<p>A quick skim of a few papers on Earth's tropical year, e.g. <a href="http://www.public.asu.edu/~mjwhite/Sidereal%20and%20tropical%20years.pdf" rel="nofollow noreferrer">M.J.White</a>, suggests to me that existing formulas are best-fit to data, not analytic.  How Saturn's ellipticity of orbit &amp; axial precession(s) play into this is beyond my learning, but I'm getting the feeling that "it just is," and that any number of cofactors such as planetary layers, densities, and so on might affect what's being observed.</p>

CC BY-SA 4.0
2018-06-11T23:06:13.090Z
Understood, but "it just is", is what I'm trying to avoid. Having said that, my assumption that the axis is smoothly precessing may be the flaw, and the current observed (instantaneous) precession might be at play.
PaulS
CC BY-SA 4.0
2018-06-11T14:38:22.643Z
2018-06-11T14:38:22.643Z
Carl Witthoft
<p>What is the value of Saturn's "tropical year" (as opposed to its "sidereal year").</p>

<p>I am seeing it listed as 10 days shorter (10,747 days) than the sidereal year (10,759.22 days).</p>

<p>How could this be if the precession of the <strong>axis</strong> (equinox) is so slow <a href="https://www.astro.umd.edu/~dphamil/research/preprints/WardHam04.pdf" rel="nofollow noreferrer">(~1.86M years)</a>? Is this an "transient/osculating" value? Frame of reference (e.g. J2000)? Am I completely missing something?</p>

<p>I <em>think</em> I can calculate the tropical year by (close to, this works for Earth):</p>

<p>$P_{tropical} = P_{sidereal} \cdot (1-\frac{P_{sidereal} }{P_{precession}})$</p>

<ul>
<li><p>Using <a href="https://nssdc.gsfc.nasa.gov/planetary/factsheet/saturnfact.html" rel="nofollow noreferrer">NASA planetary facts data on Saturn</a>
(these figures are re-used on many websites):</p>

<p>$10759.22\cdot(1-\frac{10759.22}{1.86\times10^6 \space\times\space 365.256})=10759.05$</p></li>
<li><p>Using <a href="https://ssd.jpl.nasa.gov/horizons.cgi" rel="nofollow noreferrer">JPL Horizons</a> data:</p>

<p>$10755.698\cdot(1-\frac{10755.698}{1.86\times10^6 \space\times\space 365.256})=10755.528$</p></li>
</ul>

<p>about 0.170 days less.</p>

<p>Also, interestingly, see the <em>instantaneous</em> <strong>sidereal</strong> year of Saturn from the graph of JPL Horizon's data, using the sun's centre (blue) and solar system barycentre (red). I think the <em>instantaneous</em> <strong>tropical</strong> year (at epoch) would always be 0.170 days less.</p>

<p>I think this is where the confusion is: texts have used particular values, and further the <em>same</em> text has picked a sidereal from one epoch (or average), and a tropical from another...</p>

<p><a href="https://i.sstatic.net/Y8GdD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Y8GdD.png" alt="enter image description here"></a></p>

<p>Saturn's instantaneous length of sidereal year in days, at epoch.</p>

CC BY-SA 4.0
2018-06-11T04:06:08.850Z
This was originally posted in Space SE but the OP has decided to ask here instead. It's not a deliberate cross post. Now, I hope someone here can actually answer this, otherwise we'll feel silly recommending the OP move it here! ;-)
uhoh
CC BY-SA 4.0
2018-06-11T18:38:18.527Z
The dates in this [Planetary Society blog post](http://www.planetary.org/blogs/emily-lakdawalla/2016/06031044-oppositions-conjunctions-rpx.html) suggest a tropical year length of 10754.4 days, based on whatever Saturn model is built into JPL HORIZONS.
Mike G
CC BY-SA 4.0
2018-06-14T11:10:06.497Z
How are you determining the "instantaneous" sidereal year length from HORIZONS?
Mike G
CC BY-SA 4.0
2018-06-14T12:38:38.973Z
@MikeG "osculating orbital elements".  https://ssd.jpl.nasa.gov/horizons.cgi?s_type=1 JPL Horizons: Ephemeris Type =  Orbital ELEMENTS, Body=Saturn, Centre=Sun ... Look for "PR = " numeric
PaulS
<planet><orbit><orbital-elements>
Saturn's tropical year?
PaulS
# Saturn's tropical year?

