Halo orbits — linearized motion near Lagrange points: Exercises

3.2.31 · D4 · Physics › Orbital Mechanics & Astrodynamics › Halo orbits — linearized motion near Lagrange points

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Figure — Halo orbits — linearized motion near Lagrange points

Level 1 — Recognition

Exercise 1.1

Linearized out-of-plane equation hai $\ddot{ζ} = - c ζ$ . Kuch bhi solve kiye bina, motion ka type batao aur uski frequency likho.

Recall Solution

Is equation ki form hai $\ddot{ζ} = - (positive number) \cdot ζ$ . Yeh Simple Harmonic Motion ka exact signature hai (dekho Simple Harmonic Motion): acceleration displacement ke proportional aur zero ki taraf wapas point karta hua. $\ddot{ζ} = - ω^{2} ζ$ se match karne par milta hai $ω^{2} = c$ , toh $ω_{z} = c .$ Minus sign kyun matter karta hai: negative sign "restoring" arrow hai. Agar yeh $+ c$ hota, toh force aapko door dhakelta aur aapko oscillation ki jagah exponential runaway milta.

Exercise 1.2

Eigenvalue analysis mein aapko ek $Λ_{1} > 0$ aur ek $Λ_{2} < 0$ milta hai. Dono ke liye batao ki $λ = Λ$ real hai ya imaginary, aur yeh kaisi motion produce karta hai.

Recall Solution

$Λ_{1} > 0$ : $λ = \pm Λ_{1} = \pm γ$ real hai. Real exponents $e^{+ γ t}$ (badhta hai) aur $e^{- γ t}$ (ghadta hai) dete hain → ek saddle, yaani unstable.
$Λ_{2} < 0$ : $λ = \pm Λ_{2} = \pm i ω_{x y}$ imaginary hai. Imaginary exponents $e^{\pm i ω_{x y} t} = cos \pm i sin$ dete hain → pure oscillation. Picture: figure s01 mein, red arrows door point karte hain (saddle), violet loop oscillation hai.

Level 2 — Application

Exercise 2.1

Sun–Earth $L_{1}$ ke liye ek representative value hai $c = 4.06$ . $ω_{z}$ , aur in-plane roots $Λ_{1}, Λ_{2}$ compute karo.

Recall Solution

Out-of-plane. $ω_{z} = 4.06 \approx 2.015$ . In-plane. Teen ingredients banao: $2 - c = - 2.06$ , $1 + 2 c = 9.12$ , $1 - c = - 3.06$ . Discriminant $D = (2 - c)^{2} - 4 (1 + 2 c) (1 - c) = (- 2.06)^{2} - 4 (9.12) (- 3.06) = 4.2436 + 111.63 = 115.87$ . $D \approx 10.765$ . $Λ = \frac{- ( 2 - c ) \pm D}{2} = \frac{2.06 \pm 10.765}{2} .$ $Λ_{1} = \frac{2.06 + 10.765}{2} \approx 6.41$ , $Λ_{2} = \frac{2.06 - 10.765}{2} \approx - 4.35$ . Kyun: $Λ_{1} > 0$ (saddle) aur $Λ_{2} < 0$ (oscillation) — yahi structure har collinear point par hoti hai.

Exercise 2.2

Upar wale $L_{1}$ numbers use karke, instability rate $γ$ aur in-plane frequency $ω_{x y}$ compute karo. Phir in-plane oscillation period $T = 2 π / ω_{x y}$ nikalo (dimensionless time mein).

Recall Solution

$γ = Λ_{1} = 6.41 \approx 2.532.$ $ω_{x y} = - Λ_{2} = 4.35 \approx 2.086.$ $T = \frac{2 π}{ω _{x y}} = \frac{6.2832}{2.086} \approx 3.012.$ Interpretation: unstable eigenvector ke along displacement har dimensionless time unit mein $e^{γ} \approx e^{2.53} \approx 12.6$ se multiply hoti hai — yahi reason hai ki $L_{1}$ probes ko station-keeping chahiye (Station-keeping and Orbit Maintenance).

Level 3 — Analysis

Exercise 3.1

Algebraically dikhao ki har collinear point ke liye jahan $c > 1$ ho, product $Λ_{1} Λ_{2} < 0$ hota hai (ek saddle, ek oscillation). Is fact ka use karo ki $A Λ^{2} + B Λ + C = 0$ ke liye roots ka product $C / A$ hota hai.

Recall Solution

$Λ$ mein quartic hai $Λ^{2} + (2 - c) Λ + (1 + 2 c) (1 - c) = 0$ , toh $A = 1$ , $C = (1 + 2 c) (1 - c)$ . Roots ka product $= \frac{C}{A} = (1 + 2 c) (1 - c)$ . $c > 1$ ke liye: $(1 + 2 c) > 0$ hamesha, aur $(1 - c) < 0$ . Toh product $(+) (-) < 0$ hai. Negative product ka matlab hai dono roots ke opposite signs hain → exactly ek positive $Λ_{1}$ aur ek negative $Λ_{2}$ . Yeh saddle × center structure hai, bina numbers ke prove kiya gaya.

Exercise 3.2

Sun–Earth $L_{1}$ ( $ω_{x y} = 2.086$ , $c = 4.06$ ) ke liye amplitude ratio $κ$ compute karo aur batao ki in-plane ellipse kis axis ke along stretch hoti hai.

Recall Solution

$κ = \frac{ω _{x y}^{2} + ( 1 + 2 c )}{2 ω _{x y}} = \frac{( 2.086 ) ^{2} + 9.12}{2 ( 2.086 )} = \frac{4.351 + 9.12}{4.172} = \frac{13.471}{4.172} \approx 3.229.$ Kyunki $κ > 1$ hai, $η$ (yaani $y$ ) amplitude $\approx 3.23 \times$ $ξ$ (yaani $x$ ) amplitude hai. Planar ellipse $y$ ke along stretch hoti hai — Sun–Earth line ke across elongated, uske along squashed. Figure s02 dekho: violet ellipse tall hai, round nahi.

