# How to calculate the altitude of the Moon?

With the known latitude coordinate of the observing position, how to find the altitude of the moon when it is high, i.e when it crosses the local meridian?

• May be this is out of your scope, but I saw those programs may help you. There is a cool software Digital planetarium application, free to download. And another one here but shareware Starry Night Dec 17, 2013 at 9:11
• Thanks :) But, I'm looking for a mathematical approach as it was asked in one of my astronomy homeworks.
– Ken
Dec 17, 2013 at 17:48
• If it is homework, shouldn't you be attempting to research the answer yourself, or figure it out for yourself? The point being to help you comprehend what is going on, not just obtain an answer? Feb 18, 2014 at 4:04
• @Jeremy That is exactly what I did (sort of) :) AstroFloyd's answer did help me to comprehend the concept a bit, and later I got a clarified answer from my class.
– Ken
Feb 18, 2014 at 13:12

The transit altitude of an object does not only depend on your latitude ($b$), but also on the declination of the object ($\delta$): $$h_\mathrm{tr} = \arcsin\left(\sin b \sin \delta + \cos b \cos\delta\right).$$

With a lot of difficulty if you want to do it numerically by yourself. The Moon's orbit is very complex.

Far and away the easiest method would be to use NASA's HORIZONS service for calculating ephemerides, which will get you very comprehensive and accurate information on objects at your specified times.

I don't think that HORIZONS can directly return the time that the Moon is at its maximum altitude. However, if you can write some basic computer code then it shouldn't be difficult to query the ephemerides with a precise (say 1 minute) time interval and find the maximum altitude from there.

• Thanks :) But, I'm looking for a mathematical approach as it was asked in one of my astronomy homeworks.
– Ken
Dec 17, 2013 at 17:58
• Can you please give references to explanations of the moon orbit, especially as it looks from earth? I'm puzzled that the moon and the sun have interchangable paths on the sky between winter and summer. Is it possible to easily explain this? Would this be a good question (on its own) to ask in astronomy.stackexchange.com? Many thanks! Aug 11, 2016 at 13:11
• Horizons has an "RST flag" (rise, set, transit) in the observer table options, so it can do the computation for you. Aug 3, 2023 at 4:01

This can be taken a bit further. The right-hand side of $$\sin h_{\rm tr} = \sin \phi \sin \delta + \cos \phi \cos \delta$$ is the formula for the cosine of the difference of two angles, so $$\sin h_{\rm tr} = \cos( \phi - \delta).$$ Noting that $$\sin x = \cos(90^\circ - x),$$ we have $$\cos(90^\circ - h_{\rm tr}) = \cos(\phi - \delta),$$ giving $$h_{\rm tr} = 90^\circ - \left | \phi - \delta \right | .$$ The absolute value is necessary because the cosine is an even function, i.e., $$\cos(-x) = \cos x.$$