Let's assume we have an equation of orbit which is

$$\frac{1}{r} = u_o + u_1 \cos \frac{3 \theta}{2}$$

Do I have to know the value of eccentricity of the orbit to understand the shape of the orbit? I mean if the orbit is circular or ellipse, the orbit would be closed. But my problem is that how it is different to a bound orbit.

I have found a theory that says $\Delta \phi = 2 \pi \frac{m}{n} $ can be used to find a orbit is closed or not. It says the $2 pi$ has to be a rational function. Could you explain it more?

Or it is better to use any other equation?


It's actually simpler than you might think.

If we have something like
$r=\frac{1}{\cos \left(\frac{3 \theta }{2}\right)+2}$ we get a graph like

enter image description here

But if $u_0 \leq u_1$, then the denominator will go to zero at some point, so the radius will go to infinity, and so the path won't loop back on itself.


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