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If I am given the latitude and longitude of an observatory or observation point, how can I determine whether a given object, based on its equatorial coordinates, is observable from that observatory?

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I always believe that one picture can tell more than 1000 words. So, here you have it:

Diagram

The possibility of seeing the object is called declination $\delta$. It is the angular distance from the celestial equator. Thus, red star on the diagram has $\delta=75°$ while violet star has $\delta=-30°$. For the north pole: $\delta=90°$. For the celestial equator: $\delta=0°$

We must know only the latitude $\phi$ of the observatory.

If the declination of the star is $\delta\geq 90°-\phi$, then the star is always above the horizon, no matter what time it is. This is true for red star on the diagram; it always revolves throught its two marked positions, which are both above the green line: the horizon.

If the declination of the star is $\delta\leq -90°+\phi$, then the star is always under the horizon. This is true for green star on the diagram: it is always under the horizon.

If the declination is $90°-\phi\leq\delta\geq -90°+\phi$, then the star is sometimes above the horizon and sometimes under. This is true for yellow, blue and violet star on the diagram.

Additionally, if the declination of the star is lower, it is less time above the horizon. Violet star would be just approximately 2 hours above the horizon, while yellow star is around 11 hours above it. You can use formula $\cos{h}=-\tan{\delta}\tan{\phi}$ to calculate the hour angle of the set.

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Welcome to Astronomy SE!

If the latitude φ of the observatory is equal to or less than (90° + δ), then the object is visible at least some of the time.

For example, the Trifid Nebula (M 20) is at δ = −23° 02′, so any observatory at latitudes φ ≤ [90° + (−23° 02′)], viz. φ ≤ 66° 58′ (basically, any place below the Arctic Circle), will see it at least some1 of the time.

Inversely, if the object’s declination δ ≥ (−90° + φ), it will be visible at least some of the time.

For example, Miami (Florida, USA) is at φ = +25.7°, so any object that is at δ ≥ (−90° + 25.7), viz. δ ≥ −64.3°, will be visible some of the time.

In practice, though, it’s not exactly that, because light is refracted up by the atmosphere, in variable amounts depending on atmospheric pressure and temperature, making it possible to see objects further south than the formulas would say. On the other hand, objects low in the sky have their light absorbed in greater amounts by the atmosphere, as said light needs to travel through a thicker layer of air, so objects will appear fainter. Finally, there might be trees, hills/mountains, buildings, etc., in the way.


1Here some of the time will mean both that the object will be geometrically visible for some fraction of the day and that for only some fraction of the year will the sky be dark when the object is up and therefore actually usefully viewable by the observatory's telescope.

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    $\begingroup$ That’s why I wrote some of the time… It may be in December or in June… $\endgroup$ Jun 6 '21 at 16:19
  • $\begingroup$ Ah! I took some of the time to mean only some fraction of the day. I've added an edit, please feel free to edit further or roll back, but I think something like that may be helpful to some readers. If a supernova is reported in M20 in December for example, it's simply not going to be visible from most ground telescopes meeting the latitude test alone until many months later. $\endgroup$
    – uhoh
    Jun 6 '21 at 21:45
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    $\begingroup$ Your edit is good, and you are right that precision was needed. Thank you. $\endgroup$ Jun 7 '21 at 13:56

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