5
$\begingroup$

I am approximating the azimuthal angle of Callisto from Jupiter or the Jupiter+moons barycenter using a constant angular speed circular orbit. This differs from the JPL ephemerides by only about 4 degrees, and the error has a very pleasant shape with a period of around 14 days.

However, none of the higher accuracy corrections I can find explain the 4 degree error (they seem to all be much less than a degree, and do not have the right time period). Is there a simple explanation?

I used a very simple formula:

angle = 21.57107 * time-in-days + angle-at-time-zero

This can be found from https://ssd.jpl.nasa.gov/?sat_elem or Meeus's Astronomical Algorithms (Chapter 43, satellites of Jupiter).

Meeus's book (I only have the first edition) list some corrections:

  • eccentricity
  • movement of the perijove (slow, less than 0.001 degrees per day)
  • a list of terms, the largest is 0.84 degrees, the rest 0.03 degrees and quickly decreasing in size

The eccentricity correction did not appear to help (it is listed with what appears to be an incorrect value of 21.487... instead of 21.57... but maybe this is my error). I used Kepler's equation (iterative solution, then cartesian form of ellipse, and arctan to return to the angle) to more directly handle the eccentricity, but Callisto's eccentricity of 0.0074 did not make much difference (the semi-major and semi-minor axis are the same length up to rounding on the JPL website).

Here is the graph of the error over about 60 (Earth) days time. Let me know if python code would make the question more precise. The JPL data is used to compute the position of Callisto with respect to the Jupiter system barycenter, and then I just take arctan2(y,x) to get the azimuthal angle. This agrees fairly well with the simple linear function angle = 21.57107*days, but with a +-4 degree variation depicted here:

Smooth sinusoidal graph depicting the difference between a circular approximation and the JPL data

Edit: I think I can explain away half (2 degrees worth) of the error by saying the orbit is tilted 25 degrees parallel to the equator. The reduced error still has a smooth periodic shape, but as far as I know, Callisto orbits around the equator of Jupiter. I checked that the JPL data does seem to include a 25 degree tilt.

I'm having a hard time interpreting things as my terrestrial mind has just started to grapple with the idea that a day (a full day-night cycle) is shorter near the poles of Jupiter than at the equator, and so what do we even mean by longitude. Does Jupiter have an equator? Does a barycenter have an equator? What is tilt?

$\endgroup$
6
  • $\begingroup$ I have Meeus’s second edition, if it might help. But without even looking at it, I can say/presume the following… A body on a circular orbit would travel at constant speed, yes. However, a body on an elliptical orbit (and such is the case in virtually all satellite and planetary orbits) would travel faster near periastron (wrongly written as “periapsis” in most sources), and slower near apoastron. Does your model take account of that? $\endgroup$ Nov 29 '20 at 23:28
  • 3
    $\begingroup$ @PierrePaquette Your wording is actually wrong. Periastron and apastron only refer to a central stellar object (greek. astron = star), while periapsis and apoapsis work for any kind of central object; periastron is as such a special case for orbits around a star, which can be made more special by using perihel for orbits around the Sun. Check also en.wikipedia.org/wiki/Apsis $\endgroup$ Nov 29 '20 at 23:43
  • $\begingroup$ @planetmaker “Apses” are the points closest to and furthest from a primary body—see en.wikipedia.org/wiki/Apsis, among many other sources. These points are called “peri-” and “apo-” with the suffix pertaining to the body (e.g. “-gee” for Earth, thus “perigee”). The generic word for a celestial body is ἄστρον (see en.wiktionary.org/wiki/%E1%BC%84%CF%83%CF%84%CF%81%CE%BF%CE%BD). “Periapsis” would mean “the point closest to the point closest to the body.” Nonsense. The proper word is “periastron”: “periapse” is a misconstruction. $\endgroup$ Nov 29 '20 at 23:51
  • $\begingroup$ @PierrePaquette yes. The two models, circular vs elliptic, yield nearly identical errors (at the 4 degree level). Callisto's orbit is nearly circular. The semi-major axis is 1882700km, and the semi-minor is 1882648km. On the plot, I couldn't see the difference as more than a pixel or so. $\endgroup$ Nov 30 '20 at 0:04
  • $\begingroup$ (Semi-minor computed from the eccentricity, which has two few significant figures to report any difference, but I included a few decimals where they differ so that the sentence made sense.) $\endgroup$ Nov 30 '20 at 0:06

Your Answer

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

Browse other questions tagged or ask your own question.