The true anomaly motion as I understand, is not constant in an elliptical orbit. The rate of change of true anomaly is not constant as in the case of circular orbits, in which case the true anomaly is equal to the mean anomaly, and the rate of true anomaly can be taken as the angular rate in the orbit, and the rate of rate of true anomaly can be taken as zero. For elliptical orbits, however, there should be a non-constant rate of true anomaly, and some rate of rate of true anomaly. Is there some derivation for the formula that I can refer to for understanding the motion more clearly? I have the position and velocity as cartesian state vectors in the orbit in the ECI coordinate frame.
3
-
$\begingroup$ Welcome to stackexchange! If you take the tour you'll see that it's a good idea to do at least a little research on the subject you are asking about. If you can do a little reading first and try some things, and come back here if you get stuck, that would be a better fit here. Also, you might want to look at other questions and answers right here first. You can try the orbital mechanics or orbital elements tags! $\endgroup$– uhohMay 24, 2017 at 13:45
-
1$\begingroup$ Start with the relationship between true and eccentric anomaly $\tan\frac\theta2 = \sqrt{\frac{1+e}{1-e}}\tan\frac E2$, differentiate wrt time, and clean things up. You should get a fairly simple expression expressing $\dot\theta$ as a function of $\dot E$, $e$, and $\cos \theta$. Next, from Kepler's law, $M = E - e-sin E$, you should get a fairly simple expression expressing $\dot E$ as a function of $M$, $e$, and $\cos E$. Next, combine everything using the relation $\cos E = \frac {e + \cos\theta}{1+e\cos\theta}$. You should get a fairly simple expression for $\dot\theta$. $\endgroup$– David HammenMay 24, 2017 at 17:38
-
$\begingroup$ @DavidHammen. I forgot to mention that I have the cartesian state vectors of the orbit. I know how to convert to Kepler elements, and your answer was helpful. Please also tell me how I should obtain the instantaneous angular acceleration. Should I take a first order derivative approximation, or is there any analytical solution? $\endgroup$– KJ7May 25, 2017 at 14:08
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The rate of change of the true anomaly $f$ is
$$\frac{\mathrm{d}f}{\mathrm{d}t} = \dot{f} = \frac{h}{r^2},$$
where $h = \lVert \vec{r} \, \times \vec{v} \rVert$ is the magnitude of the angular momentum and $r$ is just the radial distance from the center of the main body. Note that this is actually Kepler's second law of planetary motion.
Proof:
We're only concerned with motion in the orbital plane, therefore with bidimensional motion. The position and velocity vectors in the orbital plane are written in rotating polar coordinates as
$$ \vec{r} = r \hat{i}_r$$ $$ \vec{v} = \dot{r} \hat{i}_r + r \frac{\mathrm{d} f}{\mathrm{d} t} \hat{i}_f,$$
where $\vec{i}_r$ is the unit radius vector and $\vec{i}_f$ is the unit vector orthogonal to $\vec{i}_r$ and pointing in the same direction of velocity, $\vec{i}_f = (0,0,1) \, \times \hat{i}_r$. Evaluating the angular momentum vector gives
$$\vec{h} = \vec{r} \, \times \vec{v} = r \hat{i}_r \, \times (\dot{r} \hat{i}_r + r \frac{\mathrm{d} f}{\mathrm{d} t} \hat{i}_f) = r^2 \frac{\mathrm{d}f}{\mathrm{d}t} \hat{i}_h,$$ where $\hat{i}_h = (0, 0, 1)$. $\blacksquare$
Angular acceleration
If you want the rate of change of the rate of change of the true anomaly, i.e. the angular acceleration, differentiate the first equation to get (assuming Keplerian motion)
$$\frac{\mathrm{d}^2f}{\mathrm{d}t^2} = - \frac{2h}{r^3} \frac{\mathrm{d}r}{\mathrm{d}t} = - \frac{2h}{r^3} \hat{i}_r \cdot \vec{v}.$$
All of these derivations are contained in Chapters 2 and 3 of R. H. Battin, "An Introduction to the Mathematics and Methods of Astrodynamics," 1999.
