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Calculating Velocity at any given point in an orbit

  1. Aug 16, 2005 #1

    I'm trying to really understand orbits. I want to be able to calculate the velocity at any given point in an orbit.

    Now, parametrically an ellipse can be:

    x = a*cos(t)
    y = b*sin(t)

    If those are position, can I take the derivative to obtain velocity?

    x' = -a*sin(t)
    y' = b*cos(t)

    For the overall velocity:

    V = sqrt( (-a*sin(t))^2 + (b*cos(t))^2)

    However, there is a pesky t in there, now I use:

    x' = -a*sin(t)

    Solve for t

    -x'/a = sin(t)
    asin(-x'/a) = t

    And subsitute t back into overall equation? Does this make sense or am I just making stuff up?
  2. jcsd
  3. Aug 16, 2005 #2


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    From your equations, you have computed the speed
    V = sqrt( (-a*sin(t))^2 + (b*cos(t))^2).
    x = a*cos(t)
    y = b*sin(t)
    (x/a) = cos(t)
    (y/b) = sin(t).
    V = sqrt( (-a*(y/b))^2 + (b*(x/a))^2).
  4. Aug 16, 2005 #3
    I was going for in only for terms of x, to disclude y

    Is it valid how you subsituted the original x and y, but not the differentiated ones?
    Last edited: Aug 16, 2005
  5. Aug 16, 2005 #4


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    I believe it's fine.

    Continuing on...
    (x/a) = cos(t)
    (y/b) = sin(t)
    means that
    which can be solved for (y/b)^2.
    That expression can then be inserted in the speed expression I derived, yielding an expression for the speed in terms of x... if that's what you really want.
  6. Aug 17, 2005 #5


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    Your very first step is wrong :-( You've parameterized an ellipse, but it's not the most general possible parameterization, which is

    x = a cos(f(t))
    y = b sin(f(t))

    where f(t) can be any function.

    then dx/dt = -a sin(f(t)) df/dt, dy/dt = b cos(f(t)) df/dt

    The correct parameterization will sweep out equal areas in equal times (Kepler's law - this conserves angular momentum), so the angular velocity will be inversely proportional to the radius. Your equation has the angular velocity as being constant, which is wrong.

    You should be able to work the problem out more simply, by taking advantage of the fact that angular momentum and energy are both conserved.
  7. Aug 18, 2005 #6
    nicely done my son!
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