<blockquote>
  <p>Find the time that Arcturus passed the meridian in 29 April.</p>
</blockquote>

<p>With respect to the Sun, the Earth rotates once every 24 hours.  But because the Sun's apparent position changes in the sky due to Earth's revolution around the Sun, the Earth rotates once every <a href="http://www.universetoday.com/14700/how-long-is-a-day-on-earth/" rel="nofollow">23 hours, 56 minutes, 4 seconds with respect to the stars</a>.  Because of this 3 minutes, 56 seconds difference, all stars rise and set 3 minutes, 56 seconds earlier each succeeding day.  This applies to the time of a star passing through the meridian as well.</p>

<p>So, on the next day, April 29, Arcturus will pass the meridian at 22:56:04.</p>

<blockquote>
  <p>And what day does the Arcturus will pass the meridian about 01:00:00?</p>
</blockquote>

<p>Over the course of an entire year, about 365 days and 6 hours, these 3 minutes and 56 seconds adds up to an entire day, so that exactly one year later, the same star will pass through the meridian at the same time of day.</p>

<p>For Arcturus to pass the meridian exactly 22 hours earlier, it should be 22/24 of a year before then, or about 335 days later, or about 30 days earlier, or March 29.</p>

<blockquote>
  <p>solar time equal sidereal time minus 12hr</p>
</blockquote>

<p>Generally not correct.  Solar time is based on the average time it takes for the sun to cross the meridian (24 hours), and <a href="http://earthsky.org/astronomy-essentials/what-is-sidereal-time" rel="nofollow">sidereal time</a> is based on the time it takes for a star to cross the meridian.  Sidereal time counts 24 "hours" in 23 hours, 56 minutes, and 4 seconds, meaning that it runs slightly faster than our normal solar time.</p>

<p>There is one day in which solar and sidereal time are 12 hours off, on the northward equinox in March.  The Sun is on the northward equinox, but solar time is measured since midnight where sidereal time is measured since passing the meridian (local noon).</p>

<p>There is one day in which solar and sidereal time coincide, on the southward equinox in September.  The Sun is on the southward equinox, 180 degrees in the sky from the northward equinox, so the 12 hours off from midnight and the Sun being 12 hours off from the northward equinox cancel.</p>

CC BY-SA 3.0
2015-02-07T05:20:17.217Z
I have a question, but it was too long to post as a comment here. So I post it as the adding texts of my question instead. Could you help me please? Thank you^^
user9686
CC BY-SA 3.0
2015-02-03T01:05:40.293Z
2015-02-03T01:05:40.293Z
rgettman
<blockquote>
  <p>If the observer observes Arcturus passed through the meridian about 23:00:00 of 28 April. Find the time that Arcturus passed the meridian in 29 April. And what day does the Arcturus will pass the meridian about 01:00:00?</p>
</blockquote>

<p>I think that when Arcturus passed through the meridian, the hour angle will be 0 degree. Sidereal time equal hour angle plus right ascension. right ascention of any stars always be the same. So Sidereal time will be as same as every times while the star located at the meridian. And it doesn't matter because solar time equal sidereal time minus 12hr. Do I understand something wrong? And how do I do this question?</p>

<p>Adding question;</p>

<p>Your answer was very clear. But if I consider about the moon instead ex; the observer is at the equator and saw the moon which has a declination of 0 degree set at 11:00 PM yesterday, what time the observer will see the moon set today? I try to use the thinking process like this again. I know that the sidereal month of the moon is 27.32166 days. I calculate the solar month of the moon from 365.24219 of solar year divided by 12, and I get 30.43684 days per solar month. From subtracting both of these numbers, I get 3 days plus about 2 hours 45 minutes and 52 seconds. So to answer this question I will get 2 hours 45 minutes 52 seconds behind. My answer is that I will see the moon set at 13:45:52 PM. Do I right? I don't understand why the time I get is behind instead of earlier as the question about the Arcturus I asked before. I'm also not sure that does the declination of the moon and the position of the observer relate or not too. Thank you very much:)</p>

<time>
Calculate the time when Arcturus passed through the meridian
user9686
# Calculate the time when Arcturus passed through the meridian

(Asked by: username_0 on 2015-02-02)

> If the observer observes Arcturus passed through the meridian about 23:00:00 of 28 April. Find the time that Arcturus passed the meridian in 29 April. And what day does the Arcturus will pass the meridian about 01:00:00?

