<p>It can be done with coordinates from any two different times. A difference more than one day should give acceptable results. The Moon's ecliptic latitude is a sinusoid with a period that's well known (27.40803 days). Combining that given with the ecliptic longitudes and latitudes of the two observations the time difference between the time of the first observation and the time of passing the ascending node can be determined.  That time and the moon's known orbital angular velocity give the angular difference between the ascending node and the ecliptice longitude of the first observation. Then rearranging the first equation under &quot;Finding t1&quot; finds the Moon's orbital obliquity.</p>
<p><a href="https://i.sstatic.net/c3PEt.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/c3PEt.png" alt="enter image description here" /></a></p>
<p>All of this is an approximation that assumes a circular orbit (Moon's is pretty close to a circle). To test the process against actuals JPL Horizons was used in two different ways. It was used to provide the data that would have been gained in observation pairs. Then it was used to check the results of calculations against actuals. The time range used was May and June 2023. Virtual observations were made every day in the range, and results for longitude of the ascending node and orbital obliquity were calculated for each day and the day following it. The following chart shows the input data from Horizons. The blue trace and left axis plot the ecliptic longitude for 62 days beginning 3 May 2023. The red trace and right axis show the ecliptic latitude for the same days.</p>
<p><a href="https://i.sstatic.net/ufSvI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ufSvI.png" alt="enter image description here" /></a></p>
<p>This next chart shows the results of the calculations for the same time period:blue trace and left axis for the longitude of the ascending node and red trace, right axis for orbital obliquity.</p>
<p><a href="https://i.sstatic.net/935Hu.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/935Hu.png" alt="enter image description here" /></a></p>
<p>The chart doesn't show anything that's actually happening on any pair of days, just the results of calculations made from observations at the time. The longitude result trace shows the ongoing estimate of the ecliptic longitude at the last passing of the ascending node. The latitude trace shows the ongoing estimate its peak value from the preceding to the following upward crossings of the ecliptic.
The actual two upward crossings:
May 17, day 15, 34.05 degrees;
June 14, day 43, 33.06 degrees. The actual negative peak of the ecliptic latitude -5.2344 gegrees, positive peak 5.2038.</p>

CC BY-SA 4.0
2023-04-23T09:25:56.953Z
Your equations would be more useful in text form, rather than as a PNG image. Please consider posting them as either MathJax or (pseudo)code.
PM 2Ring
CC BY-SA 4.0
2023-04-23T09:38:50.510Z
The eccentricity of the Moon's orbit is ~0.0549, so (using the Vis-Viva equation) the ratio of its perigee speed / apogee speed is ~1.116.
PM 2Ring
CC BY-SA 4.0
2023-04-22T21:39:45.257Z
2023-06-07T17:36:41.670Z
2023-06-07T17:36:41.670Z
stretch
<p>Assume we have the ecliptic coordinate of moon; how can we calculate the ecliptic coordinate of its ascending node?
I think one approximate way is to use the moon's coordinate for two subsequent days and then we can calculate the cross product of both vectors to get the plane' normal which can be used to calculate the intersection coordinate. Is there a better accurate way?</p>

CC BY-SA 4.0
2023-04-22T04:06:00.867Z
Depends on what you want the ascending node for.  If your goal is a set of Keplerian elements for computing future positions, the results will be fairly inaccurate anyway.  The only accurate method is numerical integration, but there are a few analytic ephemeris algorithms based on those (e.g. ELP82/2000, VSOP87, JPL DE).
Greg Miller
<spherical-astronomy>
how to calculate ecliptic coordinate of moon's ascending node?
Amirhessam
# how to calculate ecliptic coordinate of moon's ascending node?

(Asked by: username_0 on 2023-04-22)

Assume we have the ecliptic coordinate of moon; how can we calculate the ecliptic coordinate of its ascending node?
I think one approximate way is to use the moon's coordinate for two subsequent days and then we can calculate the cross product of both vectors to get the plane' normal which can be used to calculate the intersection coordinate. Is there a better accurate way?

**Tags:** spherical-astronomy
> **Comments:**
> **username_1**: Depends on what you want the ascending node for.  If your goal is a set of Keplerian elements for computing future positions, the results will be fairly inaccurate anyway.  The only accurate method is numerical integration, but there are a few analytic ephemeris algorithms based on those (e.g. ELP82/2000, VSOP87, JPL DE).  

---

## 1 answers

### Answer by username_2 on 2023-04-22. Score: 0 | Accepted answer

It can be done with coordinates from any two different times. A difference more than one day should give acceptable results. The Moon's ecliptic latitude is a sinusoid with a period that's well known (27.40803 days). Combining that given with the ecliptic longitudes and latitudes of the two observations the time difference between the time of the first observation and the time of passing the ascending node can be determined. That time and the moon's known orbital angular velocity give the angular difference between the ascending node and the ecliptice longitude of the first observation. Then rearranging the first equation under "Finding t1" finds the Moon's orbital obliquity.

[![enter image description here](https://i.sstatic.net/c3PEt.png)](https://i.sstatic.net/c3PEt.png)

All of this is an approximation that assumes a circular orbit (Moon's is pretty close to a circle). To test the process against actuals JPL Horizons was used in two different ways. It was used to provide the data that would have been gained in observation pairs. Then it was used to check the results of calculations against actuals. The time range used was May and June 2023. Virtual observations were made every day in the range, and results for longitude of the ascending node and orbital obliquity were calculated for each day and the day following it. The following chart shows the input data from Horizons. The blue trace and left axis plot the ecliptic longitude for 62 days beginning 3 May 2023. The red trace and right axis show the ecliptic latitude for the same days.

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

This next chart shows the results of the calculations for the same time period:blue trace and left axis for the longitude of the ascending node and red trace, right axis for orbital obliquity.

[![enter image description here](https://i.sstatic.net/935Hu.png)](https://i.sstatic.net/935Hu.png)

The chart doesn't show anything that's actually happening on any pair of days, just the results of calculations made from observations at the time. The longitude result trace shows the ongoing estimate of the ecliptic longitude at the last passing of the ascending node. The latitude trace shows the ongoing estimate its peak value from the preceding to the following upward crossings of the ecliptic.
The actual two upward crossings:
May 17, day 15, 34.05 degrees;
June 14, day 43, 33.06 degrees. The actual negative peak of the ecliptic latitude -5.2344 gegrees, positive peak 5.2038.

> **Comments:**
> **username_3**: Your equations would be more useful in text form, rather than as a PNG image. Please consider posting them as either MathJax or (pseudo)code.  
> **username_3**: The eccentricity of the Moon's orbit is ~0.0549, so (using the Vis-Viva equation) the ratio of its perigee speed / apogee speed is ~1.116.  

---

