Halo orbits — linearized motion near Lagrange points: Foundations

3.2.31 · D1 · Physics › Orbital Mechanics & Astrodynamics › Halo orbits — linearized motion near Lagrange points

Yeh parent topic ka ground-floor page hai. Parent jo bhi arrow, letter, aur dot use karta hai, woh sab yahan absolute zero se define kiye gaye hain, us order mein jismein har ek ko pichle ki zaroorat hoti hai.

1. Position, aur letter ke upar dot ka matlab

Kuch bhi move karne se pehle, hume kehna hoga ki woh kahan hai.

Ab, motion ka matlab hai position time ke saath badalna. Physicists "rate of change" ko letter ke upar ek dot se mark karte hain — yeh Newton ka shorthand hai.

Topic ko yeh kyun chahiye: equations of motion acceleration ke baare mein rules hain — woh kehte hain "given ki aap kahan hain, yahan aapka $\overset{x}{¨}$ hai." Baad ki har cheez ek acceleration bookkeeping exercise hai.

Figure s01. Ek body ke liye time ke against teen curves jo steadily speed up karti hai: blue curve position $x$ hai (tezi se barhti), yellow line uski slope $\overset{x}{˙}$ hai (velocity, seedha barhti), aur flat red line $\overset{x}{¨}$ hai (acceleration, constant). White vertical arrows dikhate hain ki har dot ek "neeche wala kitni tezi se badal raha hai?" sawal ka jawaab deta hai.

2. Mass ratio $μ$

Do bade bodies (primaries) spacecraft par kheenchte hain. Units fix karne ya unhe rakhne se pehle, hume ek number chahiye jo describe kare ki unke masses kaise compare karte hain.

Hum $μ$ pehle define karte hain kyunki agla step (units choose karna) do masses ko $1 - μ$ aur $μ$ banaa dega.

3. Spin rate $ω$

Do primaries ek steady rate par apne common balance point ke around orbit karte hain. Us rate ka ek naam hai.

Hum $ω$ pehle define karte hain neeche normalization mein use karne se pehle, kyunki normalization ka poora point $ω = 1$ set karna hai.

4. Units normalize karna taaki $ω = 1$ ho

Koi bhi equation likhne se pehle hume apne rulers aur clocks fix karne honge. Poora CR3BP dimensionless (normalized) units mein likha jaata hai — ek bookkeeping trick jo messy constants chhupaa deti hai.

Definition Normalization convention

Hum apne units choose karte hain taaki chaar cheezein $1$ ke barabar hon:

Unit length = do primaries ke beech constant distance (toh woh fixed spots par baith jaate hain).
Unit mass = pair ki total mass, $m_{1} + m_{2}$ . Is choice se bade body ki mass $\frac{m _{1}}{m _{1} + m _{2}} = 1 - μ$ aur chhote body ki $\frac{m _{2}}{m _{1} + m _{2}} = μ$ ho jaati hai (§2 se $μ$ use karke), aur woh $1$ mein add hote hain.
Gravitational constant $G = 1$ (units mein fold kiya gaya).
Unit time = aisa choose kiya taaki pair apni orbit mein per time unit ek radian complete kare, jo spin rate $ω = 1$ force karta hai (§3 se $ω$ use karke). Problem mein har ruler aur clock ko rescale karna socho jab tak do bodies exactly ek metre apart na hon, unke beech ek kilogram weighen, aur ek radian per second par ghoomein. Is page ka har formula in units mein rehta hai.

Topic ko yeh kyun chahiye: $G$ , masses, separation, aur $ω$ sab $1$ par set hone ke saath, equations apni clutter kho deti hain aur balance points clean, fixed coordinates par baith jaate hain. Jawaab pure numbers ke roop mein aate hain (jaise "period $\approx 3.4$ ") bina kilograms ya seconds ke.

5. Cheezein kahan baithi hain: origin, axes, aur primaries ki positions

Akele units kaafi nahi hain — hume kehna hoga zero kahan hai aur $x$ kis taraf point karta hai.

Figure s05. Rotating-frame layout upar se dekha gaya. White cross origin par barycentre hai; yellow dot (bada body, mass $1 - μ$ ) $x = - μ$ par baitha hai aur red dot (chhota body, mass $μ$ ) $x = 1 - μ$ par, ek unit apart. Blue arrow positive $x$ -axis ko bade se chhote ki taraf point karte dikhata hai. Blue spacecraft dot ke paas $r_{1}$ (bade body tak) aur $r_{2}$ (chhote body tak) distances dashed lines ke roop mein hain.

Ab parent jo distances use karta hai unke explicit formulas hain.

Topic ko yeh kyun chahiye: in explicit formulas ke bina hum potential $Ω$ (§7) differentiate nahi kar sakte — slopes $\partial Ω/ \partial x$ undefined honge.

