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Does Earth's mean anomaly as given by Mathematica proceed too fast?

Posted 2 years ago

Hi everyone,

when you compute earth's mean anomaly from mathematica's astronomical data over a period of time it appears to move to fast:

FromDMS /@ 
 PlanetData["Earth", 
  EntityProperty["Planet", 
     "MeanAnomaly", {"Date" -> #}] & /@ {DateObject[{2023, 1, 1, 0, 0,
       0}, TimeZone -> "Europe/London"], 
    DateObject[{2024, 1, 1, 0, 0, 0}, TimeZone -> "Europe/London"]}]

The result given is {356.374, 358.018} (angles modulo 360), and this amounts to an elliptic precession of 1.64396 degree per year. Earth should have that much in about a century. (There is nothing special about the dates chosen, the mean anomaly increases linearly as expected, but the slope is slightly too high.)

Do I have some sort of misconception, or is there some trouble with the data?

Yours,

Bernd.

POSTED BY: Bernd Günther
2 Replies

The calendar year differs from the astronomical year.

POSTED BY: Todd Rowland
Posted 2 years ago

Certainly, it is even longer; if you move the date of comparison 6h or so forward the discrepancy will increase. Also, the differences between various types of astronomical year are much smaller than the current effect.

I got a formula for earth's mean anomaly:

M = 358.47583° + 35999.04975° TJ - 0.15° 10^(-3)TJ^2 - 0.33° 10^(-5) T_J^3

which says that M should increase by 35999 degrees in a Julian century, i.e. in 36525 days - except for small non linear perturbations (from J. Meyer on sundials, who has it from Dumoulin' book on practical astronomy, who in turn has it from Newcomb). That's what I expected.

It seems that Mathematica's data are just routed through from JPL at Caltech, and I posted my question to them. They replied rather elaborately, and I have not yet completely digested their answer, but the explanation may be that formulas as the one above do not really describe earth, but the common center of mass of the earth-moon-system. Such effects could indeed be large enough.

POSTED BY: Bernd Günther
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