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Orbital question(elliptical vs circular orbit)

  1. Sep 2, 2008 #1
    are there two points or one in an elliptical orbit where the speed is equal to the speed of a circular orbit at the same radius? if so what is the expression for this point?

    Thank you
  2. jcsd
  3. Sep 14, 2008 #2
    Hi again,

    After I picking up a pencil and paper, I was able to find the 2 points where the instantaneous speed are equal in elliptical and circular orbit at the same radius, it is actually right at the perigee and apogee in the orbit ( see the diagram below). In addition, I have find expression for elliptical an circular orbital speed; which are:

    V= SQR( GM/ R) - Circular orbit

    V= SQR(GM *( (2/R)-(1/a) ) ) Elliptical orbit

    Now, I have a doubt, how do I express these to these two velocity in terms of these point( same instantaneous speed at perigee and apogee at that point?) because when I set them both to be equal each, it actually cancelled out.

    Any ideas?


    Attached Files:

  4. Sep 14, 2008 #3

    Jonathan Scott

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    If this is a homework question, please take it to the homework and coursework area.

    In a circular orbit, the speed is the same all the way round, but in an elliptical orbit, the speed is NOT the same at perigee and apogee (those are the points of maximum and minimum speed), so if that's what you're saying, you've got something wrong.
  5. Sep 14, 2008 #4
    yes, you are right that about the speed in elliptical orbit. But from the attached diagram, the speed of the two intersection points between elliptical and circular orbit should be the same( i.e at the point of V and the opposite point of it). then if that is true, a expression of these points in terms of velocity should be able to derive.

    I hope I have made the question clear
    ( I shouldn't use the word of perigee and apogee, sorry)
  6. Sep 14, 2008 #5

    Jonathan Scott

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    Nope, still doesn't make sense. Lowest point of orbit is fastest, highest is slowest.
  7. Sep 14, 2008 #6
    Of course so (although obviously or by ODE uniqueness theorems the direction will differ), because at one apex the eccentric orbit is too fast and at the other it is too slow (IVT). Your diagram is a bit funny though, since your example circle isn't even centred upon the same point that the ellipse is orbiting.
  8. Sep 14, 2008 #7
    ok, I think I got it. so there are two different points in the diagram that have the same speed. Each of the points(i.e perigee and apogee) has a corresponding speed in elliptical and circular orbit. The lowest points of orbit is fastest , and the highest of the orbit is slowest.

    Now, I hope I have made my statement right, please allow me to ask this question again: does the statement apply to the attached diagram that for both circular and elliptical orbit have the same speed in the lowest and highest orbit?
  9. Sep 15, 2008 #8

    Vanadium 50

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    Why are you plotting position vs. time as opposed to speed vs. time? The latter plot is the one where it's easy to see that there must be at least two positions where the speeds are the same.
  10. Sep 15, 2008 #9


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    Here this might help. It shows both orbits with the focus of the orbits aligned. Where the orbits cross is where the instantaneous orbital speed will be equal for both.

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