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

We start from absolute zero: two shapes, a squash, and a scoop of area.

Step 1 — Draw the ellipse and name its bones

WHAT. An ellipse is an oval. It has one long half-width and one short half-width. We give them names:

$a$ = the semi-major axis = half of the longest diameter (the long half-width).
$b$ = the semi-minor axis = half of the shortest diameter (the short half-width).
$O$ = the centre, the exact middle of the oval.

WHY these first. Everything downstream is measured against $a$ and $b$ . You cannot talk about "how squashed" the oval is until you have the two half-widths to compare.

PICTURE. Look at the figure: the long green arrow is $a$ , the short yellow arrow is $b$ , both starting from the centre $O$ (white dot). $a > b$ always for a real ellipse — that is what makes it an oval and not a circle.

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

Step 2 — Put the Sun at the focus, a distance $a e$ off-centre

WHAT. The Sun is not at the centre. It sits at a special point called the focus $S$ , shifted along the long axis by exactly $a e$ from the centre.

WHY. Kepler's First Law says planets orbit with the Sun at a focus. That off-centre shift $a e$ is the whole reason motion looks lopsided — near $S$ the planet is close, far from $S$ it is distant. Hold onto the length $a e$ : it will reappear as the $e sin E$ term.

PICTURE. The red dot is the Sun $S$ . The red segment from $O$ to $S$ has length $a e$ . Notice $a e$ grows as $e$ grows — a rounder orbit ( $e$ small) has the Sun almost at centre; a stretched orbit has the Sun far off-centre.

Step 3 — Wrap a circle of radius $a$ around the ellipse (the auxiliary circle)

WHAT. Draw a full circle centred at $O$ with radius $a$ . Its left/right edges just touch the ellipse's tips; its top/bottom bulge above the ellipse. This is the auxiliary circle.

WHY this circle and not any other. Because an ellipse is literally a circle pressed down. If you take the auxiliary circle and squash every point's height by the factor $b / a$ , you land exactly on the ellipse. We will do our geometry on the friendly circle, then squash the answer.

PICTURE. Blue circle = auxiliary circle. Grey ellipse sits inside it, touching at the left and right tips. The two vertical green arrows show the squash: a tall circle-height $a sin E$ becomes the shorter ellipse-height $b sin E$ .

Step 4 — Define the eccentric anomaly $E$ (the helper angle)

WHAT. Pick the planet's position $P$ on the ellipse. Go straight up (or down) from $P$ until you hit the auxiliary circle at point $Q$ . The angle from the centre $O$ , measured from the perihelion direction $X$ to the line $O Q$ , is the eccentric anomaly $E$ .

WHY from the centre and not the Sun. The squash factor $b / a$ is exact only about the centre $O$ . Measure $E$ there and the coordinates stay clean: $Q = (a cos E, a sin E), P = (a cos E, b sin E) .$ Both share the same $x = a cos E$ (we moved only vertically from $Q$ to $P$ ); only the height got squashed.

PICTURE. $Q$ (blue dot) on the circle, $P$ (yellow dot) directly below it on the ellipse. The angle $E$ is the blue wedge at $O$ . The dashed vertical line links $Q$ down to $P$ — same $x$ , shorter $y$ .

Step 5 — Kepler's Second Law: area is the clock

WHAT. Kepler's Second Law says the line from the Sun to the planet sweeps out equal areas in equal times. So the fraction of the orbit's total area swept since perihelion equals the fraction of the period elapsed.

WHY area, not angle. The planet's speed changes, so the angle from the Sun does not grow steadily. But the area swept per second is a fixed constant. Area is the honest clock. We convert "time" into "area", and area we can compute geometrically.

PICTURE. Two red slivers of equal area near the Sun and far from it. Near the Sun the planet covers a wide-but-short wedge quickly; far away a thin-but-long wedge slowly. Same area, same time.

Step 6 — Squash the swept area back to the circle

WHAT. The swept area $S X P$ lives on the ellipse and is awkward. Undo the squash: every ellipse region equals $\frac{b}{a}$ times the matching region on the circle.

WHY. We already know how to measure areas on a circle (sectors and triangles). So push $S X P$ up onto the circle, measure it there, then multiply by $b / a$ to come back down.

PICTURE. The yellow ellipse region $S X P$ and its taller circle twin $S X Q$ directly above it. They are related by exactly one number: $Area (S X P) = \frac{b}{a} Area (S X Q)$ .

Step 7 — Cut the circle region into a sector minus a triangle

WHAT. On the circle, the region $S X Q$ (from Sun, along perihelion, up to $Q$ ) splits neatly: $Area (S X Q) = fan from centre sector O X Q - wedge cut by Sun offset triangle O S Q .$

WHY these two pieces. The fan $O X Q$ is a plain circular slice — easy. But we wanted the region measured from the Sun $S$ , not the centre $O$ . The Sun is offset by $a e$ (Step 2), so we subtract the little triangle $O S Q$ that the offset creates. This subtraction is where $a e$ — and therefore $e sin E$ — enters.

