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

3.2.32 · D3 · Physics › Orbital Mechanics & Astrodynamics › Three-body problem — restricted (CR3BP), characteristic equa

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Scenario matrix

Neeche har worked example us cell ke saath tagged hai jo woh fill karta hai.

Cell	Case class	Distinguishing feature	Example
A	Collinear points $L_{1}, L_{2}, L_{3}$ , generic $μ$	$U_{x y} = 0$ , $c < 0$ → saddle (TEENO)	Ex 1 (+ note on $L_{2}, L_{3}$ ), Ex 8 ( $L_{2}$ )
B	Triangular point, small $μ$	$c > 0, b^{2} - 4 c > 0$ → stable	Ex 2
C	Triangular point, large $μ$	$b^{2} - 4 c < 0$ → unstable	Ex 3
D	Degenerate boundary $μ = μ_{crit}$	$b^{2} - 4 c = 0$ (repeated root)	Ex 4
E	Limit $μ \to 0$	ek primary vanish ho jaati hai	Ex 5
F	Limit $μ \to \frac{1}{2}$	equal masses, most unstable triangular	Ex 6
G	Real-world word problem	ek real system chuno, fate decide karo	Ex 7 (Sun–Jupiter)
H	Exam twist / trap	growth rate recover karo, sirf sign nahi	Ex 8 ( $L_{2}$ )

Triangular curvatures geometry se fix hain (parent note): $L_{4}, L_{5}$ par $U_{xx} = \frac{3}{4}, U_{y y} = \frac{9}{4}, U_{x y} = \pm \frac{3 3}{4} (1 - 2 μ) .$ Isliye SABHI triangular examples ke liye: $b = 4 - \frac{3}{4} - \frac{9}{4} = 1$ , aur $c = \frac{3}{4} \cdot \frac{9}{4} - \frac{27}{16} (1 - 2 μ)^{2} = \frac{27}{16} [1 - (1 - 2 μ)^{2}] = \frac{27}{16} \cdot 4 μ (1 - μ) = \frac{27}{4} μ (1 - μ) .$ Aur stability discriminant hai $b^{2} - 4 c = 1 - 27 μ (1 - μ)$ , jo parent ki condition hai. Yeh do baatein dhyan mein rakho; triangular examples sab ek hi formula hain alag-alag $μ$ ke saath.

Neeche wali figure is poori page ka map hai: iska horizontal axis mass parameter $μ$ hai ( $0$ se $\frac{1}{2}$ tak, dimensionless), aur vertical axis stability discriminant $1 - 27 μ (1 - μ)$ hai (yeh bhi dimensionless — ek pure number). Jahan yellow curve zero ke upar hai (blue shading) triangular points stable hain; zero ke neeche (pink shading) unstable hain. Pink dotted line critical mass $μ_{crit}$ mark karti hai jahan curve zero cross karta hai, aur har coloured dot ek worked example ka $μ$ mark karta hai taaki compute karne se pehle tum dekh sako kahan har case basta hai.

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

Figure s01 — Saare $μ$ mein Triangular-point stability: discriminant curve, stable/unstable shading, $μ_{crit}$ , aur marked example cases.

Worked examples

Worked example Ex 1 — Cell A · Collinear point

$L_{1}$ (Earth–Moon), ek saddle Statement. Earth–Moon $L_{1}$ ke liye parent note quote karta hai $U_{xx} \approx + 11.3$ , $U_{y y} \approx - 4.2$ , aur $U_{x y} = 0$ (collinear points symmetry axis par baithe hain). Equilibrium classify karo aur ek growth rate $λ$ dhundho.

Forecast: $U_{y y}$ negative hai — landscape $y$ direction mein ek ridge hai. Guess: unstable. Lekin kitna unstable — kis size ka real $λ$ ?

