<p>The flaw in your logic is that "RA and L are constants". The right ascension (RA) of the object on the meridian is not constant. It changes continuously, otherwise we would not say that the sky is rising in the east and setting in the west.</p>

<p>Also, the Sun moves about 1 degree per day along the ecliptic, so the RA of the Sun is constantly changing. If the Sun is on the meridian at 12:00:00 every day, and if the RA of the Sun is changing every day, then the same RA is not on the meridian at the same time on every day. (Note: Because of the equation of time, the sun does not cross the meridian at the same time each day. For purposes of this thought experiment, pretend that it is true.)</p>

<p>These two things taken together lead to the fact that 24 hours of right ascension cross the meridian every 23 hours 56 minutes 4 seconds. And this is why the question asks for a date when the LMC culminates (crosses the meridian) at a specific time (9 pm) and a specific location.</p>

CC BY-SA 4.0
2018-09-25T17:58:07.997Z
2018-09-25T17:58:07.997Z
JohnHoltz
<p>I know that LST = HA + RA.<br>
And GST is LST + longitude(L).<br>
If a star culminates, it's HA=0. Therefore, GST = RA + L.<br>
Since RA and L are constants, GST must also be a constant. So, how can GST be different on different days of the year for the culmination of the same object for the same observer? Also if I'm given the L and RA, why am I given the GST in this question?<a href="https://i.sstatic.net/nQ93A.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/nQ93A.png" alt="enter image description here"></a></p>

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2018-09-25T18:33:22.610Z
GST = HA + RA + L, and HA is not constant.
Mike G
<coordinate><time>
Clarification on Greenwich Sidereal Time (GST)
Arnab Chowdhury
# Clarification on Greenwich Sidereal Time (GST)

(Asked by: username_0 on 2018-09-25)

I know that LST = HA + RA.  
And GST is LST + longitude(L).  
If a star culminates, it's HA=0. Therefore, GST = RA + L.  
Since RA and L are constants, GST must also be a constant. So, how can GST be different on different days of the year for the culmination of the same object for the same observer? Also if I'm given the L and RA, why am I given the GST in this question?[![enter image description here](https://i.sstatic.net/nQ93A.png)](https://i.sstatic.net/nQ93A.png)

**Tags:** coordinate, time

> **Comments:**
> **username_1**: GST = HA + RA + L, and HA is not constant.  

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## 1 answers

### Answer by username_2 on 2018-09-25. Score: 2

The flaw in your logic is that "RA and L are constants". The right ascension (RA) of the object on the meridian is not constant. It changes continuously, otherwise we would not say that the sky is rising in the east and setting in the west.

Also, the Sun moves about 1 degree per day along the ecliptic, so the RA of the Sun is constantly changing. If the Sun is on the meridian at 12:00:00 every day, and if the RA of the Sun is changing every day, then the same RA is not on the meridian at the same time on every day. (Note: Because of the equation of time, the sun does not cross the meridian at the same time each day. For purposes of this thought experiment, pretend that it is true.)

These two things taken together lead to the fact that 24 hours of right ascension cross the meridian every 23 hours 56 minutes 4 seconds. And this is why the question asks for a date when the LMC culminates (crosses the meridian) at a specific time (9 pm) and a specific location.


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