0
$\begingroup$

As the questions says, I have the celestial coordinates of a star(ra, dec). My trigonometry is not so great so I haven't managed to figure out a solution to this. I want to find a place (lat, long) and a time when the star will be right above the computed point?

I.e. star coordinates: 7h 23m 23.97s -13° 25' 2.64", where would be the point(lat, long) and time where the star would be right above?

$\endgroup$
4
$\begingroup$

Latitude is the easiest; it's the same as the star's declination: -13° 25' 2.64".

Longitude and time depend on each other. A star with right ascension 7h 23m 23.97s will be at the zenith point at 7h 23m 23.97s local sidereal time (by definition). You need to convert sidereal time to local time (at a given day), e.g. with the script described here. On September 23rd (or whatever day the Fall equinox falls, thanks @MikeG) on the Greenwich meridian, this will happen at 7:23am UTC.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Sidereal and solar times coincide at the September equinox. At the March equinox they are 12 hours apart; e.g the Sun is at RA 0h but is on the meridian at noon as usual. $\endgroup$ – Mike G Mar 17 '18 at 23:12
  • $\begingroup$ @MikeG ah, of course. Thanks, I've edited it in. $\endgroup$ – Glorfindel Mar 18 '18 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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