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

3.2.32 · Physics › Orbital Mechanics & Astrodynamics

WHY karte hain hum aisa setup?

WHAT: Full 3-body gravity ka koi closed-form solution nahi hai. Lekin agar $m_{3} \to 0$ aur primaries circles par move karein, toh problem rotating frame mein autonomous ban jaati hai — time potential se hat jaata hai.
WHY rotating frame: Inertial frame mein dono primaries ghoomte rehte hain, isliye potential time-dependent hota hai → messy. Unke saath unki orbital rate $ω$ par rotate karo aur woh ruk jaate hain → hume ek conserved energy-jaisi quantity milti hai (Jacobi constant) aur fixed equilibria milte hain.
HOW iski keemat pay karte hain: Rotating frames fictitious forces introduce karte hain — centrifugal (bahar dhakelta hai) aur Coriolis (moving bodies ko mudta hai). Yahi frozen picture ki keemat hai.

Equations setup karna (scratch se)

Har primary se distances: $r_{1} = (x + μ)^{2} + y^{2} + z^{2}, r_{2} = (x - 1 + μ)^{2} + y^{2} + z^{2} .$

Rotating frame mein Newton — accelerations derive karna

Inertial frame mein, $\ddot{r}_{in} = - \nabla V_{grav}$ . Ek aaise frame mein transform karo jo $ω = \overset{z}{^}$ ke saath rotate kare (yahan $ω = 1$ ). Kinematic identity deta hai: $\ddot{r}_{in} = \ddot{r} + 2 ω \times \dot{r} + ω \times (ω \times r) .$

Yeh terms kyu? $2 ω \times \dot{r}$ Coriolis term hai (aata hai kyunki velocity khud rotation se "twist" hoti hai); $ω \times (ω \times r)$ centrifugal term hai (rotation tumhe bahar phenk ta hai).

Rotating-frame acceleration ke liye rearrange karo aur centrifugal term ko effective potential mein move karo: $\overset{x}{¨} - 2 \overset{y}{˙} = \frac{\partial U}{\partial x}, \overset{y}{¨} + 2 \overset{x}{˙} = \frac{\partial U}{\partial y}, \overset{z}{¨} = \frac{\partial U}{\partial z}$

Lagrange points — the equilibria

Equilibrium ka matlab hai $˙ = ¨ = 0$ , toh $\nabla U = 0$ . Plane mein ( $z = 0$ ): $\frac{\partial U}{\partial x} = x - \frac{( 1 - μ ) ( x + μ )}{r _{1}^{3}} - \frac{μ ( x - 1 + μ )}{r _{2}^{3}} = 0,$ $\frac{\partial U}{\partial y} = y [1 - \frac{1 - μ}{r _{1}^{3}} - \frac{μ}{r _{2}^{3}}] = 0.$

$y = 0$ line → teen collinear points $L_{1}, L_{2}, L_{3}$ (ek quintic solve karo).
Bracket $= 0$ with $y \neq = 0$ → $r_{1} = r_{2} = 1$ → do triangular points $L_{4}, L_{5}$ jo primaries ke saath equilateral triangles banate hain.

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

Characteristic equation (stability)

Maano $x = x_{0} + ξ$ , $y = y_{0} + η$ (planar). Force mein $U$ ko first order tak Taylor-expand karo. $U_{xx} = \partial^{2} U / \partial x^{2}$ etc. ko us point par evaluate karo: $\ddot{ξ} - 2 \overset{η}{˙} = U_{xx} ξ + U_{x y} η,$ $\overset{η}{¨} + 2 \dot{ξ} = U_{x y} ξ + U_{y y} η .$

Maano $ξ, η \propto e^{λ t}$ . Substitute karne par: ${(λ^{2} - U_{xx}) ξ - (2 λ + U_{x y}) η = 0 (2 λ - U_{x y}) ξ + (λ^{2} - U_{y y}) η = 0$

Non-trivial solution ⇒ determinant $= 0$ :

Collinear points $L_{1}, L_{2}, L_{3}$

Wahan $U_{x y} = 0$ , aur paaya jaata hai ki $U_{xx} > 0$ jabki $U_{y y} < 0$ ⇒ product $U_{xx} U_{y y} - U_{x y}^{2} < 0$ . Negative product ek $Λ > 0$ force karta hai ⇒ ek real positive $λ$ ⇒ hamesha unstable (saddle-type). Isliye $L_{1} / L_{2}$ ke around halo orbits ko active station-keeping chahiye.

Triangular points $L_{4}, L_{5}$

Yahan $U_{xx} = \frac{3}{4}$ , $U_{y y} = \frac{9}{4}$ , $U_{x y} = \pm \frac{3 3}{4} (1 - 2 μ)$ . Plug karne par, roots $Λ$ real & negative hote hain (⇒ stable) tabhi jab discriminant $\geq 0$ ho: $1 - 27 μ (1 - μ) \geq 0 \Rightarrow μ \leq μ_{crit} = \frac{1}{2} (1 - \frac{23}{27}) \approx 0.03852.$

Worked examples

Common mistake Steel-manned traps

Trap A: "Coriolis force work karta hai aur energy change karta hai." Sahi lagta hai kyunki yeh equations mein ek real term hai. Fix: Coriolis $\propto ω \times \dot{r}$ hai, hamesha velocity ke ⟂, toh $F \cdot \dot{r} = 0$ — yeh speed change nahi kar sakta, sirf path bend kar sakta hai. Isliye precisely $C_{J}$ conserved hai.

