Lagrange points L1–L5 — derivation, stability: Question bank

3.2.30 · D5 · Physics › Orbital Mechanics & Astrodynamics › Lagrange points L1–L5 — derivation, stability

Shuru karne se pehle, ek ek-line reminder taaki koi bhi symbol unjustified na lage:

True or false — justify

L1, L2, L3, L4, L5 sab do bade bodies ke saath same orbital plane mein hain.

True — effective potential

Ω

sirf in-plane distances aur spin axis ke baare mein centrifugal term se bana hai, isliye paanchon equilibria usi plane mein hain; out-of-plane koi parking spot exist nahi karta.

Kisi bhi do-body system ke liye exactly paanch Lagrange points hote hain, masses chahe kuch bhi ho.

True — collinear condition hamesha 3 real roots deta hai aur triangular solution

r_{1} = r_{2} = 1

hamesha exist karta hai; count paanch par fixed hai, sirf unki stability

μ

par depend karti hai.

L4 aur L5 sirf kuch khaas mass ratios ke liye stable hain.

True — ye sirf tab stable hain jab

M_{1} / M_{2} > 24.96

(equivalently

μ < 0.0385

); us threshold se neechey Coriolis force ek drift karte particle ko trap karne mein capable nahi rehti.

Centrifugal force ek "real" force hai jo space mein ek fixed observer bhi measure kar sakta hai.

False — ye ek apparent force hai jo sirf isliye appear hoti hai kyunki humne ek rotating frame choose kiya; ek non-rotating observer sirf gravity plus body ko ek circle mein accelerate hote dekhta hai.

Kyunki L4/L5

Ω

ke maxima par baithe hain, wahan rakha gaya particle hamesha roll away karega.

False — ek static hilltop picture aisa kehta, lekin velocity-dependent Coriolis force (jo

Ω

mein invisible hai) drift ko ek chhoti loop mein curl kar deta hai, particle ko tab trap karta hai jab mass ratio kaafi chhota ho.

Lagrange points ko locate karne ke liye Coriolis force ki zaroorat hoti hai.

False — Coriolis

= - 2 ω \times v

ek resting particle ke liye zero ho jaata hai, aur Lagrange points rest positions ke roop mein define hote hain; sirf gravity + centrifugal locations set karte hain. Coriolis sirf stability analysis mein aata hai.

Agar tum dono masses ko double kar do lekin unka separation fixed rakho, toh paanchon points move karenge.

False — geometry sirf ratio

μ

par depend karti hai, absolute masses par nahi; dono ko double karne se

μ

unchanged rehta hai, toh points wahi rehte hain (haalaanki

ω

aur orbital period change hoti hai).

Ek spacecraft jo exactly L2 par baitha hai use wahan hamesha ke liye rehne ke liye zero fuel ki zaroorat hai.

False — L2 ek saddle point hai, isliye koi bhi tiny perturbation (radiation pressure, solar wind, numerical drift) exponentially grow karti hai; real missions jaise JWST chhote station-keeping burns fire karte hain.

Spot the error

"L1 symmetry se do bodies ke beech mein halfway par hai."

Galat — L1 chhote body

M_{2}

ki taraf khicha rehta hai; Sun–Earth ke liye ye Earth se sirf ~1.5 million km door hai, lagbhag raaste ka 1%, 50% nahi.

"L4 aur L5 stable hain kyunki ye effective potential ke minima hain."

Galat — ye actually

Ω

ke maxima hain; unki stability poori tarah Coriolis deflection se aati hai, jo static potential landscape nahi dikhaa sakta.

"Equilateral-triangle result

r_{1} = r_{2} = D

jo L4/L5 ke liye hai, sirf tab kaam karta hai jab

M_{1} = M_{2}

Galat — ye har mass ratio ke liye hold karta hai, kyunki

(1 - μ) + μ = 1

r_{1} = r_{2} = 1

ko dono equilibrium equations satisfy karne deta hai chahe

μ

kuch bhi ho; sirf apex ka x-shift

\frac{1}{2} - μ

masses par depend karta hai.

"Kyunki Coriolis force koi kaam nahi karta, ye stability ko affect nahi kar sakta."

Galat — ye energy ke baare mein kuch nahi badalta lekin ye motion ki direction badal deta hai, ek outward drift ko ek closed loop mein bend kar ke; yahi geometric steering L4/L5 ko stabilize karta hai.

"Trojan asteroids exactly geometric L4/L5 points par baithe hain."

Galat — kyunki L4/L5 potential maxima hain, real Trojans points ke aaspaas wide tadpole-shaped paths mein librate (swing) karte hain; ye Lagrange point ko orbit karte hain, uski par rest nahi karte.

"Hum L1, L2, L3 ko sirf formula se find kar sakte hain kyunki quintic ka koi closed form nahi hai; L4/L5 ko bhi numerics chahiye."

Galat — L4/L5 ka exact closed form

x = \frac{1}{2} - μ, y = \pm \frac{3}{2}

hai; sirf collinear points ka quintic numerical solution maangta hai.

"Rotating frame mein ek resting particle par Lagrange point par total apparent force nonzero lekin centripetal hai."