PICTURE. Blue fan = sector $O X Q$ with opening angle $E$ . Red triangle = $O S Q$ with base $O S = a e$ along the axis and height $a sin E$ (the height of $Q$ ). The area we want is fan minus triangle.

Step 8 — Assemble: Kepler's equation appears

WHAT. Multiply the circle area by $b / a$ (Step 6) to get the ellipse area, then feed it into the area-fraction / mean-anomaly relation (Step 5).

WHY. This is the moment all the pieces click: the $\frac{1}{2} a^{2}$ from geometry cancels against the $π ab$ total area, and the naked $(E - e sin E)$ is left standing.

$Area (S X P) = \frac{b}{a} \cdot \frac{1}{2} a^{2} (E - e sin E) = \frac{1}{2} ab (E - e sin E)$

$M = \frac{2 π}{π ab} \cdot \frac{1}{2} ab (E - e sin E)$

PICTURE. A cancellation cartoon: the $ab$ in numerator kills the $ab$ in denominator, the $2 π / (π) \cdot \frac{1}{2} = 1$ , leaving the boxed result.

Step 8b — Read off the distance $r$ from the same picture

WHAT. We already have the planet's coordinates $P = (a cos E, b sin E)$ and the Sun at $S = (a e, 0)$ . The distance $r$ (defined in Step 2) is just the length of the segment $S P$ by the Pythagorean theorem.

WHY here. $r$ is used in the edge-case examples below and in the summary, so we earn it now instead of quoting it. It costs only one line of algebra.

$r^{2} = (a cos E - a e)^{2} + (b sin E)^{2}$

Substitute $b^{2} = a^{2} (1 - e^{2})$ (Step 1), expand, and use $cos^{2} E + sin^{2} E = 1$ :

$r^{2} = a^{2} (cos E - e)^{2} + a^{2} (1 - e^{2}) sin^{2} E = a^{2} (1 - e cos E)^{2}$

Step 9 — The edge cases (never leave a scenario unshown)

WHAT & WHY. A derivation you can trust must survive its extremes. Check the turning points and the degenerate circle.

PICTURE. Three snapshots: perihelion ( $E = 0$ ), aphelion ( $E = π$ ), and a perfect circle ( $e = 0$ ).

The one-picture summary

Everything on one canvas: the ellipse, its auxiliary circle, the planet $P$ and its shadow $Q$ , the Sun offset $a e$ , the swept area, and the sector-minus-triangle that becomes $E - e sin E$ .

Recall Feynman retelling — the walkthrough in plain words

A planet on an oval speeds up near the Sun and dawdles far away, so "where is it at 3 o'clock?" has no easy answer. Our trick: forget the planet's real angle and use area as the clock, because area is swept at a steady rate. To measure that swept area we cheat — we blow the flat oval back up into a fair circle by un-squashing its heights (one factor $b / a$ ). On the circle, the area from perihelion up to our helper point is just a pizza slice (the sector, worth $\frac{1}{2} a^{2} E$ ) with one triangular bite removed (because the Sun sits off-centre by $a e$ , the bite is $\frac{1}{2} a^{2} e sin E$ ). Slice minus bite is $\frac{1}{2} a^{2} (E - e sin E)$ . Squash it back down, divide by the whole oval's area, scale by $2 π$ , and the clean bookkeeping leaves exactly $M = E - e sin E$ . The $E$ is the pizza-slice angle; the $- e sin E$ is the off-centre bite. At the tips the bite is zero (clock = geometry); for a circle there is no bite at all. And the distance $r$ from the Sun is just the length of the line to the planet, which the same coordinates hand us as $r = a (1 - e cos E)$ .

Recall Cloze self-test

Area is the clock because of Kepler's Second Law (equal areas in equal times).
We do geometry on the auxiliary circle (radius $a$ ) then squash by == $b / a$ == to reach the ellipse.
The sector $O X Q$ has area == $\frac{1}{2} a^{2} E$ == and needs $E$ in radians.
The triangle $O S Q$ has base == $a e$ and height $a sin E$ , area $\frac{1}{2} a^{2} e sin E$ ==.
$r$ is the Sun-to-planet distance, equal to == $a (1 - e cos E)$ ==.
Over one orbit both $E$ and $M$ run through == $0 \leq \cdot < 2 π$ ==.
Result: $M === E - e sin E ==$ and $r === a (1 - e cos E) ==$ .
Related: solving $M \to E$ needs Newton–Raphson; the parabola/hyperbola cousin is the hyperbolic Kepler equation; converting $E \to ν$ uses the true anomaly.