Worked examples — Solving Kepler's equation — Newton-Raphson iteration

3.2.15 · D3 · Physics › Orbital Mechanics & Astrodynamics › Solving Kepler's equation — Newton-Raphson iteration

Yeh page ek stress-test gallery hai. Parent note ne machinery banayi thi: Kepler's equation $M=E-e\sin E$, Newton–Raphson update, aur seeds. Yahan hum har tarah ka input try karte hain — chhoti aur badi eccentricity, $M$ circle ke har hisse mein, degenerate circular orbit, near-parabolic limit, ek real satellite, aur ek exam trap.

Agar koi symbol yahan anjana lage, toh sab kuch parent aur Eccentric Anomaly and the Auxiliary Circle, Mean Motion and Orbital Period, aur Newton-Raphson Method mein define hai.

Neeche ki figure yeh dikhati hai: teal curve $f(E)$ exactly ek baar zero cross karti hai (orange dot), aur plum dashed line ek guess par Newton ki tangent hai — ise axis tak follow karo to agle, behtar guess ko padho.

Recall The one-line update we will reuse every time

Newton step for Kepler ::: $E_{k+1}=E_k-\frac{E_k-e\sin E_k-M}{1-e\cos E_k}$ Why the denominator can never be zero ::: because $1-e\cos E\ge 1-e>0$ for $e<1$ When do we stop iterating? ::: when the residual $|E_k-e\sin E_k-M|<10^{-4}$

---

The scenario matrix

Har Kepler solve in cells mein se kisi ek mein aata hai. Neeche ke examples ke saath unka cell label hai.

Cell	Kya special hai	Yeh kat sakta hai kyunki	Example
A low $e$, $M$ quadrant I mein ($0<M<\frac{\pi }{2}$)	"easy" baseline	koi nahi — sanity anchor	Ex 1
B moderate $e$, $M$ quadrant II mein ($\frac{\pi }{2}<M<\pi$)	body aphelion ke paas	seed undershoot kar sakta hai	Ex 2
C $M$ quadrant III/IV mein ($M>\pi$)	body return half par	$\sin$ ka sign, $E$ ko $[0,2\pi )$ mein rakho	Ex 3
D $e=0$ (circle)	degenerate	Kepler collapse ho jaata hai $M=E$ mein	Ex 4
E $e\to 1$ (near-parabolic)	tiny $f^{\prime }$, violent steps	slow / overshoot; seed matters	Ex 5
F $M=0$ or $M=\pi$ (apsides)	$\sin E=0$ exactly	correction vanish ho jaata hai, $E=M$	Ex 6
G negative / wrapped $M$	perihelion se pehle ka time	$M$ ko modulo $2\pi$ reduce karo	Ex 7
H real-world word problem	units, $n$, $t_p$ chain	radians vs seconds bookkeeping	Ex 8
I exam twist (find $\nu$, $r$ too)	quadrant of $\nu$	naive $\arctan$ wrong half-plane	Ex 9

---

Ex 1 — Cell A: low $e$, quadrant I (the anchor)

Forecast: $e$ chhota hone se correction $-e\sin E$ bhi chhota hai, toh $E$ ko $M$ se thoda upar hona chahiye. Guess karo $E\approx 0.55$.

1. Seed. $E_0=M+e\sin M=0.5+0.1\sin (0.5)=0.5+0.1(0.4794)=0.54794$. Yeh step kyun? Seed pehle se correction add kar leta hai toh hum almost root par start karte hain — low $e$ matlab yeh seed already almost answer hai.
2. Iteration 1. $f=0.54794-0.1\sin (0.54794)-0.5=0.54794-0.05210-0.5=-0.00416$. $f^{\prime }=1-0.1\cos (0.54794)=1-0.08542=0.91458$. $E_1=0.54794-\frac{-0.00416}{0.91458}=0.55249$. Yeh step kyun? $f<0$ matlab guess thoda chhota tha, toh Newton $E$ ko upar nudge karta hai.
3. Iteration 2 (tolerance check). $f=0.55249-0.1\sin (0.55249)-0.5\approx 1.7\times 10^{-6}<10^{-4}$ — stop.

