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Parabolic orbit

  1. Oct 24, 2009 #1
    Hey all

    the prof derive the orbit equation for bodies in inverse square fields as:

    [tex]r=\frac{a(1-\epsilon^2)}{1+\epsilon\cos(\theta)}[/tex]

    Now, I understand how this gives an ellipse for epsilon between 0 and 1, but when epsilon is one, how does this give a parabola? Isn't the equation identically 0 if epsilon = 1?
     
  2. jcsd
  3. Oct 24, 2009 #2

    rcgldr

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    Last edited: Oct 24, 2009
  4. Oct 24, 2009 #3

    ideasrule

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    a is supposed to be the semi-major axis, but that's infinity for a parabola. The product of a and 1-e^2 is actually a constant, equal to h^2/GM where h is the angular momentum per unit mass.
     
  5. Oct 24, 2009 #4

    ideasrule

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  6. Oct 24, 2009 #5

    rcgldr

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    I corrected my post, misread the OP.
     
  7. Oct 24, 2009 #6
    Could you show me how that would be calculated? We don't have anything about a being a function of epsilon.
     
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