From Wikipedia, the True anomaly is defined as the angle between the direction of periapsis and the position of the celestial body. I'm trying to find how the true anomaly changed over time for a spacecraft over one orbit.

The following plot is a simulation ran on GMAT based on a spacecraft at 575km altitude (~5740 secs orbit period), at a 10 degree inclination. Eccentricity is set at approximately 0 (circular orbit). The initial True Anomaly is set at 0 degrees.

Simulated Results of TA over two orbits

Can anyone explain how this plot would make sense? I would expect that for a nearly circular orbit, the True Anomaly would increase rather linearly from 0 - 360 degrees, and not experience the abrupt jumps. Is my understanding of True Anomaly incorrect?

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    $\begingroup$ The plot doesn't make sense. Since you didn't show what you did, it's hard to tell what you did wrong (or what GMAT did wrong). For a nearly circular orbit the distinction between argument of periapsis and true anomaly become a bit fuzzy; different systems handle this fuzziness differently. I suggest plotting the argument of latitude (the sum of the argument of periapsis and true anomaly) versus time. $\endgroup$ Feb 11 at 9:19
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    $\begingroup$ Since being hinty didn't work...aren't these big jumps just an artifact of the plot wrapping from 0 to 360? There are not actually any "jumps" in the value, it just wraps from 360 to 0. The actual graph is a smooth sort of sine wave that just happens to be crossing the 0-to-360 barrier - see this cut and pasted version i.imgur.com/DA7DEE2.png I know nothing about the True Anomaly, but this just looks like a plotting problem. $\endgroup$ Feb 11 at 14:57
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    $\begingroup$ @OrganicMarble Mean anomaly should exhibit a sawtooth behavior, ramping linearly from 0° to 360° (or nearly linearly when faced with perturbations) over the course of an orbit. Eccentric anomaly is the solution $E$ to $M = E - e\sin E$, which for low eccentricity orbits means $E \approx M$. True anomaly, which I'll denote as $\nu$, is the angle in the plane of the orbit between periapsis and current position and is related to eccentric anomaly via $\sqrt{1-e}\tan\frac\nu2=\sqrt{1+e} \tan \frac E2$. For very small eccentricity, $\nu\approx E\approx M$. (continued) $\endgroup$ Feb 11 at 17:44
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    $\begingroup$ In other words, true anomaly is not sinusoidal. It should be more or less sawtooth. Either the OP did something wrong (e.g., plotting argument of periapsis rather than true anomaly), or GMAT did something wrong / anomalous. $\endgroup$ Feb 11 at 17:47
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    $\begingroup$ BTW, a simple way to deal with angle wrap-around is ang = (ang + 180) % 360 - 180, where % is the modulus operator. Depending on the language, you may need to do ang = (ang % 360 + 180) % 360 - 180. $\endgroup$
    – PM 2Ring
    Feb 12 at 1:22

1 Answer 1

  1. The jumps from near 0 to near 360 are just a branch cut. If the true anomaly were plotted from -180 to 180, the true anomaly would look more like a sine wave.

  2. Periapse, and therefore true anomaly, are meaningless for a perfectly circular orbit. Since nothing except for maybe a simulation is absolutely perfect, every real orbit has a periapse, but it is still ill-defined for a near-circular orbit.

This reminds me of an orbit that I was analyzing when I was a summer intern. It was for the AIM mission launched in 2007, targeting a 600km circular sun-synchronous orbit (inc~98deg). In some simulation cases, the achieved orbit was very close to circular.

In these cases, perturbations in the orbit, mostly due to the J2 effect, caused the periapse to precess around the orbit once per orbit. The simulated spacecraft never passed through periapse. One way to think about it is that the periapse itself isn't stable, and the whole ellipse shifts around at nearly the same speed as the spacecraft. If I remember, the argument of periapse shifted nearly 180deg in the ~2 minutes the spacecraft was flying over the equator, then 180deg again in a couple of minutes 48 minutes later over the other side.

This is something that definitely doesn't happen to more elliptical orbits. Another way to think about it is that the orbit was so close to circular that it wasn't well-modeled as an ellipse with one definite apoapse and periapse, but more like a peanut with two local minima and local maxima. These local minima and maxima were a couple hundred meters away from a perfect circle with a radius of about 7000km.

The point is, don't worry about true anomaly or periapse in a perfectly-circular orbit. Even if you do calculate them, they won't act like you expect. Instead, take whatever true anomaly you get and add it to the argument of periapse to get the argument of latitude. This will behave in the sawtooth manner you expect. If you have a near-equatorial orbit such that the ascending node isn't well-defined either, then add the longitude of ascending node to the argument of latitude to get the mean longitude. Argument of latitude and mean longitude are well-defined for elliptical orbits too, but are the only meaningful elements for near-circular or near-equatorial orbits.

  • $\begingroup$ To see if this is happening to your orbit, try plotting the argument of periapse vs time. For an unperturbed orbit, this would be constant. $\endgroup$
    – kwan3217
    Feb 13 at 1:41
  • $\begingroup$ Thank you for your help. Tried simulating it in a more elliptical orbit and the sawtooth trend I expected emerged. As for the AOP vs time plot for the near-circular, near-equatorial orbit as in the original post, I observed a rather sawtooth plot as well. $\endgroup$
    – Shawn Lim
    Feb 13 at 4:11

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