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

All angles are in radians unless a problem says otherwise. Recurring facts we lean on: $M = E - e sin E, M = n (t - t_{p}), n = \frac{2 π}{T} = \frac{μ}{a ^{3}}, r = a (1 - e cos E) .$

Definition The true-anomaly half-angle bridge

The parent note states, and D2 derives, the identity linking the geometric angle $E$ to the physical angle $ν$ seen from the Sun: $tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2} .$ Where it comes from (sketch): using $x = a cos E$ , $y = b sin E$ with the focus at $(a e, 0)$ gives $cos ν = \frac{cos E - e}{1 - e cos E}$ and $sin ν = \frac{1 - e ^{2} sin E}{1 - e cos E}$ ; feeding these into the half-angle identity $tan (ν /2) = \frac{sin ν}{1 + cos ν}$ and simplifying collapses to the boxed form. We use it as a given tool here. See True anomaly ν and the orbit equation r = a(1−e²)/(1+e cos ν). Why the half-angle form? Because $tan (ν /2)$ is monotonic over a full revolution $ν \in (- π, π)$ , so it removes the quadrant ambiguity that plagues $tan ν$ — $E$ and $ν$ always lie in the same half-turn, letting us pick the branch cleanly.

Level 1 — Recognition

These check that you can name and read each symbol.

L1.1

State, in one sentence each, what $M$ , $E$ , and $ν$ physically are, and from which point each is measured.

Recall Solution

$M$ (mean anomaly): the angle of a fictitious body moving at constant angular speed. It is not a real geometric angle — it is a rescaled clock, $M = n (t - t_{p})$ , where $t_{p}$ is the time of perihelion passage.
$E$ (eccentric anomaly): a geometric helper angle measured from the centre $O$ of the ellipse, on the auxiliary circle of radius $a$ .
$ν$ (true anomaly): the real angular position of the planet, measured from the focus $S$ (the Sun).

L1.2

Without solving anything, at which two points on the orbit is $M = E$ exactly? Why?

Recall Solution

At perihelion ( $E = 0$ ) and aphelion ( $E = π$ ). There $sin E = 0$ , so the correction term $e sin E$ vanishes and $M = E - 0 = E$ . Physically, these are the two turning points where the fictitious uniform body and the geometric angle momentarily line up.

L1.3

An orbit has $e = 0$ . What does Kepler's equation become, and what does that say about the motion?

Recall Solution

With $e = 0$ : $M = E - 0 \cdot sin E = E$ . The ellipse is a circle, $r = a (1 - 0) = a$ (constant radius), and the planet moves at uniform angular speed — all three anomalies coincide.

Level 2 — Application

Plug into the formulas, both directions.

L2.1 (easy direction: $E \to M \to t$ )

Given $E = π /3$ , $e = 0.4$ , period $T = 200$ days. Find $M$ and the time since perihelion $t - t_{p}$ .

Recall Solution

$M = E - e sin E = \frac{π}{3} - 0.4 sin \frac{π}{3} = 1.047198 - 0.4 (0.866025) = 1.047198 - 0.346410 = 0.700788 rad .$ Then $t - t_{p} = \frac{M}{2 π} T = \frac{0.700788}{6.283185} \times 200 = 22.306$ days.

L2.2 (radius from $E$ )

For the same orbit ( $e = 0.4$ ) with semi-major axis $a = 1.5 \times 1 0^{8}$ km and $E = π /3$ , find the distance $r$ .

Recall Solution

$r = a (1 - e cos E) = 1.5 \times 1 0^{8} (1 - 0.4 cos \frac{π}{3}) = 1.5 \times 1 0^{8} (1 - 0.4 \times 0.5) = 1.5 \times 1 0^{8} (0.8) = 1.2 \times 1 0^{8} km .$

L2.3 (hard direction: one Newton step)

For $e = 0.3$ , $M = 0.8$ rad, do one Newton–Raphson step starting from $E_{0} = M = 0.8$ . Report $E_{1}$ .

