- $\epsilon > 1$ and $E > 0$
- $0 < \epsilon < 1$ and $V_{min} < E < 0$
- $\epsilon = 0$ and $E = V_{min}$
- $\epsilon < 0$ and $E < V_{min}$
- $\epsilon = 1$ and $E = 0$

Thus, we can determine the orbit type by looking at the eccentricity alone (but we must also check the energy to make sure its non-sensical). If you use the trajectory equation, you can determine the cartesian equation form of the orbit by converting the polar form given above. The choices above correspond to the following orbits:

- $\epsilon > 1$ and $E > 0$ - hyperbola
- $0 < \epsilon < 1$ and $V_{min} < E < 0$ - ellipse
- $\epsilon = 0$ and $E = V_{min}$ - circle
- $\epsilon < 0$ and $E < V_{min}$ - impossible. $\epsilon$ is always positive (because of the square root)
- $\epsilon = 1$ and $E = 0$ - parabola

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This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.

We've forgotten the definition of eccentricity, but being good physicists, we know that Kepler's equations have three types of solutions, bound solutions ( ellipses ), unbound ( hyperbolas) and parabolas, which are a special case at the turning point.

So, we know that if $ E < 0 $ we are bound, if $ E> 0$ we are unbound, so the special case at the turning point must have $ E= 0$, neither bound nor unbound. Answer is E

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