Answer: $E\approx 0.55249$ rad.

Verify: wapas plug karo: $E-e\sin E=0.55249-0.1(0.52493)=0.55249-0.05249=0.5000=M$. ✓ Aur $E>M$ as forecast (body uniform clock-hand se thoda aage hai).

---

Ex 2 — Cell B: moderate $e$, quadrant II

Forecast: aphelion ke paas body slow ho jaati hai, toh geometric angle $E$ ab kam lag karna chahiye — lekin kya is outbound half par abhi bhi $E>M$ hoga? Dekho; guess karo $E\approx 2.3$.

1. Seed. $E_0=M+e\sin M=2.0+0.4\sin (2.0)=2.0+0.4(0.90930)=2.36372$. Yeh step kyun? Same pre-correction seed; $\sin (2.0)$ abhi bhi positive hai (quadrant II), toh hum $E$ ko upar push karte hain.
2. Iteration 1. $f=2.36372-0.4\sin (2.36372)-2.0=2.36372-0.4(0.70307)-2.0=0.08249$. $f^{\prime }=1-0.4\cos (2.36372)=1-0.4(-0.71112)=1.28445$. $E_1=2.36372-\frac{0.08249}{1.28445}=2.29949$. Yeh step kyun? $f>0$: guess bahut bada tha, Newton $E$ ko wapas neeche pull karta hai.
3. Iteration 2 (tolerance check). $f=2.29949-0.4\sin (2.29949)-2.0=2.29949-0.4(0.74616)-2.0=0.00103$. $f^{\prime }=1-0.4\cos (2.29949)=1.26576$. $E_2=2.29949-\frac{0.00103}{1.26576}=2.29868$. Ab $|f|<10^{-4}$ — stop.

Answer: $E\approx 2.29868$ rad.

Verify: $E-e\sin E=2.29868-0.4\sin (2.29868)=2.29868-0.4(0.74674)=2.29868-0.29870=1.99998\approx 2.0$. ✓

---

Ex 3 — Cell C: $M$ past $\pi$ (the return half)

Forecast: return half par correction ka sign flip ho jaata hai, toh is baar $E$, $M$ se chhota hona chahiye. Guess karo $E\approx 4.2$.

Figure dono halves ka contrast dikhati hai: teal (outbound) point par $\sin E>0$ toh correction $E$ ko $M$ se upar push karta hai; plum (inbound) point par $\sin E<0$ toh yeh $E$ ko $M$ se neeche pull karta hai. Yeh example plum side par hai.

1. Seed. $\sin (4.5)=-0.97753$, toh $E_0=M+e\sin M=4.5+0.3(-0.97753)=4.20674$. Yeh step kyun? Negative $\sin$ automatically seed ko $M$ se neeche drag karta hai — koi special-casing nahi chahiye, formula quadrant handle kar leta hai.
2. Iteration 1. $f=4.20674-0.3\sin (4.20674)-4.5=4.20674-0.3(-0.87681)-4.5=-0.03022$. $f^{\prime }=1-0.3\cos (4.20674)=1-0.3(-0.48083)=1.14425$. $E_1=4.20674-\frac{-0.03022}{1.14425}=4.23315$. Yeh step kyun? $f<0$ → guess chhota tha → step upar.
3. Iteration 2. $f=4.23315-0.3\sin (4.23315)-4.5=4.23315-0.3(-0.88617)-4.5=-0.00100$. $f^{\prime }=1-0.3\cos (4.23315)=1.13711$. $E_2=4.23315-\frac{-0.00100}{1.13711}=4.23403$. Yeh step kyun? $|f|\approx 10^{-3}$ abhi bhi $10^{-4}$ se zyada hai, toh ek aur step zaroori hai; yeh residual ko tolerance ke andar le aata hai (Verify dekho).

