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Suppose I would like to calculate the inclination of a satellite from the ecliptic. Would it be possible to do this with an amateur telescope? How would I go about doing so?


Note: A good answer should tell what kind of telescope an amateur would need, what measurements they would need to make, then what calculations they would have to perform to get the inclination (or the "instantaneous angular measurement from the ecliptic at the time of measurement").

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  • $\begingroup$ Finally found the oldest question on this site =D $\endgroup$
    – Alastor
    Nov 14, 2023 at 14:50

2 Answers 2

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Any telescope can be made to give you the information that you are looking for. The first thing that you will need to know is the location of the ecliptic which varies throughout the year. Or are you looking to find in relation to the the celestial equator?

https://en.wikipedia.org/wiki/Celestial_equator

Either way, you would start of the same by find the Declination (D) and Right Ascension (R.A) of the object in question. You do not need a telescope to find this, unless you are unable to see the object with the naked eye. If the telescope has a polar mount and is properly set up. You can read the R.A. and D off of the mount.

Otherwise it will require some trigonometry and knowing where you are (at least your latitude)

You will need to determine the Altitude and Azimuth of the object. Which is really just the direction from North and the angle from horizontal that your telescope is pointing. You could do this with a compass and protractor or even an astrolabe. Knowing this you can then convert to R.A and D with the following formulas:

$$RA = \arctan(\frac{- sin(Az) \times \cos(Alt)}{\cos(Lat) \times \sin(Alt) - \sin(Lat) \times \cos(Az) \times \cos(Alt)})$$

$$Dec = \arcsin(\sin(Lat) \times \sin(Alt) + \cos(Lat) \times \cos(Alt) \times \cos(Az))$$

You are looking for the inclination to the ecliptic, so you are mostly concerned with the declination. The ecliptic changes in declination throughout the year from 0 at the equinox to +/- 23.5 at the solstice. So your inclination from the ecliptic would be the Declination of your object +/- the Declination of the ecliptic.

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You just need a telescope with a "wedge" mount, (ie one for polar coordinates,) and not one which is simply a pan-and-tilt mount like commonly used for cameras. (BUT, it might be more fun to do it with a sextant -- see below.)

With a telescope that has a polar mount, you just need to set it up correctly. That means orienting the base of the mount in the correct north/south direction, and then setting the "wedge" angle to account for your geographic latitude. (Any amateur level scope with a polar mount will have instructions.) When you pan the scope it will sweep celestial lines of equal latitude, and when you tilt the scope it will sweep celestial lines of equal longitude. Point the scope at your target, and read the declination from the scope's mount.

The hardest part is going to be that the satellite will be moving pretty fast, and its orbit is going to have a continuously varying declination. (Unless you target a satellite in a circular, equatorial orbit. :)

Fun with a sextant

It might be easier to "shoot" the satellite with a simple (search for "Davis Mark 15" on ebay) sextant and something called an "artificial horizon." You need any liquid reflecting surface... a swimming pool or kiddie pool might work for shooting a moving satellite. (You can buy a small "artificial horizon" that's a few inches square, but you'll never catch the satellite's reflection in that.) Using the sextant you measure the angle between the satellite and its reflection, also noting the compass direction. Then you crunch a bunch of trigonometry. But this is exactly how you practice astronomical navigation star sighting on land.

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    $\begingroup$ This gives the inclination compared to the celestial equator and does not give the inclination per the ecliptic. $\endgroup$
    – Schleis
    Nov 5, 2013 at 23:09

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