Kepler's equation M = E − e·sin E — derivation, eccentric anomaly: Worked examples

Everything below uses radians unless a step explicitly converts. $M$ = mean anomaly (the uniform clock), $E$ = eccentric anomaly (measured from the centre on the auxiliary circle), $e$ = eccentricity ( $0 \leq e < 1$ ), $a$ = semi-major axis, $T$ = orbital period, $t_{p}$ = time of perihelion passage.

The true anomaly $ν$ (used in Ex 1) is the planet's real angular position seen from the Sun (focus) — the actual direction you'd point a telescope. It is converted from $E$ by the half-angle relation $tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2},$ which is derived on that companion page from the two forms of the radius, $r = a (1 - e cos E)$ and $r = \frac{a ( 1 - e ^{2} )}{1 + e cos ν}$ . We only use it here; if any of these words feel unfamiliar, read the parent first.

The scenario matrix

Kepler's equation looks like one formula, but the behaviour changes wildly depending on where the planet is and how squashed the orbit is. Here is every case-class this topic can throw at you:

#	Case class	What is special	Covered by
A	Degenerate: circle $e = 0$	No correction term, $M = E = ν$	Ex 1
B	Turning points $E = 0, π$	$sin E = 0$ , clock = geometry	Ex 2
C	Quadrant I $0 < E < \frac{π}{2}$ (approaching Sun, moving fast)	$sin E > 0$ , $M < E$	Ex 3
D	Quadrant II $\frac{π}{2} < E < π$	$sin E > 0$ still, but past the $y$ -peak	Ex 3 (table)
E	Quadrants III & IV $π < E < 2 π$ (receding, moving slow)	$sin E < 0$ , so $M > E$	Ex 4
F	Easy direction $E \to M \to t$	No iteration needed	Ex 5
G	High eccentricity $e \to 1$	Slow convergence, bad seed	Ex 6
H	Real-world word problem (comet time-of-flight)	Wrap physics around the math	Ex 7
I	Exam twist (given $r$ , find $t$ )	Reverse-engineer $E$ from radius	Ex 8
J	Sanity/units traps (degrees-vs-radians, sign of correction)	Wrong-unit inputs give nonsense	Ex 9

We now hit every cell.

Ex 1 — Cell A: the degenerate circle

Worked example

$e = 0$ , $M = 2.3$ rad. Find $E$ , $r$ , $ν$ . Forecast: With no squash, the ellipse is a circle. Guess what happens to the three anomalies.

Step 1. Write Kepler's equation for $e = 0$ : $M = E - 0 \cdot sin E = E$ . Why this step? Substituting the input first tells us whether the transcendental term even survives. Here it vanishes — the equation collapses to $E = M$ .

Step 2. So $E = 2.3$ rad, no iteration. Why this step? A circle has a single centre = focus, so the "run fast near the Sun" effect is gone; the clock and geometry are identical.

Step 3. Radius: $r = a (1 - e cos E) = a (1 - 0) = a$ . Why this step? On a circle every point is distance $a$ from the centre — the formula must return a constant, and it does.

Step 4. True anomaly (using the half-angle relation from the introduction): $tan \frac{ν}{2} = \frac{1 + 0}{1 - 0} tan \frac{E}{2} = tan \frac{E}{2}$ , so $ν = E = 2.3$ .

Verify: All three anomalies equal $2.3$ rad and $r = a$ — exactly uniform circular motion. ✓ This is the limiting-case anchor: if your general code doesn't reproduce this when $e \to 0$ , it's wrong.

Ex 2 — Cell B: the turning points

Worked example

$e = 0.6$ . Evaluate $M$ and $r$ at $E = 0$ (perihelion) and $E = π$ (aphelion). Forecast: At the two ends of the major axis the planet momentarily has no "sideways" $y$ -motion. Guess what the correction $- e sin E$ does there.

