Take the 2-minute tour ×
Astronomy Stack Exchange is a question and answer site for astronomers and astrophysicists. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 9 down vote accepted

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.

share|improve this answer
    
This isn't wrong, but it doesn't address Kepler's second law that was part of the question. –  Stan Liou Jan 1 at 23:11

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:

Vector Cross Product (Wikipedia)

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.