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Elliptical Orbit's curvature the same at opposite ends of semi-major axis?

  1. Oct 29, 2011 #1
    Let the elliptical orbit be bisected through the semi-minor axis...
    Ellipses are symetrical and both sides here are mirror images of each other, true?
    If so, then they share matching curvature at opposite ends of the semi-major axis (and opposite ends of the semi-minor axis, and reflections across both the semi-major and semi-minor axses...)

    Comparing the parts of the orbit at each end of the semi-major axis, both should have the same curvature...
    But one is close to the focus with the central mass and the other is further away.
    So at the close end the path is in a stronger gravitational acceleration, the other end in a weaker one.
    This difference appears in the orbiter at the close end moving faster than when at the far end.

    The question is: If the distances from the central mass are different, and the speeds along the orbital path are different, why do both ends of the orbit describe the identical curvature?
  2. jcsd
  3. Oct 29, 2011 #2
    It's actually very simple: They do not.

    Unless you mean something other than what I think of when I would use the word curvature: d2r/dθ2
    Last edited: Oct 29, 2011
  4. Oct 29, 2011 #3


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    Correct. If post #2 wants to invent a new defintion of "curvature", that's fine, but it's not the standard defintion. (In fact it's a rather strange definition, because if you performed a rigid body translation of a curve in space, its "curvature" as defined by #2 would change!

    If the radius of curvature is the same, but the speed is different, the acceleration towards the centre and the force required to produce the acceleration is higher if the speed is higher.
    As you said, the gravitational force IS higher when the mass is at the "close end" of the ellipse.

    For an inverse square law of gravitation, everything is the "right" proportions to give elliptical orbits.
  5. Oct 29, 2011 #4
    Thanks. It is not intuitive to me so I'm trying to think of it in terms of vectors...

    If I draw vectors for the instantaneous orbital speed (a tangent to the orbit, right?) for each of the two points where the orbit crosses the semi-major axis, the one for the closer is longer than the one for the further (its orbital speed is faster).

    If I draw vectors for the gravitational attraction to the foci with the central mass (and both point inward to the mass foci - so each is 90 degrees to the orbital speed vector, right?), then likewise the one for the orbital point closer is longer than the further.

    When I add those vectors at these two points in the orbit, the one closer with two longer vectors yields a longer vector sum, the one further with two shorter vectors yields a smaller vector sum.

    If I'm right so far, what I'm asking is how two different sized resultant vectors at orbital points each end of the semi-major axis can cause the orbiter to make the same shaped path of curvature.

    There must either be another vector in play that I'm leaving out or my model is incorrect? Something left out needs to offset the imbalance (of two different sized vectors producing the same curvature)?

    Does the solution take the form that the resultant vectors for both points must be identical with respect to those orbital point positions? So the curvature may be identical?
  6. Oct 30, 2011 #5


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    Kepler's Second law - a line connecting a planet with the sun sweeps out equal areas in equal times.
  7. Oct 30, 2011 #6
    How does Kepler's 2nd law about sweeping equal areas in equal times answer my question?

    The shapes I'm asking about are the shapes of the orbital curves (the curved lines bounding the outer areas of the swept areas at each end of the semi-major axis) which are lines, not areas. The equal time equal area sweeps have different lengths of this outer curved line, but that does not seem to be relevant. I'm looking particularly at the curvature of these two lines and asking why they are symmetrical and share the same identical curvature in spite of the orbiter at the two points where it crosses the semi-major axis having different resultant vectors comprised of different vectors for orbital speed and gravitational attraction.

    In essence, how is it that the orbiter makes the same curve at each end of the orbit when the dynamic conditions are different from each other? Or where is my error in thinking about this?

    The sweeping of areas is time based... is that a clue as to where the missing vector component of the orbiter may be found?
  8. Oct 30, 2011 #7


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    Keplers laws can be derived from what AlephZero stated. But Kepler figured out his "elliptical" orbit solution to the planets from celestial data provided by Tycho Brahe way before Newton was on the scene with his laws of gravitation. The derivation should be available somewhere on the net, or someone here might post it.
    So it probably is a clue somehow.
    I posted Kepler because these guys figured stuff out centuries ago that we still have trouble understanding.

    I am just wondering though how you can add a velocity vector to a gravitational force vector.
  9. Oct 31, 2011 #8
    you are simply forgetting that at perigee velocity is different [>] and [consequently] time is different [<] ; two factors cancel out, so curve is the same
    Last edited: Oct 31, 2011
  10. Oct 31, 2011 #9
    I wasn't forgetting the difference in speed, I was trying to understand why the result would give the same curve.

    I made a model in Excel to examine this.
    I set constants for solar gravitational constant
    inputs for the length of the semi-major axis, eccentricity
    and calculations of radius for Rap and Rper
    and calculations for orbital speed for Vap and Vper
    and calculations for the gravitational force at Rap and Rper (to get the radial vectors)

    I wanted to see if the resultant vector's angle of deflection from tangent would be the same for both points in spite of the radial and transverse vectors at both points being different.

    To really do that I needed a way to express the transverse as a vector derived from momentum, but I had no time base to work with. Also, in orbital mechanics there is not supposed to be a transverse force, so I'm not sure what to do with the momentum anyway. Newton spoke of the orbiter having an "innate velocity", which I suspect is his phrase for some constant velocity component of the orbital speed, but I'm not sure...

    I also questioned myself if a matching deflection of resultant vectors really meant much in light of the different orbital speeds. The more I looked at it the less I was convinced it had any meaning with respect to the curvature symmetry.

    I still don't have a hard answer to my question, but after a few hours of fooling around with the calculations I can see that the symmetric curves are within the possible solutions. I still don't have a handle on exactly why it is the particular solution, but I'm no longer seeing it as a strange one, just impressed that it is the correct one.

    If I had access to a dynamic model (or could better recall my third year calculus) I'm convinced the underlying reason would be clear. I'll keep looking at it.
  11. Nov 2, 2011 #10
    The reason is simply the "wonder" the ellypse works when a force obeying the inverse-square law is in one focus
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