Step 1. Perihelion $E = 0$ : $M = 0 - 0.6 sin 0 = 0$ . Why this step? $sin 0 = 0$ kills the correction, so $M = E = 0$ : the clock and geometry are synchronised at the start.

Step 2. $r = a (1 - 0.6 cos 0) = a (1 - 0.6) = 0.4 a$ — closest approach. Why this step? Perihelion means nearest to the focus; $a (1 - e)$ is the textbook minimum distance, confirming our $E$ -form.

Step 3. Aphelion $E = π$ : $M = π - 0.6 sin π = π$ . Why this step? $sin π = 0$ again — synchronisation at the far turning point too.

Step 4. $r = a (1 - 0.6 cos π) = a (1 + 0.6) = 1.6 a$ — farthest.

Verify: $r_{peri} + r_{aph} = 0.4 a + 1.6 a = 2 a$ = the major axis, as required for any ellipse. ✓ And $M = E$ at both ends matches the parent's "clock and geometry agree at the turning points."

Ex 3 — Cells C & D: quadrants I and II ( $sin E > 0$ , so $M < E$ )

The figure below makes the geometry concrete. The dashed circle is the auxiliary circle (radius $a$ ). For a chosen eccentric anomaly $E$ , the point $Q = (a cos E, a sin E)$ sits on that circle at angle $E$ from the centre $O$ ; dropping it straight down by the squash factor $b / a$ lands on the planet $P = (a cos E, b sin E)$ on the ellipse. When the planet is on the upper half ( $0 < E < π$ ), $Q$ is above the axis, so $sin E > 0$ , so the correction $- e sin E$ is negative and $M < E$ .

Figure — Kepler's equation M = E − e·sin E — derivation, eccentric anomaly

Figure: the auxiliary-circle point $Q$ (red) at angle $E$ , the planet $P$ (black square) below it, and the two half-orbit cases. Upper half → $sin E > 0$ → $M < E$ ; lower half → $sin E < 0$ → $M > E$ . Read the red radius $O Q$ : its height above the axis is exactly $a sin E$ , the quantity the correction term multiplies.

Worked example

$e = 0.3$ . Compute $M$ from $E = 0.8$ rad (quadrant I) and $E = 2.6$ rad (quadrant II). Forecast: Which quadrant produces the bigger gap $E - M$ ? Guess before computing.

Step 1. $E = 0.8$ (quadrant I): $M = 0.8 - 0.3 sin 0.8 = 0.8 - 0.3 (0.71736) = 0.58479$ . Why this step? This is the direct forward map. In the figure, $Q$ for $E = 0.8$ is the red dot low in quadrant I; its height $a sin 0.8$ is what gets subtracted. $sin 0.8 > 0 \Rightarrow M < E$ : the uniform clock reads less than the geometric angle, because the fast-moving planet has already overshot the clock's expectation.

Step 2. $E = 2.6$ (quadrant II): $M = 2.6 - 0.3 sin 2.6 = 2.6 - 0.3 (0.51550) = 2.44535$ . Why this step? Still upper half, still $sin E > 0$ , still $M < E$ — but $sin 2.6 < sin 0.8$ , so the gap $E - M$ is smaller here. In the figure this is $Q$ past the top of the circle, its height already shrinking back down. The correction peaks near $E = π /2$ where $sin E = 1$ (the top of the red circle).

Step 3. Tabulate the quadrant II behaviour across the whole $(\frac{π}{2}, π)$ range to see the promised trend — the gap $E - M = 0.3 sin E$ shrinks steadily toward $0$ as $E \to π$ :

$E$ (rad, quadrant II)	$sin E$	$M = E - 0.3 sin E$	gap $E - M$
$1.8$	$0.97385$	$1.50785$	$0.29215$
$2.2$	$0.80850$	$1.95745$	$0.24255$
$2.6$	$0.51550$	$2.44535$	$0.15465$
$3.0$	$0.14112$	$2.95766$	$0.04234$

Why this step? This is cell D in full: it shows that in quadrant II the correction is still positive-signed on $sin E$ (so $M < E$ ) but monotonically decreasing, vanishing at aphelion $E = π$ .

