical $L_1,L_2,L_3$)	Kya hamesha ek saddle + ek oscillation milta hai?
B. Out-of-plane axis	decoupled $\zeta$ equation	Kya yeh kisi bhi $c>0$ ke liye hamesha pure SHM hai?
C. Sign of $\Lambda$ roots	discriminant aur product of roots	Kaun sa root positive hai, kaun sa negative, aur kabhi do positive kyun nahi?
D. Degenerate limit $c\to 1^{+}$	boundary case	Frequencies $c=1$ par exactly kya hoti hain?
E. Large-$c$ limit	$c\gg 1$ (deep gravity well)	$\gamma$, ${\omega }_{xy}$, ${\omega }_z$ kaise scale karte hain?
F. Amplitude ratio $\kappa$	eigenvector geometry	Planar ellipse ki shape; kya hamesha $>0$ hai?
G. Real-world word problem	Sun–Earth $L_1$ (SOHO)	Dimensionless rates ko real days mein convert karo.
H. Exam twist	Lissajous closing condition	Loop halo kab hota hai aur Lissajous kab?

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Example 1 — Cell A: nominal collinear point ($c=3.19$, Earth–Moon $L_2$)

Forecast: signs ka andaza lagao — kya do oscillations milenge, do run-aways, ya ek-ek?

1. Coefficients assemble karo. $2-c=-1.19$, $1+2c=7.38$, $1-c=-2.19$. Yeh step kyun? Characteristic equation ${\Lambda }^2+(2-c)\Lambda +(1+2c)(1-c)=0$ (upar formula callout mein restate ki gayi) ko sirf yahi teen numbers chahiye aur kuch nahi.
2. Discriminant. $D=(2-c{)}^2-4(1+2c)(1-c)=1.4161-4(7.38)(-2.19)=1.4161+64.6488=66.0649$, toh $\sqrt{D}=8.1281$. Yeh step kyun? $D>0$ guarantee karta hai do real, distinct values of $\Lambda$ — koi complex $\Lambda$ nahi, isliye koi spiral nahi. Figure s01 dekho (characteristic equation ka parabola): woh zero ko do baar cross karta hai.
3. Roots. $\Lambda =\frac{-(2-c)\pm \sqrt{D}}{2}=\frac{1.19\pm 8.1281}{2}$ → ${\Lambda }_1=+4.659$, ${\Lambda }_2=-3.469$. Yeh step kyun? Ek positive, ek negative — yahi saddle × center signature hai.
4. Rates. $\gamma =\sqrt{{\Lambda }_1}=\sqrt{4.659}\approx 2.158$; ${\omega }_{xy}=\sqrt{-{\Lambda }_2}=\sqrt{3.469}\approx 1.862$. Yeh step kyun? $\Lambda >0\Rightarrow \lambda =\pm \gamma$ real (growth $e^{\gamma t}$). $\Lambda <0\Rightarrow \lambda =\pm i{\omega }_{xy}$ (oscillation).

Figure s01 — $c=3.19$ ke liye characteristic-equation parabola $f(\Lambda )$: iske do zero-crossings (amber dots) roots ${\Lambda }_1>0$ (saddle) aur ${\Lambda }_2<0$ (oscillation) hain.

Verify: product ${\Lambda }_1{\Lambda }_2=4.659\times (-3.469)\approx -16.16$, aur $(1+2c)(1-c)=7.38\times (-2.19)=-16.16$. ✓ Sum ${\Lambda }_1+{\Lambda }_2=1.190=-(2-c)$. ✓

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Example 2 — Cell B: out-of-plane axis pure SHM hai (koi bhi $c>0$)

Forecast: kya out-of-plane bob kabhi in-plane saddle ki tarah run away karta hai?

