# Parabolic orbit

1. Oct 24, 2009

### 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?

2. Oct 24, 2009

### rcgldr

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

### 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.

4. Oct 24, 2009

### ideasrule

5. Oct 24, 2009

### rcgldr

I corrected my post, misread the OP.

6. Oct 24, 2009

### KBriggs

Could you show me how that would be calculated? We don't have anything about a being a function of epsilon.