Mean longitude and mean anomaly of a body are calculated as the angle covered by the body along a fictious circular orbit, and this is ok.

But where is this orbit centered? Five different sources say five different things:

Wikipedia: circle center = ellipse focus


University of Texas: who knows?!? No ellipse at all:

U Texas diagram, white bg

University of California: who knows? Mean anomaly mentioned but not drawn; only eclipse shown, no circle:

CSU Northridge diagram, white bg

Britannica Enciclopedia: who knows? Both circle and ellipse shown, but mean anomaly not shown:


Max-Planck-Institut für Sonnensystemforschung: circle centered on ellipse center

Max Planck Inst. diagram, white bg

Theory & formulas: longitudes and anomalies summed up together, no pictures at all.

I am very confused.

  • $\begingroup$ Fitzpatrick@UTexas illustrates true and eccentric anomalies in a diagram here. $\endgroup$ – Mike G Aug 25 '20 at 15:13
  • $\begingroup$ The Wikipedia illustration has scale problems, and probably should have been removed from the Mean Anomaly article years ago. $\endgroup$ – notovny Aug 25 '20 at 15:18
  • $\begingroup$ @notovny If the image on Wikipedia is wrong, anybody (also you) can remove/fix it. $\endgroup$ – jumpjack Aug 26 '20 at 7:24

But where is this orbit centered?

The true anomaly is the angle as measured from the central body between periapsis passage and the object's current location. The orbit is an ellipse with one of the two foci at the central body. This concept is central to Kepler's laws and Newtonian mechanics.

The eccentric anomaly is the angle as measured from the center of the ellipse between periapsis passage and the projection along the minor axis of the point in question onto the minimal bounding circle that encompasses the elliptical orbit. That minimal bounding circle is necessarily centered on the center of the orbit rather than the central body.

Mean anomaly is a fictitious angle. Unlike true anomaly and eccentric anomaly, mean anomaly does not indicate where the object is. Mean anomaly instead is the angular displacement that a fictitious object in a circular orbit about the central body with the same semi-major axis length as the object in question would have passed through in the time since periapsis passage.

Aside: The image from wikipedia is flat-out wrong. The center is correct, but the size is incorrect. The circular orbit should intersect the ellipse, twice. It should be outside of the ellipse at periapsis and inside the ellipse at apoapsis.
  • $\begingroup$ This answer does not answer. Question was: where is the fictitious orbit centered? I was not asking for definitions of the quantities. All sites say "Mean anomaly instead is the angular displacement that a fictitious object in a circular orbit about the central body", but none says where the circular orbit is centered. $\endgroup$ – jumpjack Aug 26 '20 at 7:19
  • $\begingroup$ @jumpjack - I wrote "Mean anomaly instead is the angular displacement that a fictitious object in a circular orbit about the central body". I can update the answer if it isn't clear to you from what I already wrote that the central body is the center of the circular orbit. $\endgroup$ – David Hammen Aug 26 '20 at 7:24
  • $\begingroup$ I'd write "circular orbit centered on ellipse focus". "Centered on central body" is not so helpful. $\endgroup$ – jumpjack Aug 26 '20 at 7:39
  • $\begingroup$ @jumpjack - An ellipse has two foci. Writing "centered on the ellipse focus" is ambiguous because ellipses have two foci, so I am not changing my answer to your suggestion. The central body is in fact one of the two foci of a Keplerian ellipse. The other focus has no physical meaning. $\endgroup$ – David Hammen Aug 26 '20 at 7:44

Fortunately all diagrams in the question are consistent in these respects:

  • the true anomaly $\nu$ or $f$ is measured around the focus at the central body
  • the eccentric anomaly $E$ is measured around the center of the ellipse
  • both are zero at the periapsis

The mean anomaly $M$ is a pseudo-angular quantity useful in computing $E$ and $\nu$, with the same zero and period but increasing linearly with time. $M$ is not a geometric angle around an actual point. There is no right place to draw it.

You could represent $M$ with a circle around the center of the ellipse, around the focus at the central body, or even around the other focus; any such choice is arbitrary. Better yet, leave it out of the diagram to avoid suggesting a geometric meaning where there isn't one.

  • $\begingroup$ After looking at Kepler equation, I think we can say that M is centered on ellipse center: M = E - e sin(E) , being E centered on ellipse center. This means that the fictitious orbit is a circle centered on ellipse center. Given that the major semiaxis is the reference parameter of the ellipse, we could assume that this would be the radius of the fictitious orbit for e=0, hence such orbit is a circle tangent to the ellipse. Hence in the end the Max-Planck-Institut für Sonnensystemforschung picture is the right one. "Centered on ellipse center" has also geometrical sense. $\endgroup$ – jumpjack Aug 26 '20 at 16:51
  • 2
    $\begingroup$ @jumpjack M = E - e sin E expresses a numerical relation, not a geometric one. The geometric center of M is undefined. If you must define one anyway, remember that you made it up. $\endgroup$ – Mike G Aug 26 '20 at 18:46
  • $\begingroup$ yes but what is confusing is that M = E - e sin E is a numerical relation, but for M is given a geometrical definition (the angle that IN THEORY a body would cover if it orbit was elliptic), and I have also formulas which sum up an angle centered somewhere to an angle centered (apparently) somewhere else ( aa.quae.nl/en/reken/zonpositie.html#5 ) , and to a further angle (capital $ \pi $) centered on same center of both the ellipse and the circle... $\endgroup$ – jumpjack Aug 27 '20 at 10:37
  • $\begingroup$ @jumpjack The quae.nl article's $\Pi$ is the heliocentric longitude of the perihelion, reckoned from the ellipse focus where the Sun is. Their equation (7) for $\lambda$ assumes that the orbital inclination $i$ is negligible. $\endgroup$ – Mike G Aug 27 '20 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.