Solving Kepler's equation — Newton-Raphson iteration: Exercises

Before the cast, three quantities the formulas below lean on:

The three characters you will use over and over:

The Newton-Raphson recipe we hammer throughout:

$E_{k + 1} = E_{k} - \frac{E _{k} - e s i n E _{k} - M}{1 - e c o s E _{k}}, f (E) = E - e sin E - M, f^{'} (E) = 1 - e cos E .$

The picture below shows what one Newton step actually does — we will build to it in Level 2, so glance at it now and return after Exercise 2.2:

Figure — Solving Kepler's equation — Newton-Raphson iteration

The blue curve is the root function $f (E)$ ; the red dot is our starting guess (seed); the dashed green line is the tangent whose slope is $f^{'}$ . Newton slides down that tangent to where it crosses zero (green square), landing much closer to the true root (yellow dot).

Level 1 — Recognition

Exercise 1.1

State, in words, what each symbol means and give its units: $M$ , $E$ , $e$ , $n$ , $a$ , $t_{p}$ , $μ$ .

Recall Solution

$M$ — mean anomaly: a fictitious angle that grows uniformly with time. Units: radians.
$E$ — eccentric anomaly: the geometric angle at the centre of the ellipse to the point on the auxiliary circle above the body. Units: radians.
$e$ — eccentricity: dimensionless shape number, $0 \leq e < 1$ for an ellipse.
$n$ — mean motion: average angular rate, $n = μ / a^{3}$ . Units: rad/s.
$a$ — semi-major axis: half the long diameter of the ellipse. Units: metres (or any length).
$t_{p}$ — time of perihelion passage: the clock reading when the body is closest to the focus. Units: seconds.
$μ$ — gravitational parameter $= G M_{central}$ : strength of gravity of the central body. Units: $km^{3} / s^{2}$ .

Exercise 1.2

Which of these is Kepler's equation? (a) $E = M - e sin M$ (b) $M = E - e sin E$ (c) $M = E - e cos E$ .

Recall Solution

(b) $M = E - e sin E$ . Option (a) has the roles of $M$ and $E$ swapped and uses $sin M$ — that is actually the first-order seed, not the exact equation. Option (c) uses $cos$ ; the $cos$ shows up in the derivative $f^{'} = 1 - e cos E$ , not in Kepler's equation itself.

Exercise 1.3

Given the root function $f (E) = E - e sin E - M$ , write its derivative $f^{'} (E)$ .

Recall Solution

Differentiate term by term with $M, e$ constant: $\frac{d}{d E} E = 1, \frac{d}{d E} (- e sin E) = - e cos E, \frac{d}{d E} (- M) = 0.$ $f^{'} (E) = 1 - e cos E .$ Since $e < 1$ , we have $f^{'} (E) \geq 1 - e > 0$ everywhere — $f$ is strictly increasing, so the root is unique.

Level 2 — Application

Exercise 2.1

For $M = 0.8$ rad, $e = 0.3$ , compute the first-order seed $E_{0} = M + e sin M$ .

Recall Solution

$sin (0.8) = 0.71736$ . $E_{0} = 0.8 + 0.3 (0.71736) = 0.8 + 0.21521 = 1.01521 rad .$ The seed pre-corrects for the $- e sin E$ shift, so it starts us near the answer.

Exercise 2.2

Do ONE Newton-Raphson step from $E_{0} = 1.01521$ for $M = 0.8$ , $e = 0.3$ . (Match your work to the figure at the top of the page.)

Recall Solution

$f (E_{0}) = 1.01521 - 0.3 sin (1.01521) - 0.8 = 1.01521 - 0.25548 - 0.8 = - 0.04027.$ $f^{'} (E_{0}) = 1 - 0.3 cos (1.01521) = 1 - 0.15853 = 0.84147.$ $E_{1} = 1.01521 - \frac{- 0.04027}{0.84147} = 1.01521 + 0.04786 = 1.06307 rad .$ $f < 0$ told us the guess was too small, so Newton pushed $E$ up — exactly the "slide down the tangent to zero" the top figure shows. This matches Worked Example 1 in the parent note.

