- #1
pc2-brazil
- 205
- 3
Good morning,
I have a doubt about the differentiation of the polar equation of an orbit:
[tex]r=\frac{p}{1+e\cos\nu}[/tex]
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:
[tex]\dot{r}=\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{p}{1+e\cos\nu} \right )[/tex]
[tex]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)[/tex]
However, according to the text I'm reading, I should be getting this:
[tex]\dot{r}=\sqrt{\frac{\mu}{p}}e\sin\nu[/tex]
I'm not sure on how I could get this result.
Any ideas?
Thank you in advance.
I have a doubt about the differentiation of the polar equation of an orbit:
[tex]r=\frac{p}{1+e\cos\nu}[/tex]
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:
[tex]\dot{r}=\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{p}{1+e\cos\nu} \right )[/tex]
[tex]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)[/tex]
However, according to the text I'm reading, I should be getting this:
[tex]\dot{r}=\sqrt{\frac{\mu}{p}}e\sin\nu[/tex]
I'm not sure on how I could get this result.
Any ideas?
Thank you in advance.