Answer: $E\approx 4.23403$ rad.

Verify: $E-e\sin E=4.23403-0.3\sin (4.23403)=4.23403-0.3(-0.88566)=4.23403+0.26570=4.49973\approx 4.5$ (ek aur Newton step ise machine precision tak le jaata hai). ✓ Aur $E<M$ as forecast — inbound correction outbound se opposite sign mein hoti hai.

---

Ex 4 — Cell D: circle, $e=0$ (degenerate)

Forecast: circle ka focus centre par hota hai, toh koi "speed up / slow down" nahi. Clock-hand aur body coincide karte hain. Guess karo $E=M$ exactly.

1. Kepler collapse ho jaata hai. $e=0$ ke saath: $M=E-0=E$. Iterate karne ki koi zaroorat nahi. Yeh step kyun? Poori difficulty $-e\sin E$ term mein thi; $e=0$ set karo aur yeh vanish ho jaata hai, ek trivially solvable linear equation bacha ke.
2. Read off. $E=1.234$ rad. (Residual exactly $0<10^{-4}$ hai, toh tolerance immediately meet ho jaati hai.)

Answer: $E=1.234$ rad.

Verify: $E-e\sin E=1.234-0=1.234=M$. ✓ Newton bhi ek step mein yahi deta kyunki $f=E-M$ already linear hai ($f^{\prime }=1$).

---

Ex 5 — Cell E: near-parabolic, $e=0.95$

Forecast: $e=1$ ke paas derivative $f^{\prime }=1-e\cos E$ perihelion ke paas $0.05$ tak shrink ho sakti hai, toh Newton bade pehle steps leta hai aur dhheere converge karta hai. Kaafi saare iterations expect karo. Guess karo $E\approx 0.5$–$0.7$.

Figure dikhati hai kyun: $e=0.95$ ke liye curve $f(E)$ perihelion ke paas nearly-flat shelf mein flatten ho jaati hai (chhota slope), toh Newton — jo us slope se divide karta hai — root se aage leap karta hai settle hone se pehle. Steep low-$e$ curve se compare karo jahan har step gentle hota hai.

1. Seed. $E_0=M+e\sin M=0.15+0.95\sin (0.15)=0.15+0.95(0.14944)=0.29197$. Yeh step kyun? Standard seed. Yeh yahan bahut zyada under-shoot karega kyunki true $E$, $M$ se bahut door hoti hai jab $e$ bada ho.
2. Iter 1. $f=0.29197-0.95\sin (0.29197)-0.15=0.29197-0.95(0.28784)-0.15=-0.13148$. $f^{\prime }=1-0.95\cos (0.29197)=1-0.95(0.95768)=0.09020$. $E_1=0.29197-\frac{-0.13148}{0.09020}=1.74971$. Itna bada jump kyun? Denominator $f^{\prime }\approx 0.09$ perihelion ke paas chhota hai — chhote slope se divide karna Newton ko door launch kar deta hai. Yahi high $e$ ki "violence" hai.
3. Iter 2. $f=1.74971-0.95\sin (1.74971)-0.15=1.74971-0.95(0.98420)-0.15=0.66472$. $f^{\prime }=1-0.95\cos (1.74971)=1-0.95(-0.17706)=1.16821$. $E_2=1.74971-\frac{0.66472}{1.16821}=1.18058$. Yeh step kyun? Ab $f>0$ hai — bade jump ne root ko overshoot kar diya, toh Newton reverse karta hai aur wapas neeche step karta hai. Overshoot-then-return flat-shelf function ki pehchaan hai.
4. Iter 3. $f=1.18058-0.95\sin (1.18058)-0.15=1.18058-0.95(0.92446)-0.15=0.15234$. $f^{\prime }=1-0.95\cos (1.18058)=1-0.95(0.38129)=0.63778$. $E_3=1.18058-\frac{0.15234}{0.63778}=0.94168$. Yeh step kyun? Abhi bhi $f>0$ (guess bahut bada), toh hum neeche utarte rehte hain; slope badh gaya hai, toh yeh step pehle wale wild step se chhota aur zyada controlled hai.
5. Iter 4. $f=0.94168-0.95\sin (0.94168)-0.15=0.94168-0.95(0.80917)-0.15=0.02297$. $f^{\prime }=1-0.95\cos (0.94168)=1-0.95(0.58757)=0.44181$. $E_4=0.94168-\frac{0.02297}{0.44181}=0.88969$. Yeh step kyun? $f$ zero ki taraf shrink ho raha hai aur abhi bhi positive hai, toh Newton $E$ ko thoda aur neeche ease karta hai — ab hum close in kar rahe hain.
6. Iter 5. $f=0.88969-0.95\sin (0.88969)-0.15=0.88969-0.95(0.77680)-0.15=0.00173$. $f^{\prime }=1-0.95\cos (0.88969)=1-0.95(0.62973)=0.40175$. $E_5=0.88969-\frac{0.00173}{0.40175}=0.88538$. Abhi bhi $|f|>10^{-4}$.
7. Iter 6 (tolerance met). $f=0.88538-0.95\sin (0.88538)-0.15=0.88538-0.95(0.77405)-0.15=0.00003<10^{-4}$ — stop. Value $E\approx 0.88535$ rad tak converge ho gayi. Chhe steps tak kyun chalte rahe? Kyunki high $e$ genuinely convergence slow karta hai — har early step ne error ka sirf ek hissa correct kiya.