Exercise 2.3

Given $E = 1.0627$ rad, $e = 0.3$ , $a = 1$ AU, find the distance $r$ from the focus.

Recall Solution

$r = a (1 - e cos E) = 1 \cdot (1 - 0.3 cos (1.0627)) = 1 - 0.3 (0.48679) = 1 - 0.14604 = 0.85396 AU .$ Since $r < a$ , the body is on the near half of the orbit — consistent with $E < \frac{π}{2}$ .

Level 3 — Analysis

Exercise 3.1

For $M = 0.1$ rad, $e = 0.9$ , do the first Newton step from the seed $E_{0} = M + e sin M$ and explain why the step is so large in terms of $f^{'}$ .

Recall Solution

Seed: $sin (0.1) = 0.09983$ , so $E_{0} = 0.1 + 0.9 (0.09983) = 0.18985$ . Newton's step size is $- f / f^{'}$ . Here $f (E_{0}) = 0.18985 - 0.9 sin (0.18985) - 0.1 = 0.18985 - 0.16989 - 0.1 = - 0.08004,$ $f^{'} (E_{0}) = 1 - 0.9 cos (0.18985) = 1 - 0.9 (0.98204) = 1 - 0.88384 = 0.11616,$ a small denominator. Then $E_{1} = 0.18985 - \frac{- 0.08004}{0.11616} = 0.18985 + 0.68908 = 0.87893 rad .$ Because $e$ is close to $1$ , $f^{'} = 1 - e cos E$ dips near zero — the curve $f (E)$ is nearly flat there, so a tiny vertical error maps to a big horizontal correction. This is exactly why high- $e$ orbits are numerically harder: the corrections are "violent," and the iteration takes many steps to settle. See the flat region in the figure.

Exercise 3.2

Show algebraically that $r = a (1 - e cos E)$ gives perihelion distance $a (1 - e)$ and aphelion distance $a (1 + e)$ , and identify which value of $E$ produces each.

Recall Solution

$cos E$ ranges over $[- 1, 1]$ .

Perihelion (closest): $r$ minimum when $cos E$ is maximum, i.e. $cos E = 1 \Rightarrow E = 0$ . Then $r = a (1 - e \cdot 1) = a (1 - e)$ . ✓
Aphelion (farthest): $r$ maximum when $cos E = - 1 \Rightarrow E = π$ . Then $r = a (1 - e (- 1)) = a (1 + e)$ . ✓

Between them $r$ varies smoothly, matching the ellipse's near/far geometry.

Exercise 3.3

Kepler's equation $M = E - e sin E$ can also be solved by fixed-point iteration $E_{k + 1} = M + e sin E_{k}$ . State the condition on $e$ for this simpler scheme to converge, discuss where the iterate must start, and say why Newton is preferred for large $e$ .

Recall Solution

The fixed-point map is $g (E) = M + e sin E$ , with $g^{'} (E) = e cos E$ . The Banach/contraction test says the iteration converges if the map is a contraction on the interval we iterate in, i.e. $∣ g^{'} (E) ∣ \leq L < 1$ there. Since $∣ g^{'} (E) ∣ = ∣ e cos E ∣ \leq e$ for every $E$ , the bound $L = e$ holds on the whole real line, not just locally — so $g$ is a global contraction whenever $e < 1$ . That is stronger than a purely local result: any starting guess $E_{0}$ maps into a bounded region and the iterates form a Cauchy sequence converging to the unique fixed point.

Region of convergence: because the contraction bound is uniform, any $E_{0}$ works for $e < 1$ ; a natural and safe choice is $E_{0} = M$ (or $E_{0} = M + e sin M$ ).
Rate: the error shrinks by roughly the factor $e$ each step. For $e \to 1$ it barely shrinks (multiply by $0.9$ each time → very slow).