Compute $c = U_{xx} U_{y y} - U_{x y}^{2} = (11.3) (- 4.2) - 0 = - 47.46$ . Yeh step kyun? $c$ do $Λ$ -roots ka product hai; sirf iska sign saddle vs centre decide karta hai.
$c < 0$ hai, toh roots $Λ_{\pm}$ ke signs opposite hain. Compute $b = 4 - 11.3 - (- 4.2) = - 3.1$ . Yeh step kyun? Quadratic solve karne aur actual positive $Λ$ nikalane ke liye $b$ chahiye.
Solve $Λ^{2} - 3.1 Λ - 47.46 = 0$ : $Λ = \frac{3.1 \pm 3. 1 ^{2} + 4 ( 47.46 )}{2} = \frac{3.1 \pm 199.45}{2}$ . $199.45 = 14.123$ , toh $Λ_{+} = 8.611$ , $Λ_{-} = - 5.511$ . Yeh step kyun? Positive root real exponential deta hai; negative wala oscillation deta hai.
Growth rate $λ = Λ_{+} = 8.611 = 2.935$ ( $ω = 1$ clock par jo upar define ki — toh $λ$ orbital angle ke per radian measure hua hai). Yeh step kyun? Small offset $ξ \propto e^{λ t}$ follow karta hai (yaad karo $ξ$ $L_{1}$ se tiny displacement hai). $λ = 2.935$ ke saath, ek time unit ke baad — Earth–Moon orbit ka ek radian — nudge $e^{2.935} \approx 18.8 \times$ badhta hai. Yahi relentless amplification kyun hai ki $L_{1}$ ko active station-keeping chahiye.

Verify: Roots ka product $Λ_{+} Λ_{-} = 8.611 \times (- 5.511) = - 47.46 = c$ . ✓ Sum $= 8.611 - 5.511 = 3.1 = - b$ . ✓ Opposite signs saddle confirm karte hain.

Worked example Ex 2 — Cell B · Triangular

$L_{4}$ (Earth–Moon), stable Statement. Earth–Moon $μ = 0.0123$ . $L_{4}$ classify karo aur do oscillation frequencies do.

Forecast: $L_{4}$ $U$ ka ek maximum hai (ek hilltop). Naively "balls roll off hilltops" ⇒ unstable. Lekin Coriolis small- $μ$ triangular points ko bachata hai. $μ = 0.0123$ ke saath jo $0.0385$ se kaafi neeche hai, guess: stable, do distinct real oscillation frequencies.

$b = 1$ (hamesha, triangular ke liye). Compute $c = \frac{27}{4} μ (1 - μ) = \frac{27}{4} (0.0123) (0.9877) = 0.08202$ . Yeh step kyun? $c > 0$ pehla stability gate hai; yahaan pass karta hai.
Discriminant $b^{2} - 4 c = 1 - 4 (0.08202) = 1 - 0.32808 = 0.67192 > 0$ . Yeh step kyun? $1 - 27 μ (1 - μ)$ ke equivalent; positive ⇒ dono $Λ$ real. $c > 0, b > 0$ ke saath ⇒ dono negative ⇒ stable.
Roots: $Λ_{\pm} = \frac{- 1 \pm 0.67192}{2} = \frac{- 1 \pm 0.81971}{2}$ , giving $Λ_{+} = - 0.09015$ , $Λ_{-} = - 0.90985$ . Yeh step kyun? Dono negative pure imaginary $λ$ confirm karta hai.
Frequencies $ω_{1, 2} = - Λ_{\pm}$ : $ω_{1} = 0.09015 = 0.3003$ (long-period libration), $ω_{2} = 0.90985 = 0.9539$ (short-period). Yeh usi $ω = 1$ clock par measured hain, toh $ω_{2} \approx 0.95$ matlab fast wobble almost utni hi quick hai jitni ek orbital radian. Yeh step kyun? Yeh woh do libration frequencies hain jis par real Trojan-like bodies wobble karte hain.

Verify: $Λ_{+} Λ_{-} = (- 0.09015) (- 0.90985) = 0.08202 = c$ . ✓ Sum $= - 1 = - b$ . ✓ Dono negative ⇒ stable, parent ke Ex 2 se match ( $27 μ (1 - μ) = 0.328 < 1$ ).

Worked example Ex 3 — Cell C · Triangular point, mass just over critical → unstable

Statement. Ek hypothetical binary mein $μ = 0.05$ hai (above $μ_{crit} \approx 0.03852$ ). $L_{4}$ classify karo.

Forecast: $μ$ critical value se zyada hai, toh guess: discriminant flip hoga negative → complex $Λ$ → unstable (spiralling growth).

