Lagrange points L1–L5 — derivation, stability: Exercises

3.2.30 · D4 · Physics › Orbital Mechanics & Astrodynamics › Lagrange points L1–L5 — derivation, stability

Shuru karne se pehle, symbols ka ek reminder, taaki kuch bhi unexplained na rahe:

Level 1 — Recognition

Problem 1.1

Paanch saare Lagrange points ke naam batao aur, ek-ek phrase mein, kahan hain yeh dono bodies ke relative.

Recall Solution

L1: $M_{1}$ aur $M_{2}$ ke beech ki line par, unke beech mein (lighter body $M_{2}$ ke zyada kareeb).
L2: usi line par, $M_{2}$ ke bahar (chhote body se aage).
L3: usi line par, $M_{1}$ se aage, lagbhag $M_{2}$ ke opposite side par.
L4: line se alag, dono bodies ke saath ek equilateral triangle banata hua, $M_{2}$ se $6 0^{\circ}$ aage (leading).
L5: L4 ka mirror, $M_{2}$ se $6 0^{\circ}$ peeche (trailing).

Neeche ka map dekho — teen collinear points kaale dashed axis par hain, aur dono triangular points equilateral triangles ke upar hain.

Figure — Lagrange points L1–L5 — derivation, stability

Problem 1.2

Kaun se Lagrange points unstable hain, aur kaun se stable ho sakte hain? Woh ek cheez kya hai jo stable walon ko bachati hai?

Recall Solution

Unstable: L1, L2, L3 (yeh $Ω$ ke saddle points hain — ek nudge badhta jaata hai).
Stable ho sakte hain: L4, L5, chahe yeh $Ω$ ke maxima hain.
Bachaane wali cheez hai Coriolis force — yeh sirf ek moving particle par kaam karti hai, ek drifting test mass ko ek chhoti closed loop mein curve karti hai instead of use escape karne dene ke. Dekho Coriolis and Centrifugal Forces (Rotating Frames).

Problem 1.3

Sach ya jhooth: "Coriolis force ki zaroorat Lagrange points ki locations find karne mein hoti hai." Justify karo.

Recall Solution

Jhooth. Rotating frame mein rest par baitha particle ki velocity $v = 0$ hai, aur Coriolis $= - 2 ω \times v = 0$ . Toh points locate karne ke liye sirf gravity + centrifugal force chahiye. Coriolis sirf stability (dynamics) analysis mein aata hai, jab particle actually move karta hai.

Level 2 — Application

Problem 2.1

Sun–Earth system: $M_{1} / M_{2} = 3.33 \times 1 0^{5}$ , toh $μ \approx 3.0 \times 1 0^{- 6}$ . Separation hai $D = 1.50 \times 1 0^{8}$ km. Earth se L1/L2 distance $r$ estimate karo is formula se: $r \approx D (\frac{μ}{3})^{1/3} .$

Recall Solution

KYA: numbers ko Hill-radius-scale formula mein daalo. $\frac{μ}{3} = \frac{3.0 \times 1 0 ^{- 6}}{3} = 1.0 \times 1 0^{- 6}, (1.0 \times 1 0^{- 6})^{1/3} = 1.0 \times 1 0^{- 2} .$ $r \approx 1.50 \times 1 0^{8} \times 1 0^{- 2} = 1.5 \times 1 0^{6} km .$ Sanity check: SOHO L1 par hai aur JWST L2 par, dono ~1.5 million km Earth se. ✓ Dekho Halo Orbits & Station-Keeping (JWST, SOHO).

Problem 2.2

Sun–Jupiter system ke liye, $M_{1} / M_{2} = 1.047 \times 1 0^{3}$ . L4/L5 stability criterion apply karo: $\frac{M _{1}}{M _{2}} > \frac{25 + 621}{2} \approx 24.96.$ Kya Jupiter ke Trojans stable hain?

Recall Solution

$\frac{25 + 621}{2} = \frac{25 + 24.92}{2} \approx 24.96.$ Kyunki $1047 ≫ 24.96$ , criterion bahut bade margin se satisfy hota hai ⇒ Jupiter ke Trojans stable hain. Yahi wajah hai ki hazaron Trojan Asteroids Jupiter ke L4 aur L5 par cluster karte hain. ✓

Problem 2.3

Parent note ke standard units mein ( $G (M_{1} + M_{2}) = 1$ , $D = 1$ ), Kepler's third law ne $ω^{2} = G (M_{1} + M_{2}) / D^{3}$ diya tha. Dikhao ki $ω = 1$ aur batao yeh humein kya faayda deta hai.

