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Let $r=f(\theta)$ be a smooth, $2\pi$-periodic function, representing a curve star-shaped with respect to the origin. Maybe something like this:
         Polar Plot
                                        $r=f(2^\theta)-1.7$ (idea from here)

Which functions $f$ can be realized as the orbit of a particle of mass $m$ about a (much larger) mass $M$ at the origin, with the usual Newtonian gravity force, $F = G m M / r^2$, but with $M=M(t)$ an arbitrary smooth function of time $t$? In other words, if one could control variation of the origin mass, could that suffice to match any smooth $f(\theta)$?

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Basically what you're saying is that the force $F(t)$ is always directed radially, but its magnitude varies arbitrarily. Actually you didn't say $M > 0$, and in fact negative "mass" may be required in some cases. Then the answer is yes.

In polar coordinates $(r,\theta)$, your assumption says that $2 \dot{r} \dot{\theta} + r \ddot{\theta} = 0$ Basically this is conservation of angular momentum. If $r = f(\theta)$, the chain rule says $\dot{r} = f'(\theta) \dot{\theta}$, so $2 f'(\theta) \dot{\theta}^2 + f(\theta) \ddot{\theta} = 0$. For any given function $f$ with $f > 0$, a solution of the differential equation $2 f'(\theta) \dot{\theta}^2 + f(\theta) \ddot{\theta}$ with, say, $\theta(0) = 0$ and $\dot{\theta}(0)=1$ gives us a motion that starts at $\theta = 0$ and stays on the curve $r = f(\theta)$. Now the existence and uniqueness theorems for differential equations say that (if $f$ is smooth and $f > 0$ everywhere) a unique solution exists in some interval of time $t$. Moreover, the only way such a solution will stop existing is that $\theta$ goes off to $\infty$ in finite time. But we're assuming $f$ is periodic, and because of the conservation of angular momentum you have to come back to the starting position with the same speed as you started, so that won't happen.

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    Brilliant, Robert! And interesting that $M$ must be permitted to be negative. Thanks for such a clear answer!2012-09-01