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Planetary retrograde motion

  1. Jun 16, 2015 #1
    How can we relate mathematically the time interval in which a planet is in retrograde motion and the devided distances from the sun of the planets ?

    My first drawing was this:
    nothing.png Where S is thee sun , P is the planet, T si the Earth. The v's are the corespondent speeds

    If we take a look at it we can say that:

    r x cos(lambda)= ap x cos(labda2) - ao x cos(lambda1) , same for sine

    What next?

    Oh, and gamma is the vernal point
  2. jcsd
  3. Jun 16, 2015 #2
    It seems complicated. Here's what I understand from your question (correct me if I am wrong):

    You are looking for the time interval in which the planet is in retrograde motion (let's call it ##t_\text{retro}##). Retrograde motion only happens for planets farther than Earth, from the Sun. The information you have is the distances of the planets from the Sun.

    Now, will you consider your orbits to be perfectly circular, each planet has a constant velocity?
  4. Jun 16, 2015 #3


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    Here's a youtube for an animation I whipped up to show just what you are up against.

    You are basically looking for the period during which the white dot moves backward. First you are going to have to work out the geometry that give you the angle the position(as an angle) of the white dot for any given positions of the Earth and the outer planet. Then you have to incorporate how these positions change with time ( using the orbital velocities of both to work out how the position of the white dot change with time. And once you get an equation that represents that, you will need to differentiate it to find the rate at which the position of the white dot changes with time, and then from this determine for how much of one cycle this value is negative.

    I'll start you off with one hint. The angular velocity of either planet (in radian/sec) will be equal to

    [tex] \omega = \sqrt{\frac{GM}{r^3}}[/tex]

    where G is the gravitational constant, M is the mass of the Sun and r is the radius of the planet's orbit.
  5. Jun 17, 2015 #4

    Yes, they are circular, but anyway , I figured it out... You have to extract "d" from one equation in terms of the lambda's and the central distances, plug it in the other, differentiate the equation with dt. The condition at start and end of retrograde motion is d(lambda)/dt=0 , so notating d(lambda2)/dt and d(lambda1)/dt with the angular speeds omega1 and omega2 one can calculate and obtain the needed answer
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