Lagrange points L1–L5 — derivation, stability: Foundations

3.2.30 · D1 · Physics › Orbital Mechanics & Astrodynamics › Lagrange points L1–L5 — derivation, stability

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1. Mass — kitna "stuff" hai aur yeh kitna pull karta hai

Chhota number ek subscript hai — sirf ek naam ka tag, multiplication nahi. $M_{2}$ padho "mass number two", " $M$ times $2$ " nahi.

Hum $m$ (small letter) bhi use karte hain test mass ke liye — woh satellite ya dust grain jo itna tiny hai ki woh kisi ko pull nahi karta. Bade letters pull karte hain; small letter bas saath chalta hai.

2. Position, x-axis, aur $(x, y)$ — cheezein kahan hain

Hum do badi bodies ko horizontal line $y = 0$ (x-axis) ke saath line up karte hain. Sab kuch isi ek flat plane mein hota hai, isliye do numbers kaafi hain.

Figure — Lagrange points L1–L5 — derivation, stability

Figure s01 kaise padhein: baayein taraf bada lavender dot $M_{1}$ hai; daayein taraf chhota coral dot $M_{2}$ hai; upar aur daayein taraf chhota mint dot hamara test mass $m$ hai kisi $(x, y)$ par. Dashed lines ke crossing par butter dot origin $(0, 0)$ hai. Notice karo ki mint dot axis se off hai (uska $y \neq = 0$ ) — yahi off-axis position hai jo hume baad mein L4/L5 ke liye chahiye. Neeche ke do double-headed arrows dikhate hain ki $M_{1}$ origin se sirf ek chhoti distance $μ$ par hai, jabki $M_{2}$ ek badi distance $(1 - μ)$ par hai.

3. Barycenter aur mass ratio $μ$

Do bodies ek doosre ke around orbit nahi karte jaise ek nailed down ho. Dono ek shared pivot ke around circle karte hain — barycenter (centre of mass). Do bachche see-saw par imagine karo: bhaari wala middle ke paas baithta hai, halka wala door, taaki balance ho.

Pehle humein names chahiye ki har body kahan hai. $x_{1}$ ko body $M_{1}$ ka $x$ -coordinate hone do aur $x_{2}$ ko body $M_{2}$ ka $x$ -coordinate (dono axis par hain, isliye unka $y = 0$ ).

Pair kitna "lopsided" hai yeh ONE number se describe karne ke liye, mass ratio define karo:

Parent $M_{1}$ ko $x_{1} = - μ$ par aur $M_{2}$ ko $x_{2} = + (1 - μ)$ par rakhta hai. Scaled masses $M_{1} = 1 - μ$ , $M_{2} = μ$ ke saath balance rule check karo: $(1 - μ) (- μ) + μ (1 - μ) = 0$ ✓. Figure s01 mein bhaari body origin se chhoti distance $μ$ baayein baithti hai, halki body badi distance $(1 - μ)$ daayein — exactly see-saw rule.

4. Distance aur $r_{1}, r_{2}$ ka square-root formula

Hume baar baar chahiye hoga "test mass har badi body se kitni door hai?" Pehle un do distances ke naam rakh lo:

Ab hum coordinates se straight-line distance kaise compute karein? Pehle coordinate differences define karo:

Figure s02 kaise padhein: lavender dot $M_{1}$ hai, mint dot test mass $m$ hai. Coral horizontal leg $Δ x = x + μ$ hai (test mass ka $x$ minus $M_{1}$ ka $x = - μ$ ), butter vertical leg $Δ y = y$ hai, aur slate slanted line distance $r_{1}$ khud hai — hypotenuse. Corner par chhota square right angle mark karta hai jo Pythagoras apply hone deta hai.

Toh do distances jo parent likhta hai woh sirf $m$ se har body tak Pythagoras hain: $r_{1} = (x + μ)^{2} + y^{2} (to M_{1}), r_{2} = (x - (1 - μ))^{2} + y^{2} (to M_{2}) .$ $(x + μ)$ hai "kitna right $m$ hai $M_{1}$ se", kyunki $M_{1}$ $x_{1} = - μ$ par baitha hai aur negative subtract karne se add ho jaata hai.

5. Gravity — pull, aur yeh $1/ r$ ya $1/ r^{2}$ se kyun fade hoti hai

Parent force ki jagah potential energy use karta hai. Energy "stored pull" hai:

6. Rotating frames, velocity $v$ , aur centrifugal push

Hum spinning pair ke saath ride karna choose karte hain — ek rotating reference frame. Is frame mein do badi bodies move karna band kar deti hain (frozen ho jaati hain), yahi poora trick hai jo "parking spots" possible banata hai.

Figure s03 kaise padhein: centre par slate "×" spin axis hai; lavender aur coral dots $M_{1}$ aur $M_{2}$ hain, frozen kyunki hum unke saath spin kar rahe hain. Mint arrows sab axis se seedhe outward point karte hain, aur woh lambe hote jaate hain jitna door se start hon — yeh $ω^{2} ρ$ ka $ρ$ ke saath grow karna picture karta hai. Faint grey curl tumhe remind karta hai ki poora frame rate $ω$ se turn kar raha hai. Yeh outward push teesri force hai jo, dono gravities ke saath milkar, ek Lagrange point par cancel honi chahiye.

Ye dono forces properly build hain Coriolis and Centrifugal Forces (Rotating Frames) mein; poora spinning-frame setup Restricted Three-Body Problem hai.