Trap B: " $L_{4} / L_{5}$ , $U$ ke hilltops hain, toh unstable hone chahiye." Sahi lagta hai — $L_{4}, L_{5}$ maxima hain $U$ ke! Fix: Ordinary intuition ("balls roll off hilltops") Coriolis term ko ignore karta hai, jo runaway motion ko point ke around ek stable orbit mein steer karta hai — jab tak $μ < 0.0385$ . Rotation ek maximum ko stabilise karta hai.

Trap C: "Restricted ka matlab third body chhota size ka hai." Fix: Iska matlab hai chhota mass — woh primaries par negligible gravity exert karta hai. Size irrelevant hai.

Trap D: $λ^{4} + (4 - U_{xx} - U_{y y}) λ^{2} + \dots$ mein " $4$ " bhool jaana. Fix: "4" Coriolis coupling terms $2 \overset{η}{˙}$ aur $2 \dot{ξ}$ se aata hai (har factor of 2, determinant mein squared). Ise drop karo toh tum (galat) non-rotating stability criterion paaoge.

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek merry-go-round hai jisme do bhaari bachche uski par baith ke still hain (kyunki tum unke saath spin kar rahe ho). Zameen par kahin kuch special spots hain jahan ek marble sirf baith jaayega bina girhe — gravity-pull, spinning-fling, sab cancel. Kuch spots ek katori ke neeche jaisi hain: marble ko push karo aur woh wapas aa jaata hai. Baaki pahaad ki choti jaisi hain: push karo aur woh hamesha ke liye roll kar jaata hai. "Characteristic equation" hamaara chota math machine hai: hum ise kisi spot ke aas-paas ki zameen ki shape dete hain, aur woh batata hai "katori" (safe) ya "pahadi" (runaway). Surprise: yahan ek hilltop spot bhi safe ho sakta hai, kyunki spinning marble ko wapas curve karte rehti hai — yahi $L_{4}$ aur $L_{5}$ ka raaz hai.

Flashcards

CR3BP mein "restricted" ka kya matlab hai?

Teesre body ki mass negligible hai, toh woh dono primaries ko perturb nahi karta.

CR3BP mein rotating frame kyu use karte hain?

Do primaries stationary ho jaate hain aur potential time-independent ho jaata hai, jisse ek conserved Jacobi constant aur fixed equilibria milte hain.

Mass parameter

μ

define karo.

μ = m_{2} / (m_{1} + m_{2})

, chhote primary ka mass fraction (

0 < μ \leq 1/2

Effective potential

U

likho.

U = \frac{1}{2} (x^{2} + y^{2}) + \frac{1 - μ}{r _{1}} + \frac{μ}{r _{2}}

Jacobi constant kya hai?

C_{J} = 2 U - (\overset{x}{˙}^{2} + \overset{y}{˙}^{2} + \overset{z}{˙}^{2})

; CR3BP mein motion ka akela integral.

Coriolis force ke baad bhi Jacobi constant conserved kyu hai?

Coriolis force velocity ke perpendicular hai, toh woh koi work nahi karta.

(Planar) characteristic equation batao.

λ^{4} + (4 - U_{xx} - U_{y y}) λ^{2} + (U_{xx} U_{y y} - U_{x y}^{2}) = 0

Characteristic equation mein "4" kahan se aata hai?

Coriolis coupling terms

2 \overset{η}{˙}

aur

2 \dot{ξ}

se (har 2 ka factor, determinant mein squared).

Roots ke stable hone ki condition kya hai?

Dono

Λ = λ^{2}

real aur negative ⇒

λ = \pm iω

(pure oscillation).

L_{1}, L_{2}, L_{3}

hamesha unstable kyu hain?

U_{xx} U_{y y} - U_{x y}^{2} < 0

, toh ek

Λ > 0

hota hai, jo ek real positive

λ

deta hai (saddle).

L_{4}, L_{5}

ke liye stability condition kya hai?

1 - 27 μ (1 - μ) \geq 0

, yaani

μ \leq μ_{crit} \approx 0.03852

Stable

L_{4} / L_{5}

ka ek real physical example do.

Jupiter ke Trojan asteroids (

μ_{Sun-Jup} \approx 0.00095

Collinear aur triangular points physically mein kaise alag hain?

Collinear primary–primary line par hote hain (unstable saddles); triangular primaries ke saath equilateral triangles banate hain (

r_{1} = r_{2} = 1

Connections

Lagrange points — woh paanch equilibria jinhe yeh note stabilise karta hai
Jacobi constant & zero-velocity curves — conserved energy aur forbidden regions
Rotating reference frames — Coriolis & centrifugal forces
Eigenvalues & linear stability analysis — jahan se characteristic equation generally aati hai
Halo orbits & station-keeping — unstable $L_{1} / L_{2}$ ka practical use
Trojan asteroids — $L_{4} / L_{5}$ par natural bodies
Two-body problem & Kepler orbits — $μ \to 0$ limiting case