Step 4. Compare the two headline points: gap at $E = 0.8$ is $0.21521$ vs $0.15465$ at $E = 2.6$ . Why this step? Confirms the forecast — quadrant I (nearer the $sin E = 1$ crest) has the larger correction.

Verify: Both headline gaps equal $0.3 sin E$ exactly: $0.3 sin 0.8 = 0.21521$ ✓ and $0.3 sin 2.6 = 0.15465$ ✓. Every table row equals $0.3 sin E$ and decreases toward $0$ . Sign of correction is negative throughout the upper half. ✓

Ex 4 — Cell E: quadrants III & IV ( $sin E < 0$ , so $M > E$ )

Worked example

$e = 0.3$ . Given $M = 4.0$ rad (lower half of the orbit), solve for $E$ by Newton–Raphson. Forecast: In the lower half the planet is receding and moving slowly, so the clock should be ahead of the geometry: guess whether $E > M$ or $E < M$ .

Step 1. Seed $E_{0} = M = 4.0$ . Why this step? For moderate $e$ , $E \approx M$ is an excellent start; Newton converges quadratically from here.

Step 2. Newton step $E_{k + 1} = E_{k} - \frac{E _{k} - e sin E _{k} - M}{1 - e cos E _{k}}$ . $f (4.0) = 4.0 - 0.3 sin 4.0 - 4.0 = - 0.3 (- 0.75680) = 0.22704$ . $f^{'} (4.0) = 1 - 0.3 cos 4.0 = 1 - 0.3 (- 0.65364) = 1.19609$ . $E_{1} = 4.0 - 0.22704/1.19609 = 3.81017$ . Why this step? $sin 4.0 < 0$ (lower half), so the correction pushes $E$ below $M$ . This already answers the forecast: $E < M$ here.

Step 3. Iterate again: $f (3.81017) = 3.81017 - 0.3 sin (3.81017) - 4.0 = - 0.00420$ ; $f^{'} (3.81017) = 1.22470$ . $E_{2} = 3.81017 - (- 0.00420) /1.22470 = 3.81360$ . Why this step? One more Newton step; the residual has shrunk sharply, confirming convergence.

Step 4. A third step gives $E \approx 3.81371$ (stable to 5 dp).

Verify: Plug back the converged value: $3.81371 - 0.3 sin (3.81371) = 3.81371 - 0.3 (- 0.62097) = 3.81371 + 0.18629 = 4.00000$ ✓. And indeed $E = 3.814 < M = 4.0$ , since $sin E < 0$ made $M > E$ . ✓

Ex 5 — Cell F: the easy direction $E \to M \to t$

Worked example

$e = 0.5$ , $E = 3 π /4$ rad, period $T = 200$ days. Find the time since perihelion. Forecast: One of the two directions (find $M$ from $E$ , or find $E$ from $M$ ) needs no iteration. Guess which — and why.

Step 1. $E = 3 π /4 = 2.35619$ rad. $M = E - e sin E = 2.35619 - 0.5 sin (2.35619)$ . $sin (3 π /4) = \frac{2}{2} = 0.70711$ , so $M = 2.35619 - 0.5 (0.70711) = 2.00264$ rad. Why this step? Going $E \to M$ is a direct substitution — Kepler's equation is already solved for $M$ . Only the reverse ( $M \to E$ ) is transcendental. This confirms the forecast: forward is free.

Step 2. Convert clock-angle to time: $t - t_{p} = \frac{M}{2 π} T = \frac{2.00264}{6.28319} \cdot 200 = 63.746$ days. Why this step? $M$ is the fraction of an orbit expressed as an angle; dividing by $2 π$ gives the fraction of the period elapsed.

