Worked examples — Lagrange points L1–L5 — derivation, stability

3.2.30 · D3 · Physics › Orbital Mechanics & Astrodynamics › Lagrange points L1–L5 — derivation, stability

Yahan sab kuch wohi normalized units use karta hai jo parent se hain: total mass $=1$, body separation $D=1$, $G(M_1+M_2)=1$, isliye co-rotating angular speed $\omega =1$ hai. Do heavy bodies $x=-\mu$ (bada wala, $M_1$) aur $x=+(1-\mu )$ (chhota wala, $M_2$) par baithe hain, jahan $\mu =\frac{M_2}{M_1+M_2}.$

Prereqs jo aap open rakhna chahein: Restricted Three-Body Problem, Effective Potential & Jacobi Constant, Coriolis and Centrifugal Forces (Rotating Frames), Hill Sphere, Kepler's Third Law.

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Scenario matrix

Is topic ke har problem ko inhi cells mein se koi ek milta hai. Neeche ke examples us cell se tagged hain jo woh cover karte hain.

Cell	Case class	Kya special / degenerate hai	Covered by
A	Collinear L1, small $\mu$	bodies ke beech; Hill-radius approx use karo	Ex 1
B	Collinear L2, small $\mu$	chhote body ke baad; $r$ ka sign flip hota hai	Ex 2
C	Collinear L3, small $\mu$	far side, $-1$ ke paas; apna expansion chahiye	Ex 3
D	Triangular L4/L5 location	off-axis, $y\ne 0$, equilateral triangle	Ex 4
E	Stability decision, small ratio	kya $M_1/M_2>24.96$ hai?	Ex 5
F	Degenerate equal masses $\mu =\frac{1}{2}$	symmetry, L4/L5 unstable, L3 center-ish par	Ex 6
G	Real word problem (JWST at L2)	real km plug karo, units convert karo	Ex 7
H	Exam twist: Jacobi/energy sign	L1 par zero-velocity gate	Ex 8

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Example 1 — Cell A: Sun–Earth L1

Step 1 — Sahi tool chunna: Hill-radius approximation. Kyunki $\mu \ll 1$ hai, L1 chhote body ke bahut paas baith'ta hai. Poora collinear equation ek quintic hai, lekin $M_2$ ke paas yeh leading balance tak collapse ho jata hai $r\approx D{\left(\frac{\mu }{3}\right)}^{1/3}.$ Yeh step kyun? Hume exact quintic root ki zaroorat nahi; small-$\mu$ expansion sirf dominant tidal + centrifugal terms rakhti hai (woh "3" = tidal 2 + centrifugal 1). Dekho Hill Sphere.

Step 2 — Plug in karo. $r\approx 1.496\times 10^8{\left(\frac{3.0\times 10^{-6}}{3}\right)}^{1/3}=1.496\times 10^8\times (10^{-6}{)}^{1/3}.$ Yeh step kyun? $\mu /3=10^{-6}$ exactly yahan — cube root clean hai.

Step 3 — Cube root evaluate karo. $(10^{-6}{)}^{1/3}=10^{-2},r\approx 1.496\times 10^8\times 10^{-2}=1.496\times 10^6\text{ km}.$

Verify: $\approx 1.5$ million km — yahi woh jagah hai jahan SOHO rehta hai. Yeh $D$ ka $\sim 1\%$ hai, halfway nahi. Forecast "Earth se chipka hua" sahi nikla. Units: km $\times$ (dimensionless) $=$ km ✓.

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Example 2 — Cell B: Sun–Earth L2 aur sign flip kyun hota hai

Step 1 — Same leading balance, opposite side. L2 ke liye chhote body ki gravity ab Sun ki taraf point karti hai (inward) jabki centrifugal baahir point karta hai. Collinear equation ko far side par expand karne par wahi leading form milta hai $r\approx D{\left(\frac{\mu }{3}\right)}^{1/3}.$ Yeh step kyun? $\mu$ mein leading order tak, L1 aur L2 $M_2$ ke baare mein symmetric hain — dono $\pm r$ par. Antar sirf next order par aata hai.

Step 2 — Numbers leading order tak identical hain. $r\approx 1.496\times 10^6\text{ km, lekin ab anti-Sun side par.}$ Yeh step kyun? Same $\mu$, same cube root; geometry (kaun si side) flip hoti hai, magnitude nahi.

