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

3.2.32 · D1 · Physics › Orbital Mechanics & Astrodynamics › Three-body problem — restricted (CR3BP), characteristic equa

Is page pe kuch bhi assumed nahi hai. Har ek letter, har ek squiggle, har ek fancy word jis par parent topic rely karta hai, yahan build kiya gaya hai, ek aisi order mein jahan har ek brick pehle waale par tiki hai.

1. Ek vector aur uski position — cheezein kahan hain

Isko picture karo: ek table par ek pencil khadi rakho. Ek kone ko origin choose karo. Us kone se pencil ke tip tak seedha arrow hi $r$ hai. Teen numbers $(x, y, z)$ bas itna batate hain ki tum wahan pahunchne ke liye kitna aage, kitna across, aur kitna upar gaye.

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

Topic ko yeh kyun chahiye: poora problem hai "chhota body kahan hai, aur kahan ja raha hai?" Tum yeh pooch hi nahi sakte jab tak koi location ko naam dene ka tarika na ho. Woh naam hai $r = (x, y, z)$ .

Upar wali dot — $\dot{r}$ — ka matlab hai velocity (position kitni tezi se change ho rahi hai, ek arrow jis direction mein ja rahe ho). Do dots, $\ddot{r}$ , ka matlab hai acceleration (velocity khud kitni tezi se change ho rahi hai). Ek dot = change ki rate; do dots = change ki rate ki change ki rate.

Recall

$\overset{y}{˙}$ ka plain words mein kya matlab hai? ::: The rate at which the $y$ -coordinate is changing — velocity ka up/down component.

2. Mass aur mass parameter $μ$ — kaun bhaari hai

Is problem mein exactly teen masses hain: $m_{1}$ (bada primary, jaise Sun ya Earth), $m_{2}$ (chhota primary, jaise Earth ya Moon), aur $m_{3}$ (chhota teesra body — ek spacecraft ya asteroid). "Restricted" ka matlab hai $m_{3}$ itna halka hai ki woh doosron ko feel karta hai lekin unhe kabhi kheenchta nahi waapis.

Isko picture karo: dono primaries ko ek seesaw par rakho. $μ$ batata hai ki total weight kaise baanta hai. Agar $μ = 0.01$ hai, to chhota body pair ka $1%$ hai — bada wala dominate karta hai. Agar $μ = 0.5$ hai, to wo twins hain.

Topic ko yeh kyun chahiye: baad ke har equation sirf is ek ratio $μ$ par depend karti hai, raw kilograms par nahi. Sun–Jupiter aur Sun–Saturn "same shape" mein behave karte hain, sirf apne $μ$ ke zariye alag hote hain. Yeh mass pairs ke poore zoo ko ek single dial mein compress kar deta hai.

Recall

Agar $m_{1} = m_{2}$ ho, to $μ$ kya hai? ::: $μ = \frac{1}{2}$ — maximum, ek equal-mass pair.

3. Nondimensional units — rulers, clocks aur scales chunna taaki numbers gaayab ho jaayein

Hum teen independent choices karte hain, har ek basic dimension ke liye ek:

Length unit = do primaries ke beech ki doori. Yeh unka separation $= 1$ force karta hai.
Mass unit = do primaries ki total mass. Yeh $m_{1} + m_{2} = 1$ force karta hai, to seedha $m_{1} = 1 - μ$ aur $m_{2} = μ$ (mass parameter $μ = m_{2} / (m_{1} + m_{2})$ ban jaata hai chhoti mass khud).
Time unit = is tarah choose kiya jaata hai ki gravitational constant $G = 1$ ho.

To char scalings $G = 1$ , $m_{1} + m_{2} = 1$ , separation $= 1$ , aur $ω = 1$ sab ek saath consistent hain — pehle teen hamare free choices hain, aakhri unka consequence hai.

