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

[KBriggs]

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?

 

[rcgldr]

Equations at wiki:

http://en.wikipedia.org/wiki/Orbit_equation

 

[ideasrule]

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]

Your equation is for 1/r, not r. Equations at wiki:

http://en.wikipedia.org/wiki/Orbit_equation

That article shows that the OP's equation is correct.

 

[rcgldr]

That article shows that the OP's equation is correct.

I corrected my post, misread the OP.

 

[KBriggs]

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.