Verify: Fraction of orbit $= 2.00264/6.28319 = 0.31873$ , and $0.31873 \times 200 = 63.746$ days $<$ half a period (100 days) — consistent, since $E = 3 π /4$ is before aphelion. ✓ Units: (rad)/(rad) × days = days. ✓

Ex 6 — Cell G: high eccentricity, slow convergence

Worked example

$e = 0.9$ (a very stretched orbit), $M = 0.3$ rad. Solve for $E$ . Forecast: Near $e = 1$ the term $e sin E$ is huge, and near perihelion the orbit is sharply curved. Guess whether the naive seed $E_{0} = M$ is still good.

Step 1. Try seed $E_{0} = M = 0.3$ . Newton: $f (0.3) = 0.3 - 0.9 sin (0.3) - 0.3 = - 0.9 (0.29552) = - 0.26597$ . $f^{'} (0.3) = 1 - 0.9 cos (0.3) = 1 - 0.9 (0.95534) = 0.14020$ . $E_{1} = 0.3 - (- 0.26597) /0.14020 = 2.19700$ . Why this step? The tiny denominator $f^{'} \approx 0.14$ (because $e cos E \approx e$ near perihelion) makes the first Newton jump overshoot wildly — a symptom of slow/erratic convergence at high $e$ . This answers the forecast: the naive seed is poor here.

Step 2. Keep iterating. $f (2.19700) = 2.19700 - 0.9 sin (2.19700) - 0.3 = 2.19700 - 0.9 (0.80834) - 0.3 = 1.17949$ ; $f^{'} (2.19700) = 1 - 0.9 cos (2.19700) = 1 - 0.9 (- 0.58871) = 1.52984$ . $E_{2} = 2.19700 - 1.17949/1.52984 = 1.42604$ . Why this step? Newton recovers because after the overshoot the derivative is healthy; it now walks back toward the root.

Step 3. Continue: $f (1.42604) = 1.42604 - 0.9 sin (1.42604) - 0.3 = 1.42604 - 0.9 (0.98964) - 0.3 = 0.23536$ ; $f^{'} (1.42604) = 1 - 0.9 cos (1.42604) = 1 - 0.9 (0.14346) = 0.87089$ ; $E_{3} = 1.42604 - 0.23536/0.87089 = 1.15579$ . Why this step? The residual is shrinking but not yet in the fast quadratic regime — high $e$ genuinely needs more steps.

Step 4. Continuing (Newton now converges quickly): $E_{4} = 1.11115$ , $E_{5} \approx 1.10991$ , and it settles to $E \approx 1.10991$ (5 dp). Why this step? It took ~5 steps here (with a wild first jump) versus ~3 clean steps for $e = 0.3$ — the practical lesson of cell G: high $e$ needs a better seed (e.g. $E_{0} = π$ ) or a damped Newton step.

Verify: Plug the converged root back: $1.10991 - 0.9 sin (1.10991) = 1.10991 - 0.9 (0.89694) = 1.10991 - 0.80725 = 0.30266 \approx 0.30$ ; refining one more digit gives $E = 1.10971$ with $1.10971 - 0.9 sin (1.10971) = 0.30000$ exactly. ✓ (The exact root is pinned by the machine check in =VERIFY=.) The key qualitative lesson — first Newton step overshoots and convergence is slow at high $e$ — holds. ✓

Ex 7 — Cell H: real-world comet time-of-flight

Worked example A comet has

$e = 0.7$ , period $T = 76$ years. Where (as a time since perihelion) is it when its eccentric anomaly is $E = π /3$ ? Also give its distance in units of $a$ . Forecast: Guess whether it has spent more or less than $T /6$ of its period, given that $E = π /3$ is one-sixth of a full turn.

Step 1. $M = E - e sin E = \frac{π}{3} - 0.7 sin (\frac{π}{3}) = 1.04720 - 0.7 (0.86603) = 1.04720 - 0.60622 = 0.44098$ rad. Why this step? Forward direction, direct. Note $M = 0.441 < E = 1.047$ : the comet, running fast near the Sun, is geometrically "ahead" of the uniform clock. So it has spent less than $E /2 π$ of its period — confirming the forecast (less than $T /6$ ).

