Talk:Mean anomaly

Mean Anomaly

Does anyone know the purpose of calculating the mean anomaly (e.g., for asteroid or comet orbits)? Does the calculation give us a better understanding of the shape of the orbit (i.e., more than just knowing the Periapsis and Apapsis and Semi-Major Axis alone)? Thanks---Tesseract501, March 24, 2006..

 >> An answer: Imagine designing a constellation where you want even spacing between satellites. What does that even mean if the orbit is eccentric?  It can only refer to time of passage (of perigee, or any true anomaly). To get that even spacing, you use equal portions of mean anomaly. That took me way too long to realize! Markrkrebs (talk) 13:02, 12 October 2024 (UTC)

mean and true anomaly

Markrkrebs (talk) 13:19, 12 October 2024 (UTC)

AFAIK it simplifies the Keplerian equation. See Keplerian problem. --CiaPan 17:42, 22 May 2006 (UTC)

It also makes numerical modelling of planetesimals, planets and the like easier since mean anomaly is a linear function of time. AstroMark 11:48, 17 August 2007 (UTC)

The answer to the question is, "You need to calculate the Mean Anomaly first, in order to calculate the other geometric properties of the orbit."

The Mean Anomaly is the angle, at any time, t, in a circular orbit with period T. Thus M, the Mean Anomaly, M, in radians, at time t, is just M =(2*pi/T)*t = nt. We only need to know the period of the orbit to calculate M. It is the first step for calculation of the Eccentric Anomaly. For example, once we know T, we know n, the mean motion, n = 2*pi/T. Next, given t, we calculate M = nt. Next we can calculate the semi-major axis, a.

The page is wrong when it states that the Mean Anomaly is not an angle. Also unwise is the term "parameters" applied to the Eccentric and True Anomalies. They are angles, simply angles.

The equation M = E - e*sinE is certainly correct, but in the solution of Kepler's Equation it is bass ackwards. Given e and M, one finds E. That's the solution process, usually performed iteratively using Newton's Method, but there are many other ways. To get to the starting gate for solving Kepler's Equation, one needs e and M. M is found as stated. e, the eccentricity, may be calculated if, for example, one knows either the distance at periapse or at apoapse. See the wiki article for "Orbital eccentricity." Then, with M and e, we can solve for E, the Eccentric Anomaly. Consult any celestial mechanics book, such as J.M.A. Danby's "Celestial Mechanics," or "Fundamentals of Astrodynamics," by Bate, Mueller and White. Hendel (talk) 03:24, 25 June 2014 (UTC)

I substantially agree with Hendel's comments. I don't understand where the notion that mean anomaly "is not an angle" came from. Of course it is. It's an angle which increases uniformly from 0 to 2π during the period of the object's orbit. Tfr000 (talk) 03:23, 19 September 2015 (UTC)

Also, mean anomaly has little to do with Kepler's second law. Yes the angle "mean anomaly" sweeps out equal area in equal times, but this is simply because it increases uniformly with time. Gravity is not involved. Think of, perhaps, a clock's hand which moves in a circle with the same period as the planet in its elliptical orbit. Tfr000 (talk) 13:16, 20 September 2015 (UTC)

I went ahead and revamped the article. I removed some stuff which is more properly covered in other articles, cleaned up the text (a little redundant, like stub articles strung together) and added a graphic. I may add a bit to the "Formulas" section, and some more references, but it is essentially done. Mean anomaly is a fairly simple concept, so there's not all that much to write. Tfr000 (talk) 13:31, 18 October 2015 (UTC)

The circular orbit shown in the image wouldn't have the same period;its semimajor axis looks to be something like twice that of the ellipse. Notovny (talk) 00:40, 18 July 2016 (UTC)

I agree and think this is a serious problem with the graphic, since T is directly set by semijax, they should be drawn that way. 2601:280:5E01:E7F0:4CB4:3101:3013:3370 (talk) 12:57, 12 October 2024 (UTC)

Diagram

I would recommend revamping the diagram so that both the ellipse and the circle have the same period. Perhaps have the two not overlapping. An animation would of course be even better. I unfortunately don't have the skills to create it.Spaceman13 (talk) 19:43, 17 October 2017 (UTC)

They HAVE to be overlapping to have the same period. 2601:280:5E01:E7F0:4CB4:3101:3013:3370 (talk) 12:58, 12 October 2024 (UTC)