7. Kepler's Third Law → $ω = 1$ milna

Parent units choose karta hai jahan $D = 1$ aur $G (M_{1} + M_{2}) = 1$ , toh $ω^{2} = 1$ , yani $ω = 1$ . Yeh sirf rulers aur clocks ki ek smart choice hai taaki equations se clutter hat sake — koi physics change nahi hoti. Dekho Kepler's Third Law.

8. Gradient $\nabla$ aur effective potential $Ω$

9. Answers padhna: quintic, saddle, equilateral triangle, aur stability

Intuition L4/L5 stable kyun ho sakte hain even though woh hilltops hain (aur mass-ratio condition)

$Ω$ ke hilltop par baithna unstable lagta hai — ek nudge $m$ ko roll off kar dena chahiye. Lekin jab $m$ drift karna start karta hai toh woh moving hai, isliye Coriolis force ( $- 2 ω \times v$ ) kick in karta hai aur uska path ek tiny loop mein curl kar deta hai, use point ke paas trap kar ke. Yeh whirlpool rescue tabhi kaam karta hai jab spin dominant enough ho, jo tab hota hai jab do bodies kaafi lopsided hon. Classical result hai: $L4/L5 stable ⟺ μ < μ_{crit} \approx 0.0385 (equivalently \frac{M _{1}}{M _{2}} > 24.96) .$ Sun–Jupiter ( $M_{1} / M_{2} \approx 1047$ ) aur Earth–Moon ( $\approx 81$ ) easily pass karte hain, isliye real Trojan Asteroids billions of years tak survive karte hain. Full linear-stability algebra parent note mein hai.

L1/L2 ke liye use ki gayi near-body scale $r \approx D (μ /3)^{1/3}$ Hill Sphere radius hai — ek body ka gravitational control ka zone.

Prerequisite map

Equipment checklist

Right side cover karo aur khud test karo. Agar tum sab answer kar sako, tum parent note ke liye ready ho.

M_{2}

mein subscript ka matlab kya hai?

Ek naam ka tag ("mass number two"), kabhi multiplication nahi.

Mass ratio

μ

words mein kya hai?

Total mass ka woh fraction jo chhoti body mein hai,

M_{2} / (M_{1} + M_{2})

Barycenter equation mein

x_{1}

aur

x_{2}

kya represent karte hain?

Axis par bodies

M_{1}

aur

M_{2}

x

-coordinates.

Barycenter do masses ke relative kahan hota hai?

Bhaari wale ke paas, halke wale se door (see-saw balance).

Δ x

aur

Δ y

kya hain?

Do points ke beech horizontal aur vertical coordinate differences,

x - x^{'}

aur

y - y^{'}

r_{1}

aur

r_{2}

kya measure karte hain?

Test mass

m

M_{1}

aur

M_{2}

tak ki distances respectively.

r = Δ x^{2} + Δ y^{2}

kyun?

Pythagoras: yeh right triangle ka hypotenuse hai jiske legs

Δ x

aur

Δ y

hain.

Gravity

1/ r^{2}

se kyun fade hoti hai?

Pull ek sphere ki surface par spread hota hai jiska area

r^{2}

se grow karta hai, isliye yeh

1/ r^{2}

se thin hota hai.

Potential energy negative kyun hoti hai?

Gravity attractive hai; bound positions ek well mein baithte hain, deeper = zyada negative.

Rotating frame mein

v

kya hai, aur yeh zero kab hota hai?

Frame ke saath spin karne wale kisi ko dikhti

m

ki velocity; zero jab

m

Lagrange point par still baitha ho.

Centrifugal force kya hai aur kitna bada hota hai?

Spinning frame mein feel hone wala outward shove, axis se

ρ

distance par size

ω^{2} ρ

Centrifugal potential

- \frac{1}{2} ω^{2} ρ^{2}

mein

\frac{1}{2}

kahan se aata hai?

Linear force

ω^{2} ρ

integrate karne se —

ρ^{2}

ka derivative

2 ρ

hai, isliye potential

\frac{1}{2}

carry karta hai.

Coriolis force zero kab hota hai?

Jab particle rotating frame mein at rest ho (

v = 0

- G M_{1} / r_{1}

kyun

- (1 - μ) / r_{1}

ban jaata hai?

M_{1} = (1 - μ) (M_{1} + M_{2})

likho aur unit choice

G (M_{1} + M_{2}) = 1

use karo, toh prefactor

1

ban jaata hai.

\nablaΩ = 0

ka physically kya matlab hai?

Wahan landscape flat hai — koi net force nahi — ek Lagrange point.

Saddle aur maximum mein kya fark hai?

Saddle = ek taraf downhill, doosri taraf uphill (L1,L2,L3); maximum = har taraf downhill (L4,L5).

L4/L5 par

m

M_{1}

M_{2}

kaisi shape banate hain?

Ek equilateral triangle (

r_{1} = r_{2} = D

L4/L5 stable hone ki kya condition hai?

Jab

μ < μ_{crit} \approx 0.0385

, yani

M_{1} / M_{2} > 24.96

Kepler's Third Law

ω = 1

kaise deta hai?

ω^{2} = G (M_{1} + M_{2}) / D^{3}

, aur

G (M_{1} + M_{2}) = 1

D = 1

units choose karne par

ω = 1

milta hai.