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Keplerian Orbits

  1. May 1, 2009 #1
    In a Keplerian orbit (an ellipse), the acceleration vector always points to the focus containing the central mass. Since that means the acceleration vector is no longer orthogonal to the orbit (except at two points), does the relation [tex]a = \frac{v^2}{r}[/tex] still apply if we consider v to be the velocity orthogonal to the radius vector?
     
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  3. May 1, 2009 #2

    Ben Niehoff

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    There is no a priori reason that it should, although it might by coincidence.

    Keplerian orbits follow the areal law: the radius vector sweeps out equal areas in equal times.

    It should take only a bit of algebra to show how this relates to the velocity orthogonal to the radius.
     
  4. May 2, 2009 #3

    D H

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    In general, no. At two very specific locations on the orbit, yes.
     
  5. May 2, 2009 #4
    What I really require is an idea of how tangential velocity and radial acceleration are related in Keplerian orbits. Any clues?
     
  6. May 2, 2009 #5

    D H

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    Is this a homework problem, or you you simply curious?
     
  7. May 3, 2009 #6
    I am curious, and no its not homework. I'm working on a problem of my own using data from JPL Horizons and I'm a little stuck on this issue.
     
  8. May 3, 2009 #7

    D H

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    OK.

    First, some nomenclature. You mentioned tangential velocity in post #4. The velocity vector is tautologically tangent to the orbit, so the tangential velocity is just the velocity. You meant the transverse (not tangential) component of the velocity vector.

    (I'm having troubles with LaTeX, so I'll have to make do with (yuck) HTML math. Bear with me.)

    In polar coordinates, the radial vector from the Sun to a planet at a radial distance r from the Sun and a true anomaly θ is

    r = r rhat

    where rhat is the unit vector

    rhat = cosθ xhat + sinθ yhat

    Here xhat points in the direction of perihelion and yhat is normal to xhat such that the velocity vector is directed along the yhat axis when the planet is at perihelion.

    The velocity vector in polar coordinates is

    v = r θdot θhat + rdot rhat

    You are looking for the transverse component of velocity, vt = r θdot. This is just h/r, where h=r2 θdot. h is the specific orbital angular momentum (angular momentum divided by the reduced mass), which is a constant of the orbit:

    h2 = α G (M_s+M_p)

    where α is the semi-latus rectum of the orbit. The semi-latus rectum α, the semi-major axis a, and eccentricity e are related via α=a(1-e2).

    Dividing by r2,

    h2/r2 = α G(M_s+M_p)/r2

    The left hand side is the square of the transverse component of velocity while the right hand side is the product of the semi-latus rectum and the relative acceleration. In short,

    vt2 = α acc
     
  9. May 4, 2009 #8
    I need to clarify three things. In Keplerian motion:

    1. The acceleration a is the radial acceleration.

    2. The transverse velocity [tex]v_{trans}[/tex] is normal to the orbit at any point in the orbit

    3. At any point in the orbit, the fraction [tex]\frac{v_{trans}^2}{a}[/tex] is always equal to the semi-latus rectum.

    Correct?
     
  10. May 4, 2009 #9

    D H

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    The transverse velocity is the component of the velocity vector normal to the radius vector, not to the orbit. Other than that, yes, you have it correct.
     
  11. May 4, 2009 #10
    I'm just recopying your reply into tex so I can see the math in its proper form

    In polar coordinates, the radial vector from the Sun to a planet at a radial distance r from the Sun and a true anomaly θ is

    [tex]\hat r = \frac{\vec{r}}{|r|}[/tex]

    where [tex]\hat r[/tex] is the unit vector

    [tex]\hat r = \cos(\theta ) \hat x + \sin(\theta ) \hat y[/tex]

    Here [tex]\hat x[/tex] points in the direction of perihelion and [tex]\hat y[/tex] is normal to [tex]\hat x[/tex] such that the velocity vector is directed along the [tex]\hat y[/tex] axis when the planet is at perihelion.

