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

This page is a stress-test gallery. The parent note built the machinery: Kepler's equation $M=E-e\sin E$, the Newton–Raphson update, and the seeds. Here we throw every kind of input at it — small and large eccentricity, $M$ in every part of the circle, the degenerate circular orbit, the near-parabolic limit, a real satellite, and an exam trap.

If any symbol here feels unfamiliar, everything is defined in the parent and in Eccentric Anomaly and the Auxiliary Circle, Mean Motion and Orbital Period, and Newton-Raphson Method.

The figure below shows this: the teal curve $f(E)$ crosses zero exactly once (orange dot), and the plum dashed line is Newton's tangent at a guess — follow it down to the axis to read off the next, better guess.

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 $|E_k-e\sin E_k-M|<10^{-4}$

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The scenario matrix

Every Kepler solve falls into one of these cells. The examples below are labelled with the cell they cover.

Cell	What makes it special	Why it can bite you	Example
A low $e$, $M$ in quadrant I ($0<M<\frac{\pi }{2}$)	the "easy" baseline	none — sanity anchor	Ex 1
B moderate $e$, $M$ in quadrant II ($\frac{\pi }{2}<M<\pi$)	body near aphelion	seed can undershoot	Ex 2
C $M$ in quadrant III/IV ($M>\pi$)	body on the return half	sign of $\sin$, keep $E$ in $[0,2\pi )$	Ex 3
D $e=0$ (circle)	degenerate	Kepler collapses to $M=E$	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 vanishes, $E=M$	Ex 6
G negative / wrapped $M$	time before perihelion	must reduce $M$ modulo $2\pi$	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

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

Forecast: with $e$ tiny the correction $-e\sin E$ is small, so $E$ should sit just above $M$. Guess $E\approx 0.55$.

1. Seed. $E_0=M+e\sin M=0.5+0.1\sin (0.5)=0.5+0.1(0.4794)=0.54794$. Why this step? The seed pre-adds the correction so we start almost on the root — low $e$ means this seed is nearly the answer already.
2. Iteration 1. $f=0.54794-0.1\sin (0.54794)-0.5=0.54794-0.05210-0.5=-0.00416$. $f^{\prime }=1-0.1\cos (0.54794)=1-0.08542=0.91458$. $E_1=0.54794-\frac{-0.00416}{0.91458}=0.55249$. Why this step? $f<0$ means the guess was slightly too small, so Newton nudges $E$ up.
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: plug back: $E-e\sin E=0.55249-0.1(0.52493)=0.55249-0.05249=0.5000=M$. ✓ And $E>M$ as forecast (body slightly ahead of the uniform clock-hand).

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

Forecast: near aphelion the body has slowed, so the geometric angle $E$ should now lag less strongly — but still $E>M$ on this outbound half? Let's see; guess $E\approx 2.3$.

1. Seed. $E_0=M+e\sin M=2.0+0.4\sin (2.0)=2.0+0.4(0.90930)=2.36372$. Why this step? Same pre-correction seed; $\sin (2.0)$ is still positive (quadrant II), so we push $E$ up.
2. Iteration 1. $f=2.36372-0.4\sin (2.36372)-2.0=2.36372-0.4(0.70307)-2.0=0.08249$. $f^{\prime }=1-0.4\cos (2.36372)=1-0.4(-0.71112)=1.28445$. $E_1=2.36372-\frac{0.08249}{1.28445}=2.29949$. Why this step? $f>0$: guess was too big, Newton pulls $E$ back down.
3. Iteration 2 (tolerance check). $f=2.29949-0.4\sin (2.29949)-2.0=2.29949-0.4(0.74616)-2.0=0.00103$. $f^{\prime }=1-0.4\cos (2.29949)=1.26576$. $E_2=2.29949-\frac{0.00103}{1.26576}=2.29868$. Now $|f|<10^{-4}$ — stop.

Answer: $E\approx 2.29868$ rad.

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

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

Forecast: on the return half the correction flips sign, so this time $E$ should be less than $M$. Guess $E\approx 4.2$.

