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Radial velocity extremes

  1. Aug 6, 2009 #1
    Hi, I hope I chose the right forum for this topic. I'd like to show you an interesting property of orbiting bodies. I don't want you to solve it for me, I just thought the result is so beautiful that there must be simpler answer than just mathematically calculate it (I haven't managed to Google any). That's why I would like to hear your opinions.

    A body is orbiting in a radial gravitational field. And you are interested in its radial velocity (Doppler spectroscopy uses this), i.e. you wish to project the velocity to a certain direction (orange arrow, see attachment). Suppose the direction of observation lies in the orbital plane, so it becomes a 2D problem. Now where are the velocity extremes?

    The projection (dot product) is consisted of two factors: how big the velocity is and how favourably it is directed. The closer to periapsis (p), the greater the size. So if we are watching from the direction of minor semiaxes (from the bottom), one extreme surely is at periapsis, because the velocity is greatest and directed exactly towards us. But what about more general case? There are always two points of the most favourable direction (tangents to the ellipsis parallel to the direction of observation - points a,b, lines A,B), so if the velocity had constant size, the extremes would occur there. But the size isn't constant and its greatest value is at periapsis. So we expect the extremes to be somewhere between the periapsis and those points a and b. Let us advance from the point a to periapsis (p). At first the velocity diverges very slowly from the ideal direction (A), whereas the size increases significantly. But at some point the angle between the velocity and the direction of observation will be much closer to 90° and the velocity size cannot make up for it. At periapsis the velocity is huge, but its projection (pink vector) even smaller than it was at point a. Where does the diverging of the velocity "beat" the increasing of its size? It could surely be quite complicated, but surprisingly it isn't. Try it :-)

    Hint:
    Use polar coordinates with the ellipse's focus as the point of origin, express the projection of the velocity for a general situation and then put in the ellipse equation and Kepler's second law about area velocity.

    The answer:
    Make a line through the point of origin (focus) perpendicular to the direction of observation. The intersections of this line and the ellipse are where the radial velocity is greatest. So simple! Why? :-)

    Sorry for my English, if there is something unclear, just ask and I'll do my best to explain it.
     

    Attached Files:

    Last edited: Aug 6, 2009
  2. jcsd
  3. Aug 6, 2009 #2
    I don't understand what is meant by the line perpendicular to the direction of observation. There is a plane that is perpendicular. The plane would intersect the ellipse in two places.
     
  4. Aug 6, 2009 #3
    How about a diagram?
     
  5. Aug 6, 2009 #4
    Sorry, I was automatically thinking in the orbital plane. I've edited the post to clear it up. Nevertheless, both the line and the plane intersect the ellipse in two same points and these are the answer. If you want a projection to any direction outside the orbital plane, the answer is still the same, because the whole radial velocity only gets multiplied by sine of the inclination.

    junglebeast: Okay, I'll try to make some.
     
    Last edited: Aug 6, 2009
  6. Aug 8, 2009 #5
    I believe it should be clearer now.
     
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