1. SHM template se compare karo. SHM hai $\ddot{s}=-{\omega }^{2}s$. Yahan $\ddot{\zeta }=-c\zeta$, toh ${\omega }_z^2=c$. Yeh step kyun? Template se match karne par frequency bina re-solving ke pad leti hai — restoring force displacement ke proportional hai aur back point karti hai, kyunki $c>0$.
2. Frequency. ${\omega }_z=\sqrt{3.19}\approx 1.786$.
3. Period. $T_z=2\pi /{\omega }_z=6.2832/1.786\approx 3.518$ dimensionless time units. Yeh step kyun? Bob har $T_z$ pe repeat karta hai; halos ke liye yeh in-plane period ke saath line up karna zaroori hai.

Verify: ${\omega }_z^2=1.786^2=3.190=c$. ✓ $T_z{\omega }_z=3.518\times 1.786=6.283=2\pi$. ✓

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Example 3 — Cell F: in-plane amplitude ratio $\kappa$ (ellipse shape)

Forecast: kya loop $x$ ya $y$ ke along zyada wide hai?

1. Formula kahan se aata hai. Oscillatory root ek eigenvector $(A,B)$ deta hai; amplitude mein $\eta =\kappa \xi$ force karna aur pehli linearized equation use karne par milta hai $\kappa =\frac{{\omega }_{xy}^2+(1+2c)}{2{\omega }_{xy}}$. Yeh step kyun? $\kappa$ free nahi hai — yeh oscillatory root ke eigenvector se fixed hai. Unit-$\xi$ amplitude ke liye $\eta$ amplitude exactly $\kappa$ hai.
2. Plug in karo. $\kappa =\frac{{\omega }_{xy}^2+(1+2c)}{2{\omega }_{xy}}=\frac{3.469+7.38}{3.724}=\frac{10.849}{3.724}\approx 2.913$. Yeh step kyun? Humhare paas pehle se ${\omega }_{xy}^2=3.469$ hai (Example 1 se $-{\Lambda }_2$ hai) aur $1+2c=7.38$; known numbers substitute karna hi eigenvector formula ko ek concrete shape factor mein badalne ke liye bacha hua arithmetic hai.
3. Geometry interpret karo. Kyunki $\kappa \approx 2.91>1$, ellipse $y$ ke along stretch hua hai lagbhag teen ke factor se. Figure s02 dekho. Yeh step kyun? Parametric pair $\xi =-A_x\cos ,\eta =\kappa A_x\sin$ ek ellipse trace karta hai jiske semi-axes hain $A_x$ ($x$ mein) aur $\kappa A_x$ ($y$ mein).

Figure s02 — parametric in-plane ellipse $(\xi ,\eta )$: $y$ semi-axis $x$ semi-axis se $\kappa \approx 2.91$ guna hai, toh loop Lagrange point ke around vertically elongated hai.

Verify: $\kappa \times 2{\omega }_{xy}=2.913\times 3.724=10.849={\omega }_{xy}^2+1+2c$. ✓

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Example 4 — Cell D: degenerate limit $c\to 1^{+}$

Forecast: kya $c=1$ approach karte waqt instability strengthen hoti hai ya vanish?

1. $c=1$ par constant term. $(1+2c)(1-c)=3\times 0=0$. Yeh step kyun? Roots ka product yahi term hai; agar yeh zero hai, toh ek root exactly zero hai.
2. $c=1$ par solve karo. ${\Lambda }^2+(2-1)\Lambda +0=\Lambda (\Lambda +1)=0$ → ${\Lambda }_1=0,{\Lambda }_2=-1$. Yeh step kyun? Factored form seedha padhne se boundary values milti hain.
3. Interpret karo. $\gamma =\sqrt{{\Lambda }_1}=0$: exponential instability rate $c=1$ par zero ho jaata hai. Saath hi ${\omega }_{xy}=\sqrt{1}=1$ aur ${\omega }_z=\sqrt{c}=1$. Yeh step kyun? $c=1$ precisely saddle (collinear, $c>1$) aur pure-center (triangular-like, $c<1$) behaviour ke beech ki borderline hai. Figure s03 dekho: amber parabola origin par zero touch karta hai.

Figure s03 — product-of-roots curve $(1+2c)(1-c)$ vs $c$: $c>1$ ke liye negative (saddle), exactly $c=1$ par zero (marginal, $\gamma \to 0$), aur $c<1$ ke liye positive (pure oscillation).

