Solving Kepler's equation — Newton-Raphson iteration

3.2.15 · Physics › Orbital Mechanics & Astrodynamics

YEH ZAROORI KYUN HAI?

Ek time $t$ diya hai, hum jaanna chahte hain ki satellite apni ellipse par kahan hai. Clean chain yeh hai:

$time t ⟶ M ⟶ E ⟶ ν ⟶ (r, θ)$

$M$ = mean anomaly — ek kalpnik angle jo time ke saath uniformly badhta hai.
$E$ = eccentric anomaly — auxiliary circle par geometric angle.
$ν$ = true anomaly — focus se body ka actual angle.

Takleef $M$ aur $E$ ke beech mein hai.

KEPLER'S EQUATION DERIVE KAISE HOTI HAI (scratch se)?

$M = E - e sin E$ sirf quote nahi karni, hume ise earn karna hoga. Yeh Kepler's 2nd law (equal areas in equal times) se aati hai.

Setup. Ellipse (semi-axes $a, b$ ) aur uska circumscribing auxiliary circle of radius $a$ banao. Eccentric anomaly $E$ woh angle hai jo centre par body ke upar circle ke point tak banta hai.

Step 1 — Swept area time ke saath scale hoti hai. 2nd law se, perihelion se swept area elapsed time ke proportional hai: $\frac{A _{ellipse sector}}{A _{ellipse total}} = \frac{t - t _{p}}{T} = \frac{M}{2 π} .$ Kyun? Equal areas / equal times ka matlab hai fraction of area = fraction of period, aur $M /2 π$ exactly elapsed period ka fraction hai (kyunki $M = n (t - t_{p})$ aur $n T = 2 π$ ).

Step 2 — Circle ko ellipse mein squash karo. Circle par har $y$ -coordinate $b / a$ se scale hota hai ellipse par land karne ke liye. Isliye circle ki har area $b / a$ se scale hoti hai: $A_{ellipse sector} = \frac{b}{a} A_{circle sector} .$

Step 3 — Circle sector ki area. Perihelion se swept circle "sector," focus se measure karke, centre ke baare mein ek pie-slice minus ek triangle mein split hoti hai: $A_{circle sector} = pie slice, angle E \frac{1}{2} a^{2} E - triangle to focus \frac{1}{2} a^{2} e sin E .$ Triangle kyun? Focus centre se $c = a e$ offset hai. Centre, focus aur circle point se bane triangle ki base $a e$ aur height $a sin E$ hai, area $\frac{1}{2} (a e) (a sin E)$ .

Step 4 — Assemble karo. Total ellipse area $= π ab$ . Plug in karo: $\frac{\frac{b}{a} ( \frac{1}{2} a ^{2} E - \frac{1}{2} a ^{2} e s i n E )}{π ab} = \frac{M}{2 π} .$ $a$ 's aur $b$ 's khoobsoorati se cancel ho jaate hain: $\frac{\frac{1}{2} ab ( E - e s i n E )}{π ab} = \frac{E - e s i n E}{2 π} = \frac{M}{2 π} \Rightarrow M = E - e sin E .$

NEWTON-RAPHSON KYUN? (80/20 method)

Hume $M = E - e sin E$ ko $E$ ke liye solve karna hai, $M, e$ diye hue hain. Koi closed form exist nahi karta. Root function define karo:

$f (E) = E - e sin E - M .$

Hum $f (E) = 0$ chahte hain. Newton-Raphson workhorse hai: fast (quadratic convergence), sirf $f$ aur $f^{'}$ chahiye, aur yahaan $f^{'}$ trivial hai.

$f^{'} = 1 - e cos E$ kyun? $f (E) = E - e sin E - M$ ko term by term differentiate karo: $\frac{d}{d E} E = 1$ , $\frac{d}{d E} (- e sin E) = - e cos E$ , $M$ constant hai. Kyunki $e < 1$ , $f^{'} > 0$ hamesha — function strictly increasing hai, isliye root unique hai aur Newton well-behaved hai.

Figure — Solving Kepler's equation — Newton-Raphson iteration

Worked Example 1 — moderate eccentricity

Diya hai $M = 0.8$ rad, $e = 0.3$ . $E$ nikaliye.