Newton-Raphson instead converges quadratically (error squares each step) near the root, so it beats fixed-point badly once $e$ is large — at the cost of needing $f^{'}$ and a decent seed. See Fixed-Point Iteration for Kepler.

Level 4 — Synthesis

Exercise 4.1

A satellite has $a = 7000$ km, $μ = 398600 km^{3} / s^{2}$ , $e = 0.1$ . Find the mean motion $n$ and the orbital period $T$ .

Recall Solution

$n = \frac{μ}{a ^{3}} = \frac{398600}{700 0 ^{3}} = \frac{398600}{3.43 \times 1 0 ^{11}} = 1.16210 \times 1 0^{- 6} = 1.07801 \times 1 0^{- 3} rad/s .$ $T = \frac{2 π}{n} = \frac{6.28319}{1.07801 \times 1 0 ^{- 3}} = 5828.7 s \approx 97.1 min .$ A sensible low-Earth-orbit period. See Mean Motion and Orbital Period.

Exercise 4.2

For that same satellite, $t - t_{p} = 600$ s. Find $M$ , then solve for $E$ (two Newton steps from $E_{0} = M + e sin M$ ), then find the distance $r$ . Use $n = 1.07801 \times 1 0^{- 3}$ rad/s, $e = 0.1$ , $a = 7000$ km.

Recall Solution

Step 1 — mean anomaly: $M = n (t - t_{p}) = 1.07801 \times 1 0^{- 3} \times 600 = 0.64681 rad .$ Step 2 — seed: $sin (0.64681) = 0.60266$ , so $E_{0} = 0.64681 + 0.1 (0.60266) = 0.70708 rad .$ Newton step 1: $f = 0.70708 - 0.1 sin (0.70708) - 0.64681 = 0.70708 - 0.06497 - 0.64681 = - 0.00470,$ $f^{'} = 1 - 0.1 cos (0.70708) = 1 - 0.07603 = 0.92397,$ $E_{1} = 0.70708 - \frac{- 0.00470}{0.92397} = 0.70708 + 0.00509 = 0.71217 rad .$ Newton step 2: $f = 0.71217 - 0.1 sin (0.71217) - 0.64681 = 0.71217 - 0.06537 - 0.64681 = - 0.00001,$ already machine-tiny, so $E \approx 0.71218$ rad. Step 3 — distance: $r = a (1 - e cos E) = 7000 (1 - 0.1 cos (0.71218)) = 7000 (1 - 0.1 (0.75648)) = 7000 (0.92435) = 6470.5 km .$

Exercise 4.3

Using $E = 0.71218$ rad and $e = 0.1$ from Ex 4.2, find the true anomaly $ν$ (the actual focus-measured angle defined in the symbol list above).

Recall Solution

$tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2} = \frac{1.1}{0.9} tan (0.35609) .$ $1.1/0.9 = 1.22222 = 1.10554$ , and $tan (0.35609) = 0.37193$ , so $tan \frac{ν}{2} = 1.10554 \times 0.37193 = 0.41118 \Rightarrow \frac{ν}{2} = arctan (0.41118) = 0.38984,$ $ν = 0.77968 rad .$ Note $ν > E > M$ : the body has run ahead of both the geometric and the uniform angle, because it is on the fast (near-focus) part of the orbit. Since $E$ is in the first half-plane $(0, π)$ , $ν$ is too, so the plain $arctan$ is safe here — see True Anomaly and the Orbit Equation.

Level 5 — Mastery

Exercise 5.1

Prove that for $e < 1$ Kepler's equation $M = E - e sin E$ has exactly one solution $E$ , and that Newton-Raphson started from any point in the physical range cannot get stuck (its denominator never vanishes).

Recall Solution

Fix a target $M \in [0, 2 π]$ and work on the physical interval $E \in [0, 2 π]$ (all other cases reduce to this by adding a whole number of $2 π$ to both $M$ and $E$ , since $sin$ has period $2 π$ ).