Verify: $c=1$ par, roots $0$ aur $-1$; sum $=-1=-(2-c)$; product $=0=(1+2c)(1-c)$. ✓ Aur ${\omega }_z=\sqrt{1}=1$. ✓

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Example 5 — Cell E: deep-well limit $c\gg 1$

Forecast: $c$ ke saath konsa zyada tezi se badata hai — instability rate ya in-plane frequency?

1. Large $c$ ke liye roots approximate karo. $2-c\approx -c$ aur constant term $\approx (2c)(-c)=-2c^2$ ke saath: $\Lambda \approx \frac{c\pm \sqrt{c^2+8c^2}}{2}=\frac{c\pm 3c}{2}.$ Yeh step kyun? Sirf top powers of $c$ rakhne se dominant behaviour isolate hota hai; $\sqrt{c^2+8c^2}=\sqrt{9c^2}=3c$.
2. Do roots. ${\Lambda }_1\approx \frac{c+3c}{2}=2c$, ${\Lambda }_2\approx \frac{c-3c}{2}=-c$. Yeh step kyun? Ab $\gamma \approx \sqrt{2c}$ aur ${\omega }_{xy}\approx \sqrt{c}$, jabki ${\omega }_z=\sqrt{c}$ exactly.
3. $c=100$ par test karo (exact). $D=(2-100{)}^2-4(201)(-99)=9604+79596=89200$, $\sqrt{D}=298.66$. $\Lambda =\frac{98\pm 298.66}{2}$ → ${\Lambda }_1=198.33$, ${\Lambda }_2=-100.33$. Toh $\gamma =14.083$, ${\omega }_{xy}=10.017$, ${\omega }_z=10$. Yeh step kyun? Approximations se compare karo: $\sqrt{2\cdot 100}=14.14$ vs $14.08$ ✓; $\sqrt{100}=10$ vs $10.017$ ✓.

Verify: $c=100$ par: ${\Lambda }_1+{\Lambda }_2=198.33-100.33=98=-(2-c)$ ✓; product $=198.33\times (-100.33)=-19898\approx (201)(-99)=-19899$ ✓.

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Example 6 — Cell G: real-world word problem (Sun–Earth $L_1$, SOHO)

Forecast: kya ek uncontrolled SOHO weeks, months, ya years mein drift off kar jaayega?

1. Dimensionless in-plane rates. $2-c=-2.061$, $1+2c=9.122$, $1-c=-3.061$. $D=(-2.061{)}^2-4(9.122)(-3.061)=4.2477+111.69=115.94$, $\sqrt{D}=10.768$. ${\Lambda }_1=\frac{2.061+10.768}{2}=6.4145$, ${\Lambda }_2=-4.3535$. Toh $\gamma =2.5327$, ${\omega }_{xy}=2.0865$. Yeh step kyun? Example 1 jaisi hi machine, bas naya $c$.
2. Time unit. Ek full frame revolution $2\pi$ normalized time units hai aur $365.25$ real days leta hai, toh $1$ normalized time unit $=\frac{365.25}{2\pi }=58.13$ days. Yeh step kyun? Humari har computed rate per normalized time unit mein hai; days mein bolne ke liye time-units ko $58.13$ days se multiply karte hain.
3. Instability ka $e$-folding time. Unstable mode $e^{\gamma t}$ ki tarah badata hai; yeh $t=1/\gamma =1/2.5327=0.3948$ time units $=0.3948\times 58.13\approx 22.9$ days mein factor $e$ se badta hai. Yeh step kyun? $1/\gamma$ natural "main kitni tezi se run away kar raha hun" clock hai. ~23 days → SOHO ko har kuch weeks mein station-keeping chahiye.
4. In-plane period days mein. $T_{xy}=2\pi /{\omega }_{xy}=6.2832/2.0865=3.0114$ time units $=3.0114\times 58.13\approx 175.1$ days. Yeh step kyun? Wobble period ko calendar figure mein convert karta hai — in-plane loop per roughly aadha saal.