Recall

Nondimensional units mein, $m_{2}$ kiske barabar hai? ::: Exactly $μ$ — kyunki humne $m_{1} + m_{2} = 1$ set kiya, chhoti mass hi mass parameter hai.

4. Orbital plane tak restrict karna

Yahan yeh kyun matter karta hai: halanki position $r = (x, y, z)$ aur gradient puri 3D mein aaram se rehta hai, equilibria dhundne aur picture karne ke liye hum deliberately motion ko $z = 0$ tak restrict karte hain — chhota body primaries ke plane mein rehta hai. Isliye effective potential ke baad se tum $U (x, y)$ dekhoge bina $z$ ke: humne $z = 0$ set kar diya hai. (Out-of-plane $z$ direction sirf tab waapis aati hai jab vertical wobbles study karte hain; parent topic Lagrange-point hunt ke liye plane mein hi rehta hai.)

5. Barycentre — woh balance point jiske around sab kuch circle karta hai

Isko picture karo: bhaari adult aur halka bachcha seesaw par. Pivot adult ke kareeb baithta hai. Isi tarah barycentre bade primary ke kareeb baithta hai. Dono primaries is point ke around orbit karte hain, na ek doosre ke centres ke around.

In hamare units mein barycentre origin $(0, 0)$ hai. Bada primary $m_{1} = 1 - μ$ $x = - μ$ par baithta hai; chhota primary $m_{2} = μ$ $x = 1 - μ$ par baithta hai. Unka separation $(1 - μ) - (- μ) = 1$ hai, exactly waise jaise haara length unit demand karta hai.

Figure mein kya notice karna hai: bada kaala dot (mass $1 - μ$ ) origin ke sirf ek whisker left par $x = - μ$ par baitha hai, jabki chhota dot ( $μ$ ) kaafi door right par $x = 1 - μ$ par hai. Laal barycentre dot $0$ par hai — bhaari body se chipka hua, exactly seesaw pivot adult ke kareeb jaisi tarah. Apni aankhon se confirm karo ki laal dot se do dashed distances inversely weighted hain: bhaari body kareeb hai, halki body door hai.

Topic ko yeh kyun chahiye: origin ko barycentre par anchor karna rotating-frame equations ko clean aur symmetric banata hai — centrifugal "bowl" (agla section) yahan center hota hai.

Related vault reading: Two-body problem & Kepler orbits explain karta hai ki do masses ek common barycentre ke around kyun orbit karti hain.

6. Rotating reference frame — camera ko spin karna

Isko picture karo: tum ek merry-go-round par ho. Zameen par khade kisi ke liye (inertial view) ghode ghumte dikh rahe hain. Tumhare liye (rotating view) tumhare paas ka ghoda frozen laga raha hai, lekin poori baahri duniya ghoomti lagti hai.

CR3BP ka trick: ek aisa frame ride karo jo exactly $ω$ par turn karta ho, primaries ki orbital rate par. Us frame mein do primaries jagah par jamd jaati hain. Poori machinery ke liye Rotating reference frames — Coriolis & centrifugal forces dekho.

Keemat — fictitious forces. Apna viewpoint spin karna objects ko aisa lagta hai ki koi forces push kar rahe hain jo "real" pulls nahi hain:

Centrifugal — woh baahri fling jo tumhe merry-go-round par feel hoti hai; yeh spin axis se doori ke saath badhti hai.
Coriolis — rotating frame mein kisi bhi moving cheez ka sideways deflection; yeh hamesha tumhari velocity ke perpendicular dhakelta hai, isliye yeh tumhara path mod deta hai bina kabhi speed badhaye ya ghataye.

Figure mein kya notice karna hai: body (kaala dot) kaale velocity arrow ke saath move kar raha hai, lekin laal Coriolis arrow ek perfect right angle par point karta hai. Dono arrows ko apni ungli se trace karo — wo $9 0^{\circ}$ par milte hain. Woh right angle hi poori wajah hai ki Coriolis ek path steer kar sakta hai bina kabhi speed add ya remove kiye.