Recall Solution

$ω^{2} = \frac{G ( M _{1} + M _{2} )}{D ^{3}} = \frac{1}{1 ^{3}} = 1 ⟹ ω = 1.$ Kyun helpful hai: $ω = 1$ ke saath centrifugal potential $- \frac{1}{2} ω^{2} (x^{2} + y^{2})$ apna clutter kho deta hai aur simply $- \frac{1}{2} (x^{2} + y^{2})$ ban jaata hai, toh $Ω$ differentiate karna clean ho jaata hai. Yeh hai Kepler's Third Law dono bade bodies par apply kiya hua.

Level 3 — Analysis

Problem 3.1

$M_{1}$ ko $x = - μ$ par aur $M_{2}$ ko $x = + (1 - μ)$ par rakho. Verify karo ki yeh choices barycenter (center of mass) ko origin par rakhti hain.

Recall Solution

KYA: barycenter $x_{cm} = \frac{M _{1} x _{1} + M _{2} x _{2}}{M _{1} + M _{2}}$ par hai. Hamare units mein $M_{1} = 1 - μ$ aur $M_{2} = μ$ . $M_{1} x_{1} + M_{2} x_{2} = (1 - μ) (- μ) + μ (1 - μ) = - μ (1 - μ) + μ (1 - μ) = 0.$ Toh $x_{cm} = 0$ ✓ — dono contributions bilkul cancel ho jaate hain, jo barycenter ki defining property hai.

Problem 3.2

Dikhao ki triangular points $r_{1} = r_{2} = 1$ satisfy karte hain jab $(x, y) = (\frac{1}{2} - μ, \frac{3}{2})$ par rakhe jaayein.

Recall Solution

KYA: dono distances compute karo. $r_{1}^{2} = (x + μ)^{2} + y^{2} = (\frac{1}{2} - μ + μ)^{2} + (\frac{3}{2})^{2} = \frac{1}{4} + \frac{3}{4} = 1.$ $r_{2}^{2} = (x - (1 - μ))^{2} + y^{2} = (\frac{1}{2} - μ - 1 + μ)^{2} + \frac{3}{4} = (- \frac{1}{2})^{2} + \frac{3}{4} = 1.$ Dono se $r_{1} = r_{2} = 1 = D$ milta hai. DEKHNE MEIN KAISA LAGTA HAI: ek equilateral triangle — teeno sides barabar length $1$ (figure s01). Isliye L4/L5 ko triangular points kehte hain.

Problem 3.3

Axis $y = 0$ par, collinear condition hai: $- \frac{1 - μ}{∣ x + μ ∣ ^{3}} (x + μ) - \frac{μ}{∣ x - ( 1 - μ ) ∣ ^{3}} (x - (1 - μ)) + x = 0.$ Sign-by-sign explain karo kyun L1 (bodies ke beech) ke liye do gravity terms ko partly oppose karna padta hai, jabki L2 (bodies ke bahar) ke liye unhe add karna padta hai.

Recall Solution

Har region ke signs set karo (test mass position $x$ par):

L1 par ( $- μ < x < 1 - μ$ ): $x + μ > 0$ (mass $M_{1}$ ke daayein taraf hai, toh $M_{1}$ use baayein / negative- $x$ direction mein kheenchti hai), aur $x - (1 - μ) < 0$ (mass $M_{2}$ ke baayein taraf hai, toh $M_{2}$ use daayein / positive- $x$ direction mein kheenchti hai). Dono gravities axis ke saath opposite directions mein point karti hain → yeh partly cancel ho jaati hain, aur bacha hua plus centrifugal $+ x$ zero tak balance ho jaata hai.
L2 par ( $x > 1 - μ$ ): ab $x + μ > 0$ aur $x - (1 - μ) > 0$ . Dono gravities mass ko bodies ki taraf wapas kheenchti hain (negative- $x$ ) → yeh add ho jaati hain. Akele centrifugal $+ x$ (outward) dono ko balance karna padta hai. Isliye L2 ek naive single-body estimate se zyada door hai.

Dekhne mein kaisa lagta hai: figure s02 mein L1 par do gravity arrows ek doosre ki taraf point karte hain; L2 par yeh ek hi direction mein point karte hain (inward), aur centrifugal arrow (outward) unhe balance karta hai.

Level 4 — Synthesis

Problem 4.1

Small- $μ$ L1/L2 distance $r \approx D (μ /3)^{1/3}$ scratch se derive karo. Explicitly dikhao "3" kahan se aata hai.