The figure contrasts the two halves: on the teal (outbound) point $\sin E>0$ so the correction pushes $E$ above $M$; on the plum (inbound) point $\sin E<0$ so it pulls $E$ below $M$. This example lives on the plum side.

1. Seed. $\sin (4.5)=-0.97753$, so $E_0=M+e\sin M=4.5+0.3(-0.97753)=4.20674$. Why this step? The negative $\sin$ automatically drags the seed below $M$ — no special-casing needed, the formula handles the quadrant.
2. Iteration 1. $f=4.20674-0.3\sin (4.20674)-4.5=4.20674-0.3(-0.87681)-4.5=-0.03022$. $f^{\prime }=1-0.3\cos (4.20674)=1-0.3(-0.48083)=1.14425$. $E_1=4.20674-\frac{-0.03022}{1.14425}=4.23315$. Why this step? $f<0$ → guess too small → step up.
3. Iteration 2. $f=4.23315-0.3\sin (4.23315)-4.5=4.23315-0.3(-0.88617)-4.5=-0.00100$. $f^{\prime }=1-0.3\cos (4.23315)=1.13711$. $E_2=4.23315-\frac{-0.00100}{1.13711}=4.23403$. Why this step? $|f|\approx 10^{-3}$ still exceeds $10^{-4}$, so one more step is warranted; it brings the residual under tolerance (see Verify).

Answer: $E\approx 4.23403$ rad.

Verify: $E-e\sin E=4.23403-0.3\sin (4.23403)=4.23403-0.3(-0.88566)=4.23403+0.26570=4.49973\approx 4.5$ (one more Newton step nails it to machine precision). ✓ And $E<M$ as forecast — the inbound correction is opposite in sign to the outbound one.

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

Forecast: a circle has its focus at the centre, so there is no "speed up / slow down". The clock-hand and the body coincide. Guess $E=M$ exactly.

1. Kepler collapses. With $e=0$: $M=E-0=E$. There is nothing to iterate. Why this step? The whole difficulty was the $-e\sin E$ term; set $e=0$ and it vanishes, leaving a trivially solvable linear equation.
2. Read off. $E=1.234$ rad. (The residual is exactly $0<10^{-4}$, so the tolerance is met immediately.)

Answer: $E=1.234$ rad.

Verify: $E-e\sin E=1.234-0=1.234=M$. ✓ Newton would also give this in one step since $f=E-M$ is already linear ($f^{\prime }=1$).

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

Forecast: near $e=1$ the derivative $f^{\prime }=1-e\cos E$ can shrink toward $0.05$ near perihelion, so Newton takes large first steps and converges slowly. Expect several iterations. Guess $E\approx 0.5$–$0.7$.

The figure shows why: for $e=0.95$ the curve $f(E)$ flattens into a shelf near perihelion (small slope), so Newton — which divides by that slope — leaps violently before settling. Compare the steep low-$e$ curve where each step is gentle.