Seed: $E_{0} = M + e sin M = 0.8 + 0.3 (0.7174) = 1.0152$ . Yeh seed kyun? Yeh $- e sin E$ term ke liye pre-correct karta hai, hume root ke paas land karta hai taaki kam iterations lagein.

Iteration 1: $f = 1.0152 - 0.3 sin (1.0152) - 0.8 = 1.0152 - 0.2555 - 0.8 = - 0.0403$ $f^{'} = 1 - 0.3 cos (1.0152) = 1 - 0.1585 = 0.8415$ $E_{1} = 1.0152 - \frac{- 0.0403}{0.8415} = 1.0631$ Yeh step kyun? Hum $E$ ko upar move karte hain kyunki $f < 0$ ka matlab tha hamara guess bahut chhota tha.

Iteration 2: $f = 1.0631 - 0.3 sin (1.0631) - 0.8 = 1.0631 - 0.2626 - 0.8 = 0.0005$ $E_{2} = 1.0631 - \frac{0.0005}{0.8558} \approx 1.06266$

Pehle hi $∣ f ∣ \sim 1 0^{- 4}$ ; ek aur step machine precision deta hai. $E \approx 1.0627$ rad. Dhyan do $E > M$ — body uniform angle se aage nikal gayi hai, jaisa expect tha.

Worked Example 2 — high eccentricity (jahaan seeds matter karte hain)

Diya hai $M = 0.1$ rad, $e = 0.9$ .

Seed choice — kyun parwah karein? High $e$ ke liye, $E_{0} = M$ slowly converge ho sakta hai ya oscillate kar sakta hai. $E_{0} = π$ sirf tab use karo jab $M$ $π$ ke paas ho; yahaan $M$ chhota hai toh seed $E_{0} = M + e sin M = 0.1 + 0.9 (0.0998) = 0.1898$ .

Iter 1: $f = 0.1898 - 0.9 sin (0.1898) - 0.1 = 0.1898 - 0.1700 - 0.1 = - 0.0802$ $f^{'} = 1 - 0.9 cos (0.1898) = 1 - 0.8838 = 0.1162$ $E_{1} = 0.1898 - (- 0.0802/0.1162) = 0.8801$

Itna bada jump kyun? High $e$ ke paas, $f^{'}$ chhota hota hai (denominator $\approx 0.12$ ), isliye Newton bada step leta hai. Yehi reason hai ki high- $e$ orbits mein dhyan chahiye — correction violent hoti hai.

Iter 2: $f = 0.8801 - 0.9 sin (0.8801) - 0.1 = 0.8801 - 0.6909 - 0.1 = 0.0892$ $f^{'} = 1 - 0.9 cos (0.8801) = 1 - 0.5731 = 0.4269$ $E_{2} = 0.8801 - 0.0892/0.4269 = 0.6712$

Iter 3: $f = 0.6712 - 0.9 sin (0.6712) - 0.1 = 0.6712 - 0.5595 - 0.1 = 0.0117$ ; $f^{'} = 0.2914$ ; $E_{3} = 0.6310$ Continue karne par converge hota hai $E \approx 0.6249$ rad par. Dheera — high $e$ genuinely mushkil hai.

$E$ se true anomaly $ν$ tak

Jab $E$ mil jaaye: $tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2}, r = a (1 - e cos E) .$ $r = a (1 - e cos E)$ kyun? Geometry project karo: focus se body ki distance, eccentric anomaly ke through express karke, yeh compact form milti hai (aphelion par max $E = π$ : $r = a (1 + e)$ ; perihelion par min $E = 0$ : $r = a (1 - e)$ — check karo).

Common mistake Classic errors ko steel-man karna

1. "Hamesha $E_{0} = M$ use karo." Sahi lagta hai kyunki chhothe $e$ ke liye theek hai aur simple hai. Fix: $e ≳ 0.8$ ke liye yeh seed bahut dheere converge ho sakti hai ya pehla step overshoot kar sakta hai; $E_{0} = M + e sin M$ ya apside ke paas $E_{0} = π$ use karo.

