6
$\begingroup$

When viewing star maps in Stellarium, I noticed that the meridians of the equatorial grid were denoted in hours and not degrees, and that always puzzled me. Why did astronomers choose this system? I understand it has something to do with being relative to the Earth's rotation which takes 24 hours, but I can't figure out exactly what.

$\endgroup$
5
$\begingroup$

In order to know when a star will be above horizon, you'll need an equation with times, not "celestial latitudes".

So you end up adding RA and sidereal time.

EDIT

In order to find out the position of a star you need to find out how its Hour Angle varies with time.

The Hour Angle is how far the star is from the observer's meridian:

Diagram of angles

In this image, it is the angle between the site's celestial meridian (North Celestial P ole-Z enith-S outh horizon) and the star's celestial meridian (P-X-Y). It is the same angle as between where X culminates and X.

So it comes than $HA=ST-RA$ and since you may not have a "local sidereal time" clock at hand, you can use Greenwich's Sidereal Time: $HA=GST-Lat-RA$

  • $HA$ (local) Hour Angle of a star
  • $ST$ Local Sidereal Time
  • $GST$ Greenwich Sidereal Time
  • $RA$ Right Ascension of a star
  • $Lat$ (local) Geographical Latitude
$\endgroup$
8
  • $\begingroup$ This is actually not correct! The Sidereal Time is not a time but an angle just as right ascension is. It is the angle between the local meridian and the vernal equinox. For instance, you cannot use sidereal time as the time of an event because the same sidereal time occurs every 23h56m... The strict interpretation of time includes the date. $\endgroup$
    – Dieudonné
    Jul 9 '14 at 13:38
  • 1
    $\begingroup$ @Dieudonné You are wrong. Sidereal time is a time, measured in Sidereal seconds, not in degrees.Please check e.g. Wikipedia: en.wikipedia.org/wiki/Sidereal_time Local Sidereal Time at any locality differs from the Greenwich Sidereal Time value of the same moment, by an amount that is proportional to the longitude of the locality. When one moves eastward 15° in longitude, sidereal time is larger by one sidereal hour (note that it wraps around at 24 hours). I have actually used sidereal time clocks in my observations. $\endgroup$
    – Envite
    Jul 9 '14 at 21:57
  • $\begingroup$ To me, time can be used to identify a specific moment (in time), and should include a date. A time therefore, should not occur again. You cannot use sidereal time to identify a specific moment in time because 23h56m later it will occur again. I would call sidereal time an angle because it specifies the right ascension of objects which lie on the local meridian (i.e. the line from south to zenith in the northern hemisphere). Sirius, for instance, will be on the meridian when the sidereal time is 6h45m. $\endgroup$
    – Dieudonné
    Jul 10 '14 at 8:01
  • $\begingroup$ As you say, sidereal time is measured in sidereal seconds and not in degrees. But a sidereal second is not necessarily a unit of time. A second is precisely defined with the number of cycles of radiation of cesium 133. A sidereal second cannot be defined so exactly. The duration of a sidereal second in real seconds in January will be different from the duration in July because of variations in the rotation of the Earth. The practical use of sidereal time is as the right ascension of the local meridian at a specific time. And right ascension is an angle. $\endgroup$
    – Dieudonné
    Jul 10 '14 at 8:12
  • 1
    $\begingroup$ @Dieudonné No, you can not define a time unit as an angle, they are different magnitudes. You can, instead, define a time unit by using an angle: a sidereal second is the time the Earth rotates 360/(24*60*60) degrees in reference to the fixed stars in average. $\endgroup$
    – Envite
    Jul 11 '14 at 10:28
2
$\begingroup$

Astronomers use this system simply because it is (was) convenient. The sky shifts about 15° every hour or a full circle, 360° every 24 hours (actually 23h56m). So if you want to know where a star is in an hour, simply add an hour to the local hour angle.

$\endgroup$

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.