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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?

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    $\begingroup$ This question is somewhat similar, but I was wondering more specifically about declination and orbital inclination. $\endgroup$ Dec 16, 2022 at 8:21
  • $\begingroup$ 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. $\endgroup$
    – stretch
    Dec 19, 2022 at 4:50
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    $\begingroup$ @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)? $\endgroup$ Dec 19, 2022 at 14:56

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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:

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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:

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T orbital period - angles in degrees, time in years

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

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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:

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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.

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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.

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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:

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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

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