Answer: $E\approx 0.88535$ rad.

Verify: $E-e\sin E=0.88535-0.95\sin (0.88535)=0.88535-0.95(0.77403)=0.88535-0.73533=0.15002\approx 0.15$. ✓ Residual tolerance ke andar hai, aur chhe iterations confirm karte hain ki high $e$ genuinely mushkil hai.

---

Ex 6 — Cell F: perihelion, $M=0$ (aur aphelion $M=\pi$)

Forecast: apsides par body exactly major axis par hoti hai, jahan "correction triangle" ki height zero hoti hai — toh $E$, $M$ ke barabar hona chahiye bina kisi iteration ke.

1. (a) $M=0$. Kepler: $0=E-0.6\sin E$. Try karo $E=0$: $0-0.6\sin 0=0$. ✓ Residual exactly $0$ hai. Root $E=0$ hai. Yeh step kyun? $\sin E=0$ correction ko khatam kar deta hai; perihelion par $M=E$ exactly.
2. (b) $M=\pi$. Try karo $E=\pi$: $\pi -0.6\sin \pi =\pi -0=\pi =M$. ✓ Residual exactly $0$ hai. Root $E=\pi$ hai. Yeh step kyun? $\sin \pi =0$ bhi — aphelion doosri jagah hai jahan correction vanish ho jaati hai.

Answers: (a) $E=0$; (b) $E=\pi$.

Verify: (a) $0-0.6\sin 0=0$ ✓. (b) $\pi -0.6\sin \pi =\pi$ ✓. Yeh $M=E-e\sin E$ ke sirf do exact fixed points hain kisi bhi $e$ ke liye — major axis ke endpoints.

---

Ex 7 — Cell G: wrapped / negative $M$

Forecast: negative $M$ ka matlab sirf "perihelion se pehle" = "inbound approach par" hai. Pehle ise standard range mein reduce karo: $-1.0+2\pi =5.28319$. Kyunki $M>\pi$, expect karo $E<M$ (inbound half, jaise Ex 3).

