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?

  • $\begingroup$ Note: I corrected the expression for the vis viva equation. $\endgroup$ Jun 7, 2015 at 9:15

1 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.

  • $\begingroup$ In this system, in which direction is the "angular speed"? Perpendicular to r it seems? (But that's of little geometrical meaning I think)? $\endgroup$
    – paul23
    Jun 7, 2015 at 9:33
  • $\begingroup$ @paul23 It will in deed be perpendicular to r, but also in the plane of the orbit, which is defined by the inclination, longitude of ascending node and argument of periapsis and reference direction, see orbital elements. $\endgroup$
    – fibonatic
    Jun 7, 2015 at 13:15
  • $\begingroup$ @paul23 However you only look in two dimensions, thus the orbital plane will be defined by that. The only thing that remains is whether the orbit is clockwise or anticlockwise, which can be changed by changing the sign of the angular velocity. $\endgroup$
    – fibonatic
    Jun 7, 2015 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.