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## Thursday, 4 August 2011

### Physics GRE - #3

In Kepler's problem of planetary motion, various values of the eccentricity and hence the energy $E$ classify orbits according to conic sections. What value of the eccentricity $\epsilon$ and the energy E belongs to a parabolic orbit?

• $\epsilon > 1$ 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 $E = 0$
Choice number 5 is the answer. Recall that the eccentricity is defined by $\epsilon = \sqrt{1+\frac{E}{V_{min}}}.$ Additionally, we also have the corresponding trajectory equation: $\frac{p}{r}=1+\epsilon \cos{\phi}$ where $p$ is the radius of the circular orbit (when $\epsilon = 0$).

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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