Verify: JWST aur Gaia Sun–Earth L2 ke around $\approx 1.5$ million km anti-Sunward orbit karte hain. Real mission se consistent hai (Halo Orbits & Station-Keeping (JWST, SOHO)). ✓

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Example 3 — Cell C: Sun–Earth L3 (far side)

Step 1 — Alag expansion. L3 $x=-1$ ke paas hai, tiny body se door. Yahan small-$\mu$ result yeh hai $x_{L3}\approx -1-\frac{5}{12}\mu .$ Yeh step kyun? Sun ke paas chhota body ek door perturber hai, isliye hum quintic ko $x=-1$ ke baare mein expand karte hain, $M_2$ ke baare mein nahi. Coefficient $5/12$ standard hai.

Step 2 — $\mu =3.0\times 10^{-6}$ plug karo. $\delta =\frac{5}{12}\mu =\frac{5}{12}(3.0\times 10^{-6})=1.25\times 10^{-6}.$ Toh $x_{L3}\approx -1.00000125$. Yeh step kyun? Offset $\delta$ normalized units mein antipode ke baad shift hai.

Verify: $\delta \times D=1.25\times 10^{-6}\times 1.496\times 10^8\approx 187$ km. Toh L3 exact opposite point se $\sim 190$ km aage hai — ek sach mein tiny nudge, jaise forecast tha. ✓

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Example 4 — Cell D: L4/L5 geometry

Step 1 — Exact triangular solution use karo. Parent ne dikhaya ki $r_1=r_2=1$ dono ${\partial }_x\Omega =0$ aur ${\partial }_y\Omega =0$ solve karta hai, jisse milta hai $x=\frac{1}{2}-\mu ,y=+\frac{\sqrt{3}}{2}.$ Yeh step kyun? Koi approximation zaroorat nahi — L4/L5 exact hain chahe $\mu$ kuch bhi ho. Equilateral triangle ek geometric identity hai, small-$\mu$ trick nahi.

Step 2 — $\mu =9.54\times 10^{-4}$ plug karo. $x=0.5-0.000954=0.499046,y=\mathrm{0.866025.}$

Step 3 — Check karo ki dono distances 1 ke barabar hain. Distance to $M_1$ at $(-\mu ,0)$: $r_1=\sqrt{(x+\mu {)}^2+y^2}=\sqrt{0.5^2+(\sqrt{3}/2{)}^2}=\sqrt{0.25+0.75}=1$. Distance to $M_2$ at $(1-\mu ,0)$: $r_2=\sqrt{(x-(1-\mu ){)}^2+y^2}=\sqrt{(-0.5{)}^2+0.75}=1$. Yeh step kyun? Equilateral claim confirm karta hai: dono legs $=1=$ base $D$.

Verify: $r_1=r_2=D=1$ ✓ — equilateral triangle confirmed. Yahi woh jagah hai jahan Trojan Asteroids cluster karte hain.

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Example 5 — Cell E: Stability decision

Step 1 — Criterion state karo. L4/L5 stable hain iff $\frac{M_1}{M_2}>\frac{25+\sqrt{621}}{2}\approx \mathrm{24.9599.}$ Yeh step kyun? Yeh characteristic equation ${\lambda }^4+{\lambda }^2+\frac{27}{4}\mu (1-\mu )=0$ se aata hai jisme $s={\lambda }^2$ mein real-negative roots chahiye — Coriolis force (dekho Coriolis and Centrifugal Forces (Rotating Frames)) itni strong honi chahiye ki drift karte mass ko wapas ghuma sake.

Step 2 — Threshold compute karo. $\sqrt{621}=24.9199,\frac{25+24.9199}{2}=\mathrm{24.9599.}$

Step 3 — Compare karo. Sun–Jupiter: $M_1/M_2\approx 1047.6>24.96$ → stable. Earth–Moon: $M_1/M_2\approx 81.3>24.96$ → stable (weakly; Kordylewski dust clouds). Yeh step kyun? Dono ratios bar ko kaafi margin se cross karte hain.

