<p>Orbital inclination can be estimated fairly well from two observations made when the planet is in opposition to the Sun. If the observation means/method gives equatorial coordinates they should be converted to geocentric ecliptic. To convert:</p>
<p><a href="https://i.sstatic.net/yuju4.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/yuju4.png" alt="enter image description here" /></a></p>
<p>A planet is in opposition when its geocentric ecliptic longitude and the Sun's are  180 degrees apart.  The Sun's geocentric ecliptic coordinates are available online. JPL Horizons is a convenient and accurate source. The reason for using ecliptic coordinates at opposition is that at opposition geocentric and heliocentric ecliptic longitudes are the same. The geocentric latitude is somewhat greater than the heliocentric, but can be adjusted.</p>
<p>After the second observation, it's possible to calculate the planet's orbital period assuming a circular orbit. Planet orbits are all elliptical but their eccentricities typically are small enough to get reasonable orbit parameters with the circular orbit assumption.</p>
<p>Definitions:</p>
<p><a href="https://i.sstatic.net/xA9nO.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xA9nO.png" alt="enter image description here" /></a></p>
<p>T orbital period - angles in degrees, time in years</p>
<p><a href="https://i.sstatic.net/TiMT1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/TiMT1.png" alt="enter image description here" /></a></p>
<p>Radius of its assumed circular orbit, from Kepler's third law:</p>
<p><a href="https://i.sstatic.net/btTFf.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/btTFf.png" alt="enter image description here" /></a></p>
<p><strong>Curve fitting - Longitude of the Ascending Node</strong></p>
<p>Over the course of its period a planet's angular distance from the ecliptic is a single period of a quasi-sinusoid: &quot;quasi&quot; because of its small departures from perfect sinusoid, which come from its orbit being elliptical. Knowing two points on a sinusoid and its period all of its other points can be calculated.</p>
<p>It can be shown that:</p>
<p><a href="https://i.sstatic.net/V3jVF.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/V3jVF.png" alt="enter image description here" /></a></p>
<p>The equation finds the elapsed time from the last passing of the ascending node to the first observation. It gives the wrong answer half the time, because the arccotangent function, over the expected range of values, starts over again at the descending node. Whenever the ecliptic latitude of the first observation is negative its necessary to add half the orbital period to the return.</p>
<p>Knowing the time of the ascending node passage, its longitude can
be found.</p>
<p><a href="https://i.sstatic.net/dIVUY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dIVUY.png" alt="enter image description here" /></a></p>
<p>The orbital inclinatiion as seen from Earth is equal the peak of the geocentric ecliptic latitude sinusoid. It can be found using time and latitude values from either observation.</p>
<p><a href="https://i.sstatic.net/jFY30.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/jFY30.png" alt="enter image description here" /></a></p>
<p>That value will be slightly greater than the heliocentric peak ecliptic latitude for a typical planet. Solving the triangle with vertices at Earth, Sun and planet finds the angle at the Sun which is the corrected value.</p>
<p>Bhpk = Bgpk - asin(sin(Bgpk)/R)</p>
<p><strong>Saturn test case</strong></p>
<p>Opposition times, values 2022, 2023 - from JPL Horizons - and first calculations:</p>
<p><a href="https://i.sstatic.net/3nZtj.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3nZtj.png" alt="enter image description here" /></a></p>
<p>Using the equation for the time elapsed since passing the ascending node gives 2.426 years. Considering that the ecliptic latitude seen in the first observation is negative, add half the orbital period to get the corrected value: 18.02 years.</p>
<p>That leads to finding the ecliptic longitude of the ascending node: 113.894 degrees.</p>
<p>Then, find the maximum value of the ecliptic latitude, as seen from Earth: 2.774 degrees.</p>
<p>Then adjust for Earth being closer than the Sun: 2.494 degrees.</p>
<p><strong>Comparison, calculated values to online, trusted values:</strong></p>
<p><a href="https://i.sstatic.net/9YIqX.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9YIqX.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-01-09T03:58:37.070Z
2023-06-01T22:50:45.083Z
2023-06-01T22:50:45.083Z
stretch
<p>I wanted to ask if it's possible to establish a correlation between the observed equatorial coordinates of a Solar System planet, especially <strong>declination</strong>, and the planet's <strong>orbital inclination</strong> with respect to the ecliptic.</p>
<p>In case there is such correlation, would it be possible, for example, to derive the orbital inclination of an outer planet by comparing its declination angles observed at different times?</p>

CC BY-SA 4.0
2022-12-16T08:21:30.563Z
[This question](https://astronomy.stackexchange.com/questions/18316/calculating-orbits-using-observational-data) is somewhat similar, but I was wondering more specifically about declination and orbital inclination.
user3764418
CC BY-SA 4.0
2022-12-19T04:50:14.313Z
Orbital inclination can be determined from a number of right ascension, declination observations. It's been done for all the planets in the solar system. The best way to do it depends on how far off the ecliptic the hypothetical planet's orbital plane might be and the accuracy needed. There's no simple, short answer.
stretch
CC BY-SA 4.0
2022-12-19T14:56:02.383Z
@stretch Just for the sake of argument: could two distinct observations of the planet over a years-long timespan be sufficient in principle to determine the orbital plane (if only low accuracy is needed)?
user3764418
<orbit><coordinate><inclination>
Deriving orbital inclination from equatorial coordinates?
user3764418
# Deriving orbital inclination from equatorial coordinates?