$c = \frac{27}{4} (0.05) (0.95) = 0.320625$ ; $b = 1$ . Yeh step kyun? $Λ$ mein quadratic set up karo.
Discriminant $b^{2} - 4 c = 1 - 1.2825 = - 0.2825 < 0$ . Yeh step kyun? Negative discriminant ⇒ $Λ$ complex conjugates hain, real nahi. Tab $λ = Λ$ ka nonzero real part hoga ⇒ growth.
Complex roots $Λ = \frac{- 1 \pm i 0.2825}{2} = - 0.5 \pm 0.26575 i$ . Growing direction hamesha kyun hoti hai? Ek complex $Λ$ ko polar form mein likho $Λ = R e^{i θ}$ jahan $R > 0$ aur angle $- π < θ < π$ ho. Iske do square roots hain $λ = \pm R e^{i θ /2}$ . Unke real parts hain $\pm R cos (θ /2)$ , aur $- \frac{π}{2} < θ /2 < \frac{π}{2}$ hone se $cos (θ /2) > 0$ — toh ek root ka real part strictly positive hota hai jab bhi $θ \neq = 0$ (yaani jab bhi $Λ$ real nahi hai). Yahi positive real part exponential growth hai. Yeh oscillatory instability hai: body spiral karke bahar jaati hai, collinear points ke pure exponential saddle se alag.

Verify: discriminant sign check: $27 μ (1 - μ) = 27 (0.05) (0.95) = 1.2825 > 1$ , toh $1 - 27 μ (1 - μ) = - 0.2825 < 0$ . ✓ Unstable confirmed.

Worked example Ex 4 — Cell D · Degenerate boundary

$μ = μ_{crit}$ Statement. Exactly $μ_{crit} = \frac{1}{2} (1 - 23/27)$ par discriminant zero hai. $μ_{crit}$ numerically dhundho aur (repeated) root $Λ$ nikalo.

Forecast: Stable aur unstable ke beech boundary — expect karo ek repeated real root, marginal case. Guess $μ_{crit} \approx 0.0385$ .

Solve $1 - 27 μ (1 - μ) = 0$ , i.e. $27 μ^{2} - 27 μ + 1 = 0$ . Tab $μ = \frac{27 - 2 7 ^{2} - 4 \cdot 27}{2 \cdot 27} = \frac{27 - 621}{54}$ . Yeh step kyun? Stability discriminant, zero set karke, $μ$ mein ek plain quadratic hai.
$621 = 24.9199$ , toh $μ_{crit} = \frac{27 - 24.9199}{54} = \frac{2.0801}{54} = 0.038520$ . Yeh step kyun? Parent ka $0.03852$ confirm karta hai aur algebraic form $\frac{1}{2} (1 - 23/27)$ identical dikhata hai (kyunki $621 = 2 7^{2} - 108$ aur $23/27 = 621/729$ ).
Boundary par $b^{2} - 4 c = 0$ , toh $Λ = - b /2 = - 1/2$ (double root). Tab $λ = \pm i / 2$ (repeated). Yeh step kyun? Ek repeated imaginary pair matlab do libration frequencies merge ho jaate hain; is $μ$ se aage wo complex pair mein split ho jaate hain aur destabilise karte hain. Yeh coalescence stability boundary ka mathematical signature hai.

Verify: $\frac{1}{2} (1 - 23/27) = \frac{1}{2} (1 - 0.922958) = 0.038521$ . ✓ step 2 se match karta hai. Aur $27 μ_{crit} (1 - μ_{crit}) = 1$ . ✓

Worked example Ex 5 — Cell E · Limit

$μ \to 0$ (second primary vanish ho jaati hai) Statement. $μ \to 0^{+}$ hone do. $c$ ka aur triangular-point stability ka kya hoga? Physically interpret karo.

Forecast: $m_{2} \to 0$ ke saath hum (almost) two-body / one-body situation mein wapas hain. Guess: $L_{4}$ marginally, phir trivially stable ho jaata hai; libration frequencies $0$ aur $1$ approach karti hain.

$c = \frac{27}{4} μ (1 - μ) \to 0$ as $μ \to 0$ . Yeh step kyun? $c$ roots ka product hai; agar yeh $\to 0$ hai, toh ek root $Λ \to 0$ .
$b = 1$ aur $c \to 0$ ke saath: roots $Λ_{+} \to 0$ aur $Λ_{-} \to - 1$ . Frequencies $ω_{1} = - Λ_{+} \to 0$ , $ω_{2} = - Λ_{-} \to 1$ . Yeh step kyun? $ω_{2} \to 1$ orbital frequency hi hai ( $ω = 1$ clock par) — tiny body akele primary ke around mean motion par orbit karta hai. $ω_{1} \to 0$ matlab long-period libration infinitely slow ho jaati hai (doosra mass hold karne ke liye nahi hai).
Discriminant $1 - 27 μ (1 - μ) \to 1 > 0$ : stable region ke andar deeply. Yeh step kyun? Confirm karta hai ki tiny secondary wale har real planetary system mein stable Trojans hote hain.

