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

WHY do we even need this?

For a circle, position is trivial: angle $= \omega t$. For an ellipse, the planet sweeps equal areas in equal times, not equal angles. The angle you actually want (the true anomaly $\nu$, measured from the Sun) changes non-uniformly with time. We need a chain of three angles to untangle this.

HOW we build it: the auxiliary circle

Draw the ellipse (semi-major $a$, semi-minor $b$, eccentricity $e$). Around it draw a circle of radius $a$ centred at the ellipse centre $O$. The Sun sits at focus $S$, a distance $ae$ from $O$.

For a planet at point $P$ on the ellipse, drop a vertical line through $P$; it hits the circle at point $Q$. The angle $\angle QOX$ (from centre, $X$ = perihelion direction) is the eccentric anomaly $E$.

The magic property of an ellipse: it is a circle squashed in $y$ by factor $b/a$. So if $Q=(a\cos E,\; a\sin E)$ on the circle, the planet is $x = a\cos E, \qquad y = b\sin E.$

DERIVATION 1 — Kepler's equation from areas (first principles)

Kepler's 2nd law: the radius from the focus sweeps area at a constant rate. So the fraction of orbital period elapsed equals the fraction of total ellipse area swept since perihelion:

$\frac{t - t_p}{T} = \frac{\text{Area swept } SXP}{\text{total ellipse area } \pi a b}.$

The left side, times $2\pi$, is by definition $M$: M = \frac{2\pi}{T}(t-t_p) = \frac{2\pi \cdot (\text{Area } SXP)}{\pi a b}. \tag{1}

Step: compute Area $SXP$. Use the squash. Every $y$-coordinate on the ellipse is $b/a$ times the corresponding $y$ on the circle. Area scales the same way, so any region on the ellipse $= \frac{b}{a}\times$ (its image region on the circle).

Why this step? It turns a messy elliptical area into a circular-sector area, which we can compute with simple geometry.

$\text{Area } SXP_{\text{ellipse}} = \frac{b}{a}\,\text{Area } SXQ_{\text{circle}}.$

Step: Area $SXQ$ on the circle = (circular sector $OXQ$) − (triangle $OSQ$).

Why? $SXQ$ is the sector from perihelion swept about the centre minus the wedge cut off by the focus offset.

• Sector $OXQ$: a slice of a circle radius $a$, angle $E$: area $= \tfrac12 a^2 E$.
• Triangle $OSQ$: base $OS = ae$ along the axis, height $= y_Q = a\sin E$: $\text{area} = \tfrac12 (ae)(a\sin E) = \tfrac12 a^2 e\sin E.$

So $\text{Area } SXQ_{\text{circle}} = \tfrac12 a^2 E - \tfrac12 a^2 e\sin E = \tfrac12 a^2 (E - e\sin E).$

Step: convert to ellipse and plug into (1). $\text{Area } SXP = \frac{b}{a}\cdot \tfrac12 a^2 (E - e\sin E) = \tfrac12 ab\,(E - e\sin E).$

$M = \frac{2\pi}{\pi ab}\cdot \tfrac12 ab\,(E - e\sin E).$

DERIVATION 2 — radius and true anomaly in terms of $E$

From the focus $S=(ae,0)$ to planet $P=(a\cos E, b\sin E)$: $r^2 = (a\cos E - ae)^2 + (b\sin E)^2.$ Use $b^2 = a^2(1-e^2)$: $r^2 = a^2(\cos E - e)^2 + a^2(1-e^2)\sin^2 E.$ Expand and use $\cos^2 E + \sin^2 E = 1$: $r^2 = a^2\big[\cos^2E - 2e\cos E + e^2 + \sin^2E - e^2\sin^2E\big] = a^2(1 - e\cos E)^2.$ $\boxed{r = a(1 - e\cos E)}$ Why useful? Once you solve for $E$, you instantly get the distance $r$, and the components $x=a\cos E,\ y=b\sin E$ give full position. The true anomaly follows from $\tan\frac{\nu}{2} = \sqrt{\frac{1+e}{1-e}}\,\tan\frac{E}{2}.$

HOW to actually solve it (it's transcendental!)

$M = E - e\sin E$ cannot be inverted algebraically for $E(M)$. We iterate (Newton–Raphson): $E_{k+1} = E_k - \frac{E_k - e\sin E_k - M}{1 - e\cos E_k}, \qquad E_0 = M.$

Common mistakes (Steel-manned)

Quick recall

Flashcards

Connections

• True anomaly ν and the orbit equation r = a(1−e²)/(1+e cos ν)
• Kepler's Second Law — equal areas in equal times
• Kepler's Third Law — T² ∝ a³ and mean motion n
• Eccentricity and ellipse geometry (a, b, ae, b²=a²(1−e²))
• Numerical root-finding — Newton–Raphson
• Time of flight and orbit propagation
• Hyperbolic Kepler equation M = e·sinh F − F

Concept Map

Hinglish (regional understanding)