Selected Questions - Answers From All Experts Astronomy Forum (Sidereal time)
Question: Can you explain what is sidereal time and how one can find it? - Marge M., Montreal
Answer:
"Sidereal time" basically means time by the stars, i.e. using star positions to obtain the time at one's location. The diagram below is useful as a reference:
Here, the observer ('Obs') at longitude L with respect to Greenwich (G, or the Greenwich meridian at 0 degrees). The L.S.T. or local sidereal time is 6h 00m as denoted by the fact that 6h Right Ascension (RA) is on the observer's meridian. Then the hour angle (HA) for the star will be as shown. (Note: the view shown is a POLAR one, i.e. the Earth as seen from above Earth's north pole).
The crucial thing to note is how the sidereal time is related to RA as well as hour angle. So to find it one usually needs to know - or find- these measures as well.
The Right Ascension (RA) is measured in units (hours, minutes, seconds) associated with time, and the Right Ascension of the star is clearly equal to the local sidereal time (L.S.T.) plus the hour angle (which in this case is negative). Thus, we can write:
HA = RA of observer meridian - RA of object
Again, the hour angle is the angular distance in hours separating your local meridian and the RA of the object. It increases westwards, unlike RA - which increases eastwards. As may be seen, the RA of the observer's meridian (6 h in Figure 1) is none other than the local sidereal time. If you know what Right Ascension is on your meridian you know the L.S.T.
Another useful expression used often is:
Local time of transit (L.T.T.) = Star's RA - Sun's RA
Where L.T.T. is the local mean solar time of transit (or L.M.T.) of the star or other object and Sun's RA is the Sun's Right Ascension on the particular date. A simple interpolation method is adequate (for most purposes) to obtain the latter, given the base data for the specific dates of the solstices and equinoxes below:
Date: ----- March 21---------June 22-----------Sept. 23--------Dec. 22
Sun R.A. ----( 0 hr.)--------(6 hr.)-----------(12 hr.)----------(18 hr.)
March 21 is just the (approx.) date of the Vernal Equinox, June 22 is the summer solstice, Sept. 23 the autumnal equinox and Dec. 22 the winter solstice. Since the Earth complete one circuit of the Sun in 365 days, the Sun (as seen from the Earth) will be apparently displaced along the ecliptic (the plane of the Earth's orbit projected onto the celestial sphere) by about one degree per day or equivalent to a 4 minute time difference. (Since 24 hours of Right Ascension corresponds to a 360 deg angular difference).
Thus, using either of the four known RA dates for the Sun - and knowing the Sun's RA changes by 4 mins/day, the new solar RA on any day can be found.
Example (1)
Find the Right Ascension of the Sun on July 1st.
We note that 8 days separates July 1st from June 22 when the Sun's R.A. is known to be 6h 00m.
The time difference is: 8 days x (4 min/day) = 32 mins.
So the Sun's RA on July 1st = 6h 00m + 32 min = 6h 32 m
Example (2):
Aldebaran (RA = 4h 34m) is found to be 45 degrees east of your meridian on a particular date. What is your L.S.T.?
Solution
First, find the hour angle (HA) of Aldebaran. We have 45 deg = 45 deg/(15 deg/h) = 3 h. But since the HA is measured east of the meridian then one must use the negative value (-3 h) so:
24 h + (- 3h) = 21 h 00m
We know:
HA = RA of the meridian - RA (object)
and RA of the meridian = L.S.T.
Therefore:
L..S.T.= HA + RA of object
L.S.T. = 21h 00m + 4 h 34 m = 25h 34m = 1h 34m (25h 34m - 24h 00m)
A nice additional problem for practice, to see if you understand sidereal time, is given below:
The star Canopus (RA = 6h 20m) is observed to have a local hour angle = 45 deg on Feb. 10th for a given location.
What is the local sidereal time ?
Answer:
"Sidereal time" basically means time by the stars, i.e. using star positions to obtain the time at one's location. The diagram below is useful as a reference:
Here, the observer ('Obs') at longitude L with respect to Greenwich (G, or the Greenwich meridian at 0 degrees). The L.S.T. or local sidereal time is 6h 00m as denoted by the fact that 6h Right Ascension (RA) is on the observer's meridian. Then the hour angle (HA) for the star will be as shown. (Note: the view shown is a POLAR one, i.e. the Earth as seen from above Earth's north pole).
The crucial thing to note is how the sidereal time is related to RA as well as hour angle. So to find it one usually needs to know - or find- these measures as well.
The Right Ascension (RA) is measured in units (hours, minutes, seconds) associated with time, and the Right Ascension of the star is clearly equal to the local sidereal time (L.S.T.) plus the hour angle (which in this case is negative). Thus, we can write:
HA = RA of observer meridian - RA of object
Again, the hour angle is the angular distance in hours separating your local meridian and the RA of the object. It increases westwards, unlike RA - which increases eastwards. As may be seen, the RA of the observer's meridian (6 h in Figure 1) is none other than the local sidereal time. If you know what Right Ascension is on your meridian you know the L.S.T.
Another useful expression used often is:
Local time of transit (L.T.T.) = Star's RA - Sun's RA
Where L.T.T. is the local mean solar time of transit (or L.M.T.) of the star or other object and Sun's RA is the Sun's Right Ascension on the particular date. A simple interpolation method is adequate (for most purposes) to obtain the latter, given the base data for the specific dates of the solstices and equinoxes below:
Date: ----- March 21---------June 22-----------Sept. 23--------Dec. 22
Sun R.A. ----( 0 hr.)--------(6 hr.)-----------(12 hr.)----------(18 hr.)
March 21 is just the (approx.) date of the Vernal Equinox, June 22 is the summer solstice, Sept. 23 the autumnal equinox and Dec. 22 the winter solstice. Since the Earth complete one circuit of the Sun in 365 days, the Sun (as seen from the Earth) will be apparently displaced along the ecliptic (the plane of the Earth's orbit projected onto the celestial sphere) by about one degree per day or equivalent to a 4 minute time difference. (Since 24 hours of Right Ascension corresponds to a 360 deg angular difference).
Thus, using either of the four known RA dates for the Sun - and knowing the Sun's RA changes by 4 mins/day, the new solar RA on any day can be found.
Example (1)
Find the Right Ascension of the Sun on July 1st.
We note that 8 days separates July 1st from June 22 when the Sun's R.A. is known to be 6h 00m.
The time difference is: 8 days x (4 min/day) = 32 mins.
So the Sun's RA on July 1st = 6h 00m + 32 min = 6h 32 m
Example (2):
Aldebaran (RA = 4h 34m) is found to be 45 degrees east of your meridian on a particular date. What is your L.S.T.?
Solution
First, find the hour angle (HA) of Aldebaran. We have 45 deg = 45 deg/(15 deg/h) = 3 h. But since the HA is measured east of the meridian then one must use the negative value (-3 h) so:
24 h + (- 3h) = 21 h 00m
We know:
HA = RA of the meridian - RA (object)
and RA of the meridian = L.S.T.
Therefore:
L..S.T.= HA + RA of object
L.S.T. = 21h 00m + 4 h 34 m = 25h 34m = 1h 34m (25h 34m - 24h 00m)
A nice additional problem for practice, to see if you understand sidereal time, is given below:
The star Canopus (RA = 6h 20m) is observed to have a local hour angle = 45 deg on Feb. 10th for a given location.
What is the local sidereal time ?
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