Verify: $μ = 0$ par, $c = 0$ , roots $0$ aur $- 1$ ; $ω_{2} = 1 = ω$ (orbital rate). ✓

Worked example Ex 6 — Cell F · Limit

$μ \to \frac{1}{2}$ (equal masses) Statement. Equal-mass binary, $μ = 0.5$ . $L_{4}$ classify karo aur complex $Λ$ dhundho.

Forecast: Do equal stars — sabse symmetric, aur ( $0.5 > μ_{crit}$ hone se) most unstable triangular case. Guess: strongly complex $Λ$ .

$c = \frac{27}{4} (0.5) (0.5) = \frac{27}{16} = 1.6875$ ; $b = 1$ . Yeh step kyun? $c$ yahaan maximum possible hai, kyunki $μ (1 - μ)$ $μ = \frac{1}{2}$ par peak karta hai.
Discriminant $1 - 4 (1.6875) = 1 - 6.75 = - 5.75 < 0$ ⇒ complex conjugate $Λ$ . Yeh step kyun? Strong oscillatory instability confirm karta hai; note karo $U_{x y} = \frac{3 3}{4} (1 - 2 μ) = 0$ yahaan, toh twist vanish ho jaata hai phir bhi instability rehti hai — instability twist term ke baare mein nahi hai.
$Λ = \frac{- 1 \pm i 5.75}{2} = - 0.5 \pm 1.19896 i$ . Yeh step kyun? $λ = Λ$ mein nonzero real part ⇒ runaway (Ex 3 ke polar-form argument se). Equal-mass binaries mein koi stable Trojan points nahi hote.

Verify: $∣ Λ ∣ = 0. 5^{2} + 1.1989 6^{2} = 0.25 + 1.4375 = 1.6875 = c$ . ✓ (ek conjugate pair ka product $= ∣Λ ∣^{2} = c = 1.6875$ .)

Worked example Ex 7 — Cell G · Real-world word problem: Sun–Jupiter Trojans

Statement. Jupiter Sun ki mass ka $9.55 \times 1 0^{- 4}$ hai. Real asteroids (Trojans) $L_{4}, L_{5}$ par baithe hain. Dikhao ki theory unhe allow karta hai aur do libration periods nikalo (Jupiter ke orbital period ke units mein).

Forecast: $μ \approx 0.00095 ≪ 0.0385$ , toh stable — observation se match karta hai. Expect karo ek bahut long-period aur ek nearly-orbital-period libration.

$μ = \frac{9.55 \times 1 0 ^{- 4}}{1 + 9.55 \times 1 0 ^{- 4}} = 9.541 \times 1 0^{- 4}$ . Yeh step kyun? Mass ratio ko mass parameter $μ = m_{2} / (m_{1} + m_{2})$ mein convert karo.
$c = \frac{27}{4} μ (1 - μ) = \frac{27}{4} (9.541 \times 1 0^{- 4}) (0.99905) = 6.4327 \times 1 0^{- 3}$ ; $b = 1$ . Yeh step kyun? Tiny $c$ ⇒ ek root near $0$ , ek near $- 1$ .
Roots $Λ_{\pm} = \frac{- 1 \pm 1 - 4 c}{2} = \frac{- 1 \pm 0.974269}{2}$ . $0.974269 = 0.987051$ . Toh $Λ_{+} = - 0.0064746$ , $Λ_{-} = - 0.993525$ . Yeh step kyun? Dono negative ⇒ stable; frequencies mein feed karo.
Frequencies $ω_{1} = - Λ_{+} = 0.0064746 = 0.080465$ , $ω_{2} = - Λ_{-} = 0.993525 = 0.996757$ ( $ω = 1$ clock par). Yeh step kyun? Yeh do libration rates hain; ab periods mein convert karo.
$ω = 1$ clock par primaries ka orbital period $T_{orb} = 2 π / ω = 2 π$ hai. Ek libration period $T = 2 π / ω_{i}$ hai, toh orbital period se uska ratio $T / T_{orb} = ω / ω_{i} = 1/ ω_{i}$ hai. Isliye long-period libration $1/ ω_{1} = 1/0.080465 = 12.43$ Jupiter orbits chalta hai, aur short wala $1/ ω_{2} = 1/0.996757 = 1.003$ Jupiter orbits. Yeh step kyun? ~12.4-orbit wali long libration real Trojans ki observed slow "tadpole" wobble se match karti hai.