Recall Solution

Newton step: $E_{1} = E_{0} - \frac{E _{0} - e sin E _{0} - M}{1 - e cos E _{0}}$ . (All trig values below are carried to 6 decimal places; that is the source of any tiny rounding in the last digit.)

$f (E_{0}) = 0.8 - 0.3 sin (0.8) - 0.8 = - 0.3 \times 0.717356 = - 0.215207.$
$f^{'} (E_{0}) = 1 - 0.3 cos (0.8) = 1 - 0.3 \times 0.696707 = 0.790988.$
$E_{1} = 0.8 - \frac{- 0.215207}{0.790988} = 0.8 + 0.272073 = 1.072073.$

Level 3 — Analysis

Now explain why, and probe signs and quadrants.

L3.1 (all quadrants — sign of the correction)

For $e = 0.5$ , compute $M$ for $E = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ (one in each quadrant). In which quadrants is $M < E$ , and in which is $M > E$ ? Explain physically.

How to read the figure below. The horizontal axis is the eccentric anomaly $E$ (one full revolution, $0$ to $2 π$ ); the vertical axis is the mean anomaly $M$ . The dashed slate line is the reference $M = E$ (what you'd get with no correction). The coral curve is the actual $M = E - 0.5 sin E$ . Where the coral curve dips below the dashed line, $M < E$ (lavender shading, the outbound half $0 < E < π$ ); where it rises above, $M > E$ (mint shading, the inbound half $π < E < 2 π$ ). The four butter dots are the sample points you compute.

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

Recall Solution

$M = E - 0.5 sin E$ .

$E = \frac{π}{4} = 0.785398$ : $sin E = + 0.707107$ , $M = 0.785398 - 0.353553 = 0.431845$ . $M < E$ .
$E = \frac{3 π}{4} = 2.356194$ : $sin E = + 0.707107$ , $M = 2.356194 - 0.353553 = 2.002641$ . $M < E$ .
$E = \frac{5 π}{4} = 3.926991$ : $sin E = - 0.707107$ , $M = 3.926991 + 0.353553 = 4.280544$ . $M > E$ .
$E = \frac{7 π}{4} = 5.497787$ : $sin E = - 0.707107$ , $M = 5.497787 + 0.353553 = 5.851340$ . $M > E$ .

Pattern: where $sin E > 0$ (upper half, quadrants I–II, the outbound leg from perihelion), $M < E$ ; where $sin E < 0$ (lower half, quadrants III–IV, inbound to perihelion), $M > E$ . Physically: on the way out the planet has sprinted through perihelion, so real geometry ( $E$ ) has run ahead of the uniform clock ( $M$ ); on the way back in, it lags and the clock catches up. In the figure the coral curve sits below the dashed $M = E$ line on the left half (lavender) and above it on the right half (mint).

L3.2 (monotonicity — why Newton never divides by zero)

Show that $\frac{d M}{d E} = 1 - e cos E > 0$ for all $E$ when $0 \leq e < 1$ . Why does this guarantee a unique $E$ for each $M$ , and why is Newton's denominator safe?

Recall Solution

Differentiate $M = E - e sin E$ with respect to $E$ : $\frac{d M}{d E} = 1 - e cos E$ . Since $- 1 \leq cos E \leq 1$ and $0 \leq e < 1$ , we have $e cos E \leq e < 1$ , so $1 - e cos E \geq 1 - e > 0$ . It is strictly positive everywhere.

Uniqueness: $M (E)$ is strictly increasing, hence one-to-one — each $M$ maps to exactly one $E$ .
Newton safety: the denominator in the Newton step is $f^{'} (E) = 1 - e cos E$ , exactly this quantity, which is bounded below by $1 - e > 0$ . It can never be zero, so no division-by-zero blowup.

