Take the 2-minute tour ×
Astronomy Stack Exchange is a question and answer site for astronomers and astrophysicists. It's 100% free, no registration required.

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?

share|improve this question
1  
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 –  Ahmed Hamdy Dec 17 '13 at 9:11
    
Thanks :) But, I'm looking for a mathematical approach as it was asked in one of my astronomy homeworks. –  Karthikeyan KC Dec 17 '13 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? –  Jeremy Feb 18 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. –  Karthikeyan KC Feb 18 at 13:12
add comment

2 Answers 2

up vote 2 down vote accepted

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

share|improve this answer
add comment

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.

share|improve this answer
    
Thanks :) But, I'm looking for a mathematical approach as it was asked in one of my astronomy homeworks. –  Karthikeyan KC Dec 17 '13 at 17:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.