During each day, every celestial objectappears to move along a circular path on thecelestial sphere due to theEarth's rotation creating two moments when it crosses the meridian.[2][3] Except at thegeographic poles, any celestial object passing through the meridian has anupper culmination, when it reaches its highest point (the moment when it is nearest to thezenith), and nearly twelve hours later, is followed by alower culmination, when it reaches its lowest point (nearest to thenadir). The time ofculmination (when the object culminates) is often used to mean upper culmination.[2][3][4]
An object'saltitude (A) in degrees at its upper culmination is equal to 90 minus the observer'slatitude (L) plus the object'sdeclination (δ):
The object is above thehorizon even at its lower culmination; i.e. if|δ +L | > 90° (i.e. if inabsolute value the declination is more than the colatitude, in the corresponding hemisphere)
The object is below the horizon even at its upper culmination; i.e. if|δ −L | > 90° (i.e. if in absolute value the declination is more than the colatitude, in the opposite hemisphere)
The upper culmination is above and the lower below the horizon, so the body is observed to rise and set daily; in the other cases (i.e. if in absolute value the declination is less than thecolatitude)
The third case applies for objects in a part of the full sky equal to thecosine of the latitude (at the equator it applies for all objects, because the sky turns around the horizontal north–south line; at the poles it applies for none, because the sky turns around the vertical line). The first and second case each apply for half of the remaining sky.[citation needed]
The period between a culmination and the next is asidereal day, which is exactly 24sidereal hours and 4 minutes less than 24 commonsolar hours, while the period between an upper culmination and a lower one is 12 sidereal hours. The period between successive day to day (rotational) culminations is effected mainly byEarth's orbitalproper motion, which produces the different lengths between thesolar day (the interval between culminations of the Sun) and the sidereal day (the interval between culminations of anyreference star) or the slightly more precise,precession unaffected,stellar day.[5] This results in culminations occurring every solar day at different times, taking asidereal year (366.3 days), a year that is one day longer than thesolar year, for a culmination to reoccur. Therefore, only once every 366.3 solar days the culmination reoccurs at the same time of a solar day, while reoccurring every sidereal day.[6] The remaining small changes in the culmination period time from sidereal year to sidereal year is on the other hand mainly caused bynutation (with a 18.6 years cycle), resulting in the longer time scaleaxial precession of Earth (with a 26,000 years cycle),[7][8] whileapsidal precession and other mechanics have a much smaller impact on sidereal observation, impacting Earth's climate through theMilankovitch cycles significantly more. Though at such timescales stars themself change position, particularly those stars which have, as viewed from theSolar System, ahigh proper motion.
Stellar parallax appears to be a similar motion like all these apparent movements, but has only from non-averaged sidereal day to sidereal day a slight effect, returning to its original apparent position, completing a cycle every orbit, with a slight additional lasting change to the position due to the precessions. This phenomenon results from Earth changing position on its orbital path.
The time of day at a location on Earth (except at the poles) can be inferred from the culmination of the Moon in the sky and its phase: each lunar phase culminates closest to thezenith (being exactly south or north of it, crossing themeridian) in the sky at a specific daytime, as marked in the diagram, rising (east) and setting (west) during the time of the day preceding and succeeding the culmination.[9]
Earth'ssubsolar point occurs at the point where the upper culmination of the Sun reaches the point'szenith. At this point, which moves around thetropics throughout the year, the Sun is perceived to be directly overhead.
We apply the previous equation,A = 90° −L +δ, in the following examples.
Supposing that thedeclination of the Sun is +20° when it crosses the local meridian, then thecomplementary angle of 70° (from the Sun to the pole) is added to and subtracted from the observer'slatitude to find the solar altitudes at upper and lower culminations, respectively.
From52° north, the upper culmination is at 58° above the horizon due south, while the lower is at 18° below the horizon due north. This is calculated as 52° + 70° = 122° (thesupplementary angle being 58°) for the upper, and 52° − 70° = −18° for the lower.
From80° north, the upper culmination is at 30° above the horizon due south, while the lower is at 10° above the horizon (midnight sun) due north.