    The velocity vector in polar coordinates is

    [tex]\vec{v} = \vec{r}\dot{\theta} \cdot \hat{\theta} + \vec{r} \cdot \hat r[/tex]

    You are looking for the transverse component of velocity, [tex]v_t = \vec{r} \dot{\theta}[/tex]. This is just [tex]h/r[/tex], where [tex]h=r^2 \dot{\theta}[/tex]. h is the specific orbital angular momentum (angular momentum divided by the reduced mass), which is a constant of the orbit:

    [tex]h^2 = \alpha G (M_s+M_p)[/tex]

    where [tex]\alpha[/tex] is the semi-latus rectum of the orbit. The semi-latus rectum [tex]\alpha[/tex], the semi-major axis a, and eccentricity e are related via [tex]\alpha=a(1-e^2)[/tex].

    Dividing by [tex]r^2[/tex],

    [tex]\frac{h^2}{r^2} = \alpha \frac{G(M_s+M_p)}{r^2}[/tex]

    The left hand side is the square of the transverse component of velocity while the right hand side is the product of the semi-latus rectum and the relative acceleration. In short,

    [tex]v_t^2 = \alpha a[/tex]

    [Hopefully this has been transcribed correctly! Please check!]
     
    Last edited: May 4, 2009
  12. May 4, 2009 #11
    Doh! Because the velocity normal to the velocity vector would be zero

    Let me capture your reply in vector notation

    [tex]v_t = \vec{v} \times \hat r[/tex]
     
    Last edited: May 4, 2009
  13. May 4, 2009 #12

    Sorry this last part should be

    [tex]v_t^2 = \alpha a_{cc}[/tex]
     
  14. May 4, 2009 #13

    D H

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    As written, this defines the unit vector [tex]rhat[/tex]. You are given r and θ. You don't know the vector [tex]\vec r[/tex]. This equation is much better expressed as

    [tex]\vec r = r \hat r[/tex]

    No. [tex]\hat{\theta}[/tex] is a vector. That should be a scalar r coupled with [tex]\hat{\theta}[/tex], not the vector [tex]\vec r[/tex]. This got you in trouble in several places. Note that it is customary to denote the scalar magnitude of some vector [tex]\vec q[/tex] with the corresponding symbol sans the vector glyph: [tex]q\equiv||\vec q||[/tex]. That is what is being done here with r.

    The second term is even worse. It has units of length, not velocity.

    Apply the product rule to [tex]\vec r = r \hat r[/tex]. Doing so yields

    [tex]\vec v = \dot r \hat r + r\frac{d}{dt}(\hat r)[/tex]

    Now differentiate the unit vector rhat:

    [tex]\frac{d}{dt}\hat r =
    -\sin\theta\,\dot{\theta}\hat x + \cos\theta\,\dot{\theta}\hat y
    = \dot{\theta}\hat\theta[/tex]

    Collecting terms yields the expression for the velocity vector in polar coordinates:

    [tex]\vec{v} = r\dot{\theta} \hat \theta + \dot r \hat r[/tex]


    Finally, you used the symbol a to represent both the semi-major axis and the acceleration. Using one symbol to represent two different things is never a good idea.


    No!

    That vector is normal to the orbital plane. Think of it this way: The specific angular momentum vector is

    [tex]\mathbf h = \mathbf r \times \mathbf v[/tex]


    To get the component of some vector q normal to some unit vector u you need to subtract the projection of q onto u:

    [tex]
    \mathbf q_{\perp} =
    \mathbf q - (\mathbf q\cdot \hat{\mathbf u}) \hat{\mathbf u}
    [/tex]

    Alternatively, you can use the vector triple product:

    [tex]
    \mathbf q_{\perp} =
    \hat{\mathbf u} \times (\mathbf q \times \hat{\mathbf u}) =
    (\hat{\mathbf u} \times \mathbf q) \times \hat{\mathbf u}
    [/tex]

    Note that in general the vector triple product is not associative. It is in this case because the unit vector uhat appears twice.