Centrifugal term kahan se aata hai. Rotating frame mein spin axis se $ρ = x^{2} + y^{2}$ doori par mass ke per unit outward centrifugal force $ω^{2} ρ$ hai, baahri direction mein point karta hua. Ek force jo kisi potential ka (positive) gradient hai, $\frac{1}{2} ω^{2} ρ^{2} = \frac{1}{2} (x^{2} + y^{2})$ se aana chahiye ( $ω = 1$ use karke), kyunki $\frac{1}{2} (x^{2} + y^{2})$ ko $x$ ke respect mein differentiate karne se $x$ waapis milta hai, yani outward push. Isliye exactly $\frac{1}{2} (x^{2} + y^{2})$ term neeche effective potential mein appear karta hai — yeh centrifugal force hai ek landscape ki tarah likha hua.

Recall

Coriolis force kisi body ki speed kyun nahi badal sakta? ::: Yeh hamesha velocity ke perpendicular point karta hai, isliye yeh zero work karta hai — sirf path ko modta hai.

7. Gradient $\nabla$ aur partial derivatives — ek landscape ki slope

Isko picture karo: ek hillside par khado. Bilkul east ki taraf face karo aur poochho ki zameen us direction mein kitni steep hai — yeh $\partial U / \partial x$ hai. North ki taraf face karne ke liye ghoom jao — yeh $\partial U / \partial y$ hai. Har partial derivative ek chosen direction mein slope hai.

Topic ko yeh kyun chahiye: is convention ke saath, force per mass $\nabla U$ hai — objects landscape ki slope ke along higher $U$ ki taraf push hote hain. Jahan zameen flat ho, $\nabla U = 0$ : woh ek equilibrium hai, ek possible resting spot. Woh flat spots Lagrange points hain.

Double subscripts — $U_{xx} = \partial^{2} U / \partial x^{2}$ , $U_{x y} = \partial^{2} U / \partial x \partial y$ — second derivatives hain: landscape ki curvature. $U_{xx} > 0$ = $x$ ke along upar curve karta hai; $U_{xx} < 0$ = neeche curve karta hai. Curvature batati hai ki ek flat spot bowl hai ya saddle.

Recall

$\nabla U = 0$ physically kya signal karta hai? ::: Landscape locally flat hai — koi net force nahi — ek equilibrium point.

8. Distances $r_{1}, r_{2}$ aur effective potential $U$ — frozen landscape khud

$U$ likhne se pehle humein woh do distances chahiye jinpar yeh depend karta hai. Chhote body se $(x, y)$ par (hum plane mein hain, $z = 0$ ) har primary tak:

Gravity term ko consistently padhna. Parent ke flipped sign convention ke saath (§7 dekho), gravity $U$ uthata hai jab tum ek primary ke kareeb jaate ho: jaise $r_{1} \to 0$ , $(1 - μ) / r_{1} \to + \infty$ . To is landscape mein har primary ek infinitely tall spike hai, ek well nahi — objects upar primary ki taraf push hote hain kyunki yahi real gravitational pull ki direction hai. Metaphor ko clean rakhne ke liye hum consistently inhe gravity spikes (tall peaks) kahenge, "wells" kabhi nahi. Is topic mein sab kuch upar ki taraf = stronger attraction ki taraf point karta hai.

"Effective" kyun: Coriolis force ko landscape mein fold nahi kiya ja sakta (yeh velocity par depend karta hai, position par nahi), isliye $U$ sirf gravity + centrifugal capture karta hai. Coriolis motion equations mein ek separate term ki tarah rehta hai. Yeh landscape exactly wahi hai jiske Lagrange points flat spots hain — aur Jacobi constant & zero-velocity curves dekho ki kaise $2 U$ ke level sets un walls ban jaate hain jinhein spacecraft cross nahi kar sakta.

Downstream yeh directly Lagrange points aur Eigenvalues & linear stability analysis mein feed hota hai.