Galat — defining condition

\nablaΩ = 0

hai, yaani net apparent force exactly zero hai; isliye particle rest mein rehta hai.

Why questions

Hum rotating frame mein jaane ki takleef kyun uthate hain?

Kyunki usme do bade bodies frozen hain, isliye "unke relative equilibrium" ek genuine rest condition ban jaata hai; ek fixed frame mein pura configuration spin karta hai aur kuch bhi kabhi rest mein nahi hota.

Centrifugal potential

- \frac{1}{2} ω^{2} (x^{2} + y^{2})

ke roop mein minus sign ke saath kyun appear karta hai?

Taaki iska (negative) gradient baahir ki taraf point kare,

+ ω^{2} ρ \overset{ρ}{^}

, outward centrifugal force ko reproduce karte hue; sign isliye choose kiya gaya hai taaki derivative correct push de.

Hill-radius formula

r \approx D (μ /3)^{1/3}

mein "3" itna specific kyun hai?

Ye collinear equation ko chhote body ke paas expand karne par tidal-stretch factor of 2 ko centrifugal factor of 1 se add karne se aata hai; wo do effects milke balance length scale set karte hain.

L1 aur L2 Earth ke opposite sides par hone ke bawajood almost same distance par kyun baithe hain?

Dono Earth ki gravity aur combined tidal + centrifugal gradient ke beech same local balance se governed hain, jo leading order mein

\pm r

mein symmetric hai, almost equal distances

\sim D (μ /3)^{1/3}

dete hue.

Trojan asteroids L4/L5 par billions of years tak survive kar sakte hain jabki L1/L2 par telescopes ko constant correction chahiye — kyun?

L4/L5 (chhote mass ratio ke saath) genuinely stable hain — Coriolis kisi bhi drift ko return karta hai — jabki L1/L2 saddle points hain jahan perturbations grow karti hain, isliye wahan kuch bhi actively hold kiya jaana chahiye.

Stability threshold mass ratio par depend karta hai na ki individual masses par — kyun?

Linearized equations of motion ek characteristic equation mein reduce hoti hain jo sirf product

μ (1 - μ)

par depend karti hai; absolute masses length aur time units ke saath scale out ho jaati hain.

Paanchon mein se L3 real missions ke liye sabse kam useful kyun hai?

M_{1}

ke opposite side par door baitha hai, larger body ke peeche permanently chhupa hua, aur ek unstable saddle hai — pahunchna mushkil, pakde rakhna mushkil, aur communication ke liye nazar se ooha.

Edge cases

μ \to 0

hone par triangular points ka kya hota hai (test body negligible, ek mass dominate karti hai)?

Apex

x = \frac{1}{2} - μ \to \frac{1}{2}

ho jaata hai aur triangle almost central primary ke baare mein equilateral ban jaata hai; stability guaranteed hai kyunki

μ ≪ 0.0385

Exact threshold

μ = μ_{c} \approx 0.0385

par kya hota hai?

Discriminant

1 - 27 μ (1 - μ)

zero ho jaata hai, do oscillation frequencies merge ho jaati hain, aur stability marginal hoti hai — perturbations clearly grow bhi nahi karti aur clearly decay bhi nahi karti, toh point borderline hai.

Equal masses

M_{1} = M_{2}

(toh

μ = \frac{1}{2}

) ke liye, kya L4/L5 stable hain?

Nahi —

μ = \frac{1}{2}

μ_{c} \approx 0.0385

se kaafi zyada hai, isliye Coriolis force drift ko trap karne ke liye kaafi weak hai aur triangular points unstable ho jaate hain.

Limit mein

ω

kya hai jab do bodies bahut door hain (

D

large)?

Kepler's third law se

ω^{2} = G (M_{1} + M_{2}) / D^{3}

, toh

ω \to 0

; system kabhi dheeray spin karta hai, aur centrifugal term accordingly shrink karta hai, haalaanki paanchon points abhi bhi exist karte hain.

Agar test mass negligible nahi hoti (comparable to

M_{2}

), kya ye paanch points abhi bhi exact equilibria honge?

Nahi — restricted-three-body assumption (

m

kuch affect nahi karta) essential hai; ek heavy third body primaries ki orbit ko perturb karta aur clean five-point structure break down ho jaata.

Ek bade body ki location par effective potential kaisa dikhta hai (

r_{1} \to 0

r_{2} \to 0

Gravity term

- \frac{1 - μ}{r _{1}}

- \frac{μ}{r _{2}}

- \infty

tak diverge karta hai, infinitely deep well banaate hue; Lagrange points in wells ke saddles aur ridges par baithe hain, kabhi bottom par nahi.

Recall Ek-line summary jo tumhe recite karni chahiye

Paanch points, hamesha; teen collinear saddles (unstable, fuel chahiye), do triangular maxima (stable sirf tab jab $M_{1} / M_{2} > 24.96$ , Coriolis se bache hue); Coriolis stability set karta hai location nahi.

Related: Restricted Three-Body Problem · Effective Potential & Jacobi Constant · Coriolis and Centrifugal Forces (Rotating Frames) · Hill Sphere · Trojan Asteroids · Kepler's Third Law · Halo Orbits & Station-Keeping (JWST, SOHO) · Roche Limit