If I observe the moon pass my celestial meridian at a specific north polar clock time will I observe every subsequent passage at ${\rm current\,time} + 13.2^\circ \times ({\rm days\,passed}) \mod 360^\circ$?

Does the moon pass celestial meridian at exactly the same time if observed from every point on this longitude?


1 Answer 1


Your formula is a crude approximation. It's ok if you just want a rough idea of the number of orbits the Moon makes over a long period of time, but it's not useful for calculating the daily motion of the Moon.

The Moon's orbit is fairly eccentric (about 0.0549), so its daily speed varies quite a bit. The Babylonians discovered that the mean daily motion is approximately 13° 10' 35", but that it can range from 11° 6' 35" to 15° 14' 35". The Moon's motion is rather complex, the Wikipedia article on lunar theory is a good introduction, with a rundown of the history, and quite a few relevant formulae.

The fact that the Moon's motion is so hard to compute accurately (compared to the motions of the Sun and the visible planets) is a mixed blessing. On the one hand, it would have been great for navigators if it had been easier to prepare Moon position tables. On the other hand, it made it rather obvious that early models of the celestial motions were incorrect. No matter what they tried, the early cosmologists weren't able to produce a model of the Moon's motions that was consistent with observation. Eg, in the best lunar theory of Ptolemy's model of the cosmos, which was good enough at estimating the Moon's position to use for eclipse prediction, the apparent size of the Moon ought to change by around 100% every month, which it obviously doesn't do. :) So if it weren't for the Moon's tricky orbit we might still be using Ptolemy's geocentric model of the cosmos today, and the whole scientific revolution may never have happened.


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