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Calculating elliptical orbits

  1. Apr 25, 2009 #1
    I've been doing some calculations but am obviously too dumb to work out something which should be straightforward (and no, this isn't homework)

    In a circular orbit, the acceleration is related to the tangential velocity and the radius via

    a=v^2/r

    In an elliptical orbit, both the velocity and the acceleration are constantly changing. At two points in the orbit, the radial acceleration is zero, for example.

    Q: Is it possible to infer the distance to the center of mass, knowing the 2-velocity and 2-acceleration at any instance of time?
     
  2. jcsd
  3. Apr 25, 2009 #2
    what do you mean by" 2-velocity and 2-acceleration"
     
  4. Apr 25, 2009 #3
    OK- 3-acceleration and 3-velocity then.

    I only mention 2-velocity and 2-accleration because in an elliptical orbit (like a planet around the sun) the orbit is executed on a plane.
     
  5. Apr 25, 2009 #4

    Nabeshin

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    I think kof was more confused by your wording, as am I, than anything else.

    Do you mean if you make 2 (or 3) measurements of velocity/acceleration of a satellite, you can infer the orbit? (P.S all orbits are executed in a plane, not just ellipses)
     
  6. Apr 25, 2009 #5
    NO.

    3-acceleration and 3-velocity means that you can take measurements of acceleration and velocity in 3 orthogonal directions.
     
  7. Apr 25, 2009 #6

    Nabeshin

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    Well, if you knew the full acceleration vector, and knowing it's due completely to gravity, a simple newton's 2nd law and universal law of gravitation should yield the distance to the COM.
     
  8. Apr 25, 2009 #7
    Prove it.
     
  9. Apr 25, 2009 #8

    Nabeshin

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    I don't know if latex is working...

    F=m2a=Gm1m2/r^2

    r=(Gm1/a)^(1/2)

    That's all there is to that one. The thing is, you don't usually have an acceleration measurement.
     
  10. Apr 25, 2009 #9
    Actually, you can calculate the entire elliptical orbit from just the state vectors, ie. the position and velocity vectors. I can't remember how I did it just now, but a few years ago I made a 2d gravitational simulation program which did just that.

    Cheers,
    Mike
     
  11. Apr 25, 2009 #10
    Not what I'm looking for: the G and M1 data are not known accurately enough for my purposes.

    What I need is a vector equation relating velocity, acceleration and distance to center of mass in the case of an elliptical orbit.
     
  12. Apr 26, 2009 #11

    Nabeshin

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    Umm, G is known to pretty high precision so you must be looking for a damn good measurement. Like h4tt3n said, you can calculate the elliptical orbit from the state vectors, but this involves the position vector which it sounds like you don't have.

    Off the top of my head I can't think of any vector equations relating the three quantities you mentioned (Measurements of these values are probably going to be a lot more imprecise than the measurement of G though...). I'll think about it more in the morning, but if you could describe what you need this for, that information might prove useful.
     
  13. Apr 26, 2009 #12

    tony873004

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    You can compute your semi-major axis with
    a = (2 / R - V ^ 2 / Mu) ^ -1
    where mu is G*M, G is the gravitational constant, and M is the mass of the sun.
    where R is sqrt(Rx^2+Ry^2+Rz^2), where R is the magnitude of your position vector
    where V is sqrt(Vx^2+Vy^2+Vz^2), where V is the magnitude of your velocity vector

    You can then compute your eccentricity in the following manner:

    Hx = Ry * Vz - Rz * Vy
    Hy = Rz * Vx - Rx * Vz
    Hz = Rx * Vy - Ry * Vx
    H = Sqr(Hx ^ 2 + Hy ^ 2 + Hz ^ 2)

    p = H ^ 2 / Mu
    q = Rx * Vx + Ry * Vy + Rz * Vz ' dot product of r*v

    E = Sqr(1 - p / a) ' eccentricity

    With your SMA and your eccentricity, you should be able to answer your question.
     
  14. Apr 27, 2009 #13
    But I'm not trying to calculate the eccentricity and I don't know what the value of the semi-major axis is!

    I'm trying to work out how far the centre of mass is given the velocity and acceleration at a point in time and knowing that the orbit is elliptical!
     
  15. Apr 27, 2009 #14

    Chronos

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    Keplers law still works pretty well.
     
  16. Apr 27, 2009 #15

    D H

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    G is one of the least well-known physical constants: four decimal places of accuracy. We know the fine structure constant to eleven digits or so. Astronomers rarely use G; they use μ=G*M instead because the standard gravitational parameter μ can be measured independently of either G or M and can be measured to a high degree of accuracy. We know μ for the Sun to ten places of accuracy, for example.

    You can't if all you know is the velocity and acceleration. Think of it this way: The gravitational acceleration toward an object of mass M and distance r is -GM/r2, directed toward the mass. Double the distance and quadruple the mass and you get exactly the same acceleration, but a very different orbit.
     
  17. May 8, 2009 #16
    Hello Diamondgeezer
    I wanted to react on your first alinea, but now I have read the second it has become more complicated. I have to do that in short time because if I take to long I am asked to login again and after that my quote has dissapeared, a very boresome technical problem I have encounted earlier but have to live with it.
     
  18. May 8, 2009 #17
    OK to your first alinea I wanted to say it is a beginvalue problem then, but that you state in your second alinea. The velocity known is OK but the acceleration has two parts Newtonian attraction and centrifugal acceleration. The remark of Chronos about Kepler is correct... So I would say you need at least the surface of the ellipse. Or could you add the actual distance to the sun to your list of beginvalues. Knowing that it is an ellipse is a limiting value and not a number. From that you only know that the total energy = potential energy + kinetic energy is below zero
    greetings Janm
     
  19. May 10, 2009 #18
    You can use Kepler's law p^2=a^3.

    The period squared is equal to the semi-major axis cubed.
     
  20. May 10, 2009 #19

    D H

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    There's only one problem with using Kepler's law (well, two problems). He doesn't know the period and he doesn't know the semi-major axis.
     
  21. May 10, 2009 #20
    Hello tony873004
    Somewhere here must lie the anwer. I have problems with the unities a=>L/T , R=>L , Mu=>L^3/(WT^2), V^2=>L^2/T^2
    so V^2/Mu=>W/L, with L=lenght T=time and W=Weight
    with this formula correct R can be calculated from V and a.

    greetings Janm
     
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