Verify: $Λ_{+} Λ_{-} = (- 0.0064746) (- 0.993525) = 6.4327 \times 1 0^{- 3} = c$ . ✓ Dono roots negative ⇒ stable, $μ ≪ μ_{crit}$ . ✓ Long-period ratio $1/ ω_{1} \approx 12.43$ . ✓

Worked example Ex 8 — Cell H (aur cell A, point

$L_{2}$ ) · Exam twist: $L_{2}$ par e-foldings ki number nikalo Statement. Sun–Earth $L_{2}$ (space telescopes ka ghar) mein effective curvatures hain $U_{xx} \approx + 7.38$ , $U_{y y} \approx - 2.19$ , $U_{x y} = 0$ . Ek spacecraft ko nudge kiya jaata hai aur uncontrolled chhod diya jaata hai. Ek dimensionless time unit baad ( $t = 1$ , yaani ek orbital radian) displacement kitne factor se badh jaayegi?

Forecast: $L_{2}$ collinear hai ⇒ (cell A se) unstable ⇒ real positive $λ$ . Trap yeh hai: bahut se log "unstable" par ruk jaate hain. Exam growth factor $e^{λ}$ chahta hai.

$c = U_{xx} U_{y y} - U_{x y}^{2} = (7.38) (- 2.19) - 0 = - 16.1622 < 0$ ⇒ saddle, exactly baaki collinear points ki tarah. Yeh step kyun? Confirm karta hai ki ek real positive $λ$ exist karta hai (cell A behaviour $L_{2}$ ke liye).
$b = 4 - 7.38 - (- 2.19) = - 1.19$ . Solve $Λ^{2} - 1.19Λ - 16.1622 = 0$ : $Λ_{+} = \frac{1.19 + 1.4161 + 64.6488}{2} = \frac{1.19 + 66.0649}{2} = \frac{1.19 + 8.1281}{2} = 4.6591$ . Yeh step kyun? Positive root unstable direction ki rate deta hai.
$λ = 4.6591 = 2.1585$ . Offset $ξ$ ka $t = 1$ baad growth factor: $e^{λ \cdot 1} = e^{2.1585} = 8.658$ . Yeh step kyun? Ek time unit mein (Sun–Earth orbit ka ek radian, ~58 s-scaled din) ek unmanaged deviation ~8.7× badhti hai. Yahi quantitative reason hai ki station-keeping burns roughly har kuch hafte hoti hain.

Verify: $Λ_{+} = 4.6591$ ; check $Λ_{+}^{2} - 1.19 Λ_{+} + (- 16.1622) \approx 0$ : $21.707 - 5.544 - 16.162 = 0.001 \approx 0$ . ✓ Growth factor $e^{2.1585} = 8.66$ . ✓

Recall Self-test (guess karne ke baad reveal karo)

Collinear points $L_{1}, L_{2}, L_{3}$ hamesha hote hain ::: unstable saddles, kyunki $U_{x y} = 0$ with $U_{xx} > 0, U_{y y} < 0$ har ek par $c < 0$ deta hai. Triangular points stable hain iff ::: $μ \leq μ_{crit} \approx 0.03852$ , equivalently $27 μ (1 - μ) \leq 1$ . Ek collinear point ke liye saddle vs centre decide karne wali ek quantity kya hai? ::: $c = U_{xx} U_{y y}$ ka sign ( $Λ$ -roots ka product). Equal-mass binaries mein stable Trojans kyun nahi hote? ::: $μ = 0.5 > μ_{crit}$ , toh discriminant $1 - 27 μ (1 - μ) = - 5.75 < 0$ , complex $Λ$ aur oscillatory runaway deta hai. Characteristic equation mein "4" aata hai ::: Coriolis coupling terms $2 \overset{η}{˙}, 2 \dot{ξ}$ se jo determinant mein squared hote hain. "Ek time unit" ka matlab hai ::: primaries ke orbital angle ka ek radian (woh clock jahan $ω = 1$ ).

Yeh bhi dekho: Lagrange points, Jacobi constant & zero-velocity curves, Rotating reference frames — Coriolis & centrifugal forces.