<p>When one asks the question, "How is a year defined?", there are actually a few different answers one can provide. What you're implicitly asking, even if you don't know, is how one defines a <a href="https://en.wikipedia.org/wiki/Sidereal_year" rel="nofollow noreferrer">Sidereal Year</a>. In effect, the Sidereal Year is the time necessary to complete one, full, 360 degree revolution around the Sun. This is opposed to the <a href="https://en.wikipedia.org/wiki/Tropical_year" rel="nofollow noreferrer">Tropical Year</a> (or else the Solar Year) which is the time necessary to travel just slightly less than a full 360 degrees (due to things like precession and the definition of the Tropical Year).</p>

<p>So your question boils down to, how is a Sidereal year determined? In practice, this is often done by choosing a specific starting point and noting where the Sun is with respect to the stars in the Sky. Then, it will be an entire sidereal year (and a 360 degree revolution) before the Sun returns to the same position as previously observed. Once you've observed the Sun in the <em>exact same position</em> you know a full revolution has occurred of the Earth around the Sun. It should be noted that the motion of the stars over a year is almost completely inconsequential. The stars are so immensely far away compared to their own motion, that their apparent motion (otherwise known as the proper motion) is basically nothing over a year. In this way, we consider the stars "fixed".</p>

CC BY-SA 3.0
2018-04-23T15:43:31.647Z
Thanks very much for this answer. Should I take it that the movement of the stars we compare the sun's position to, is completely irrelevant - or do we do (or are able to do) additional computation that takes this into account?
Bilkokuya
CC BY-SA 3.0
2018-04-23T15:47:44.407Z
@Bilkokuya Pretty much yes. The stars are so immensely far away, that any motion they have in the sky is virtually unnoticeable and irrelevant over the course of the Year. [Barnard's Star](https://en.wikipedia.org/wiki/Barnard%27s_Star) is the star with the fastest proper motion and even that is still pretty minuscule after only a year. For any other star the proper motion is inconsequential.
zephyr
CC BY-SA 3.0
2018-04-23T15:48:26.303Z
[USNO](http://asa.usno.navy.mil/SecM/Glossary.html#_Y) says the Gregorian calendar is based on the tropical year.
Mike G
CC BY-SA 3.0
2018-04-23T15:53:52.897Z
@zephyr Thanks so much for the clarification - I think if you edited in the comment (regarding the proper motion being almost non-existant), it would answer this question perfectly (it was the main stumbling block that was confusing me).
Bilkokuya
CC BY-SA 3.0
2018-04-23T15:41:38.273Z
2018-04-23T15:55:19.483Z
2018-04-23T15:55:19.483Z
zephyr
<p>The start and end of a single orbit are when they cross a specified, albeit arbitrary, point.  For satellites, the start of an orbit or rev is when it crosses a boundary like the equator. </p>

CC BY-SA 3.0
2018-04-25T19:29:12.833Z
2018-04-25T19:29:12.833Z
Kurt Heckman
<p><em>This is being asked from a lay-person point of view. However, please do not hesitate to give more detailed answers that require prior knowledge - if these answer the question more accurately.</em></p>

<p>My confusion currently comes from the fact that everything in the universe is moving in some way or another - so any reference point we use would also be moving/not completely reliable.</p>

<p>As such, how can we accurately determine that we have completed an exact 360 degree rotation around the sun (and as such, measure exactly how long it took)?</p>

<p><em>To be clear, this is asking about how we can determine that we have completed a single 360 orbit of our sun, and not the timing aspect of how we know how long a second is. If this situation is different for measuring the orbits of other bodies around their star - it would be great to expand on those differences as well.</em></p>

CC BY-SA 3.0
2018-04-23T14:16:32.003Z
*Apologies for any misuse of terms, or bad tagging etc. Please do not hesitate to edit directly, or correct me on anything. Many thanks in advance.*
Bilkokuya
<orbit>
How do we measure one complete orbit?
Bilkokuya
# How do we measure one complete orbit?

(Asked by: username_0 on 2018-04-23)

*This is being asked from a lay-person point of view. However, please do not hesitate to give more detailed answers that require prior knowledge - if these answer the question more accurately.*

My confusion currently comes from the fact that everything in the universe is moving in some way or another - so any reference point we use would also be moving/not completely reliable.

As such, how can we accurately determine that we have completed an exact 360 degree rotation around the sun (and as such, measure exactly how long it took)?

*To be clear, this is asking about how we can determine that we have completed a single 360 orbit of our sun, and not the timing aspect of how we know how long a second is. If this situation is different for measuring the orbits of other bodies around their star - it would be great to expand on those differences as well.*

**Tags:** orbit

> **Comments:**
> **username_0**: *Apologies for any misuse of terms, or bad tagging etc. Please do not hesitate to edit directly, or correct me on anything. Many thanks in advance.*  

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## 2 answers

### Answer by username_1 on 2018-04-23. Score: 3 | Accepted answer

When one asks the question, "How is a year defined?", there are actually a few different answers one can provide. What you're implicitly asking, even if you don't know, is how one defines a [Sidereal Year](https://en.wikipedia.org/wiki/Sidereal_year). In effect, the Sidereal Year is the time necessary to complete one, full, 360 degree revolution around the Sun. This is opposed to the [Tropical Year](https://en.wikipedia.org/wiki/Tropical_year) (or else the Solar Year) which is the time necessary to travel just slightly less than a full 360 degrees (due to things like precession and the definition of the Tropical Year).

So your question boils down to, how is a Sidereal year determined? In practice, this is often done by choosing a specific starting point and noting where the Sun is with respect to the stars in the Sky. Then, it will be an entire sidereal year (and a 360 degree revolution) before the Sun returns to the same position as previously observed. Once you've observed the Sun in the *exact same position* you know a full revolution has occurred of the Earth around the Sun. It should be noted that the motion of the stars over a year is almost completely inconsequential. The stars are so immensely far away compared to their own motion, that their apparent motion (otherwise known as the proper motion) is basically nothing over a year. In this way, we consider the stars "fixed".

> **Comments:**
> **username_0**: Thanks very much for this answer. Should I take it that the movement of the stars we compare the sun's position to, is completely irrelevant - or do we do (or are able to do) additional computation that takes this into account?  
> **username_1**: @username_0 Pretty much yes. The stars are so immensely far away, that any motion they have in the sky is virtually unnoticeable and irrelevant over the course of the Year. [Barnard's Star](https://en.wikipedia.org/wiki/Barnard%27s_Star) is the star with the fastest proper motion and even that is still pretty minuscule after only a year. For any other star the proper motion is inconsequential.  
> **username_2**: [USNO](http://asa.usno.navy.mil/SecM/Glossary.html#_Y) says the Gregorian calendar is based on the tropical year.  
> **username_0**: @username_1 Thanks so much for the clarification - I think if you edited in the comment (regarding the proper motion being almost non-existant), it would answer this question perfectly (it was the main stumbling block that was confusing me).  

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### Answer by username_3 on 2018-04-25. Score: 0

The start and end of a single orbit are when they cross a specified, albeit arbitrary, point. For satellites, the start of an orbit or rev is when it crosses a boundary like the equator.


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