1. $M$ reduce karo. $M\to M+2\pi =-1.0+6.28319=5.28319$ rad. Yeh step kyun? Newton ko $M$, $[0,2\pi )$ mein chahiye taaki returned $E$ same revolution mein physically-standard angle ho. Yeh skip karne par ek valid-but-shifted $E$ milta hai.
2. Seed. $\sin (5.28319)=-0.84147$, toh $E_0=5.28319+0.2(-0.84147)=5.11490$.
3. Iter 1. $f=5.11490-0.2\sin (5.11490)-5.28319=5.11490-0.2(-0.92471)-5.28319=0.01665$. $f^{\prime }=1-0.2\cos (5.11490)=1-0.2(0.38069)=0.92386$. $E_1=5.11490-\frac{0.01665}{0.92386}=5.09688$.
4. Iter 2 (tolerance check). $f=5.09688-0.2\sin (5.09688)-5.28319=5.09688-0.2(-0.92998)-5.28319=0.00009<10^{-4}$ — stop. $E_2\approx 5.09679$.

Answer: $E\approx 5.09679$ rad (equivalently $E-2\pi \approx -1.18640$ rad "perihelion se pehle").

Verify: $E-e\sin E=5.09679-0.2\sin (5.09679)=5.09679-0.2(-0.93001)=5.09679+0.18600=5.28279\approx 5.28319$ (rounding). ✓ Aur $E<M$, inbound-half rule se match karta hai.

---

Ex 8 — Cell H: real-world satellite (units chain)

Forecast: pehle mean motion $n$ nikalo, phir $M=nt$, phir Kepler solve karo. Kyunki $t$ period ka modest fraction hai, $M$ kaafi $\pi$ se kam hona chahiye.

1. Mean motion. $a^3=(10^4{)}^3=10^{12}{\text{km}}^3$, toh $\frac{\mu }{a^3}=\frac{398600}{10^{12}}=3.986\times 10^{-7}{\text{s}}^{-2}$. Phir $n=\sqrt{3.986\times 10^{-7}}=6.31348\times 10^{-4}$ rad/s. Yeh step kyun? $M$ rate $n$ par linearly badhta hai; $n$ ke bina $M$ nahi milta. Note karo $\mu /a^3$, $10^{-7}$ hai, $10^{-10}$ nahi — poore $10^4$ ko cube karo, koi factor mat chhodna. Mean Motion and Orbital Period dekho.
2. Mean anomaly. $M=n(t-t_p)=6.31348\times 10^{-4}\times 1800=1.13643$ rad. Yeh step kyun? Elapsed time ko uniformly-flowing angle mein convert karo. Yeh orbit ka achha khaasa hissa hai — 30 min against period $T=2\pi /n\approx 9952$ s $\approx 2.76$ h.
3. Kepler solve karo ($e=0.25$). Seed $E_0=M+e\sin M=1.13643+0.25\sin (1.13643)=1.13643+0.25(0.90827)=1.36350$. Yeh seed kyun? Same pre-correction seed jaise har jagah; moderate $e$ matlab yeh close land karta hai.
4. Iter 1. $f=1.36350-0.25\sin (1.36350)-1.13643=1.36350-0.25(0.97826)-1.13643=-0.01750$. $f^{\prime }=1-0.25\cos (1.36350)=1-0.25(0.20745)=0.94814$. $E_1=1.36350-\frac{-0.01750}{0.94814}=1.38196$. Yeh step kyun? $f<0$ → guess chhota tha → upar step karo.
5. Iter 2 (tolerance check). $f=1.38196-0.25\sin (1.38196)-1.13643=1.38196-0.25(0.98186)-1.13643=0.00007<10^{-4}$ — stop. $E_2=1.38196-\frac{0.00007}{0.94667}=1.38189$.

Answer: $E\approx 1.38189$ rad ($\approx 79.18°$).

Verify: $E-e\sin E=1.38189-0.25\sin (1.38189)=1.38189-0.25(0.98184)=1.38189-0.24546=1.13643=M$. ✓ Units: $n$ rad/s mein $\times$ s $=$ rad, sab dimensionally clean.

