Lagrange points L1–L5 — derivation, stability: Visual walkthrough

3.2.30 · D2 · Physics › Orbital Mechanics & Astrodynamics › Lagrange points L1–L5 — derivation, stability

Hum Restricted Three-Body Problem ke setup aur Effective Potential & Jacobi Constant ke landscape idea par build kar rahe hain. Spinning-frame forces Coriolis and Centrifugal Forces (Rotating Frames) se aate hain.

Step 1 — Stage: do bhaari bodies aur ek spinning turntable

Turntable kyun? Agar hum still rehte toh sab kuch ghoomta dikhta — koi "parking spot" dhundna impossible hota. Do bodies ko freeze karna ek motion problem ko "flat spot dhundo" wale problem mein badal deta hai. Yahi rotating frame ka poora trick hai.

PICTURE. Neeche, $M_{1}$ left mein hai, $M_{2}$ right mein, barycenter (black dot) dono ke beech mein hai. Curved arrow dikhata hai ki poori picture saath mein rate $ω$ par spin kar rahi hai (Greek "omega", hamara symbol kitni tezi se turntable ghoomta hai ke liye, radians per second mein).

Figure — Lagrange points L1–L5 — derivation, stability

Step 2 — Rulers chunna taaki numbers clean rahein

$ω = 1$ kyun? Kyunki Kepler's Third Law spin ko gravity aur distance se link karta hai: do bodies ke ek doosre ke around chakkar lagate rehne ke liye, gravity ko centripetal pull supply karna padega, jo force karta hai $ω^{2} = \frac{G ( M _{1} + M _{2} )}{D ^{3}} .$ Term by term: $G (M_{1} + M_{2})$ total gravitational strength hai, $D^{3}$ separation ka cube hai, aur unka ratio hi spin-squared hai. Units choose karo jahan $G (M_{1} + M_{2}) = 1$ aur $D = 1$ ho toh right side $1$ ban jaata hai, toh $ω = 1$ . Ek symbol, khatam.

PICTURE. Hum $M_{1}$ ko $x = - μ$ par rakhte hain aur $M_{2}$ ko $x = + (1 - μ)$ par, jahan $μ$ (Greek "mu") chhote body mein mass ka fraction hai: $μ = \frac{M _{2}}{M _{1} + M _{2}} .$ Yahan $μ$ yeh answer deta hai ki "total weight ka kitna hissa $M_{2}$ mein hai?" — $0$ aur $\frac{1}{2}$ ke beech ka number. Barycenter exactly $x = 0$ par land karta hai.

Recall

$M_{1}$ $- μ$ par kyun hai, $- 1/2$ par kyun nahi? ::: Kyunki bhaari body balance point ke zyada paas baitta hai. Lever law $M_{1} \cdot (arm_{1}) = M_{2} \cdot (arm_{2})$ barycenter ko bhaari side ki taraf khiinchta hai.

Step 3 — Do forces jo ek still marble feel karta hai

Sirf do kyun? Coriolis force $- 2 ω \times v$ hai: ise velocity $v$ chahiye. Rest mein marble ka $v = 0$ hai, toh Coriolis exactly zero hai. Parking spots dhundne ke liye sirf gravity + centrifugal chahiye. (Coriolis Step 8 mein jaag jaata hai, stability ke liye.)

PICTURE. $(x, y)$ par marble ke liye hum do distances measure karte hain: $r_{1} = (x + μ)^{2} + y^{2}, r_{2} = (x - (1 - μ))^{2} + y^{2} .$ Padhke: $r_{1}$ $M_{1}$ tak straight-line gap hai (jo $- μ$ par baitha hai), $r_{2}$ $M_{2}$ tak gap hai ( $1 - μ$ par). Magenta aur violet arrows do gravity pulls hain; orange arrow centrifugal push hai.

Step 4 — Saari forces ko ek landscape $Ω$ mein pack karna

Potential kyun? Ek force jo "kisi height map ke neeche ki taraf" likha ja sake uske baare mein reason karna kaafi aasaan hai: vectors balance karne ke bajay hum dekhte hain kahan zameen level hai. Gravity aur centrifugal force dono is tarah ke hain. Yeh effective potential hai.