6. Rotating frame aur uski do extra forces

Sun aur ek planet (ya Earth aur Moon) ek doosre ke around ghoomte hain. Bahar se dekhne ki jagah, hum saath ride karte hain, $ω = 1$ rate par unke saath spin karte hue, toh do bade bodies hamare view mein still baith jaate hain (exactly §5 ki fixed positions).

Topic ko yeh kyun chahiye: ek still (non-rotating) frame mein do bade bodies chalti rehti hain, aur koi fixed balance point nahi hai jiske paas hover kiya ja sake. Unke saath spin karna unhe jagah par freeze kar deta hai, jo ki ek hi tarika hai "hamesha ke liye $L_{1}$ ke paas baithe rehna" ka matlab bhi banta hai.

Lekin merry-go-round par ride karne ki ek keemat hai: do extra apparent (fictitious) forces appear hoti hain.

Definition Coriolis effect aur factor of

$2$ kahan se aata hai Ek sideways deflection jo sirf un cheezon par act karta hai jo rotating frame mein move kar rahi hain. Ek frame ke liye fictitious-force law jo $ω$ par spin karta hai: $a_{Coriolis} = - 2 ω \times v,$ jahan $v = (\overset{x}{˙}, \overset{y}{˙}, \overset{z}{˙})$ velocity hai aur $\times$ cross product hai (ek rule jo dono inputs ke perpendicular vector produce karta hai). Factor $2$ koi fudge nahi hai — yeh acceleration ko rotating frame mein transform karne ke calculus se drop out hota hai (ek rotating position vector ko do baar differentiate karne par rotation rate do baar milti hai). $ω = (0, 0, 1)$ ke saath ( $z$ par unit rate par spin), cross product deta hai $- 2 (0, 0, 1) \times (\overset{x}{˙}, \overset{y}{˙}, \overset{z}{˙}) = - 2 (- \overset{y}{˙}, \overset{x}{˙}, 0) = (2 \overset{y}{˙}, - 2 \overset{x}{˙}, 0) .$ Inhe "acceleration = force" ke left-hand side par move karne se unke signs flip ho jaate hain, exactly parent ke $- 2 \overset{y}{˙}$ $x$ -line mein aur $+ 2 \overset{x}{˙}$ $y$ -line mein produce karte hain. $z$ -line ko kuch nahi milta kyunki spin axis $z$ ke along hai: $z$ ke around rotating karna vertical motion ko deflect nahi kar sakta.

Figure s02. Upar se dekha gaya ek spinning disk. Blue dot spacecraft hai, uska blue arrow woh batata hai ki woh kaise move kar raha hai. Red arrow (centrifugal) hamesha yellow spin axis se seedha bahar point karta hai aur sirf kahan body hai chahiye. Green arrow (Coriolis) motion ke sideways point karta hai aur sirf isliye aata hai kyunki body move kar rahi hai. Yeh do force families ki visual root hai.

7. Effective potential $Ω$ aur uska gradient $\nablaΩ$

Hum ab "centrifugal + dono gravities" ko ek landscape mein package karte hain.

Slopes ke baare mein baat karne ke liye hume gradient chahiye.

"Partial" kyun aur ordinary derivative kyun nahi? Kyunki $Ω$ ek saath teen variables par depend karta hai. Yeh isolate karne ke liye ki woh unme se ek ke saath kaise badalta hai, hum baakiyon ko freeze karte hain. Curly $\partial$ flag hai jo kehta hai "baaki held fixed."

Ab hum finally parent ki central equations likh sakte hain — rotating frame mein acceleration ke rules.

Figure s03. $x$ – $y$ plane mein $Ω$ ki contour lines. Yellow dot bada body hai, red dot chhota body — har ek rings ke tight nest mein baitha hai ( $Ω$ ka tall spike). Contours wahan crowd karte hain jahan landscape steep hai. Green $\times$ ek Lagrange point mark karta hai: woh ek spot jahan rings khul jaate hain aur ground momentarily flat hai ( $\nablaΩ = 0$ ) — woh balance jiske paas hum hover karte hain.

8. Displacement $(ξ, η, ζ)$ — "zoom-in" coordinates

Hume spacecraft ki absolute position ki parwah nahi; hume parwah hai ki woh balance point se kitna stray karta hai.

Topic ko yeh kyun chahiye: chhoti quantities woh hain jo hume linearize karne deti hain (agla section). Agar hum full nonlinear $Ω$ rakhen, toh hum solve nahi kar sakte.

9. Taylor expansion aur "linear"

Yeh woh engine hai jo ek unsolvable problem ko solvable banata hai.