L3.3 (limiting behaviour — high eccentricity)

As $e \to 1^{-}$ , what happens to $d M / d E$ near perihelion ( $E = 0$ )? What does that imply about how fast $E$ changes with the clock there?

Recall Solution

At $E = 0$ : $d M / d E = 1 - e cos 0 = 1 - e$ . As $e \to 1^{-}$ , this $\to 0^{+}$ . Since $d E / d M = 1/ (1 - e cos E)$ , near perihelion $d E / d M = 1/ (1 - e) \to \infty$ . So a tiny change in the clock $M$ produces a huge change in $E$ : the planet whips through perihelion extremely fast for near-parabolic orbits. This is exactly why high- $e$ cases are numerically stiff and need a better seed than $E_{0} = M$ .

Level 4 — Synthesis

Chain several relations together.

L4.1 ( $t \to M \to E \to r \to ν$ , the full propagation)

An orbit has $e = 0.2$ , $a = 2$ AU, $T = 1000$ days. At $t - t_{p} = 100$ days, find (a) $M$ , (b) $E$ (Newton to 5 dp), (c) $r$ , (d) the true anomaly $ν$ .

Recall Solution

(a) Clock $\to$ mean anomaly. $n = \frac{2 π}{T} = \frac{6.283185}{1000} = 0.00628319$ rad/day. $M = n (t - t_{p}) = 0.00628319 \times 100 = 0.628319 rad .$ (b) Solve Kepler by Newton, $E_{0} = M = 0.628319$ , $e = 0.2$ . Stopping rule: stop when $∣ E_{k + 1} - E_{k} ∣ < 1 0^{- 5}$ , i.e. the 5th decimal has settled.

$f_{0} = 0.628319 - 0.2 sin (0.628319) - 0.628319 = - 0.2 \times 0.587785 = - 0.117557$ ; $f_{0}^{'} = 1 - 0.2 cos (0.628319) = 1 - 0.2 \times 0.809017 = 0.838197$ . $E_{1} = 0.628319 + 0.140250 = 0.768569$ . (step size $0.140250$ )
$f_{1} = 0.768569 - 0.2 sin (0.768569) - 0.628319 = 0.768569 - 0.138931 - 0.628319 = 0.001319$ ; $f_{1}^{'} = 1 - 0.2 cos (0.768569) = 0.857157$ . $E_{2} = 0.768569 - 0.001539 = 0.767030$ . (step size $0.001539$ )
$f_{2} = 0.767030 - 0.2 sin (0.767030) - 0.628319 = 0.767030 - 0.138710 - 0.628319 = 0.000001$ ; $f_{2}^{'} = 0.857005$ . $E_{3} = 0.767030 - 0.0000012 = 0.767029$ . (step size $\approx 1.2 \times 1 0^{- 6} < 1 0^{- 5}$ → stop.)
Converged: $E \approx 0.767029$ rad.

(c) Radius. $r = a (1 - e cos E) = 2 (1 - 0.2 cos 0.767029) = 2 (1 - 0.2 \times 0.719963) = 2 (0.856007) = 1.712014 AU .$

(d) True anomaly. $tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2} = \frac{1.2}{0.8} tan (0.383515) = 1.224745 \times 0.403553 = 0.494292.$ $\frac{ν}{2} = arctan (0.494292) = 0.457546$ , so $ν = 0.915093 rad \approx 52.4 3^{\circ} .$ Quadrant check: here $E = 0.767$ lies in the first half-turn ( $0 < E < π$ ), and the half-angle form guarantees $ν$ lies in the same half-turn, so $ν /2 \in (0, π /2)$ and the principal $arctan$ branch is correct — no $π$ needs to be added. (Rule of thumb: if $E \in (π, 2 π)$ , add $2 π$ to the $ν$ you get from doubling, or equivalently take $ν /2$ in the matching half so $ν$ shares $E$ 's revolution.)