Level 4 — Synthesis

Exercise 4.1

Sun–Earth $L_{1}$ ( $ω_{x y} \approx 2.086$ , $ω_{z} \approx 2.015$ ) ke numbers use karke explain karo ki linear theory ek Lissajous kyun produce karta hai, halo nahi, aur quantify karo ki dono frequencies percentage mein kitni mismatched hain.

Recall Solution

In-plane wobble har $T_{x y} = 2 π / ω_{x y}$ pe repeat hoti hai aur out-of-plane bob har $T_{z} = 2 π / ω_{z}$ pe. Ek path ek single 3-D loop mein tabhi close hota hai jab ye dono periods commensurate hon — simple halo ke liye hum chahte hain 1:1 resonance $ω_{x y} = ω_{z}$ . Mismatch: $\frac{ω _{x y} - ω _{z}}{ω _{z}} = \frac{2.086 - 2.015}{2.015} \approx 0.0352 = 3.52%$ . Kyunki $ω_{x y} \neq = ω_{z}$ hai (dono $\approx 3.5%$ se different hain), in-plane aur out-of-plane motions phase se drift ho jaati hain aur trajectory ek torus-jaisi band fill karta hai → ek Lissajous orbit, quasi-periodic, kabhi close nahi hota. Figure s03 dekho: dono frequencies ek non-closing ribbon weave karti hain. Kya fix karta hai: nonlinear frequency-amplitude corrections (Richardson ka third-order expansion). Ek specific amplitude par corrected frequencies $ω_{x y} = ω_{z}$ satisfy karti hain, aur loop snap karke ek halo mein close ho jaata hai. Linear theory ingredients deta hai ( $ω_{x y}, ω_{z}, κ$ , ellipse shape); closing condition nonlinear hai.

Exercise 4.2

Ek designer chahta hai ki out-of-plane bob in-plane wobble ke relative jitna possible ho utna slow ho, taaki ek bahut flat Lissajous bane. Sun–Earth $L_{1}$ ( $c = 4.06$ ) aur Earth–Moon $L_{2}$ ( $c = 3.19$ ) ke beech, kaun sa smaller ratio $ω_{z} / ω_{x y}$ deta hai? Dono compute karo.

Recall Solution

Sun–Earth $L_{1}$ : $ω_{z} = 2.015$ , $ω_{x y} = 2.086$ (Ex 2.2/2.1). Ratio $= 2.015/2.086 \approx 0.9660$ . Earth–Moon $L_{2}$ : parent worked example se, $ω_{z} = 3.19 \approx 1.786$ , $ω_{x y} = 3.47 \approx 1.863$ . Ratio $= 1.786/1.863 \approx 0.9587$ . Earth–Moon $L_{2}$ ka smaller ratio hai ( $0.959 < 0.966$ ), toh uska out-of-plane bob proportionally sabse slow hai — dono mein se flatter Lissajous.

Level 5 — Mastery

Exercise 5.1

Ek collinear point ke liye stability index ko loosely define karo per in-plane period exponential growth factor ke roop mein, $Σ \equiv e^{γ T_{x y}}$ jahan $T_{x y} = 2 π / ω_{x y}$ . Earth–Moon $L_{2}$ ( $γ \approx 2.16$ , $ω_{x y} \approx 1.863$ ) ke liye $Σ$ compute karo aur station-keeping cadence ke liye interpret karo.

Recall Solution

$T_{x y} = \frac{2 π}{ω _{x y}} = \frac{6.2832}{1.863} \approx 3.372.$ $Σ = e^{γ T_{x y}} = e^{2.16 \times 3.372} = e^{7.284} \approx 1454.$ Matlab: unstable eigenvector ke along koi bhi error ek single in-plane period mein $\approx 1450$ ke factor se badhta hai. 1 km error ek cycle mein $\sim$ 1450 km ban jaata hai. Yahi reason hai ki maintenance burns ko orbit mein kai baar schedule karna padta hai, sirf ek baar nahi — manifold structure (Invariant Manifolds and Low-Energy Transfers) jo aapko door le jaata hai wahi leverage hai jo station-keeping saste mein correct karne ke liye use karta hai.

Exercise 5.2

Dikhao ki eigenvector amplitude ratio $κ$ ko $ω_{x y}$ ki jagah $Λ_{2}$ use karke likha ja sakta hai, aur Earth–Moon $L_{2}$ ke liye numerically verify karo ki dono forms agree karti hain. Yaad karo $ω_{x y}^{2} = - Λ_{2}$ .

Recall Solution

Start karo $κ = \frac{ω _{x y}^{2} + ( 1 + 2 c )}{2 ω _{x y}}$ se. Substitute karo $ω_{x y}^{2} = - Λ_{2}$ aur $ω_{x y} = - Λ_{2}$ : $κ = \frac{- Λ _{2} + ( 1 + 2 c )}{2 - Λ _{2}} .$ Verify karo (Earth–Moon $L_{2}$ , $c = 3.19$ , $Λ_{2} = - 3.47$ ): Numerator $= 3.47 + 7.38 = 10.85$ . Denominator $= 2 3.47 = 2 (1.863) = 3.726$ . $κ = 10.85/3.726 \approx 2.912$ — parent note ke $κ \approx 2.92$ se match karta hai. ✓ Yeh kyun matter karta hai: yeh aapko $κ$ directly eigenvalue $Λ_{2}$ se lene deta hai bina $ω_{x y}$ ke liye alag se square root liye — design pipeline mein kam rounding steps.