Existence and uniqueness. Define $f (E) = E - e sin E - M$ , continuous on $[0, 2 π]$ . Its derivative is $f^{'} (E) = 1 - e cos E$ , and because $∣ cos E ∣ \leq 1$ with $e < 1$ , $f^{'} (E) \geq 1 - e > 0 for all E .$ So $f$ is strictly increasing on $[0, 2 π]$ . Check the two endpoints: $f (0) = 0 - 0 - M = - M \leq 0, f (2 π) = 2 π - 0 - M = 2 π - M \geq 0,$ using $0 \leq M \leq 2 π$ . A continuous, strictly increasing function that is $\leq 0$ at the left end and $\geq 0$ at the right end crosses zero exactly once on $[0, 2 π]$ — this is the Intermediate Value Theorem applied inside the physical domain, and strict monotonicity forbids a second crossing. Hence a unique $E$ .

Newton is never stuck. Newton's update divides by $f^{'} (E_{k}) = 1 - e cos E_{k}$ . Because $∣ cos E_{k} ∣ \leq 1$ and $e < 1$ , we always have $f^{'} (E_{k}) \geq 1 - e > 0$ , so the denominator is bounded strictly away from zero and can never vanish — the update is well-defined for any starting $E_{0}$ . Since $f$ is monotone with no stationary points, the iterates cannot become trapped, and Newton-Raphson is guaranteed to home in on the single root. This completes the proof. See Newton-Raphson Method.

Exercise 5.2

A comet has $e = 0.967$ (Halley-like). At $M = 0.2$ rad, perform Newton-Raphson from $E_{0} = M + e sin M$ until $∣ f ∣ < 1 0^{- 6}$ . Report the number of steps and final $E$ . Comment on the speed compared with the $e = 0.3$ case.

Recall Solution

Seed: $sin (0.2) = 0.19867$ , $E_{0} = 0.2 + 0.967 (0.19867) = 0.39211$ .

$k$	$E_{k}$	$f (E_{k})$	$f^{'} (E_{k})$
0	0.39211	$- 0.17646$	$0.10787$
1	$0.39211 + 0.17646/0.10787 = 2.02790$	$0.96906$	$1.43046$
2	$2.02790 - 0.96906/1.43046 = 1.35544$	$0.20915$	$1.20441$
3	$1.35544 - 0.20915/1.20441 = 1.18178$	$0.02010$	$1.36135$
4	$1.18178 - 0.02010/1.36135 = 1.16702$	$0.00013$	$1.37262$
5	$1.16702 - 0.00013/1.37262 = 1.16693$	$\sim 4 \times 1 0^{- 8}$ ✓	—

Answer: about 5–6 steps, $E \approx 1.16693$ rad (a sixth step polishes $∣ f ∣$ below $1 0^{- 6}$ ). The very small $f^{'} \approx 0.108$ at the seed threw the first step wildly past the root to $2.03$ , then it clawed back over the next steps. Contrast $e = 0.3$ in Ex 2.2, which was essentially converged in 2 steps. High eccentricity = flat $f$ near the root = slow, dramatic convergence, exactly as predicted at L3.

Exercise 5.3

Sketch the full time-to-position pipeline as a flow of quantities and name the tool used at each arrow.

Recall Solution

$t \to M$ : trivial multiply by mean motion $n$ (linear, instant).
$M \to E$ : the hard step — transcendental, needs iteration (Newton-Raphson).
$E \to ν$ : half-angle tangent identity, watch the quadrant.
$E \to r$ : direct plug-in $r = a (1 - e cos E)$ .

Recall One-line self-test before you leave

The single genuinely hard arrow in the whole pipeline is which one, and which method conquers it? The $M \to E$ arrow (solving $M = E - e sin E$ ) ::: it is transcendental, so we use Newton-Raphson: $E_{k + 1} = E_{k} - \frac{E _{k} - e sin E _{k} - M}{1 - e cos E _{k}}$ .