1. Seed. $E_0=M+e\sin M=0.15+0.95\sin (0.15)=0.15+0.95(0.14944)=0.29197$. Why this step? Standard seed. It will under-shoot badly here because the true $E$ is far from $M$ when $e$ is huge.
2. Iter 1. $f=0.29197-0.95\sin (0.29197)-0.15=0.29197-0.95(0.28784)-0.15=-0.13148$. $f^{\prime }=1-0.95\cos (0.29197)=1-0.95(0.95768)=0.09020$. $E_1=0.29197-\frac{-0.13148}{0.09020}=1.74971$. Why the huge jump? The denominator $f^{\prime }\approx 0.09$ is tiny near perihelion — dividing by a small slope launches Newton far. This is the "violence" of high $e$.
3. Iter 2. $f=1.74971-0.95\sin (1.74971)-0.15=1.74971-0.95(0.98420)-0.15=0.66472$. $f^{\prime }=1-0.95\cos (1.74971)=1-0.95(-0.17706)=1.16821$. $E_2=1.74971-\frac{0.66472}{1.16821}=1.18058$. Why this step? $f>0$ now — the big jump overshot the root, so Newton reverses and steps back down. Overshoot-then-return is the signature of a flat-shelf function.
4. Iter 3. $f=1.18058-0.95\sin (1.18058)-0.15=1.18058-0.95(0.92446)-0.15=0.15234$. $f^{\prime }=1-0.95\cos (1.18058)=1-0.95(0.38129)=0.63778$. $E_3=1.18058-\frac{0.15234}{0.63778}=0.94168$. Why this step? Still $f>0$ (guess too big), so we keep descending; the slope has grown, so this step is smaller and more controlled than the wild first one.
5. Iter 4. $f=0.94168-0.95\sin (0.94168)-0.15=0.94168-0.95(0.80917)-0.15=0.02297$. $f^{\prime }=1-0.95\cos (0.94168)=1-0.95(0.58757)=0.44181$. $E_4=0.94168-\frac{0.02297}{0.44181}=0.88969$. Why this step? $f$ is shrinking toward zero and still positive, so Newton eases $E$ down a little more — we are now closing in.
6. Iter 5. $f=0.88969-0.95\sin (0.88969)-0.15=0.88969-0.95(0.77680)-0.15=0.00173$. $f^{\prime }=1-0.95\cos (0.88969)=1-0.95(0.62973)=0.40175$. $E_5=0.88969-\frac{0.00173}{0.40175}=0.88538$. Still $|f|>10^{-4}$.
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. The value has converged to $E\approx 0.88535$ rad. Why keep going through six steps? Because high $e$ genuinely slows convergence — each early step corrected only part of the error.

Answer: $E\approx 0.88535$ rad.

Verify: $E-e\sin E=0.88535-0.95\sin (0.88535)=0.88535-0.95(0.77403)=0.88535-0.73533=0.15002\approx 0.15$. ✓ The residual is within tolerance, and the six iterations confirm high $e$ is genuinely harder.

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

Forecast: at the apsides the body sits exactly on the major axis, where the "correction triangle" has zero height — so $E$ should equal $M$ with no iteration.

1. (a) $M=0$. Kepler: $0=E-0.6\sin E$. Try $E=0$: $0-0.6\sin 0=0$. ✓ Residual is exactly $0$. Root is $E=0$. Why this step? $\sin E=0$ kills the correction; $M=E$ exactly at perihelion.
2. (b) $M=\pi$. Try $E=\pi$: $\pi -0.6\sin \pi =\pi -0=\pi =M$. ✓ Residual is exactly $0$. Root is $E=\pi$. Why this step? $\sin \pi =0$ too — aphelion is the other place the correction vanishes.

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

Verify: (a) $0-0.6\sin 0=0$ ✓. (b) $\pi -0.6\sin \pi =\pi$ ✓. These are the only two exact fixed points of $M=E-e\sin E$ for any $e$ — the endpoints of the major axis.

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Ex 7 — Cell G: wrapped / negative $M$

Forecast: negative $M$ just means "before perihelion" = "on the inbound approach". Reduce it to the standard range first: $-1.0+2\pi =5.28319$. Since $M>\pi$, expect $E<M$ (inbound half, like Ex 3).

1. Reduce $M$. $M\to M+2\pi =-1.0+6.28319=5.28319$ rad. Why this step? Newton needs $M$ in $[0,2\pi )$ so that the returned $E$ is the physically-standard angle in the same revolution. Skipping this gives a valid-but-shifted $E$.
2. Seed. $\sin (5.28319)=-0.84147$, so $E_0=5.28319+0.2(-0.84147)=5.11490$.
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 "before perihelion").

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). ✓ And $E<M$, matching the inbound-half rule.

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

Forecast: first find the mean motion $n$, then $M=nt$, then solve Kepler. Since $t$ is a modest fraction of the period, $M$ should be well under $\pi$.