L4.2 (period unknown — combining Kepler's 3rd law)

A planet orbits a star of mass giving $μ = 1.327 \times 1 0^{20} m^{3} / s^{2}$ (Sun) with $a = 1.496 \times 1 0^{11}$ m. At $E = 1.2$ rad with $e = 0.3$ , how many days have elapsed since perihelion?

Recall Solution

Mean motion via Kepler's 3rd law: $n = \frac{μ}{a ^{3}}$ . $a^{3} = (1.496 \times 1 0^{11})^{3} = 3.348071 \times 1 0^{33} m^{3}, \frac{μ}{a ^{3}} = \frac{1.327 \times 1 0 ^{20}}{3.348071 \times 1 0 ^{33}} = 3.963779 \times 1 0^{- 14} s^{- 2} .$ $n = 3.963779 \times 1 0^{- 14} = 1.990924 \times 1 0^{- 7} rad/s .$ Mean anomaly: $M = E - e sin E = 1.2 - 0.3 sin (1.2) = 1.2 - 0.3 \times 0.932039 = 1.2 - 0.279612 = 0.920388$ rad. Time: $t - t_{p} = \frac{M}{n} = \frac{0.920388}{1.990924 \times 1 0 ^{- 7}} = 4.622882 \times 1 0^{6} s .$ In days: $\frac{4.622882 \times 1 0 ^{6}}{86400} = 53.506$ days.

Level 5 — Mastery

Edge cases, stiff numerics, and reverse engineering.

L5.1 (high eccentricity — bad seed vs good seed)

For $e = 0.9$ , $M = 0.3$ rad, iterate Kepler's equation. (a) Do two fixed-point iterations $E_{k + 1} = M + e sin E_{k}$ from $E_{0} = M$ . (b) Do two Newton iterations from $E_{0} = π$ . Which converges faster toward the true root $E \approx 1.16227$ ?

Recall Solution

(a) Fixed point, $E_{0} = 0.3$ :

$E_{1} = 0.3 + 0.9 sin (0.3) = 0.3 + 0.9 \times 0.295520 = 0.565968.$
$E_{2} = 0.3 + 0.9 sin (0.565968) = 0.3 + 0.9 \times 0.535896 = 0.782306.$ Still far from $1.16227$ — fixed point crawls because $e$ is large (its convergence factor is $\approx e cos E$ , near $0.9$ ).

(b) Newton, $E_{0} = π = 3.141593$ :

$f_{0} = 3.141593 - 0.9 sin (3.141593) - 0.3 = 3.141593 - 0.9 \times 2.65 \times 1 0^{- 6} - 0.3 = 2.841590$ ; $f_{0}^{'} = 1 - 0.9 cos (3.141593) = 1 + 0.9 = 1.900000$ . $E_{1} = 3.141593 - 1.495574 = 1.646019.$
$f_{1} = 1.646019 - 0.9 sin (1.646019) - 0.3 = 1.646019 - 0.9 \times 0.997191 - 0.3 = 0.448547$ ; $f_{1}^{'} = 1 - 0.9 cos (1.646019) = 1 - 0.9 \times (- 0.074937) = 1.067443$ . $E_{2} = 1.646019 - 0.420210 = 1.225809.$ Already within $0.06$ of the root after two Newton steps, versus fixed-point still stuck near $0.78$ . Newton wins decisively for large $e$ ; the parent's default seed $E_{0} = M$ is poor here, and $E_{0} = π$ (or $M + e sign$ ) is a common robust choice.

L5.2 (degenerate case — $e$ just below 1, perihelion speed)

For $e = 0.99$ , compare $d E / d M$ at perihelion ( $E = 0$ ) and at aphelion ( $E = π$ ). By what factor does the eccentric anomaly respond faster to the clock at perihelion?

Recall Solution

$\frac{d E}{d M} = \frac{1}{1 - e cos E}$ .