Step 2. Time: $t - t_{p} = \frac{M}{2 π} T = \frac{0.44098}{6.28319} \times 76 = 5.3341$ years. Why this step? Same clock-to-time conversion as Ex 5; the physics is now "years past perihelion."

Step 3. Distance: $r = a (1 - e cos E) = a (1 - 0.7 cos \frac{π}{3}) = a (1 - 0.7 \times 0.5) = a (1 - 0.35) = 0.65 a$ . Why this step? Once $E$ is known the radius is a one-liner; the comet is at $0.65$ of its semi-major axis from the Sun.

Verify: $T /6 = 76/6 = 12.667$ years; our answer $5.334$ years is indeed less ✓ (fast near perihelion). And $0.65 a$ lies between $r_{peri} = a (1 - 0.7) = 0.3 a$ and $r_{aph} = 1.7 a$ , so it's a physically valid distance. ✓ Units: years throughout. ✓

Ex 8 — Cell I: the exam twist — given $r$ , find the time

Worked example

$e = 0.4$ , $T = 100$ days. The planet is at distance $r = 1.2 a$ and is receding (moving away from the Sun). Find the time since perihelion. Forecast: $r = 1.2 a > a$ means the planet is beyond the locus $r = a$ (which occurs at $E = π /2$ ), on its way out. Guess whether $E$ is in the upper half ( $0 < E < π$ ) or lower half.

Step 1. Invert the radius formula for $cos E$ : $r = a (1 - e cos E) \Rightarrow cos E = \frac{1 - r / a}{e} = \frac{1 - 1.2}{0.4} = \frac{- 0.2}{0.4} = - 0.5$ . Why this step? $r$ depends only on $cos E$ , and $cos$ is an even function: $cos E = cos (2 π - E)$ . So $arccos (- 0.5)$ has two solutions in one orbit, $E = 2 π /3$ (upper half) and $E = 4 π /3$ (lower half). The radius alone cannot tell them apart.

Step 2. Resolve the branch using "receding." Here is the rule, stated cleanly: for $0 < E < π$ (upper half) the planet is moving from perihelion toward aphelion, so $r$ is increasing — it is receding. For $π < E < 2 π$ (lower half) it is moving from aphelion back to perihelion, so $r$ is decreasing — it is approaching. The problem says receding, so we take the upper-half root: $E = 2 π /3 = 2.09440$ rad. Why this step? This is the whole point of the exam trap: $cos E = - 0.5$ is ambiguous, and the extra physical word ("receding") selects the arccos branch. Approaching would force $E = 4 π /3$ instead.

Step 3. $M = E - e sin E = 2.09440 - 0.4 sin (2 π /3) = 2.09440 - 0.4 (0.86603) = 2.09440 - 0.34641 = 1.74799$ rad. Why this step? Forward map to the clock.

Step 4. $t - t_{p} = \frac{M}{2 π} T = \frac{1.74799}{6.28319} \times 100 = 27.819$ days. Why this step? Final clock-to-time conversion.

Verify: Check the radius: $a (1 - 0.4 cos (2 π /3)) = a (1 - 0.4 (- 0.5)) = a (1 + 0.2) = 1.2 a$ ✓. Had we (wrongly) taken the approaching branch $E = 4 π /3$ we'd get $M = 4.18879 - 0.4 (- 0.86603) = 4.53520$ and $t = 72.18$ days — which contradicts "receding." Branch choice matters. ✓

Ex 9 — Cell J: the degrees-vs-radians and sign trap

Worked example A student is given

$M = 3 0^{\circ}$ , $e = 0.2$ and blindly writes $E = M + e sin E$ with the calculator in degree mode, getting $E \approx 30 + 0.2 sin (3 0^{\circ}) = 30. 1^{\circ}$ . What is wrong, and what is the correct $E$ ? Forecast: Kepler's equation was derived from an area ( $\frac{1}{2} a^{2} E$ ), where the angle must be in radians. Guess how badly degree-mode corrupts the answer.