1. Mean motion. $a^3=(10^4{)}^3=10^{12}{\text{km}}^3$, so $\frac{\mu }{a^3}=\frac{398600}{10^{12}}=3.986\times 10^{-7}{\text{s}}^{-2}$. Then $n=\sqrt{3.986\times 10^{-7}}=6.31348\times 10^{-4}$ rad/s. Why this step? $M$ grows linearly at rate $n$; we can't get $M$ without it. Note $\mu /a^3$ is $10^{-7}$, not $10^{-10}$ — cube the full $10^4$, don't drop a factor. See Mean Motion and Orbital Period.
2. Mean anomaly. $M=n(t-t_p)=6.31348\times 10^{-4}\times 1800=1.13643$ rad. Why this step? Convert elapsed time into the uniformly-flowing angle. This is a fair chunk of the orbit — 30 min against a period $T=2\pi /n\approx 9952$ s $\approx 2.76$ h.
3. Solve Kepler ($e=0.25$). Seed $E_0=M+e\sin M=1.13643+0.25\sin (1.13643)=1.13643+0.25(0.90827)=1.36350$. Why this seed? Same pre-correction seed as everywhere; moderate $e$ means it lands close.
4. Iter 1. $f=1.36350-0.25\sin (1.36350)-1.13643=1.36350-0.25(0.97826)-1.13643=-0.01750$. $f^{\prime }=1-0.25\cos (1.36350)=1-0.25(0.20745)=0.94814$. $E_1=1.36350-\frac{-0.01750}{0.94814}=1.38196$. Why this step? $f<0$ → guess too small → step up.
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: $n$ in rad/s $\times$ s $=$ rad, everything dimensionally clean.

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

Forecast: $E$ is in quadrant II (between $\frac{\pi }{2}$ and $\pi$), and $\nu$ always lives in the same half-plane as $E$. So $\nu$ should also be in quadrant II, a little larger than $E$. See True Anomaly and the Orbit Equation.

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 $=1.52753\times 2.20390=3.36655$. Why this step? The half-angle form avoids the sign ambiguity of $\cos \nu$ — because $E/2$ and $\nu /2$ both sit in $(0,\frac{\pi }{2})$ for $E\in (0,\pi )$, their tangents are cleanly positive.
2. Recover $\nu$. $\frac{\nu }{2}=\arctan (3.36655)=1.28281$, so $\nu =2.56562$ rad. Why this step? Because $E<\pi$ we take the principal $\arctan$; $\nu /2$ stays in the first quadrant, no atan2 gymnastics needed. If $E>\pi$ we would add $2\pi$ to keep $\nu$ in the same revolution.
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. Why this step? $r=a(1-e\cos E)$ turns the geometric angle straight into distance — no need to go through $\nu$.

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

Verify (three ways):

• Quadrant: $\nu =2.566$ rad is in quadrant II, same half-plane as $E$, and $\nu >E$ — matches forecast. ✓
• Radius bounds: $r$ must lie in $[a(1-e),a(1+e)]=[4200,9800]$ km; $8862.5$ km sits inside, near the aphelion side. ✓
• 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 — agrees to rounding. ✓

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Recall Matrix self-test

Which cell has $\sin E=0$ at the solution? ::: Cell F (apsides $M=0$ or $M=\pi$) — the correction term vanishes and $E=M$ exactly Why does Cell E (high $e$) need many iterations? ::: $f^{\prime }=1-e\cos E$ shrinks near perihelion, so Newton takes violent, overshooting steps before settling What must you do to a wrapped/negative $M$ before iterating (Cell G)? ::: reduce it modulo $2\pi$ into $[0,2\pi )$ On the inbound half ($M>\pi$, Cell C), is $E$ bigger or smaller than $M$? ::: smaller — $\sin M<0$ flips the correction sign Which formula avoids the $\nu$ quadrant trap (Cell I)? ::: the half-angle $\tan \frac{\nu }{2}=\sqrt{\frac{1+e}{1-e}}\tan \frac{E}{2}$ When do we stop iterating on any example? ::: when the residual $|E-e\sin E-M|<10^{-4}$