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Kepler's 2nd.law and calculation

  1. Dec 20, 2013 #1
    Hello,

    While dealing with Kepler's 2nd.law, when calculating the area of the triangle which sweeps out:


    Area of a triangle = 1/2 base x height

    Now, we know dθ = dx/r (where dx = arc length, r=radius)

    So, dx= rdθ

    So, area of a triangle = 1/2 base(rdθ) x height(h)

    Is this correct?

    Thanks
     
  2. jcsd
  3. Dec 20, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    If dx is defined to be something like an "arc length in tangential direction", yes. It is not the actual arc length if the motion is not circular.
     
  4. Dec 20, 2013 #3

    K^2

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    Science Advisor

    Keep in mind that Kepler's Second Law is just another way of saying that angular momentum is conserved. In other words, ##\omega r^2## is a constant. Keeping in mind that for a small section h = r, this is in agreement with what you wrote.
     
  5. Dec 20, 2013 #4
    Thank you very much for letting me clear the confusion.
     
  6. Dec 22, 2013 #5

    Philip Wood

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    Gold Member

    Hope I'm not going to regenerate the confusion, but for an elliptical orbit there are just two places where the triangle area is given by [itex]\frac{1}{2} r\ ds = \frac{1}{2} rv\ dt[/itex], and those are the two ends of the major axis. [I'm using ds as infinitesimal portion of arc length, r as distance from focus and v as speed of orbiting body.]

    At every other point, r and ds are not perpendicular to each other, so a sine or cosine factor has to be inserted.
     
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