2. " $M = E - e sin E$ degrees mein kaam karta hai." Sahi lagta hai — angles toh angles hain. Fix: equation sirf radians mein exact hai, kyunki $sin E$ ko $E$ mein hi add kiya jaata hai; degrees mix karne se balance toot jaata hai. Pehle convert karo.

3. " $f^{'} = 1 - e sin E$ ." Sahi lagta hai — tune $sin$ Kepler's equation se copy kiya. Fix: $- e sin E$ ka derivative $- e cos E$ hai, isliye $f^{'} = 1 - e cos E$ . Galat $f^{'}$ kabhi kabhi (sahi root par!) converge bhi ho jaata hai lekin quadratic speed barbad kar deta hai.

4. $ν$ aur $E$ dono same half-plane mein hain, yeh bhuulna. Naively $arctan$ use karne se $ν$ galat quadrant mein aa jaata hai. Fix: two-argument tangent (atan2) use karo ya half-angle sign track karo.

Recall Ise 12-saal ke bachche ko explain karo

Socho ek runner ek squashed circle jaisi racetrack par daudh raha hai. Ek ghadi ki sui (woh hai $M$ ) bilkul steady speed se ghoomti rehti hai. Lekin asli runner andar ke post ke paas speed up karta hai aur door slow down karta hai. Toh runner aur ghadi ki sui match nahi karte — aur koi magic formula nahi hai jo ghadi ke time se runner ki jagah pal bhar mein bata de. Toh hum guess karte hain, check karte hain kitna galat hai, slope use karke smarter guess banate hain, aur repeat karte hain. 3-4 guesses ke baad hum bilkul sahi hote hain. Woh guessing recipe hai Newton-Raphson!

Flashcards

Kepler's equation kya hai?

M = E - e sin E

, mean anomaly

M

, eccentric anomaly

E

, eccentricity

e

ko link karta hai.

Kepler's equation algebraically

E

ke liye solve kyun nahi ho sakti?

Yeh transcendental hai —

E

linearly bhi appear karta hai aur

sin E

ke andar bhi, isliye koi closed-form inverse exist nahi karta; iterate karna padta hai.

Kepler's equation ke liye Newton-Raphson update kya hai?

E_{k + 1} = E_{k} - \frac{E _{k} - e sin E _{k} - M}{1 - e cos E _{k}}

Yahaan use kiya gaya

f (E)

aur

f^{'} (E)

kya hai?

f (E) = E - e sin E - M

aur

f^{'} (E) = 1 - e cos E

Root unique kyun hai?

Kyunki

e < 1 \Rightarrow f^{'} (E) = 1 - e cos E > 0

har jagah, isliye

f

strictly increasing hai.

Moderate

e

ke liye achha initial guess?

E_{0} = M + e sin M

(first-order correction), ya chhothe

e

ke liye

E_{0} = M

Newton-Raphson root ke paas kaunsi convergence rate deta hai?

Quadratic — error har iteration mein roughly square ho jaati hai.

E

ν

kaise nikaalte hain?

tan (ν /2) = \frac{1 + e}{1 - e} tan (E /2)

E

ke terms mein focus se distance?

r = a (1 - e cos E)

; min

a (1 - e)

E = 0

par, max

a (1 + e)

E = π

par.

Kepler's equation radians kyun use karni chahiye?

Kyunki

E

aur

sin E

seedha add hote hain; equality sirf consistent radian measure ke saath hold karti hai.

Kepler's equation kis physical law se derive hoti hai?

Kepler's 2nd law (equal areas swept in equal times), auxiliary circle ke through.

Mean motion formula?

n = μ / a^{3}

, jisse

M = n (t - t_{p})

milta hai.

Connections

Kepler's Laws of Planetary Motion — 2nd law is equation ko janam deta hai.
Eccentric Anomaly and the Auxiliary Circle — $E$ ka geometric meaning.
True Anomaly and the Orbit Equation — $E$ solve karne ke baad ka agla step.
Newton-Raphson Method — general root-finding, convergence conditions.
Mean Motion and Orbital Period — jahaan se $n$ aur $M$ aate hain.
Fixed-Point Iteration for Kepler — simpler (slower) alternative $E_{k + 1} = M + e sin E_{k}$ .