9. Exponentials $e^{λ t}$ aur eigenvalues $λ$ — grow karo, oscillate karo, ya decay karo?

Isko picture karo: ek bowl mein marble bottom ke around oscillate karta hai (imaginary $λ$ ). Ek dome par balanced marble tezi se roll off karta hai (real positive $λ$ ).

Parent ki characteristic equation $λ^{4} + (4 - U_{xx} - U_{y y}) λ^{2} + (U_{xx} U_{y y} - U_{x y}^{2}) = 0$ bas yeh hai ki "kaun se growth-rates $λ$ $U$ ki curvature plus Coriolis coupling (akela $4$ ) ke saath consistent hain?"

Recall

Ek Lagrange point real positive $λ$ deta hai. Stable hai ya unstable? ::: Unstable — ek nudge $e^{λ t}$ ki tarah grow karta hai aur bhaag jaata hai.

Prerequisite map

Ise upar se neeche padho: kahan aur kitni tezi se ka naam lena pehle aata hai; units chunna aur plane tak restrict karna stage set karta hai; camera spin karna scene freeze karta hai; frozen landscape $U$ ke flat spots hain (Lagrange points); unhe nudge karna aur $e^{λ t}$ se test karna characteristic equation deta hai.

Equipment checklist

Main describe kar sakta hoon ki arrow $r$ aur uske dots $\dot{r}, \ddot{r}$ ka kya matlab hai ::: $r$ = origin se position arrow; ek dot = velocity; do dots = acceleration.
Main do masses se $μ$ compute kar sakta hoon ::: $μ = m_{2} / (m_{1} + m_{2})$ , chhote body ka pair mein fraction.
Main teen unit choices aur unke force karne wali quantity list kar sakta hoon ::: Length = separation (→1), mass = total (→1), time taaki $G = 1$ ; mil ke yeh Kepler's third law ke zariye $ω = 1$ force karte hain.
Main dono primaries ko nondimensional units mein place kar sakta hoon ::: $m_{1} = 1 - μ$ at $x = - μ$ , $m_{2} = μ$ at $x = 1 - μ$ , separation $1$ , barycentre at origin.
Main jaanta hoon ki parent ke analysis mein $U$ mein $z$ kyun nahi hai ::: Hum motion ko primaries ke orbital plane, $z = 0$ , tak restrict karte hain.
Main $(x, y)$ se $r_{1}$ aur $r_{2}$ likh sakta hoon ::: $r_{1} = (x + μ)^{2} + y^{2}$ , $r_{2} = (x - (1 - μ))^{2} + y^{2}$ .
Main bata sakta hoon ki centrifugal term $\frac{1}{2} (x^{2} + y^{2})$ kahan se aata hai ::: Yeh woh potential hai jiska gradient $ω = 1$ ke saath outward centrifugal force $ω^{2} ρ$ deta hai.
Main yahan $U$ ke sign convention ko jaanta hoon ::: Yeh topic acceleration $= + \nabla U$ use karta hai (push higher $U$ ki taraf), usual $F = - \nabla V$ ka ulta; isliye primaries spikes hain, wells nahi.
Main do fictitious forces ko naam de sakta hoon aur har ek ke baare mein ek key fact ::: Centrifugal baahri ki taraf fling karta hai (doori ke saath badhta hai); Coriolis moving bodies ko velocity ke perpendicular deflect karta hai (koi work nahi karta).
Main $\partial U / \partial x$ aur $U_{xx}$ padh sakta hoon ::: First partial = $x$ ke along slope; second $U_{xx}$ = curvature.
Main jaanta hoon ki $\nabla U = 0$ kya signal karta hai ::: Ek flat spot — koi net force nahi — ek equilibrium (Lagrange point).
Main $e^{λ t}$ mein $λ$ ke sign/type ko interpret kar sakta hoon ::: Real positive → bhaag jaata hai (unstable); pure imaginary → oscillate karta hai (stable).