I think that when Arcturus passed through the meridian, the hour angle will be 0 degree. Sidereal time equal hour angle plus right ascension. right ascention of any stars always be the same. So Sidereal time will be as same as every times while the star located at the meridian. And it doesn't matter because solar time equal sidereal time minus 12hr. Do I understand something wrong? And how do I do this question?

Adding question;

Your answer was very clear. But if I consider about the moon instead ex; the observer is at the equator and saw the moon which has a declination of 0 degree set at 11:00 PM yesterday, what time the observer will see the moon set today? I try to use the thinking process like this again. I know that the sidereal month of the moon is 27.32166 days. I calculate the solar month of the moon from 365.24219 of solar year divided by 12, and I get 30.43684 days per solar month. From subtracting both of these numbers, I get 3 days plus about 2 hours 45 minutes and 52 seconds. So to answer this question I will get 2 hours 45 minutes 52 seconds behind. My answer is that I will see the moon set at 13:45:52 PM. Do I right? I don't understand why the time I get is behind instead of earlier as the question about the Arcturus I asked before. I'm also not sure that does the declination of the moon and the position of the observer relate or not too. Thank you very much:)

**Tags:** time


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## 1 answers

### Answer by username_1 on 2015-02-03. Score: 2 | Accepted answer

> Find the time that Arcturus passed the meridian in 29 April.

With respect to the Sun, the Earth rotates once every 24 hours. But because the Sun's apparent position changes in the sky due to Earth's revolution around the Sun, the Earth rotates once every [23 hours, 56 minutes, 4 seconds with respect to the stars](http://www.universetoday.com/14700/how-long-is-a-day-on-earth/). Because of this 3 minutes, 56 seconds difference, all stars rise and set 3 minutes, 56 seconds earlier each succeeding day. This applies to the time of a star passing through the meridian as well.

So, on the next day, April 29, Arcturus will pass the meridian at 22:56:04.

> And what day does the Arcturus will pass the meridian about 01:00:00?

Over the course of an entire year, about 365 days and 6 hours, these 3 minutes and 56 seconds adds up to an entire day, so that exactly one year later, the same star will pass through the meridian at the same time of day.

For Arcturus to pass the meridian exactly 22 hours earlier, it should be 22/24 of a year before then, or about 335 days later, or about 30 days earlier, or March 29.

> solar time equal sidereal time minus 12hr

Generally not correct. Solar time is based on the average time it takes for the sun to cross the meridian (24 hours), and [sidereal time](http://earthsky.org/astronomy-essentials/what-is-sidereal-time) is based on the time it takes for a star to cross the meridian. Sidereal time counts 24 "hours" in 23 hours, 56 minutes, and 4 seconds, meaning that it runs slightly faster than our normal solar time.

There is one day in which solar and sidereal time are 12 hours off, on the northward equinox in March. The Sun is on the northward equinox, but solar time is measured since midnight where sidereal time is measured since passing the meridian (local noon).

There is one day in which solar and sidereal time coincide, on the southward equinox in September. The Sun is on the southward equinox, 180 degrees in the sky from the northward equinox, so the 12 hours off from midnight and the Sun being 12 hours off from the northward equinox cancel.

> **Comments:**
> **username_0**: I have a question, but it was too long to post as a comment here. So I post it as the adding texts of my question instead. Could you help me please? Thank you^^  

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