Verify: Threshold $=24.9599$; dono $1047.6$ aur $81.3$ ise exceed karte hain. Forecast "yes/yes" sahi nikla. ✓

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Example 6 — Cell F: Degenerate equal masses $\mu =\frac{1}{2}$

Step 1 — L4/L5 location. $x=\frac{1}{2}-\mu =\frac{1}{2}-\frac{1}{2}=0$, $y=\pm \frac{\sqrt{3}}{2}$. Yeh step kyun? Equal masses ke saath triangle apex exactly $y$-axis par baith'ta hai (barycenter ke upar). Equilateral geometry phir bhi hold karti hai — woh kabhi $\mu$ par depend nahi karti.

Step 2 — Stability check. $\mu <{\mu }_c=\frac{1}{2}(1-\sqrt{23/27})\approx 0.0385$ chahiye. Yahan $\mu =0.5\gg 0.0385$. Yeh step kyun? Coriolis-strength condition badly fail karti hai; ek nudge bina bound ke grow karta hai.

Verify: ${\mu }_c\approx 0.0385$ aur $0.5>0.0385$ → unstable. Equal-mass binaries L4/L5 par Trojans trap nahi kar sakti. Woh symmetry jo points create karti hai unhe stabilize nahi karti. ✓

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Example 7 — Cell G: Real word problem — JWST ki station-keeping region size karna

Step 1 — Hill radius. $R_H=D{\left(\frac{\mu }{3}\right)}^{1/3}=1.496\times 10^8\times (10^{-6}{)}^{1/3}=1.496\times 10^8\times 10^{-2}.$ Yeh step kyun? Hill radius (dekho Hill Sphere) L1 aur L2 dono ka scale set karta hai small $\mu$ ke liye — woh essentially Earth se $\pm R_H$ par baith'te hain.

Step 2 — Evaluate karo. $R_H\approx 1.496\times 10^6\text{ km}.$

Verify: $R_H\approx 1.5\times 10^6$ km $\approx$ L2 ki actual distance. JWST roughly is scale ke andar ek loop karta hai. Units: km ✓. Ex 1/Ex 2 se match karta hai jaise karna chahiye (L1, L2, aur $R_H$ leading order share karte hain). ✓

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Example 8 — Cell H: Exam twist — Jacobi/zero-velocity gate

Step 1 — Potential yaad karo. $\Omega (x,0)=-\frac{1-\mu }{|x+\mu |}-\frac{\mu }{|x-(1-\mu )|}-\frac{1}{2}{x}^2.$ Yeh step kyun? Axis par $y=0$ hai, toh $r_1=|x+\mu |$, $r_2=|x-(1-\mu )|$. Yeh Effective Potential & Jacobi Constant line tak restricted hai.

Step 2 — $\mu =0.1$ ke liye L1 aur L2 locate karo. Collinear quintic numerically solve karne par approximately milta hai $x_{L1}\approx 0.6090,x_{L2}\approx \mathrm{1.2668.}$ Yeh step kyun? Pehle roots chahiye taki pata chale $\Omega$ kahan evaluate karna hai.

Step 3 — $\Omega$ evaluate karo. $\Omega (x_{L1})\approx -1.8871,\Omega (x_{L2})\approx -\mathrm{1.8541.}$ Yeh step kyun? Zyada $\Omega$ = zero-velocity picture mein lower "barrier" ($C_J=-2\Omega$ chhota hota hai). Inner gate (L1) zyada negative hai → baad mein khulta hai; L2 par outer neck L1 se pehle khulta hai yahan.

Verify: Numerically $\Omega (L2)>\Omega (L1)$ (dono $-1.85$ se $-1.89$ ke paas). Jaise jaise energy badhti hai pehle khulne wala neck wohi hota hai jiska sabse bada $\Omega$ ho; in dono mein woh L2 hai. Yahi exact reason hai ki ek spacecraft L1 par inward cross karne se pehle L2 ke baad leak ho sakta hai. ✓

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Recall Har example kaun sa cell cover karta hai?

Ex 1 collinear L1 small $\mu$ ::: Cell A Ex 2 collinear L2 small $\mu$ (sign flip) ::: Cell B Ex 3 collinear L3 far side ::: Cell C Ex 4 triangular L4/L5 geometry ::: Cell D Ex 5 stability decision ::: Cell E Ex 6 degenerate equal masses ::: Cell F Ex 7 real word problem (JWST/Hill) ::: Cell G Ex 8 exam twist (Jacobi gate) ::: Cell H