Kyunki balance point flat hai ( $\nablaΩ = 0$ ), expansion ka constant term zero hai, aur pehla surviving term linear wala hai — $(ξ, η, ζ)$ ke proportional.

10. $Ω$ se $U$ tak: second derivatives aur constant $c$

Hamare "spring" ki stiffness is baat se set hoti hai ki slope khud kitni tezi se badlata hai — yeh ek second derivative hai.

Formula

$U_{xx} = 1 + 2 c, U_{y y} = 1 - c, U_{z z} = - c$ ki full derivation Hum $Ω = \frac{1}{2} (x^{2} + y^{2}) + \frac{1 - μ}{r _{1}} + \frac{μ}{r _{2}}$ ko do baar differentiate karte hain aur phir $x$ -axis par evaluate karte hain, jahan $y = z = 0$ toh dono bodies point ke saath $x$ ke along line up hote hain.

Piece A — centrifugal bowl $\frac{1}{2} (x^{2} + y^{2})$ .

\frac{\partial^2}{\partial y^2}[\cdots] = 1,\quad \frac{\partial^2}{\partial z^2}[\cdots] = 0$$ (isme $z$ nahi hai). Contribution: $(+1, +1, 0)$. **Piece B — ek gravity well $\frac{m}{r}$** jahan $r = \sqrt{(x-x_0)^2 + y^2 + z^2}$. Inverse-distance potential ke liye ek standard result yeh hai ki uske second derivatives, source ke through axis par evaluate kiye (toh offset purely $x$ ke along hai), hain $$\frac{\partial^2}{\partial x^2}\!\left(\frac{m}{r}\right) = +\frac{2m}{r^3}\ \text{(body tak line ke along)},\qquad \frac{\partial^2}{\partial y^2}\!\left(\frac{m}{r}\right) = \frac{\partial^2}{\partial z^2}\!\left(\frac{m}{r}\right) = -\frac{m}{r^3}\ \text{(perpendicular)}.$$ Line ke along $+2$ aur perpendicular $-1$ classic **tidal** pattern hai (mass ki taraf line ke along stretch, us par across squeeze), aur inhe zero mein sum karna hoga — $2 - 1 - 1 = 0$ — kyunki $1/r$ potential mass se door Laplace's equation satisfy karta hai. **Dono bodies par sum.** Masses $1-\mu$ ($r_1$ par) aur $\mu$ ($r_2$ par) ke saath, har curvature mein gravity contribution hai: - $x$ ke along: $+\frac{2(1-\mu)}{r_1^3} + \frac{2\mu}{r_2^3} = +2c$, - $y$ ke along: $-\frac{1-\mu}{r_1^3} - \frac{\mu}{r_2^3} = -c$, - $z$ ke along: $-\frac{1-\mu}{r_1^3} - \frac{\mu}{r_2^3} = -c$, definition $c = \frac{1-\mu}{r_1^3} + \frac{\mu}{r_2^3}$ use karke. **Piece A + Piece B add karo:** $$U_{xx} = 1 + 2c,\qquad U_{yy} = 1 - c,\qquad U_{zz} = 0 + (-c) = -c.$$ Kyunki $c>1$, sirf $U_{xx}$ positive hai ($x$ mein valley/spring); $U_{yy}$ aur $U_{zz}$ negative hain (hills/anti-springs). Yahi mixed sign hai jo point ko **saddle** banata hai.

11. Exponentials, sinusoids, $λ$ , $γ$ , $ω_{x y}$ , $ω_{z}$

Jab hum spring-like equation ka solution guess karte hain, hum $e^{λ t}$ try karte hain.

Figure s04. Do displacement-versus-time curves dikhate hain ki $λ$ kya decide karta hai. Red curve $e^{γ t}$ hai real positive $λ$ ke liye — yeh bina bound ke fly away karta hai (instability). Green curve imaginary $λ$ ke liye wobble hai — ek bounded sine jo hamesha frequency $ω$ par oscillate karta hai. Collinear Lagrange points dono ek saath carry karte hain, jo saddle signature hai.

12. Amplitudes, $κ$ , Lissajous vs halo

Prerequisite map

graph TD
  A["Position x y z and dots"] --> V["Vector r and velocity v"]
  V --> MU["Mass ratio mu"]
  MU --> W["Spin rate omega"]
  W --> N["Normalized units omega equals 1"]
  N --> O["Origin at barycentre, x axis big to small"]
  O --> R["Distances r1 and r2 with formulas"]
  R --> B["Rotating frame"]
  B --> C["Centrifugal and Coriolis minus 2 omega cross v"]
  C --> E["Effective potential Omega"]
  E --> EQ["Equations of motion"]
  EQ --> F["Gradient equals zero = Lagrange point"]
  F --> G["Displacement xi eta zeta"]
  G --> H["Taylor expand to linear"]
  H --> I["