According to Kepler's second law: A planet sweeps out equal areas in equal times (A simple definition). This means that if the orbit is somewhat eccentric, the planet will move faster when on the closest approach to it's parent star, and slowest when at aphelion. So why does this happen, is the curvature of space-time greater closer to the star?
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2 Answers
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Velocity is a form of kinetic energy, while height within a gravity well is a form of potential energy. For an orbiting body, conservation of energy will keep the total energy constant.
So as a planet moves away from the parent star, it loses velocity and gains potential energy. As it moves closer, it trades the potential energy back for velocity. The point of lowest potential has the highest velocity and the point of highest potential has the lowest velocity.
answered Oct 7, 2013 at 19:58
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$\begingroup$ This isn't wrong, but it doesn't address Kepler's second law that was part of the question. $\endgroup$ Commented Jan 1, 2014 at 23:11
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1$\begingroup$ @StanLiou No, but it does address the reason why Kepler's second law holds. $\endgroup$– anonCommented Jun 25, 2015 at 16:32
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$\begingroup$ @QPaysTaxes no, it does not. Conservation of energy is generated by time symmetry, while conservation of angular momentum by rotational symmetry. Nothing in this answer addresses Kepler's second law at all. Those things are logically independent of one another, and indeed it is possible to imagine a conservative force without angular momentum conservation, e.g., any non-radial conservative force will do. $\endgroup$ Commented Jun 25, 2015 at 21:36
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$\begingroup$ @StanLiou ...Well, there goes that bit of physics knowledge. I was taught wrong, I guess. Apologies. $\endgroup$– anonCommented Jun 25, 2015 at 21:37
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The reason is that gravity is a radial force, and so conserves angular momentum $\mathbf{L} = \mathbf{r}\times m\mathbf{v}$, where $\mathbf{r}$ is the vector to the planet from the star and $\mathbf{v}$ is its velocity.
First, remember that the cross product of two vectors $\mathbf{a}$ and $\mathbf{b}$ has a magnitude that's equal to the area of a parallelogram with those sides, and hence is twice the area of the triangle with those sides:
If the planet is at position $\mathbf{r}$ and you wait a small amount of time $\mathrm{d}t$, in that time it will be displaced by $\mathrm{d}\mathbf{r} = \mathbf{v}\,\mathrm{d}t$. Think of the area $\mathrm{d}A$ it sweeps out in that time as the area of the triangle defined by its old position $\mathbf{r}$ and its displacement $\mathrm{d}\mathbf{r}$, and therefore its rate of change of the swept area is: $$\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{1}{2}\left|\mathbf{r}\times\mathbf{v}\right| = \frac{1}{2m}|\mathbf{L}|\text{,}$$ which is constant by conservation of angular momentum. Since the rate of increase swept area does not change, a line joining the star and planet sweeps out equal areas in equal times.
Thus, Kepler's second law is exactly equivalent to the statement that the magnitude of the angular momentum of the planet is conserved. Of course, if you know that the orbit is planar, the direction of the angular momentum must be conserved as well.
Conversely, conservation of angular momentum implies that the orbit is planar and that Kepler's second law holds.