(Asked by: username_0 on 2018-06-11)

What is the value of Saturn's "tropical year" (as opposed to its "sidereal year").

I am seeing it listed as 10 days shorter (10,747 days) than the sidereal year (10,759.22 days).

How could this be if the precession of the **axis** (equinox) is so slow [(~1.86M years)](https://www.astro.umd.edu/~dphamil/research/preprints/WardHam04.pdf)? Is this an "transient/osculating" value? Frame of reference (e.g. J2000)? Am I completely missing something?

I *think* I can calculate the tropical year by (close to, this works for Earth):

$P\_{tropical} = P\_{sidereal} \cdot (1-\frac{P\_{sidereal} }{P\_{precession}})$

* Using [NASA planetary facts data on Saturn](https://nssdc.gsfc.nasa.gov/planetary/factsheet/saturnfact.html)
  (these figures are re-used on many websites):

  $10759.22\cdot(1-\frac{10759.22}{1.86\times10^6 \space\times\space 365.256})=10759.05$
* Using [JPL Horizons](https://ssd.jpl.nasa.gov/horizons.cgi) data:

  $10755.698\cdot(1-\frac{10755.698}{1.86\times10^6 \space\times\space 365.256})=10755.528$

about 0.170 days less.

Also, interestingly, see the *instantaneous* **sidereal** year of Saturn from the graph of JPL Horizon's data, using the sun's centre (blue) and solar system barycentre (red). I think the *instantaneous* **tropical** year (at epoch) would always be 0.170 days less.

I think this is where the confusion is: texts have used particular values, and further the *same* text has picked a sidereal from one epoch (or average), and a tropical from another...

[![enter image description here](https://i.sstatic.net/Y8GdD.png)](https://i.sstatic.net/Y8GdD.png)

Saturn's instantaneous length of sidereal year in days, at epoch.

**Tags:** planet, orbit, orbital-elements
> **Comments:**
> **username_1**: This was originally posted in Space SE but the OP has decided to ask here instead. It's not a deliberate cross post. Now, I hope someone here can actually answer this, otherwise we'll feel silly recommending the OP move it here! ;-)  
> **username_2**: The dates in this [Planetary Society blog post](http://www.planetary.org/blogs/emily-lakdawalla/2016/06031044-oppositions-conjunctions-rpx.html) suggest a tropical year length of 10754.4 days, based on whatever Saturn model is built into JPL HORIZONS.  
> **username_2**: How are you determining the "instantaneous" sidereal year length from HORIZONS?  
> **username_0**: @MikeG "osculating orbital elements".  https://ssd.jpl.nasa.gov/horizons.cgi?s_type=1 JPL Horizons: Ephemeris Type =  Orbital ELEMENTS, Body=Saturn, Centre=Sun ... Look for "PR = " numeric  

---

## 1 answers

### Answer by username_3 on 2018-06-11. Score: 1

A quick skim of a few papers on Earth's tropical year, e.g. [M.J.White](http://www.public.asu.edu/~mjwhite/Sidereal%20and%20tropical%20years.pdf), suggests to me that existing formulas are best-fit to data, not analytic. How Saturn's ellipticity of orbit & axial precession(s) play into this is beyond my learning, but I'm getting the feeling that "it just is," and that any number of cofactors such as planetary layers, densities, and so on might affect what's being observed.

> **Comments:**
> **username_0**: Understood, but "it just is", is what I'm trying to avoid. Having said that, my assumption that the axis is smoothly precessing may be the flaw, and the current observed (instantaneous) precession might be at play.  

---

