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I think you are overlooking these factors: Iowa State University is at 42 degrees North latitude, so the Sun and celestial equator that they are showing in their figures are always to the south at t …

Positions are usually given relative to the observer, not relative to the object. So instead of saying "I am under the star X", you would say "star X is overhead". The word for an object directly over …

The heliocentric coordinates then get converted to other coordinate systems. … Geo longitude (Lng) is the ecliptic longitude as seen from the Earth (Geo), but otherwise, the coordinate system is the same. …

The coordinate system for an object in the sky measured relative to the horizon is known as altitude-azimuth. … See Horizontal Coordinate System on Wikipedia …

The flaw in your logic is that "RA and L are constants". The right ascension (RA) of the object on the meridian is not constant. It changes continuously, otherwise we would not say that the sky is ris …

Yes, the orientation of the Sun will be different from the Earth's northern and southern hemisphere, just like your example of the Moon. I would not say that a sunspot in the "northern" hemisphere wo …

It may be easier to visualize the formula by calculating the altitude based on the observer's latitude (lat) and the object's declination ($\delta$), and then re-arranging the equation to find the obs …

The problem is not that right ascension goes from 0 to 24 hours (or 0 to any other number). The problem is that the Earth is revolving around the Sun, and our time system is based on the Sun. Because …

You can think of the position given in your title as the coordinates of the star that is directly over the Sun's north pole. (Similar to how Polaris is almost directly over the Earth's north pole.) Li …

You are correct that topocentric coordinates are for the position of "close" objects, corrected for observing from the Earth's surface instead of the theoretical center. The topocentric and geocentri …

The position angle P of a body ($\alpha_1, \delta_1$) with respect to another body ($\alpha_2, \delta_2$) can be calculated from $$tan(P)={sin(\Delta\alpha)\over cos(\delta_2)tan(\delta_1)-sin(\delta_ …

Given a date and time, the position of the Moon can be calculated to provide the declination and right ascension. The sub-point of the Moon (the point on the Earth at which the Moon is at the zenith) …