Why is Mean Anomaly treated as an angle?

[Kaleb]

So I've been refreshing my grasp of orbital mechanics recently, and reminding myself of the whole true/eccentric/mean anomaly thing. I think I've got a pretty good grasp of them:

- True Anomaly (f) is the actual angle from the focus of the orbit to the position of orbiting body (starting at periapsis).

- Eccentric Anomaly (E) is a geometrical convenience: the angle from the center of a circumscribed circle (not the orbit's focus) to the projection of the orbiting body onto said circle (again, starting at periapsis).

- Mean Anomaly (M) is a "fictional" angle that varies linearly in time from 0 to 2À during the course of a period. I *think* it represents the angle where the satellite would be at a given time, if it were on a circular orbit around the focus. But wikipedia is clear that it is actually a parameter, not an angle.

The mean anomaly (M) can be found from the eccentric anomaly (E) and the eccentricity (e) via Kepler's Equation (which has no closed-form inverse): M = E - e*sin(E). The mean anomaly is useful for finding the time at a certain point in an orbit, since the time of a given mean anomaly is just t = M/n (assuming M = 0 at t = 0), where n is called the mean motion, and is related to the orbital period (P) by n = 2À/P. The mean motion is the constant angular velocity of this fictitious angle M.

From what I've seen so far, it occurs to me that there's not really any advantage to treating M like an angle. It doesn't represent a physical angle, or interact with other physical angles, and it isn't involved directly in any trigonometric relations. Since it's just a parameter, what makes sense to me is to normalize it, by dividing by 2À, so that it varies from 0 to 1 during the course of an orbital period. Similar to the way that Plank's constant was normalized to create h-bar by dividing by 2À, we could define M-bar = M/2À.

Then, to get the time at a location in orbit, instead of dividing by a fictional angular velocity (t = M/n), we simply multiply by the orbital period (t = M-bar * P). This shows that M-bar, rather than being a fictional quantity, is merely the fraction of the orbital period completed thus far (t/P). Furthermore, we don't even need to define mean motion at all. We've "refactored" to eliminate two intermediate fictitious variables, and replaced them with a simple fraction of an existing physical variable. This seems much simpler and more intuitive to me than a non-existent pseudo-angle and its pseudo-velocity. (For example, saying M-bar = .25 makes it instantly clear that an object is a quarter of the time into its orbit, whereas M = 1.5707 may not be immediately recognizable as À/2, and therefore one quarter of 2À. And that's with just a simple fraction.)

So my question is, why don't we do this? I know its just a simple matter of scaling, but what is the point of scaling M to the range of 0...2À rather than 0...1? Is this merely a case of "this is the way we've always done it, and it works well enough", or is there some benefit somewhere to keeping M as an angle? Maybe a trigonometric relation that I'm unaware of? I assume Kepler originally derived it in terms of angles, but that doesn't mean we need to stick with that notation if there's a cleaner, functionally equivalent notation. Or maybe people already do this, and I'm just not aware of it?

Just my random thoughts.

BTW - If you've ever heard of the "Tau Manifesto" and "Pi is Wrong" movement... I had to resist the urge to replace all the 2À's in this post with Ä's. I figured one unconventional notation was enough for the time being.

Edited July 13, 2013 by Kaleb

[Yourself]

So my question is, why don't we do this?

Because that normalization factor is going to leak out somewhere. In this case, it alters Kepler's equation:

Mbar = ( E - e * sin(E) ) / ( 2 À )

You're also missing that the equation for the mean motion looks like this:

n = sqrt( μ / a³ )

This is because the orbital period already has a 2À in it:

P = 2 À sqrt( a³ / μ )

So actually computing the mean motion in the range 0-2À doesn't really involve ever using À at all:

M = t sqrt( μ / a³ )

So I would be able to compute the eccentric anomaly without ever having to deal with a À term in any of my equations.

BTW - If you've ever heard of the "Tau Manifesto" and "Pi is Wrong" movement... I had to resist the urge to replace all the 2À's in this post with Ä's. I figured one unconventional notation was enough for the time being.

I have heard of it and I think it's stupid. This article adequately sums up why I think that: http://www.thepimanifesto.com/

[Kaleb]

Because that normalization factor is going to leak out somewhere. In this case, it alters Kepler's equation

Correct, the 2À would appear there instead. But that formula is already computing a parameter (M) from an angle (E), so I don't see any problem with introducing a normalization factor there. And the reverse formula has to be solved iteratively anyway.

So actually computing the mean motion in the range 0-2À doesn't really involve ever using À at all:

M = t sqrt( μ / a³ )

Granted, this is maybe a good point, but it assumes you don't already have the period computed. If you do, there's no reason to recompute that square root -- it's just as easy to say

M = 2À t / P

But either way, since the resulting variable is in the range 0-2À, you'd have have to divide by 2À if you actually wanted to know how far into the orbit something was.