(Asked by: username_0 on 2022-12-16)

I wanted to ask if it's possible to establish a correlation between the observed equatorial coordinates of a Solar System planet, especially **declination**, and the planet's **orbital inclination** with respect to the ecliptic.

In case there is such correlation, would it be possible, for example, to derive the orbital inclination of an outer planet by comparing its declination angles observed at different times?

**Tags:** orbit, coordinate, inclination
> **Comments:**
> **username_0**: [This question](https://astronomy.stackexchange.com/questions/18316/calculating-orbits-using-observational-data) is somewhat similar, but I was wondering more specifically about declination and orbital inclination.  
> **username_1**: Orbital inclination can be determined from a number of right ascension, declination observations. It's been done for all the planets in the solar system. The best way to do it depends on how far off the ecliptic the hypothetical planet's orbital plane might be and the accuracy needed. There's no simple, short answer.  
> **username_0**: @stretch Just for the sake of argument: could two distinct observations of the planet over a years-long timespan be sufficient in principle to determine the orbital plane (if only low accuracy is needed)?  

---

## 1 answers

### Answer by username_1 on 2023-01-09. Score: 1

Orbital inclination can be estimated fairly well from two observations made when the planet is in opposition to the Sun. If the observation means/method gives equatorial coordinates they should be converted to geocentric ecliptic. To convert:

[![enter image description here](https://i.sstatic.net/yuju4.png)](https://i.sstatic.net/yuju4.png)

A planet is in opposition when its geocentric ecliptic longitude and the Sun's are 180 degrees apart. The Sun's geocentric ecliptic coordinates are available online. JPL Horizons is a convenient and accurate source. The reason for using ecliptic coordinates at opposition is that at opposition geocentric and heliocentric ecliptic longitudes are the same. The geocentric latitude is somewhat greater than the heliocentric, but can be adjusted.

After the second observation, it's possible to calculate the planet's orbital period assuming a circular orbit. Planet orbits are all elliptical but their eccentricities typically are small enough to get reasonable orbit parameters with the circular orbit assumption.

Definitions:

[![enter image description here](https://i.sstatic.net/xA9nO.png)](https://i.sstatic.net/xA9nO.png)

T orbital period - angles in degrees, time in years

[![enter image description here](https://i.sstatic.net/TiMT1.png)](https://i.sstatic.net/TiMT1.png)

Radius of its assumed circular orbit, from Kepler's third law:

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

**Curve fitting - Longitude of the Ascending Node**

Over the course of its period a planet's angular distance from the ecliptic is a single period of a quasi-sinusoid: "quasi" because of its small departures from perfect sinusoid, which come from its orbit being elliptical. Knowing two points on a sinusoid and its period all of its other points can be calculated.

It can be shown that:

[![enter image description here](https://i.sstatic.net/V3jVF.png)](https://i.sstatic.net/V3jVF.png)

The equation finds the elapsed time from the last passing of the ascending node to the first observation. It gives the wrong answer half the time, because the arccotangent function, over the expected range of values, starts over again at the descending node. Whenever the ecliptic latitude of the first observation is negative its necessary to add half the orbital period to the return.

Knowing the time of the ascending node passage, its longitude can
be found.

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

The orbital inclinatiion as seen from Earth is equal the peak of the geocentric ecliptic latitude sinusoid. It can be found using time and latitude values from either observation.

[![enter image description here](https://i.sstatic.net/jFY30.png)](https://i.sstatic.net/jFY30.png)

That value will be slightly greater than the heliocentric peak ecliptic latitude for a typical planet. Solving the triangle with vertices at Earth, Sun and planet finds the angle at the Sun which is the corrected value.

Bhpk = Bgpk - asin(sin(Bgpk)/R)

**Saturn test case**

Opposition times, values 2022, 2023 - from JPL Horizons - and first calculations:

[![enter image description here](https://i.sstatic.net/3nZtj.png)](https://i.sstatic.net/3nZtj.png)

Using the equation for the time elapsed since passing the ascending node gives 2.426 years. Considering that the ecliptic latitude seen in the first observation is negative, add half the orbital period to get the corrected value: 18.02 years.

That leads to finding the ecliptic longitude of the ascending node: 113.894 degrees.

Then, find the maximum value of the ecliptic latitude, as seen from Earth: 2.774 degrees.

Then adjust for Earth being closer than the Sun: 2.494 degrees.

**Comparison, calculated values to online, trusted values:**

[![enter image description here](https://i.sstatic.net/9YIqX.png)](https://i.sstatic.net/9YIqX.png)


---

