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Vector confusion

  1. Jul 10, 2006 #1


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    I'm trying to compute Mean Anomoly. Mean Anomoly is an angle that describes where in an orbit an object is relative to its periapsis. The formula according to Wikipedia is

    M = E - e \cdot \sin E
    where e is eccentricity (which I know), and E is Eccentric Anomoly.

    According to Wikipedia, Eccentric Anomoly is
    E = \arccos \frac{{1 - \left| r \right|/a}}{e}
    where a is Semi-Major Axis (which I know), e is eccentricity (which I know), and r is an orbiting body's position vector. If I solve for r, I'm home free.

    According to Wikipedia, The orbital position vector r is a cartesian vector describing the position of the orbiting body. Wouldn't that mean that r would be express as something like 1.0i + 2.0j+3.0k? But I can't plug something like that into the formula for Eccentric Anomoly. I could use Pothagorean Theorum to solve for distance and use this number. But that turns the vector into a scalar, and I doubt that's the right way to do it. Direction is pretty important here.

    I know the object's x, y, and z position wrt the reference frame. How do I turn r into something I can use in these formulas?
  2. jcsd
  3. Jul 10, 2006 #2
    The formula involves only the modulus of the vector r, doesn't it?
  4. Jul 10, 2006 #3


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    Modulus = absolute value??

    In my example with 1.0i + 2.0j + 3.0k, I could do r=sqrt(1^2+2^2+3^2). But there are two points on the orbit that would have this same value for r. So I wouldn't know which one I computed.
  5. Jul 10, 2006 #4

    IIRC, you are the author of an orbit simulation program, right? I'm a [exaggeration]novice[/exaggeration] when it comes at programming, so I don't understand the exact nature of your problem. Sorry. :redface:
    Last edited: Jul 10, 2006
  6. Jul 10, 2006 #5


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    Yes :biggrin: . And if you use the program, you'll notice on the orbital elements box I leave out Mean Anomoly, simply because I don't get it right. I see many references that give the same formulas as Wikipedia for Mean Anomoly. And they all have the variable r in them. r is a vector, not a scalar, which is why I'm confused as to how to turn an x,y,& z position into a single number that I can assign to r that conveys both distance and direction. From physics classes, I remember that any answer that is a vector is stated as a scalar plus either an angle, or with the ijk-hat method. But of course any programming language doesn't want to see hats in the code. And E=Arcos((1-ABS(15, northwest)/a)/(e)) will generate a syntax error.

    Thanks for your help thus far :smile:
  7. Jul 10, 2006 #6
    I have used your program before, and had fun generating all those patterns. :biggrin: But to fully appreciate a program such as this, I think I will need a decent understanding of celestial mechancis (I know it's not a necessity, but still I prefer it that way :) ). Keep up the good work, and I hope you get this problem solved.
  8. Jul 10, 2006 #7
    I could be wrong here, because I have no clue about oribtal mechanics (if that's what it's called :smile:)

    But, maybe this will help:
    So you are trying to calculuate [itex] E [/itex]

    http://en.wikipedia.org/wiki/Eccentric_anomaly" [Broken]by the following:

    [tex] E = cos^{-1} \frac{1-\frac{| \vec r |}{a}}{e}[/tex]

    Looking up the definition of [itex] r [/itex] and [itex] e [/itex] we have:

    http://en.wikipedia.org/wiki/Orbital_position_vector" [Broken]:
    [itex] \vec r [/itex]

    Since you are taking the modulus of this, you will return a scaler:
    ie) [tex] | \vec r | = \sqrt{x^2+y^2+z^2} [/tex]

    This is the length of the vector. So assuming my geometry is correct (I haven't really worked with ellipses), you will have at least two spots where [tex] | \vec r | [/tex] can be equal. For example if the ellipse was on 2d plane.

    [tex] | \vec r_1 | = |\vec r_2 | [/tex]
    [tex] \vec r_1 = (a,0,0) [/tex]
    [tex] \vec r_2 = (-a,0,0) [/tex]

    However, you are worried about a point on an ellipse, so you need some more information to make the angle unique (since [itex] \vec r [/itex] by itself is not doing it).

    Now if we look at [itex] e [/itex],
    "[URL [Broken]
    Orbital Eccentricity[/URL] ([itex] e [/itex])

    [tex] e = | \vec e | [/tex]

    http://en.wikipedia.org/wiki/Eccentricity_vector" [Broken]([tex] \vec e [/tex])

    [tex] \vec e = \frac{\vec v \times \vec h}{\vec \mu}-\frac{\vec r}{| \vec r |} [/tex]
    [tex] \vec v = [/tex] orbital velocity vector
    [tex] \vec h = [/tex] orbital angular momentum
    [tex] \vec r = [/tex] orbital position vector
    [tex] \vec \mu = [/tex] standard gravitational parameter

    It would seem that at the point in question, [itex] \vec v [/itex] would be unique, thus causing [itex] e = | \vec e | [/itex] to be the unique quanitity that is dependent on position.

    :blushing: Like I said, I could be way off here.

    EDIT: I don't know what's up with the LaTeX... I guess I have an error somewhere. I hope this is somewhat readable.
    Last edited by a moderator: May 2, 2017
  9. Jul 10, 2006 #8


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    The LaTeX is fine. It doesn't work in preview mode though. Thanks. I think I'm getting a grip on it now. Using
    [tex] | \vec r | = \sqrt{x^2+y^2+z^2} [/tex]

    I'm getting good results. But I would expect the angle to increase from 0 to 2pi over the course of an orbit. Instead, it increases from 0 to pi and then back to 0. If I subtract this angle from 2pi anytime the object is in the 3rd or 4th quadrant (negative y value), then all is well.

    It just seems from the referenced formulas that this type of a piecewise answer is not necessary. But since taking the absolute value of the vector will yield the same answer for two different positions, I guess it is.

    One silly mistake I was making was forgetting that the angle is measured from periapsis (closest point) rather than as a longitude (angle from a fixed direction, usually the Vernal equinox). But in a circular orbit, where the only hint of eccentricity comes 5 places to the right of the decimal, this point is not well defined and it drifts all over the place, causing Mean Anomoly to do the same. Once given a little eccentricity, things begin to work a lot better.
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