Orbital motion (in a plane) Speed at a given mean anomaly

Question

For a small simulation I'm working I'm in need to provide an instantaneous speed (and acceleration).

The system is a basic "2 dimensional" orbit system. The orbits follow kepler's law of orbits. Which can be described in a polar system $(\mathbf{r}, \boldsymbol{\nu})$ as:

$$\mathbf{r(\nu)} = \frac{a (1 - e^2)}{1 + e \cos(\nu)} \hat{r}$$ $r$ is the radius, $a$ the semimajor axis $e$ the eccentricity $\nu$ the true anomaly

Using kepler's second law this can be "solved" to simple linear function by introducing a special angle called the "mean anomaly" ($M$) - $E$ is the eccentric anomaly: $$M = E + \sin(E)$$

$$M(t) = M_0 + nt $$ $$n = \sqrt{\frac{GM}{a^3}}$$ With $G$ = gravitational constant, $M$ = "solar" mass.

Now the Eccentric motion needs to be calculated numerically, but newton's method is converting fast enough for my purposes. And more importantly: using this structure there is no growing truncation error.

Anyways, I wonder how to come from this to the speed a certain point? A speed vector to be exact ($\dot{r}(t), \dot{\nu}(t)$). And the accelerataion vector.
I'm kind of stuck in the method of progression, I'd really prefer something better than just inserting two times, and calculate the $\Delta E$ between those two numerically.

I could use the vis-viva equation:

$$v^2 = GM \left(\frac 2 r - \frac 1 a \right)$$

But that only gives the magnitude and seems ugly to then use geometry to calculate the vector?

Answer 1

You can use other parameters, which define an orbit, such as angular momentum,

$$ h = \omega r^2 = \sqrt{GMa(1-e^2)}, $$

where $\omega$ is the angular velocity, thus $\dot{\nu}$. By rewriting this equation to $\omega$ and substituting in the expression for $r(\nu)$ you obtain,

$$ \omega(\nu) = \sqrt{\frac{GM}{a^3(1-e^2)^3}}(1+e\cos(\nu))^2. $$

The angular component of the velocity vector will be equal to $\omega r$. The radial component can then also be found using Pythagoras and the magnitude of the velocity from the vis-viva equation.

In order to find the values for theses time derivatives at a given time you first have to find the corresponding eccentric anomaly, $E$, for example by using the method you stated in your question, convert this to the true anomaly, $\nu$, which can then be used in the equations.