PICTURE + equation, term by term: $Ω (x, y) = well around M_{1} - \frac{1 - μ}{r _{1}} well around M_{2} - \frac{μ}{r _{2}} centrifugal dome - \frac{1}{2} (x^{2} + y^{2}) .$

$- \frac{1 - μ}{r _{1}}$ : bade body ke around deep funnel; $1 - μ = M_{1} / (M_{1} + M_{2})$ uska weight-share hai, aur $1/ r_{1}$ paas aane par plunge karta hai.
$- \frac{μ}{r _{2}}$ : $M_{2}$ ke around chhota funnel (chhota share $μ$ ).
$- \frac{1}{2} (x^{2} + y^{2})$ : ulta bowl. Iska slope bahar point karta hai, exactly centrifugal push $ω^{2} ρ$ reproduce karta hai ( $ω = 1$ ke saath). $\frac{1}{2}$ isliye hai taaki $x^{2}$ differentiate karne par clean $x$ mile.

Flat spot rule: Lagrange point woh jagah hai jahan zameen har direction mein tilt karna band kar de: $\frac{\partial Ω}{\partial x} = 0 aur \frac{\partial Ω}{\partial y} = 0.$ Symbol $\partial / \partial x$ ("partial derivative") bas yeh poochta hai: agar main $x$ ko thoda nudge karun aur $y$ fixed rakhuun, toh kya height badlegi? Zero matlab "is taraf koi tilt nahi".

Step 5 — Sideways-slope equation ( $\partial Ω/ \partial y = 0$ )

$y$ se kyun start karein? Kyunki do bade bodies dono $x$ -axis par baithe hain, $y$ -direction symmetric hai aur sabse saaf condition deta hai — yeh humein bataayega ki gravity strengths ka sum centrifugal strength ke barabar hona chahiye.

Derivative, term by term. $Ω$ ke har piece ko $y$ ke saath differentiate karte hain: $\frac{\partial Ω}{\partial y} = (\frac{1 - μ}{r _{1}^{3}}) y + (\frac{μ}{r _{2}^{3}}) y - y = 0.$

$\frac{1 - μ}{r _{1}^{3}} y$ : $M_{1}$ ke well ka $y$ -slope. $r_{1}^{3}$ (cube) isliye aata hai kyunki $1/ r_{1}$ differentiate karna jahan $r_{1} = \dots + y^{2}$ ho, ek $y / r_{1}^{3}$ le aata hai.
$\frac{μ}{r _{2}^{3}} y$ : $M_{2}$ ke well ke liye same story.
$- y$ : centrifugal dome ka $y$ -slope.

Ab $y \neq = 0$ use karo (hum off-axis point chahte hain). Poori equation ko $y$ se divide karo: $\frac{1 - μ}{r _{1}^{3}} + \frac{μ}{r _{2}^{3}} = 1.$ Padho: do gravity strengths, mass-share aur $1/ r^{3}$ se weighted, exactly $1$ (centrifugal strength) add up karni chahiye. Yeh box yaad rakho — ise Equation (Y) bolte hain.

Step 6 — Along-axis slope $r_{1} = r_{2}$ force karta hai

Equal distances kyun? Intuitively: $y$ -condition ne pehle se total gravity strength fix kar di. $x$ -condition tab left–right components balance karta hai, aur barycenter ke lever ko honest rakhne ka ek hi tarika hai ki dono bodies marble se same distance par hon.

$x$ -derivative, term by term: $\frac{\partial Ω}{\partial x} = \frac{1 - μ}{r _{1}^{3}} (x + μ) + \frac{μ}{r _{2}^{3}} (x - (1 - μ)) - x = 0.$

$(x + μ)$ : hum $M_{1}$ se kitne east hain.
$(x - (1 - μ))$ : hum $M_{2}$ se kitne east hain.

Equation (Y) substitute karo $x$ eliminate karne ke liye. " $- x$ " ko $x = x \cdot 1 = x (\frac{1 - μ}{r _{1}^{3}} + \frac{μ}{r _{2}^{3}})$ se replace karke aur simplify karne par, har $x$ cancel ho jaata hai aur tumhare paas bachta hai: $(1 - μ) μ (\frac{1}{r _{1}^{3}} - \frac{1}{r _{2}^{3}}) = 0.$ Kyunki $μ \neq = 0$ aur $1 - μ \neq = 0$ (do real bodies), ek hi raasta bachta hai: $r_{1} = r_{2} .$

PICTURE. Marble ab dono bodies se equidistant hai — woh $M_{1} M_{2}$ line ke perpendicular bisector par hai (dashed vertical line $x = \frac{1}{2} - μ$ par).

