# Doubt about the polar equation of a Kepler orbit

1. Jul 23, 2010

### pc2-brazil

Good morning,

I have a doubt about the differentiation of the polar equation of an orbit:
$$r=\frac{p}{1+e\cos\nu}$$
It represents the relative position of a planet with respect to the central body.
Here, p is the parameter, e is the eccentricity and r is the radius of the planet measured from the focus (where the central body is located).
If I differentiate it, I obtain the following result:
$$\dot{r}=\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{p}{1+e\cos\nu} \right )$$
$$p \left(\frac{e\dot{\nu}\sin \nu}{(1+e\cos\nu)^2}\right) = \frac{h^2}{\mu} \left(\frac{e\dot{\nu}\sin \nu}{(1+e\cos\nu)^2}\right)$$
However, according to the text I'm reading, I should be getting this:
$$\dot{r}=\sqrt{\frac{\mu}{p}}e\sin\nu$$
I'm not sure on how I could get this result.
Any ideas?

2. Jul 24, 2010

### Filip Larsen

From Kepler's second law you have the relationship $r^2 \dot{\nu} = h$ which you can insert into your equation to get rid of $\dot{\nu}$.

3. Jul 24, 2010

### pc2-brazil

OK, I tried it and it led me to the correct result.