Perihelion $E = 0$ : $\frac{1}{1 - 0.99 \times 1} = \frac{1}{0.01} = 100.$
Aphelion $E = π$ : $\frac{1}{1 - 0.99 \times ( - 1 )} = \frac{1}{1.99} = 0.502513.$ Ratio $= \frac{100}{0.502513} = 199.0$ . So $E$ (and hence the planet) sweeps roughly 199 times more responsively to the clock at perihelion than at aphelion — a vivid picture of the near-parabolic sprint-and-crawl.

L5.3 (reverse — find $e$ from an observation)

At $E = π /2$ a planet is measured at distance $r = 0.85 a$ . Find the eccentricity $e$ , then the mean anomaly $M$ at that instant.

Recall Solution

Use $r = a (1 - e cos E)$ with $E = π /2$ , so $cos E = 0$ : $0.85 a = a (1 - e \times 0) = a \Rightarrow contradiction if taken literally!$ This is the lesson: at $E = π /2$ the radius is $a (1 - e cos (π /2)) = a (1 - 0) = a$ for every eccentricity, so $r$ at $E = π /2$ carries no information about $e$ . The stated datum $r = 0.85 a$ at $E = π /2$ is therefore impossible — a great trap. Corrected version: suppose instead $r = 0.85 a$ occurs at $E = π /3$ . Then $0.85 = 1 - e cos \frac{π}{3} = 1 - 0.5 e \Rightarrow e = \frac{1 - 0.85}{0.5} = 0.30.$ Then $M = E - e sin E = \frac{π}{3} - 0.30 sin \frac{π}{3} = 1.047198 - 0.30 \times 0.866025 = 1.047198 - 0.259808 = 0.787390$ rad.

L5.4 (bridge to the hyperbolic case)

For an unbound trajectory the ellipse is replaced by a hyperbola and the equation becomes $M = e sinh F - F$ (see Hyperbolic Kepler equation M = e·sinh F − F). Given $e = 1.5$ , $F = 1.0$ , find $M$ . Contrast the sign structure with the elliptic case.

Recall Solution

What is $F$ ? For a hyperbolic (escape) orbit, $e > 1$ and there is no auxiliary circle; instead the geometry is captured by the hyperbolic eccentric anomaly $F$ , a real parameter defined through hyperbolic functions ( $cosh F, sinh F$ ) exactly as $E$ was defined through $cos E, sin E$ for the ellipse. It plays the same "geometric helper" role, but grows without bound rather than cycling. $M = e sinh F - F = 1.5 sinh (1.0) - 1.0 = 1.5 \times 1.175201 - 1.0 = 1.762802 - 1.0 = 0.762802.$ Contrast (complete):

Elliptic: $M = E - e sin E$ . The $sin E$ term is bounded and oscillates; the leading sign is on $E$ (plus) with the correction minus.
Hyperbolic: $M = e sinh F - F$ . The $sinh F$ term is unbounded and grows exponentially; the $e$ -term now carries the plus sign and dominates, while the lone $F$ is subtracted.
Same idea: in both, $M$ is still the uniform clock $M = n (t - t_{p})$ (with $n = μ /∣ a ∣^{3}$ for the hyperbola), and in both, solving $M \to E$ (or $M \to F$ ) is transcendental and needs Newton iteration. Only the geometry — circle vs. hyperbola, $sin$ vs. $sinh$ — differs. This is why the two equations are studied side by side.

L5.4 trap (steel-manned) and the L5 trap below are separate cautions.

Recall One-line recap of the ladder

Recognise the anomalies ::: L1 Plug both directions ( $E \to M$ easy, $M \to E$ needs Newton) ::: L2 Sign of $e sin E$ flips by hemisphere; $d M / d E = 1 - e cos E > 0$ ::: L3 Full chain $t \to M \to E \to r \to ν$ , with $n = μ / a^{3}$ ::: L4 High- $e$ stiffness, degenerate $cos E = 0$ , hyperbolic bridge ::: L5