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I am running an animation of a satellite in an elliptic orbit (defined by a parametric equation for $x$ and $y$ as a function of $t$) and want to make sure the spacecraft is traveling at the right speeds at different points in its orbit. That is, it should go slower at is apoapsis and much faster at its periapsis. I can easily calculate the tangential speed of the satellite using this equation:

$v=\sqrt{GM(\cfrac{2}{r}-\cfrac{1}{a})}$

How do I convert this to the angular speed of the satellite at this point?

I've done extensive research (hours and hours) but haven't found anything of value. The closest thing was this expression of Kepler's Third Law:

$\cfrac{dA}{dt}=\cfrac{1}{2}r^2\omega$

Since $\cfrac{dA}{dt}$ is a rate (area swept out per second) I rewrote this equation as

$\cfrac{A}{P}=\cfrac{1}{2}r^2\omega$

where $A$ is the area of the elliptic orbit (given by $A=\pi ab$ where $a$ and $b$ are the semi-major and semi-minor axes of the ellipse, respectively), and $P$ is the period of the elliptic orbit (given by $P=2 \pi \sqrt{\cfrac{a^3}{GM}}$). Solving this for $\omega$ yields:

$\omega=\cfrac{2A}{Pr^2}$

For each time step in my simulation I use the satellite's current position in this equation to compute $\omega$ and then use the result to update the current $\theta$. This updated $\theta$ is then plugged into the parametric equation mentioned above to get the satellite's $x$ and $y$ position.

I can't find my mistake anywhere and would really appreciate it if someone could point it out to me.

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  • $\begingroup$ Am I missing something? Angular speed is tangential speed divided by the instantaneous radius -- or arctan if you get picky. Is that all you wanted? $\endgroup$ – Carl Witthoft Jan 23 '14 at 20:41
  • $\begingroup$ @CarlWitthoft I tried that, too, but it still didn't work. Could I be seeing the lack of accuracy because I am using the Euler integration method (with a fairly large time step in seconds)? $\endgroup$ – Matthew R. Jan 23 '14 at 20:48
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    $\begingroup$ Carl is close, but his formula is seen as incorrect for the case of an object heading straight toward the center of the orbit. To correct the formula, you need to use the component of the velocity perpendicular to the radius instead of the angular speed. You will need to do some geometry to get the angle between the velocity and the radius. $\endgroup$ – Brian Moths Jan 23 '14 at 21:19
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs I think I understand. In a circular orbit the velocity given by the equation I'm using is always perpendicular to the radius, but that is not the case, apparently, for elliptic orbits. The equation I am using does give me tangential velocity, but not the velocity perpendicular to the radius. However, I don't think that should matter when using the equation derived from Kepler's Third Law. $\endgroup$ – Matthew R. Jan 23 '14 at 21:55
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs OK, I tried projecting the tangential velocity vector onto the unit vector perpendicular to the radius and using that magnitude for velocity and dividing by the radius (as Carl suggested), but still no luck. Any other ideas? $\endgroup$ – Matthew R. Jan 23 '14 at 22:35
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The formula $$ \dot{\theta} = \omega = \frac{2A}{Pr^2} $$ is correct; it can also be derived from the specific angular momentum $h$: $$ h = r^2\omega = \sqrt{GMa(1-e^2)} = b\sqrt{\frac{GM}{a}}, $$ with $e = \sqrt{(a^2-b^2)/a^2}$ the orbital eccentricity. However, this doesn't solve the Kepler problem, because both $\omega$ and $r$ depend on $t$ in a complicated way, which isn't specified by the above formula. In other words, the above formula gives you $\omega(r)$, but not $\omega(t)$ and $r(t)$.

Also, note that $\theta$ is the true anomaly, which is the angle between the direction of periapsis and the current position of the body, as seen from the main focal point (where the attracting body is). And $r$ is the distance between the current position and the focal point. If you want to use cartesian coordinates $(x,y)$, it is better to parametrize them using the eccentric anomaly $E$: $$ x = a\cos E,\qquad y = b\sin E. $$

enter image description here

So how to find $E(t)$? For this, we need to introduce another parameter, called the mean anomaly $M$. The mean anomaly increases linearly with time: $$ M(t) = \frac{2\pi}{P}t = \sqrt{\frac{GM}{a^3}}t. $$ From $M(t)$, we can calculate the eccentric anomaly $E(t)$ using Kepler's equation $$ M = E - e\sin E, $$ which you have to solve numerically. Once $E(t)$ is known, $(x(t),y(t))$ follow immediately.

For completeness, the true anomaly can be calculated from the eccentric anomaly: $$ \tan\frac{\theta}{2} = \sqrt{\frac{1+e}{1-e}}\tan\frac{E}{2}, $$ and the distance $r$ to the attracting body at the focal point is $$ r = \frac{a(1-e^2)}{1+e\cos\theta}. $$

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  • $\begingroup$ Thank you! I actually only needed that last equation, but that was an awesome, concise write-up! Turns out I was calculating the delta true anomaly but was converting my theta to (x,y) using the parametric equation for eccentric anomaly. Thanks again, works like a charm! $\endgroup$ – Matthew R. Jan 23 '14 at 22:58
  • $\begingroup$ @MatthewR. You're welcome. $\endgroup$ – Pulsar Jan 23 '14 at 23:12

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