Hi Tom, thanks for the kind comment! I think this is the best place to communicate feedback, although email [email protected] would work too.

I also appreciate the suggestion. If you'd like to make that improvement, feel free to edit the relevant pages and submit a pull request!

Best, Bryan

Suggested text:

While the mean anomaly can be (as we have above) introduced simply as stand-in for the time, it can be seen more physically as the fractional area of the ellipse that has been swept out since the last perihelion passage. Kepler’s second law guarantees that the rate of change of the mean anomaly is constant, so the mean anomaly is proportional to time.

As shown above it is not easy to directly translate time or the mean anomaly area directly to the true anomaly angle so an intermediate quantity the eccentric anomaly is also used. The position in the orbit on the ellipse is referred to a nominal position on a path in a circle circumscribing the actual orbit. It is easy to transform the area representing the mean anomaly into the equivalent area in the circumscribing circle, and similarly to transform the true anomaly to the matching angle on the circle.

For a circle, angles and swept out areas are proportional to one another which allows us to use this eccentric anomaly in/on this circle as a link between the mean and true anomalies. With the eccentric anomaly in hand, we can transform from mean anomaly area in the ellipse to eccentric anomaly area in the circle, immediately get the corresponding angle on the circle, and then transform that angle back to true anomaly angle on the ellipse. Or vice versa.

For unbound orbits a similar process is used: the intermediate quantity is usually called hyperbolic anomaly, and instead of a circle a hyperbola with eccentricity sqrt(2) is the reference figure.

Here's a very rough version of how I'd update the suggested text:

Due to Kepler's second law, the area swept inside the ellipse during equal times is equal. Therefore, we can define a quantity which represents the area swept out in one arbitrary time unit. Let's call this the mean anomaly. Since this is changing at the same rate for all time, it provides a simple connection between the swept area and the time. Shown below in the figure.

However, this area doesn't directly translate to calculating the true anomaly at that time, which is really what we're interested in to know the exact position of an object at any time or vice versa. So we need to define something to translate between the mean anomaly and the true anomaly. In terms of areas, we can think of stretching the ellipse so that it becomes a circle.

Circles are nice because the sector area for a given angle is also directly proportional to the angle. So let's draw a line through the true anomaly and up to the circle. A bunch of trig and we find that this angle in terms of the true anomaly is E = 2 atan(...). Then the sector area is E/2 * a2. Now draw a line from the focus to the eccentric point and subtract the area of the triangle from the origin, you get 1/2 a2 (E - e sin(E)). Scale back to the ellipse and you get 1/2 a * b * (E - e sin(E)).

Finally, relate this to the mean anomaly area. The mean anomaly area is 1/2 a * b * M. Since this area in terms of mean anomaly must equal the area in terms of eccentric anomaly, M = E - e sin(E) or Kepler's equation.

Reference: https://math.stackexchange.com/questions/388134/how-to-calculate-ellipse-sector-area-from-a-focus