Verify: $1/\gamma =0.3948$ tu; $\times 58.13=22.95$ days. $T_{xy}=3.0114$ tu $\times 58.13=175.05$ days. $58.13\times 2\pi =365.25$ ✓.

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Example 7 — Cell H: Lissajous vs Halo (exam twist)

Forecast: kya do frequencies equal hain? Agar nahi, toh kya path close hota hai?

1. Frequencies compare karo. ${\omega }_{xy}=1.862$ (in-plane wobble) aur ${\omega }_z=1.786$ (out-of-plane bob). Ratio ${\omega }_{xy}/{\omega }_z=1.862/1.786=1.0426\ne 1$. Yeh step kyun? Ek 3-D loop sirf tab khud ko retrace karta hai jab do frequencies commensurate hon; sabse clean closure 1:1 ratio hai. Kyunki $1.0426$ na $1$ hai (na koi simple ratio), periods line up nahi karte.
2. Mismatch dekhne ke liye dono periods compute karo. $T_{xy}=2\pi /{\omega }_{xy}=3.375$ tu, $T_z=2\pi /{\omega }_z=3.518$ tu. Woh har loop $0.143$ tu differ karte hain. Yeh step kyun? Explicit periods drift ko concrete banate hain: ek in-plane loop ke baad out-of-plane bob abhi apne start par wapas nahi aaya, toh 3-D curve har turn ek naye point par land karta hai.
3. Resulting motion ko naam do. Kyunki curve kabhi exactly apni starting state par return nahi karta, yeh slowly precess karta hai aur ek torus par ek band densely fill karta hai — ek quasi-periodic Lissajous orbit, closed halo nahi. Figure s04 dekho. Yeh step kyun? Yeh unequal, incommensurate frequencies ka direct geometric consequence hai.
4. Closing condition state karo. Genuine halo ke liye $\boxed{{\omega }_{xy}={\omega }_z}$ (ek 1:1 resonance) chahiye. Linear theory in dono frequencies ko unequal values par fix karta hai, toh woh condition meet nahi kar sakta. Sirf nonlinear, amplitude-dependent frequency corrections (Richardson's third-order expansion) ${\omega }_{xy}$ aur ${\omega }_z$ ko shift karti hain jab tak, ek special amplitude $A_z$ par, woh coincide nahi karte aur loop snap shut nahi hota. Yeh step kyun? Yahi is topic ka poora moral hai — linear theory ingredients supply karta hai (frequencies, $\kappa$, ellipse shape); halo ka existence ek nonlinear amplitude constraint hai.

Figure s04 — kaafi periods tak trace ki gayi $(\eta ,\zeta )$ projection: kyunki ${\omega }_{xy}\ne {\omega }_z$ hai, curve kabhi khud ko retrace nahi karta aur densely ek band fill karta hai — Lissajous orbit ki signature, closed halo ki nahi.

Verify: ${\omega }_{xy}/{\omega }_z=1.862/1.786=1.0426\ne 1$; $T_{xy}=3.375$ tu, $T_z=3.518$ tu, difference $0.143$ tu — periods genuinely unequal, toh koi closure nahi. ✓

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Recall check

Recall

Do in-plane roots ka product kaun si quantity ke barabar hai, aur collinear point ke liye kya sign? ::: $(1+2c)(1-c)$; negative kyunki $c>1$, forcing one saddle + one center. Out-of-plane motion kin values of $c$ ke liye stable hai? ::: Sabhi $c>0$ ke liye, kyunki $U_{zz}=-c<0$ pure SHM deta hai ${\omega }_z=\sqrt{c}$ ke saath. Linear theory kya FAIL karta hai produce karne mein, aur kya fix karta hai? ::: Ek closed halo; fix hai nonlinear amplitude tuning jo ${\omega }_{xy}={\omega }_z$ force karti hai.

See also: Circular Restricted Three-Body Problem · Lagrange Points L1–L5 · Jacobi Integral and Zero-Velocity Curves · Invariant Manifolds and Low-Energy Transfers · Station-keeping and Orbit Maintenance · Simple Harmonic Motion