Recall Solution

Setup. Test mass ko $M_{2}$ se $r$ distance par line par rakho, $r ≪ D$ ke saath. Real units mein kaam karo, $D$ general. Teen forces line ke saath, toward $M_{1}$ (inward) ko positive maante hue:

$M_{2}$ ki gravity (magnitude $G M_{2} / r^{2}$ ) mass ko $M_{2}$ ki taraf kheenchti hai — yaani L2 ke liye outward, toh yeh $- G M_{2} / r^{2}$ ke roop mein enter hoti hai.
$M_{1}$ ki tidal gravity. $M_{2}$ ki location par mass $G M_{1} / D^{2}$ feel karta; $r$ kadam aur bahar jaane par yeh gradient $\frac{d}{d R} (\frac{G M _{1}}{R ^{2}}) = - \frac{2 G M _{1}}{R ^{3}}$ se badalta hai. Distance $r$ par: magnitude $\frac{2 G M _{1} r}{D ^{3}}$ ka ek extra inward change... lekin $M_{2}$ ke circular motion ke relative sirf difference rakhte hain.
Centrifugal radius $\approx D + r$ par: $ω^{2} (D + r)$ jahan $ω^{2} = G M_{1} / D^{3}$ (dominant body). $M_{2}$ ki orbit ke relative yeh $+ ω^{2} r = \frac{G M _{1} r}{D ^{3}}$ outward contribute karta hai.

Dono $M_{1}$ terms combine karo (tidal $2$ + centrifugal $1$ ): large body ke environment se net restoring pull hai $(2 + 1) \frac{G M _{1} r}{D ^{3}} = \frac{3 G M _{1} r}{D ^{3}} .$ Yahi "3" ka source hai — tidal gradient $2$ contribute karta hai, centrifugal $1$ contribute karta hai.

Balance $M_{2}$ ki local gravity ke against: $\frac{G M _{2}}{r ^{2}} = \frac{3 G M _{1} r}{D ^{3}} ⟹ r^{3} = \frac{M _{2}}{3 M _{1}} D^{3} ⟹ r = D (\frac{M _{2}}{3 M _{1}})^{1/3} .$ Small $μ$ ke liye, $M_{2} / M_{1} \approx μ$ , toh $r \approx D (\frac{μ}{3})^{1/3} .$ Yeh Hill Sphere radius ke same length scale hai — koi coincidence nahi, dono tidal + centrifugal balance se aate hain.

Problem 4.2

L4/L5 stability characteristic equation hai: $λ^{4} + λ^{2} + \frac{27}{4} μ (1 - μ) = 0.$ Dikhao ki bounded (oscillatory) motion ke liye $27 μ (1 - μ) \leq 1$ chahiye, aur critical mass ratio $μ_{c}$ solve karo.

Recall Solution

KYA: $s = λ^{2}$ substitute karo quartic ko quadratic mein badalne ke liye: $s^{2} + s + \frac{27}{4} μ (1 - μ) = 0.$ KYUN: motion bounded tabhi rehta hai jab har $λ$ purely imaginary ho, yaani har $s = λ^{2}$ real aur negative ho (toh $λ = \pm i ∣ s ∣$ , pure oscillation). $s$ ke real hone ke liye, quadratic ka discriminant non-negative hona chahiye: $Δ = 1 - 4 \cdot \frac{27}{4} μ (1 - μ) = 1 - 27 μ (1 - μ) \geq 0.$ Yahi required condition hai: $27 μ (1 - μ) \leq 1$ . Boundary solve karo $27 μ (1 - μ) = 1$ , yaani $27 μ^{2} - 27 μ + 1 = 0$ : $μ = \frac{27 \pm 2 7 ^{2} - 4 \cdot 27}{2 \cdot 27} = \frac{27 \pm 621}{54} = \frac{1}{2} (1 \pm \frac{621}{729}) = \frac{1}{2} (1 \pm \frac{23}{27}) .$ Physically relevant (chhota) root hai: $μ_{c} = \frac{1}{2} (1 - \frac{23}{27}) \approx 0.0385.$ Stability $μ < μ_{c}$ ke liye hold karti hai. (Jab $s$ real-negative ho toh humein dono roots negative bhi chahiye — sum $= - 1 < 0$ aur product $= \frac{27}{4} μ (1 - μ) > 0$ yeh guarantee karte hain. ✓)

Level 5 — Mastery

Problem 5.1

Criterion $μ < μ_{c} \approx 0.0385$ ko familiar mass-ratio form $M_{1} / M_{2} > \frac{25 + 621}{2}$ mein convert karo. Phir decide karo ki Earth–Moon ( $M_{1} / M_{2} = 81.3$ ) stable Trojans host kar sakta hai ya nahi.