Which is easier to understand?

M = 0.78539816

or

M-bar = 0.125

I'd contend that the later is more recognizable as 1/8th of an orbit than the first one.

I think that's the core of my problem with M -- as a standalone variable, it doesn't seem to represent anything particularly useful, but if you rescale it, it does.

So I would be able to compute the eccentric anomaly without ever having to deal with a À term in any of my equations.

Did you mean mean anomaly here? Or are you talking about iteratively solving Kepler's equation for E? If the later, again, I don't see what the problem is with adding an extra normalization term into an equation that converts from a parameter to an angle. Especially since it already requires numerical methods to solve. If the former, I'm still not seeing the usefulness of calculating M on the range 0-2À.

This article adequately sums up why I think that

I'll have to give it a look. Thanks!

[Yourself]

But either way, since the resulting variable is in the range 0-2À, you'd have have to divide by 2À if you actually wanted to know how far into the orbit something was.

And, except in a few circumstances, knowing how far into the orbit you are as a fraction of the total orbit doesn't actually tell you where you are. If you want to know where you actually are (the true anomaly), you're going to end up using the mean anomaly as it's currently defined (see below). There's really no way around that.

I'd contend that the later is more recognizable as 1/8th of an orbit than the first one.

Okay, but why do we need to do that? If I'm computing the mean anomaly, it's because I intend to use it to compute the eccentric anomaly (and ultimately the true anomaly), so I don't much care what it's actual value is.

I think that's the core of my problem with M -- as a standalone variable, it doesn't seem to represent anything particularly useful, but if you rescale it, it does.

Not everything needs to represent a physical quantity in a natural way. Consider also that the mean anomaly can be used with hyperbolic orbits as well, which don't have an orbital period at all. However, they still have a mean motion: n = sqrt( μ / (-a)³ ) and, thus, the mean anomaly is still given by M = n t (more generally we can define the mean motion for any orbit with e != 1 to simply be n = sqrt( μ / |a|³ )). In this case, I end up solving for a different form of the eccentric anomaly:

M = e sinh(F) - F

From which I can get the true anomaly via:

cosh(F) = ( e + cos(θ) ) / ( 1 + e cos(θ) )

Or, solving for cos(θ):

cos(θ) = ( e - cosh(F) ) / ( e cosh(F) - 1 )

Normalizing the mean anomaly here would definitely be useless because it doesn't represent the fraction of an orbital period, because the orbital period is undefined. The current definition of the mean motion is simply the most natural definition.

If the later, again, I don't see what the problem is with adding an extra normalization term into an equation that converts from a parameter to an angle. Especially since it already requires numerical methods to solve. If the former, I'm still not seeing the usefulness of calculating M on the range 0-2À.

Because then iteratively solving Kepler's equation will have extra terms in it, terms that typically aren't necessary. For example, the typical iterative solution using Newton's method ends up looking like this:

E[n+1] = E[n] - ( E[n] - e cos( E[n] ) - M ) / ( 1 - e cos( E[n] ) )

Introducing that normalization factor modifies this slightly:

E[n+1] = E[n] - ( E[n] - e cos( E[n] ) - 2 À Mbar ) / ( 1 - e cos( E[n] ) )

That multiplication by 2 À is largely unnecessary. Basically you end up taking the normalized mean anomaly and putting it right back into the typical form regardless. From a numerical standpoint this is undesirable since the extra operations can introduce errors.

[Kaleb]

If I'm computing the mean anomaly, it's because I intend to use it to compute the eccentric anomaly (and ultimately the true anomaly), so I don't much care what it's actual value is.

Aha! I think this is where our perspectives differ: I don't care so much about *where* as *when*. I've only been going the other way in my computations: I know certain positions along the orbit, and I need to know the time when the object will reach that position, in order to set up a rendezvous. Expressing those times as fractions of the orbital period is useful, because then I can compute transfer orbits with a period that will create an intersection at the proper location and time. Once I know the ratio between orbital periods, it's a simple deal to find the ratio of the semi-major axes, and so on.

Consider also that the mean anomaly can be used with hyperbolic orbits as well, which don't have an orbital period at all.

Interesting... Point conceded (tentatively). I hadn't considered the generalization to e>1.

Because then iteratively solving Kepler's equation will have extra terms in it, terms that typically aren't necessary.

Yeah, in that case I would expect M-bar to be converted to M first, and then solve the equation. You wouldn't want those extra operations every iteration. Like I said, I've (so far) always been going the other way, from true anomaly to mean anomaly.

Edited July 16, 2013 by Kaleb