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Area for an ellipse

  1. Aug 23, 2014 #1
    Hello everybody,

    I'm trying to know, in a keplerian orbit, how to calculate the area of a swaped area; since the Sun is at one of the focus, I wish to calculate given an angle measured from focus to the orbiting body, the area swaped.
    I dont know if I'm explaning this right...Hope so.

    Kind regards,

    CPtolemy
     
  2. jcsd
  3. Aug 23, 2014 #2

    Mark44

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    Staff: Mentor

    swaped = ? What is this word?

    Calculating the area of an ellipse is pretty straightforward. There are several formulas if you know the equation of the ellipse. See http://en.wikipedia.org/wiki/Ellipse#Area.
     
  4. Aug 23, 2014 #3
    Hi,

    I mean swept. Sorry for my english... :(

    I don't want to know the entire area of the ellipse - just the swept area by the body.

    Regards,

    CPtolemy
     
  5. Aug 23, 2014 #4

    HallsofIvy

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    Staff Emeritus
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    I think he meant the area "swept out" by the planets motion- the area inside the elliptic orbit.

    cptolemy, the area of an ellipse with major and minor axes of lengths a and b is [itex]\pi ab[/itex].
     
  6. Aug 24, 2014 #5
    In Fundamentals of Astrodynamics by Bate, Mueller and White, ISBN 0-486-60061-0, I can see the following equation:

    dt = 2/h dA

    h is the specific angular momentum, given by h = r v sin(γ), where γ is the flight path angle, i.e. the angle between the r and v vectors. This is consistent with Kepler's second law as h is a constant for a given orbit.
     
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