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

False. $M$ is the angle of a fictitious uniformly-moving body, $M=n(t-t_p)$; the real angle from the Sun is the true anomaly $\nu$. They coincide only at perihelion and aphelion.

True. At perihelion $E=0$, so $\sin E=0$ and $M=E-e\sin E=0$; and $\nu =0$ points along the same perihelion axis. All three agree at the turning points.

False. $E$ is measured from the centre $O$ of the ellipse on the auxiliary circle of radius $a$. Measuring from the centre is what makes the clean $x=a\cos E,y=b\sin E$ squash exact.

True. The correction is directly proportional to $e$, so a larger $e$ means the planet's speed varies more, and the gap between the uniform clock $M$ and the geometric angle $E$ grows.

True. With $e=0$ the $e\sin E$ term vanishes, so $M=E$; the motion is uniform and mean, eccentric, and true anomalies all coincide — see ellipse geometry.

False. For $e>1$ the geometry uses the hyperbolic form $M=e\sinh F-F$ with the hyperbolic sine and eccentric anomaly $F$; ordinary $\sin$ and the circle only apply to ellipses ($0\le e<1$).

True. $\cos E=-1$ there gives $r=a(1+e)$, the maximum distance. At $E=0$, $\cos E=1$ gives $r=a(1-e)$, the minimum (perihelion).

False. Equal increments of $M$ mean equal time and equal swept area (Kepler's 2nd law), but $\nu$ jumps fast near perihelion and slowly near aphelion — see Kepler's Second Law — equal areas in equal times.

False. The equation was derived from areas/arcs where the angle enters in radians by construction. The $E$ term is a raw angle-as-arc, not a $\sin$-only quantity, so degrees break it. Only radians work.

The rearranged form still has $E$ on both sides — it is a fixed-point/transcendental equation, not a closed formula. You must iterate (fixed-point or Newton–Raphson).

The $y$-coordinate is squashed by $b/a$: it is $y=b\sin E$, not $a\sin E$. Using $a\sin E$ describes the point $Q$ on the auxiliary circle, not the planet $P$ on the ellipse.

The focus $S$ is at distance $ae$ from the centre $O$, not $a$. That $ae$ base (from $O$ to $S$) is exactly what generates the $\frac{1}{2}{a}^{2}e\sin E$ triangle area, hence the $e\sin E$ term.

Wrong sign. The correct $E$-form is $r=a(1-e\cos E)$ with a minus. Check: $E=0$ must give perihelion $a(1-e)$; the plus version wrongly gives $a(1+e)$, the aphelion value.

For small-to-moderate $e$, the natural seed is $E_0=M$, since $E\approx M$ when $e$ is small. Starting at $0$ wastes iterations; $M$ converges fastest.

Equal areas in equal times does not mean equal arc length in equal times. Near perihelion the radius is short, so the planet must move faster to sweep the same area; speed is not constant.

The auxiliary circle circumscribes the ellipse with radius $a$ (semi-major axis). Squashing that radius-$a$ circle by $b/a$ produces the ellipse.

Because the ellipse is exactly a circle squashed vertically by $b/a$ about the centre, so measuring from the centre keeps the clean parametrization $x=a\cos E,y=b\sin E$. The focus offset $ae$ is then handled separately and becomes the $e\sin E$ term.

On the auxiliary circle you subtract triangle $OSQ$ (corners centre $O$, Sun $S$, circle-point $Q$; base $OS=ae$, height $a\sin E$) from the circular sector, giving a $\frac{1}{2}{a}^{2}e\sin E$ area. Multiplying the whole circular figure by the vertical squash factor $b/a$ to get the ellipse area carries that term through as $-e\sin E$ — the physical correction for the planet running fast near the Sun.

$E$ appears both linearly and inside $\sin E$, mixing an algebraic and a transcendental term — no finite combination of elementary functions unmixes them, so we must solve numerically.

Because $\nu$ changes non-uniformly with time, but $M=n(t-t_p)$ is a perfectly linear clock. Kepler's equation then bridges the easy clock $M$ to the geometric angle $E$, from which $\nu$ and $r$ follow.

Larger $e$ makes the $-e\sin E$ correction bigger and the motion more non-uniform, so $E$ departs further from $M$; the starting guess $E_0=M$ is worse and iterations converge more slowly.

The mean motion $n$ sets how fast $M$ ticks; via Kepler's third law, $n=2\pi /T=\sqrt{\mu /a^3}$ with $\mu =GM$ the gravitational parameter, so the orbit's size $a$ and the central mass alone fix the clock rate. This links $M=n(t-t_p)$ to physical time.

Because every $y$-coordinate on the ellipse is $b/a$ times the corresponding $y$ on the auxiliary circle, and area scales by that same vertical factor. This converts the messy elliptical region into an easy circular sector minus triangle.

It collapses to $M=E$, describing uniform circular motion; the ellipse becomes a circle and all three anomalies coincide.

The perihelion distance $a(1-e)$ shrinks toward $0$ while aphelion $a(1+e)$ grows toward $2a$; the orbit becomes extremely elongated and the speed contrast between perihelion and aphelion becomes extreme.

Because $\sin 0=\sin \pi =0$, the correction term vanishes at both the perihelion and aphelion turning points, so the clock and geometry momentarily agree.

Reduce $M$ modulo $2\pi$ into $[0,2\pi )$ first, since the geometry repeats every revolution; then solve for $E$ in that principal range.

$M=n(t-t_p)$ becomes negative, giving a negative $E$ by the odd symmetry $\sin (-E)=-\sin E$; the planet is simply approaching perihelion rather than receding from it.

No — $e=1$ is a parabolic escape trajectory, not a bound ellipse, so the circular auxiliary construction and $M=E-e\sin E$ no longer apply; a separate parabolic (Barker) treatment is needed, and $e>1$ uses the hyperbolic form.

For $0\le e<1$ the derivative $1-e\cos E$ is always strictly positive (never zero), so the iteration is well-behaved; with seed $E_0=M$ it converges reliably, only slowing as $e$ nears $1$.