# Second Kepler law for hyperbolas

1. Jul 23, 2013

### jaumzaum

A body can describe 3 types of orbits around another (considering only the gravitational force), a elipse, a hyperbole or a parabola.
Does the second kepler law (area law) work for hyperboles or parabolas too?

2. Jul 23, 2013

### WannabeNewton

Yes. Kepler's 2nd law holds for any central force motion. Here is one proof:

Consider a particle moving under a central force $F = f(r)\hat{r}$. We see immediately that $\frac{\mathrm{d} L}{\mathrm{d} t} = r\times F = 0$ hence $L = r\times p$ is a constant of motion. Now consider the position of the particle at an instant $t$ and another instant $t + \Delta t$ so that the position of the particle is given in polar coordinates by $(r,\theta)$ and $(r + \Delta r, \theta + \Delta \theta)$ respectively. For small $\Delta \theta$, the area swept out during this interval will be given by the triangle swept out of side length $r + \Delta r$ and height $r\Delta \theta$ so $\Delta A = \frac{1}{2}r^{2}\Delta \theta + \frac{1}{2}r\Delta r \Delta \theta$ hence $\frac{\Delta A}{\Delta t} = \frac{1}{2}r^{2}\frac{\Delta \theta}{\Delta t} + \frac{1}{2}r\frac{\Delta r \Delta \theta }{\Delta t}$ so $\frac{dA}{dt} = \lim_{\Delta t\rightarrow 0 }\frac{\Delta A}{\Delta t} = \frac{1}{2}r^{2}\frac{d\theta}{d t} = \text{const.}$ as desired.

3. Jul 23, 2013

### mathman

Spelling error. You are talking about hyperbolas. Hyperbole is a completely different thing.