How Does Epsilon Equal One Result in a Parabola in Orbital Equations?

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Discussion Overview

The discussion revolves around the orbital equation for bodies in inverse square fields, specifically examining the implications of the eccentricity parameter (epsilon) on the shape of the orbit. Participants explore how the equation produces different conic sections, particularly focusing on the case when epsilon equals one and its relation to parabolic orbits.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions how the orbit equation results in a parabola when epsilon equals one, suggesting that the equation becomes identically zero in that case.
  • Another participant notes that the semi-major axis (a) is considered to be infinite for a parabolic trajectory, which raises questions about the relationship between a and epsilon.
  • There is a reference to the product of a and (1-e^2) being a constant related to angular momentum, indicating a deeper connection between these parameters.
  • One participant corrects their earlier misunderstanding regarding the equation's form, affirming that the original equation presented by the first participant is indeed correct.
  • A request is made for clarification on how to calculate the relationship between a and epsilon, highlighting a lack of information on this aspect in the current discussion.

Areas of Agreement / Disagreement

The discussion contains multiple competing views, particularly regarding the interpretation of the orbital equation at epsilon equals one and the implications for the semi-major axis. There is no consensus reached on these points.

Contextual Notes

Participants express uncertainty about the definitions and relationships between the parameters involved, particularly concerning the nature of the semi-major axis in relation to parabolic orbits.

KBriggs
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Hey all

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

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

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?
 
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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.
 
ideasrule said:
That article shows that the OP's equation is correct.
I corrected my post, misread the OP.
 
ideasrule said:
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.

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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