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Does Kepler’s second law hold true if the shape of the orbit is not closed? Let’s say, for example, a spiral?

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Kepler's second law states that the curve $\mathbf{r}(t)$ followed by a planet sweeps out equal areas in equal times. That is $$ \frac{\text{d}A}{\text{d}t} = \text{constant.} $$ The infinitesimal area element swept out in a time $\text{d}t$ by a planet travelling on such a path is $$ \text{d}A = \frac{1}{2}|\mathbf{r}\times\text{d}\mathbf{r}| $$ so that $$ \frac{\text{d}A}{\text{d}t} = \frac{1}{2}\left|\mathbf{r}\times\frac{\text{d}\mathbf{r}}{\text{d}t}\right| = \frac{1}{2}|\mathbf{r}\times\mathbf{v}| = \frac{1}{2m}|\mathbf{J}| $$ where $\mathbf{J}$ is the angular momentum. This means that Kepler's second law holds for any path where the magnitude of its angular momentum is constant. For any path in space it is possible to find a parametrisation such that angular momentum is constant (since the speed can be chosen arbitrarily at each point). It is therefore possible for Kepler's second law to hold for any curve, but the force required for it to do so may be very complicated for complex curves.

EDIT: My comment in response to Emilio Pisanty was getting too long, so I'll put this here instead.

Firstly, the path being planar is not necessary - it is entirely possible to define differential areas for non planar paths, which is exactly what the above equation for $\text{d}A$ does.

Secondly, it is true that the particle may not reverse its direction on the path as this would cause $|\mathbf{v}|=0$ for some point on the curve, making $|\mathbf{J}|$ non-constant. However, it is possible to approximate such a path arbitrarily well by extending the section of reversed path to an arbitrarily small loop. To illustrate what I mean, consider this picture of three curves: Cycloids

A particle following a common cycloidal curve reverses its direction every time it reaches the $x$ axis, so has a point where the velocity vanishes. A particle on a prolate cycloid has no such point. Further, the prolate can be made to approximate the common as well as one pleases. However, in the limit this does break down.

Finally, there is another constraint: at no point can the curve move directly away from the origin as in this case the time derivative of area is zero. More rigorously, $\mathbf{r}||\mathbf{v}$ implies $\mathbf{r}\times\mathbf{v}=\mathbf{0}$ and it is not possible to choose a speed which keeps the angular momentum constant.

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  • $\begingroup$ "For any path in space it is possible to find a parametrisation such that angular momentum is constant" - so long as the path is planar and it doesn't turn around in its tracks, that is. $\endgroup$ – Emilio Pisanty Dec 3 '19 at 22:40

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