Recall Solution

$μ$ se $M_{1} / M_{2}$ tak. Kyunki $μ = \frac{M _{2}}{M _{1} + M _{2}}$ , likho $q = M_{1} / M_{2}$ . Toh $μ = \frac{1}{q + 1} .$ Boundary $27 μ (1 - μ) = 1$ , $μ = 1/ (q + 1)$ aur $1 - μ = q / (q + 1)$ ke saath: $27 \cdot \frac{1}{q + 1} \cdot \frac{q}{q + 1} = 1 ⟹ 27 q = (q + 1)^{2} ⟹ q^{2} - 25 q + 1 = 0.$ $q = \frac{25 \pm 625 - 4}{2} = \frac{25 \pm 621}{2} .$ Stable regime bada root hai: $q > \frac{25 + 621}{2} \approx 24.96.$

Earth–Moon: $q = 81.3 > 24.96$ ✓ ⇒ Moon ke L4/L5 (weakly) stable hain. Yeh theoretikal ghar hai faint Kordylewski dust clouds ka.

Problem 5.2 (Degenerate / limiting cases)

Stability criterion ke dono extremes investigate karo: (a) $μ \to 0$ (test-body limit, jaise ek akele star ke paas ek grain). Kya L4 stable hai? (b) $μ = \frac{1}{2}$ (equal masses). Kya L4 stable hai? Physically interpret karo.

Recall Solution

$f (μ) = 27 μ (1 - μ)$ evaluate karo aur $1$ se compare karo (stable jab $f \leq 1$ ).

(a) $μ \to 0$ : $f \to 0 \leq 1$ ✓ — stable, bahut bade margin ke saath. Mantab hai: ek vanishingly light second body equilateral point ko barely perturb karta hai, aur Coriolis grain ko aasaani se trap kar leti hai.
(b) $μ = \frac{1}{2}$ : $f = 27 \cdot \frac{1}{2} \cdot \frac{1}{2} = 6.75 > 1$ ✗ — unstable. Do equal masses ke saath, triangular point par perturbing pulls itni strong hain ki Coriolis unhe tame nahi kar sakta; point kisi body ko hold nahi kar sakta.
Boundary check: stability kahin beech mein lost hoti hai, precisely $μ_{c} \approx 0.0385$ par (Problem 4.2 se). Toh stable window hai $0 \leq μ < 0.0385$ — sirf lopsided pairs (ek body bahut heavier) ko stable Trojans milte hain. Yeh reality se match karta hai: Sun–Jupiter, Sun–Neptune, Earth–Moon sabka $μ$ chhota hai.

Problem 5.3 (Synthesis with tidal physics)

L1/L2 scale $r = D (μ /3)^{1/3}$ aur Roche Limit dono tidal balance se aate hain. Ek paragraph mein contrast karo kya toota har jagah: Hill/L1 scale par physically kya hota hai versus Roche limit par kya hota hai?

Recall Solution

Dono scales ek body ki apni gravity aur ek bade neighbour ki tidal + centrifugal stretch ke competition se set hote hain.

Hill/L1 scale $r \sim D (μ /3)^{1/3}$ par, sawaal ek teesre chhote body ki orbit ke baare mein hai: is radius ke andar $M_{2}$ ek satellite ko bound orbit mein rakh sakta hai; $r$ par (L1 gate) satellite $M_{1}$ ki taraf leak ho sakta hai. Yeh gravitational ownership ki edge mark karta hai.
Roche limit par, sawaal ek fluid ya rubble-pile satellite ki apni cohesion ke baare mein hai: jab satellite ke across tidal stretch uski self-gravity se zyada ho jaati hai, satellite toot jaata hai (rings bante hain). Roche limit scale karta hai $R_{Roche} \sim R_{M_{2}} (2 ρ_{1} / ρ_{2})^{1/3}$ — directly mass ratio par nahi balki densities par. Contrast: Hill/L1 poochta hai "kya $M_{2}$ ek orbiting companion rakh sakta hai?"; Roche poochta hai "kya ek companion intact survive karega?". Same tidal machinery, alag cheez weigh ki ja rahi hai — orbit stability versus material strength.

Recall One-line self-test recap

Kaun se do Lagrange points stable ho sakte hain, aur kyun chahe yeh potential maxima hain? ::: L4 aur L5; velocity-dependent Coriolis force ek drifting mass ko ek bounded loop mein curve karti hai, lekin sirf tab jab $M_{1} / M_{2} > 24.96$ (equivalently $μ < 0.0385$ ). $r = D (μ /3)^{1/3}$ mein "3" kahan se aata hai? ::: Tidal gradient $2$ contribute karta hai, centrifugal $1$ contribute karta hai, sum $3$ hota hai. Kya L4 equal masses ke liye stable hai ( $μ = \frac{1}{2}$ )? ::: Nahi — $27 μ (1 - μ) = 6.75 > 1$ , toh motion unbounded hai.