    Using the above, the transverse component of the velocity vector is

    [tex]
    \aligned
    \mathbf v_{\text{trans}} &=
    \mathbf v - (\mathbf v\cdot \hat{\mathbf r}) \hat{\mathbf r} \\
    &= \hat{\mathbf r} \times (\mathbf v \times \hat{\mathbf r}) \\
    &= (\hat{\mathbf r} \times \mathbf v) \times \hat{\mathbf r}
    \endaligned
    [/tex]
     
  15. May 4, 2009 #14
    I'm glad I copied and asked, otherwise I'd have been going in the wrong direction (or lots of them).

    The vector triple product (perversely) makes more sense to me than "subtracting the projection of...". I must make more use of the right hand rule to prevent mistakes!

    I know about the confusing of the symbols for semi-major axis and acceleration - hence my later post.

    I must admit that my knowledge of Keplerian orbits is pretty rudimentary up to this point, so there's plenty more to learn.

    DH - this is plenty for me to digest and work on. In case this thread goes a little quiet from here on in, can I thank you for your attention to my problem and for your prompt responses?
     
  16. May 4, 2009 #15

    D H

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    That is kinda perverse. Look at it from the perspective of computational cost. "Subtracting the projection of ..." involves an inner product (3 multiplies, 2 adds), a scalar times a vector (3 multiplies), and a vector subtraction (3 adds), for a total of 6 multiplies and 5 adds. One cross product requires 6 multiplies and 3 adds. The vector triple product involves two cross products or 12 multiplies and 6 adds. The projection method, with half the number of multiplication operations as the vector triple product method, is nearly twice as fast as using the vector triple product -- and is also a lot less prone to numerical error.


    You're welcome! Glad to help.
     
  17. May 5, 2009 #16
    DH

    Sorry to bother but I need some clarification. I'm using MathCAD (hence I can crunch vector data quickly) but still need brains to do the calculations correctly.

    I'm using data from JPL Horizons for the planet Mercury (as an example)

    My calculations run like this:

    At a point xi,yi,zi from the center of the Sun the velocity vectors for Mercury are vxi, vyi, vzi

    Let [tex]\vec{r_i}=\vec{x_i},\vec{y_i},\vec{z_i}[/tex] and [tex]\vec{v_i}=\vec{vx_i},\vec{vy_i},\vec{vz_i}[/tex]

    Therefore the unit radius vector is [tex]\hat r_i = \frac{r_i}{|r_i|}[/tex]

    This means that the radial velocity is [tex]vrad_i= \hat r_i \cdot \vec{v_i}[/tex]

    To get the vector normal to the plane of the orbit:

    [tex]\vec{vn_i}= \hat r_i \times \vec{v_i}[/tex] (This should be zero)

    Then the transverse vector should be [tex]\vec{vt_i}= ( \hat r_i \times \vec{v_i}) \times \hat r_i [/tex] ?

    In which case I need the magnitude of [tex]\vec{vt_i}[/tex], square it, divide it by the radial acceleration and that should be the length of the semi-latus rectum (ie a constant)?
     
    Last edited: May 5, 2009
  18. May 5, 2009 #17

    D H

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    Correct.

    That this cross product is normal to the instantaneous orbital plane is correct. That this should be zero is absolutely incorrect. This is the specific angular momentum vector. If the planets truly did obey Kepler's laws (and they don't), this vector would be a constant.

    Correct.

    That would be a constant if the planets truly did obey Kepler's laws. Simple Keplerian orbits can be characterized by six orbital elements at some epoch time. Only of of these six elements, the true anomaly, changes with respect to time. The other orbital elements are all constants of the orbit.

    The planets do not exactly obey Kepler's laws. Kepler's laws are only approximately correct. The planets perturb each others' orbits. In addition, even Newton's law of gravitation is not quite correct; general relativity yields a better (but much, much hairier) model. The deviation between general relativity and Newtonian gravity is the greatest for Mercury because of Mercury's relatively high orbital velocity.

    The argument of perhielion is one of those constant orbital elements in a simple Keplerian orbit. Mercury's perihelion advances by 531 arcseconds per century because of the influence of the other planets and by 43 arcseconds per century because of general relativistic effects.
     
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