3.2.15 · D5Orbital Mechanics & Astrodynamics
Question bank — Solving Kepler's equation — Newton-Raphson iteration
The three angles you must keep straight throughout:
- = mean anomaly — a fake angle that grows perfectly uniformly with time.
- = eccentric anomaly — the geometric angle at the centre of the ellipse, read on the auxiliary circle.
- = true anomaly — the real angle of the body seen from the focus.
See Eccentric Anomaly and the Auxiliary Circle and True Anomaly and the Orbit Equation if any of these feels shaky.
True or false — justify
Kepler's equation has a unique solution for every .
True — because has derivative for all (since ), so is strictly increasing and crosses zero exactly once.
For a circular orbit () the mean and eccentric anomalies are identical.
True — with the equation collapses to , meaning the body genuinely moves uniformly, so there is nothing to iterate.
You can compute from instantly, but not from .
True — going is just plug-and-evaluate, but going requires solving a transcendental equation where appears both alone and inside , which cannot be algebraically isolated.
gives the correct answer whether you work in degrees or radians.
False — it is exact only in radians, because (a pure number) is added directly to (an angle), and that balance only holds when the angle is measured in radians.
For the eccentric anomaly is always larger than the mean anomaly .
True — on the way from perihelion to aphelion , so ; physically the body has sped past the uniform clock-hand near the focus.
The correction term vanishes at both perihelion and aphelion.
True — at perihelion and at aphelion , and , so exactly at both apsides.
Newton-Raphson always converges to the correct root regardless of the seed .
False — for very high a poor seed can overshoot or converge painfully slowly because becomes tiny near perihelion, making early steps violent; the seed choice genuinely matters.
The distance is largest when .
True — gives , the aphelion distance, which is indeed the maximum since is minimised there.
Doubling the eccentricity always doubles the number of Newton iterations needed.
False — iteration count grows non-linearly and depends on both and where sits; near-parabolic () orbits blow up in difficulty far faster than any linear rule predicts.
The true anomaly and eccentric anomaly are always in the same half of the orbit.
True — both are measured from perihelion in the same rotational sense, so they cross , , and together and never sit in opposite half-planes.
Spot the error
"The derivative of is ."
Wrong — you copied the from Kepler's equation instead of differentiating it; , so the correct slope is .
"Since is constant here, Newton-Raphson is just a single division."
Wrong — depends on , so it must be re-evaluated at every guess ; treating it as constant breaks the update entirely.
"I'll seed with and iterate in degrees."
Two errors — the equation must be in radians (convert first), and although is a fine seed here, the degree mistake will make every term nonsense.
"Because at my guess, the true root must be smaller, so I decrease ."
Backwards — is increasing, so means the guess is below the root; Newton's automatically adds a positive amount and moves up.
"I solved with , so is done."
Incomplete — plain returns a value in and drops the quadrant, so you must use the half-angle sign (or
atan2) to place in the same half-plane as ."My wrong still converged to the right answer, so the formula must be fine."
Misleading — a wrong-but-positive slope can still limp toward the correct root, but you lose the quadratic convergence that makes Newton worth using; correctness of the limit hides a broken speed.
" is the fraction of the orbit's angle swept from perihelion."
Wrong — by Kepler's 2nd law is the fraction of the period (and hence of the swept area) elapsed, not the fraction of geometric angle; the angle sweeps non-uniformly.
Why questions
Why do we invent the fictitious mean anomaly at all?
Because time and geometric angle are not proportional on an ellipse; is defined to grow linearly with time (), giving us a clean uniform handle that we can later convert to the real angle.
Why is Newton-Raphson preferred over plain fixed-point iteration ?
Newton converges quadratically (roughly doubling correct digits each step), whereas fixed-point converges only linearly and slows to a crawl as ; Newton uses the slope to aim, not just repeat.
Why does the seed work better than ?
It already applies one first-order correction for the term, landing the initial guess near the root so fewer iterations are needed — it is essentially one manual step of the correction built in.
Why is the root of Kepler's equation guaranteed unique?
Because makes strictly positive everywhere, so never turns around; a strictly monotonic function can cross zero at most once.
Why does the term physically exist?
It corrects for the focus being offset from the ellipse's centre by ; the equal-areas law measured from the focus produces exactly this centre-to-focus triangle term.
Why does high eccentricity make the iteration "violent"?
Near perihelion , so becomes tiny; dividing by a small slope produces enormous first steps that can overshoot the root.
Why must be obtained via a half-angle formula and not simply set equal to ?
is measured from the centre and from the focus, so they differ except at the apsides; the half-angle tangent relation converts one geometric viewpoint to the other.
Edge cases
What happens to Kepler's equation when ?
It becomes : the orbit is circular, motion is uniform, and no iteration is required at all.
What happens as (near-parabolic)?
can approach zero near perihelion, so Newton becomes unstable there; a robust seed like near the apside, or a specialised near-parabolic formulation, is needed.
What is when ?
— perihelion, where exactly because , so no correction and no iteration are needed.
What is when ?
— aphelion, again exactly since ; the two symmetric apsides are the only places and coincide for .
For just past (e.g. ), is the correction positive or negative?
On the return leg gives , so and now — the body lags the uniform clock-hand while far from the focus.
If your computed comes out negative for a small negative , is that valid?
Yes — Kepler's equation is odd-symmetric ( gives ), so negative anomalies simply describe the pre-perihelion half of the orbit and are perfectly legitimate.
What if is already below your tolerance at the seed?
Then you accept immediately with zero iterations — the seed happened to land on the root, which is common for small near an apside.
Recall One-line self-test
Cover everything above and answer aloud: Why can't we solve algebraically, and what single property guarantees Newton will find the one right ? Answer ::: appears both bare and inside so it can't be isolated (transcendental); and for all guarantees a strictly increasing with exactly one root.