# How to calculate when an object will be directly

## Main Question or Discussion Point

I'm having some trouble determining when a celestial object will be overhead (in terms of right ascension, not declination). Since we use solar time, it wouldn't be correct to simply use the right ascension of the object since each day the sidereal and solar time become four minutes different. So, given the celestial coordinates of an object, latitude and longitude of a location on earth, how do I find what time the object will be overhead? I've been grappling with this for awhile now and could really use some help instead of being asked more questions.

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I've actually just made some progress on the problem from my book. I needed to determine the local sidereal time over Cuba at approximately 6 pm in late September. Using the GMT, and assuming (to make it easier) that this event happened on the equinox, I converted Cuba's longitude to hours and subtracted them from the GMT to find the approximate local sidereal time. Is there an easier way to do this? I feel like it would be an absolute pain to calculate LST when not near the equinox.

I've actually just made some progress on the problem from my book. I needed to determine the local sidereal time over Cuba at approximately 6 pm in late September. Using the GMT, and assuming (to make it easier) that this event happened on the equinox, I converted Cuba's longitude to hours and subtracted them from the GMT to find the approximate local sidereal time. Is there an easier way to do this? I feel like it would be an absolute pain to calculate LST when not near the equinox.
I wouldn't want to do it in my head, but here's a pretty simple formula to get the sidereal time at Greenwich for any date and time. From there, you just adjust for your longitude, as you did above.

If you don't understand it, just google the various terms such as "Julian date." You might also consider googling for "practical astronomy" or "ephemeris calculations" to get a more comprehensive tutorial on time conversion.

If you just want the answer without doing the calculation yourself, there are lots of sites that will do it for you, e.g.
http://www.jgiesen.de/astro/astroJS/siderealClock/

Yes you would need time but you would also need position. Azimuth and Altitude must be calculated from the Hour Angle, Latitude and Declination with Altitude to be 90 degrees.

HA = f(gst, time and Longitude minus the Right Ascension)
AZI = f(HA, Latitude, Declination)
ALT = f(HA, Latitude, Declination)

Last edited:
BobG
Homework Helper
I've actually just made some progress on the problem from my book. I needed to determine the local sidereal time over Cuba at approximately 6 pm in late September. Using the GMT, and assuming (to make it easier) that this event happened on the equinox, I converted Cuba's longitude to hours and subtracted them from the GMT to find the approximate local sidereal time. Is there an easier way to do this? I feel like it would be an absolute pain to calculate LST when not near the equinox.
Local sidereal time (which is the same as right ascension) is the Greenwich Mean Sidereal Time plus the longitude.

You can look up the GMST at midnight for a particular date in an Astronomical Almanac, and then add the current Greenwich Mean Time (solar) to get your current GMST.

Or you can do the easy thing and download a spreadsheet that calculates GMST for any time you want at any given longitude just by setting the azimuth to 180. Or, if your real goal is correlate your location and your telescope's az/el, it will calculate the RA/dec of the object you're looking at directly instead of you having to figure it out manually. (You should do all of this manually at least once, just because, but I don't think I know anyone that would want to do this manually all the time.)