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Units problem in Keplerian equation

  1. Feb 2, 2012 #1
    See the arrow in the jpg file. There must be a unit missing somewhere, since a is distance, and is used as the sqrt(a).
     

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  2. jcsd
  3. Feb 2, 2012 #2

    Simon Bridge

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    prev chapter - derivation of Kepler's formula.
    Go carefully through what everything is.
     
  4. Feb 2, 2012 #3
    According the the previous chapter, a = b/sqrt(1-e**2), where b is the the semi-minor axis, which is a distance. r is a distance. E' presumably is a time derivative of the eccentric anomaly, which is presumably dependent upon time (degrees/sec or whatever). e is dimensionless. This is described in Herbet on page 31 in deriving the elements. The figure it is derived from shows an ellipse, parabola, E, v (true anomaly), etc. There is little mention of units. Distance is distance, meters, feet, AU, whatever.

    I'm converting an old computer program for meteor analysis according to a 60's book that has the code in it. This equation is not mentioned in the book, but a colleague working on it with me brought up the equation. Herget gives 9 equations, while the book shows 9 for the equatorial elements, and five more for the 3 elements on the ecliptic. Herget's chapter is on the Laplace method, while the book may be using some other part of Herget. So maybe I'm chasing a red herring. Neither of us have training in orbit calculations, but I am enjoying learning about them. This is a private effort.

    Are you familiar with the Herget book?
     
  5. Feb 2, 2012 #4

    Simon Bridge

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    I'd have expected E to be dimentionless too - anything inside a trig function needs the same dimentions as angle - eg. none.

    "But," says someone from the cheap-seats, "Can't I measure x meters and then take the cosine of that so the thing inside the trig function has dimentions of length?"

    Ah, but then you are taking cos(kx): k=1m-1 ... which means the wavelength is twice pi in meters. Thus kx is the phase angle in radiens. So maybe the E you measure has dimensions - but it's appearance as the argument in a trig function implies it is multiplied by 1 with dimensions [E]-1

    Bearing this example in mind - see if you can derive the last formula from Kepler's equation. You are correct - some extra thing has been implied just like k was implied above. The derivation will tell you where.

    This is somtheing you learn to watch for when you are reading technical papers.
     
  6. Feb 4, 2012 #5
    I think one of my fellow workers figured it out. I don't have time now to demonstrate how, but he started with the second law, and differentiated it, then made a substitution. One really has to pay attention to units.
     
  7. Feb 5, 2012 #6

    Chronos

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    I could tell you the sad story about an undergrad who calculated the mass of Jupiter exceeded that of the sun due to a perfectly innocent units error. It was only two orders of magnitude, for goodness sake. Not like something huge. It could happen to anybody.
     
  8. Feb 5, 2012 #7
    Good quotes.
     
  9. Feb 8, 2012 #8
    In a book on Celestial Mechanics by J. Tatum, I find this for a circular orbit:

    "if we suppose that a planet of mass m is in a circular orbit of radius a around a Sun of mass M, M being supposed to be so much larger than m that the Sun can be regarded as stationary, we can just equate the product of mass and centripetal acceleration of the planet, maw**2, to the gravitational force between planet and Sun, GMm/a**2; and, with the period being given by P = 2*pi/w, we immediately obtain the third law:" (w is omega, angular speed, w=2*pi/P)

    P**2 = ((4*pi*2)*a**3)/GM (eq 9.2.1).

    P is in unit of time. GM=mu with units of Km**3/s**2, where s is seconds.

    As an exercise, he asks:

    Exercise. Express the period in terms of a, the radius of the planet’s circular orbit
    around the centre of mass.

    Re-arranging, a**3 = (P**2)*(GM)/(4*pi**2)

    As I see it, the units are: a**3 = ((s**2)*(Km**3)/s**2) [4*pi** unitless]

    So a**3 units are : Km**3. I have no idea how this relates to the (sqrt(a**2) = P in my original post. Apparently, one needs to turn to an elliptical orbit and examine the relationships in the same way. Perhaps the author has more to say on that the elliptical orbits.
     
  10. Feb 8, 2012 #9

    Simon Bridge

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    ... looks awkward doesn't it? Is that supposed to be a pi-squared in there?
    I take it that "**" indicates "raised to the power of" as in p**2 == p^2 = p2?
    Try this [qute me to see what I did]:
    [tex]p^2 = \frac{\;\; 4\pi^2a^3}{GM}[/tex]... clearer? Now: what was the question?
    ... perhaps indeed.
     
  11. Feb 15, 2012 #10

    BobG

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    There's more than one way to skin a cat, and that particular page seems to describe one of them.

    I'd use https://www.amazon.com/Fundamentals-Astrodynamics-Applications-David-Vallado/dp/1881883140, simply because other people use it (makes communication easier).

    The author of your book also has a very stilted way of speaking. He must be very full of himself. Books written for, or at least funded or inspired by, the military are a little easier to read, since the military likes people to speak clearly and be understood by others.

    But it is pretty cool to see someone use Kepler's equation in degrees. The book was obviously written in the slide rule era.

    But was there really a need to list the number of degrees in one radian to 9 significant digits in the 60's? I just don't believe his observations were accurate enough to worry about that many significant digits.
     
    Last edited by a moderator: May 5, 2017
  12. Feb 15, 2012 #11

    BobG

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    I think that's written in Fortran and the ** does indeed represent "raised to the power of".
     
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