Step 1. Spot the unit error. In $M = E - e sin E$ , the $E$ outside $sin$ is an arc/area angle and must be in radians; the $sin E$ term expects its argument in radians too. Mixing " $30$ " (a degree number) with $sin (3 0^{\circ}) = 0.5$ silently uses two different angle systems. Why this step? The parent note's derivation used sector area $\frac{1}{2} a^{2} E$ — that formula is only true with $E$ in radians. Degrees break the very identity.

Step 2. Convert the input first: $M = 3 0^{\circ} = 30 \times \frac{π}{180} = 0.52360$ rad. Why this step? Always move to radians before touching Kepler's equation, so every term speaks the same language.

Step 3. Solve $M = E - 0.2 sin E$ with $M = 0.52360$ by Newton, seed $E_{0} = 0.52360$ : $f (0.52360) = 0.52360 - 0.2 sin (0.52360) - 0.52360 = - 0.2 (0.5) = - 0.1$ ; $f^{'} (0.52360) = 1 - 0.2 cos (0.52360) = 1 - 0.2 (0.86603) = 0.82679$ . $E_{1} = 0.52360 + 0.1/0.82679 = 0.64453$ . One more step: $f (0.64453) = 0.64453 - 0.2 sin (0.64453) - 0.52360 = - 0.00090$ ; $f^{'} = 0.84058$ ; $E_{2} = 0.64453 - (- 0.00090) /0.84058 = 0.64560$ , converged. Why this step? This is the correct transcendental solve; note $E > M$ ? No — here $sin E > 0$ so $M < E$ , i.e. $E = 0.6456 > M = 0.5236$ , the upper-half sign rule. And $- e sin E$ is a subtraction, so writing $E = M + e sin E$ is a fixed-point rearrangement, not "add the correction once."

Step 4. Compare: correct $E = 0.64560$ rad $= 36.9 9^{\circ}$ . The student's degree-mode " $30. 1^{\circ}$ " is off by nearly $7^{\circ}$ . Why this step? Quantifies the damage: the unit error is not tiny — it lost the entire correction.

Verify: Plug back in radians: $0.64560 - 0.2 sin (0.64560) = 0.64560 - 0.2 (0.60181) = 0.64560 - 0.12036 = 0.52524 \approx 0.52360$ ; refining gives $E = 0.64494$ rad with residual $0$ to 5 dp. ✓ And $0.64494$ rad $= 36.9 5^{\circ}$ , confirming the student's $30. 1^{\circ}$ was wrong by unit confusion. ✓

Quick recall

Recall Which direction needs iteration?

$E \to M$ (and hence $E \to t$ ) is a direct substitution; $M \to E$ is transcendental and needs Newton–Raphson. ::: Forward is free, inverse is iterative.

Sign of the correction $- e sin E$ in the upper half ( $0 < E < π$ )? ::: Negative, so $M < E$ .

Given only $r$ , why can't you find $E$ uniquely? ::: $r = a (1 - e cos E)$ depends on $cos E$ (even), giving two branches $E$ and $2 π - E$ ; you need extra info (approaching vs receding) to choose.

Which units must Kepler's equation use, and why? ::: Radians — it was derived from sector area $\frac{1}{2} a^{2} E$ , valid only in radians.

See also: Numerical root-finding — Newton–Raphson, Time of flight and orbit propagation, True anomaly ν and the orbit equation r = a(1−e²)/(1+e cos ν), Eccentricity and ellipse geometry (a, b, ae, b²=a²(1−e²)), and the Hyperbolic Kepler equation M = e·sinh F − F for the $e > 1$ case.