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

<p>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 <a href="https://en.wikipedia.org/wiki/Sidereal_time" rel="nofollow noreferrer">sidereal time</a> (by definition). You need to convert sidereal time to local time (at a given day), e.g. with the script described <a href="https://astronomy.stackexchange.com/q/20296/7472">here</a>. On September 23rd (or whatever day the Fall equinox falls, thanks @MikeG) on the Greenwich meridian, this will happen at 7:23am UTC.</p>

CC BY-SA 3.0
2018-03-17T23:12:54.003Z
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.
Mike G
CC BY-SA 3.0
2018-03-18T07:52:53.223Z
@MikeG ah, of course. Thanks, I've edited it in.
Glorfindel
CC BY-SA 3.0
2018-03-17T17:26:20.027Z
2018-03-18T07:52:06.023Z
2018-03-18T07:52:06.023Z
Glorfindel
<p>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?</p>

<p>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?</p>

<star><coordinate>
Given a star's ra, dec how can I compute one pair of lat, long and time where the star would be at zenith?
Cristi Berceanu
# Given a star's ra, dec how can I compute one pair of lat, long and time where the star would be at zenith?

(Asked by: username_0 on 2018-03-17)

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?

**Tags:** star, coordinate


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## 1 answers

### Answer by username_1 on 2018-03-17. Score: 6 | Accepted answer

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](https://en.wikipedia.org/wiki/Sidereal_time) (by definition). You need to convert sidereal time to local time (at a given day), e.g. with the script described [here](https://astronomy.stackexchange.com/q/20296/7472). On September 23rd (or whatever day the Fall equinox falls, thanks @MikeG) on the Greenwich meridian, this will happen at 7:23am UTC.

> **Comments:**
> **username_2**: 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.  
> **username_1**: @MikeG ah, of course. Thanks, I've edited it in.  

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