A question about atronomy?

Question

Describe the path that a star on the celestial equator follows from the time it rises until it sets for a person at a latitude of 60 degrees N and a person at the equator

I guess noone knows then?

Answer 1

Lots of people know.

It is just difficult to answer what is most probably a homework question, without writing the actual homework for you.

However, be it known that any celestial object located on the equator will always rise due East and always set due west. This is valid for any latitude (except the poles).

As for how high above the horizon when it passes the observer's meridian, consider that the latitude zero (equator) on Earth is defined as being directly underneath the celestial equator. A person on Earth's equator must look directly at the zenith to see where the celestial equator crosses the meridian.

The horizon is 90 degrees from the zenith (celestial mechanics definition; your personal horizon may vary but for math questions in astronomy, we use 90 degrees all the time for the distance from the zenith to the horizon).

Take your observer from the equator, move him 10 degrees north (to latitude 10 N); the celestial equator will appear to have moved 10 degrees south of the zenith. How high is the top of the equator now, over the horizon?

Keep going until the observer reaches 60N

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If you really want to be technical about it, you can use the celestial mechanics equations:

Sin(H) = Sin(L)*SIn(D) + Cos(L)*Cos(D)*Cos(P)
where H is the height above the horizon (also called altitude), L is latitude, D is declination of the object and P is the hour angle.
For an object on the celestial equator, D = 0.
Sin(0)=0 and Cos(0)=1

For any object, the polar angle at the time of transit (maximum altitude) is 0. Cos(0)=1

So we are left with Sin(H) = Cos(L).

For a star on the celestial equator, the polar angle at "star rise" is 90 degrees (Cos(90)=0) to the East and, at "star set" it is also 90 degrees (but to the west).

For fixed stars, the angle P changes by 15.041 degrees per hour.

The length of our day (and the divisions of our clock) is set by the Sun; the angle P changes by 15 degrees per hour for the Sun.

Once you have the altitude (and the polar angle P):
Sin(Z) = Cos(D)*Sin(P) / Cos(H)
where Z is the azimuth (direction along the horizon that one must face in order to face the celestial object), D is declination (zero) and H is the altitude, as before.

You will see that the observer at the equator sees the star at an altitude equal to the complement of the polar angle
(when P = 90, H = 0; when P = 30, H = 60; etc.).

And the sine of an angle is equal to the cosine of the complement. So the equation will always give Sin(Z) = 1, giving an angle Z = 90 (due East before meridian transit; due West after).

This shortcut does not work for the observer at 60 N.