Step 7 — Triangle band karna: $r_{1} = r_{2} = D = 1$

$r_{1} = r_{2} = r$ ko (Y) mein plug karo: $\frac{1 - μ}{r ^{3}} + \frac{μ}{r ^{3}} = \frac{( 1 - μ ) + μ}{r ^{3}} = \frac{1}{r ^{3}} = 1 ⟹ r^{3} = 1 ⟹ r = 1 = D .$ Har symbol apni jagah earn karta hai: numerator collapse ho jaata hai kyunki mass-shares $1$ mein add hote hain, aur $r^{3} = 1$ force karta hai $r = 1$ , jo exactly separation $D$ hai.

Yeh equilateral triangle kyun hai. Dono bade bodies $D$ distance par hain, aur marble dono se $D$ distance par hai. Teen points, saare pairwise distances $D$ ke barabar — yahi equilateral triangle ki definition hai.

PICTURE + coordinates. " $M_{1}$ se $- μ$ aur $M_{2}$ se $1 - μ$ distance $1$ " solve karte hain: $x = \frac{1}{2} - μ, y = \pm \frac{3}{2} .$

$x = \frac{1}{2} - μ$ : bodies ke beech mein (Step 6 se bisector).
$y = + \frac{3}{2}$ L4 hai ( $M_{2}$ se $6 0^{\circ}$ aage); $y = - \frac{3}{2}$ L5 hai ( $6 0^{\circ}$ peeche). $3 /2$ bas unit equilateral triangle ki height hai.

Step 8 — Twist: L4/L5 hilltops hain, phir bhi stable

Woh kyun hold karte hain. Jis pal marble slide karna shuru karta hai, woh move kar raha hai — aur ab so raha Coriolis force jaag jaata hai. Coriolis, $- 2 ω \times v$ , drifting marble ko sideways bend karta hai, uski escape ko hilltop ke around ek tiny loop mein curl karta hai. Marble peak ke around endlessly circle karta hai instead of jaane ke.

Catch (masses par ek condition). Coriolis tabhi jeetta hai jab spin, hill ki steepness ke relative kaafi fast ho — yaani agar $M_{2}$ kaafi chhota ho. Motion ko linearise karne par characteristic equation milti hai $λ^{4} + λ^{2} + \frac{27}{4} μ (1 - μ) = 0,$ aur bounded (oscillating) motion ke liye uska discriminant $\geq 0$ chahiye: $27 μ (1 - μ) \leq 1 ⟺ μ < μ_{c} = \frac{1}{2} (1 - \frac{23}{27}) \approx 0.0385,$ equivalently $M_{1} / M_{2} > \frac{25 + 621}{2} \approx 24.96$ .

PICTURE. Left: hilltop jahan marble spiral kar raha hai (Coriolis-trapped). Right: zyada bhaari $M_{2}$ ke liye loop unwind ho jaata hai aur marble escape kar jaata hai.

Ek-picture summary

Poori derivation ek canvas par: do bodies turntable par frozen → height map $Ω$ banao → "north tilt nahi" Equation (Y) deta hai → "east tilt nahi" $r_{1} = r_{2}$ force karta hai → saath mein triangle $r_{1} = r_{2} = D$ par band ho jaata hai → Coriolis decide karta hai ki hilltop trap karega ya release.

Recall Feynman retelling — walkthrough plain words mein

Hum ek merry-go-round par chad gaye jo Sun aur Jupiter ke saath spin kar raha tha, toh dono hamare liye still baithe rahe. Ride par rakha ek crumb do tugs feel karta hai (dono se gravity) plus ek outward fling (centrifugal) — aur jabtak woh still baitha hai, sideways Coriolis nudge so raha hai. Humne hills aur valleys ka ek landscape draw kiya jiska slope woh total tug hai, toh parking spot bas zameen ka ek flat patch hai. "Kya zameen north–south level hai?" poochne par pata chala ki do gravity tugs ko fling mein add hona chahiye. "East–west level hai?" poochne par pata chala ki crumb dono bodies se equally far hona chahiye. Inhe saath rakho aur distances dono Sun–Jupiter gap par lock ho jaate hain — teen points saari distances equal, ek perfect triangle. Woh tips L4 aur L5 hain, ek $6 0^{\circ}$ Jupiter se aage aur ek $6 0^{\circ}$ peeche. Funny part: woh hilltops hain, valleys nahi, toh crumb roll off hona chahiye — lekin jis pal woh move karta hai, so raha Coriolis nudge use ek little loop mein curl kar deta hai peak ke around, use forever trap karta hai, jab tak doosri body zyada bhaari na ho. Jupiter Sun ke samne featherweight hai, toh uske Trojan asteroids wahan tab se parked hain jab se Solar System bana tha.

Hill Sphere · Halo Orbits & Station-Keeping (JWST, SOHO) · Roche Limit · Kepler's Third Law · Restricted Three-Body Problem