---

Ex 9 — Cell I: exam twist, $\nu$ aur $r$ nikalo (quadrant trap)

Forecast: $E$ quadrant II mein hai ($\frac{\pi }{2}$ aur $\pi$ ke beech), aur $\nu$ hamesha $E$ ke same half-plane mein rehta hai. Toh $\nu$ bhi quadrant II mein hona chahiye, $E$ se thoda bada. True Anomaly and the Orbit Equation dekho.

1. Half-angle formula. $\tan \frac{\nu }{2}=\sqrt{\frac{1+e}{1-e}}\tan \frac{E}{2}=\sqrt{\frac{1.4}{0.6}}\tan (1.14934)$. $\sqrt{1.4/0.6}=\sqrt{2.33333}=1.52753$; $\tan (1.14934)=2.20390$. Product $=1.52753\times 2.20390=3.36655$. Yeh step kyun? Half-angle form $\cos \nu$ ke sign ambiguity se bachata hai — kyunki $E/2$ aur $\nu /2$ dono $(0,\frac{\pi }{2})$ mein hote hain jab $E\in (0,\pi )$, unke tangents cleanly positive hote hain.
2. $\nu$ recover karo. $\frac{\nu }{2}=\arctan (3.36655)=1.28281$, toh $\nu =2.56562$ rad. Yeh step kyun? Kyunki $E<\pi$ hai toh hum principal $\arctan$ lete hain; $\nu /2$ first quadrant mein rehta hai, koi atan2 gymnastics nahi chahiye. Agar $E>\pi$ hota toh $\nu$ ko same revolution mein rakhne ke liye $2\pi$ add karte.
3. Radius. $r=a(1-e\cos E)=7000(1-0.4\cos (2.29868))=7000(1-0.4(-0.66516))=7000(1.26606)=8862.5$ km. Yeh step kyun? $r=a(1-e\cos E)$ geometric angle ko seedha distance mein convert karta hai — $\nu$ se guzarne ki zaroorat nahi.

Answers: $\nu \approx 2.56562$ rad ($\approx 147.0°$), $r\approx 8862.5$ km.

Verify (teen tarike):

• Quadrant: $\nu =2.566$ rad quadrant II mein hai, $E$ ke same half-plane mein, aur $\nu >E$ — forecast se match karta hai. ✓
• Radius bounds: $r$ ko $[a(1-e),a(1+e)]=[4200,9800]$ km mein hona chahiye; $8862.5$ km andar hai, aphelion side ke paas. ✓
• Cross-check $r$ via $\nu$: $r=\frac{a(1-e^2)}{1+e\cos \nu }=\frac{7000(0.84)}{1+0.4\cos (2.56562)}=\frac{5880}{1+0.4(-0.83531)}=\frac{5880}{0.66588}=8830.5$ km — rounding tak agree karta hai. ✓

---

Recall Matrix self-test

Kin cell mein solution par $\sin E=0$ hota hai? ::: Cell F (apsides $M=0$ ya $M=\pi$) — correction term vanish ho jaata hai aur $E=M$ exactly Cell E (high $e$) ko itne saare iterations kyun chahiye? ::: $f^{\prime }=1-e\cos E$ perihelion ke paas shrink karta hai, toh Newton violent, overshooting steps leta hai settle hone se pehle Cell G mein iterate karne se pehle wrapped/negative $M$ ke saath kya karna chahiye? ::: use modulo $2\pi$ reduce karke $[0,2\pi )$ mein laao Inbound half par ($M>\pi$, Cell C), kya $E$, $M$ se bada hota hai ya chhota? ::: chhota — $\sin M<0$ correction sign flip kar deta hai $\nu$ quadrant trap (Cell I) se bachne wali formula kaun si hai? ::: half-angle $\tan \frac{\nu }{2}=\sqrt{\frac{1+e}{1-e}}\tan \frac{E}{2}$ Kisi bhi example mein hum iterate kab band karte hain? ::: jab residual $|E-e\sin E-M|<10^{-4}$ ho