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Positional Astronomy: Zenith, celestial poles and sidereal time

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kr75
#1
Jul11-13, 12:43 AM
P: 16
I have a little trouble with positional astronomy so I wanted to clarify a few things to get my understanding right. The imaginary celestial poles are located directly above (or below) the poles of the Earth, right? And the Zenith is the position in the sky that a person would look if one was looking straight up from local position. So, if a person were at either pole of the Earth and looking straight up, that would be a celestial pole?

I have not clearly understood the concepts of azimuth, declination (what is declination?) and the zenith and how to use them to find the position of a star in the night sky at a given time of the year from a particular location on Earth. I know that sidereal time and local time are also used in this case. Could someone explain this?
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collinsmark
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Jul24-13, 08:35 PM
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Quote Quote by kr75 View Post
I have a little trouble with positional astronomy so I wanted to clarify a few things to get my understanding right. The imaginary celestial poles are located directly above (or below) the poles of the Earth, right?
Yes. That is to say that the true poles of the Earth (not the magnetic poles, but the true poles: true north and true south), each point toward their respective celestial pole by extension.

And the Zenith is the position in the sky that a person would look if one was looking straight up from local position.
Correct again.

Let me get back to the celestial poles for a moment with an example. Suppose that you were at some location in the northern hemisphere with a latitude of 40o during a clear night. If you look directly up you will see the zenith (which is dependent upon your location). Now face north, and decrease your neck/back angle by 40o (so that you are looking 90o - 40o ≈ 50o above the horizon [depending on your horizon], facing North)*. You are now looking toward celestial North.

*(This is the same thing as 50o altitude by the way [as measured from your particular celestial horizon]. More on that later.)

There's another term I would like to introduce: meridian. That's an imaginary line -- actually a circle, that intersects the celestial poles and the zenith from your particular location. (In geography, "meridians" are lines of equal longitude, and there are many of them. In astronomy the definition is a little different in that there is only one meridian, and it passes through the zenith.)

So, if a person were at either pole of the Earth and looking straight up, that would be a celestial pole?
Yes, that's right.

I have not clearly understood the concepts of azimuth, declination (what is declination?) and the zenith and how to use them to find the position of a star in the night sky at a given time of the year from a particular location on Earth.
There are two ways to specify celestial coordinates: Each contain two parameters that go together.
Altitude and azimuth.
Declination and right ascension.

Altitude and azimuth are always measured in respect to your particular location. If you look up 50o above the horizon*, that's the altitude: 50o. Altitude is a measure of how much "up" something is. Now onto azimuth. Azimuth is a measure of how far East of North something is. If you look North, you azimuth is 0o. If you look due East your azimuth is 90o. South is 180o and West is 270o azimuth.

*(Technically, your celestial horizon. The celestial horizon is any direction 90o from the zenith.) It's essentially the same thing as the normal horizon, if you are right on the ground and the earth is an ideal sphere without anything obstructing the view such as hills or trees.)

[Edit: In case you haven't figured this out on your own, "azimuth" is pretty meaningless if you are physically located at either of the Earth's poles.]

Declination and right ascension are measurements of the night sky. Declination and right ascension stay with the stars, not your location. Stars can be mapped and specified in the night sky in terms of their declination and right ascension. Celestial north has a declination of +90o and celestial south -90o. Right ascension is measured from the vernal equinox or the first point of Aries, which is the place on the celestial sphere where the Sun crosses the celestial equator from south to north at the March equinox and is located in the constellation Pisces (that's just an arbitrary convention, by the way). The units of right ascension are traditionally hours, where the full circle of right ascension is 24 hr. Each hour is then divided into 60 arc minutes, and each arc minute into 60 arc seconds.

I know that sidereal time and local time are also used in this case. Could someone explain this?
  • 24 hours of sidereal time is the time it takes for the Earth to rotate with respect to the fixed, background stars.
  • 24 hours of solar time is the time it takes for the Earth to rotate with respect to the sun.

A mean sidereal day is about 23 hours, 56 minutes, 4.0916 seconds of solar time. That's roughly 4 minutes less than a solar day. Over the course of a year, that difference adds up to 1 day. (One day per year difference means one extra rotation per year [i.e. one rotation per revolution] difference.)

Why one extra rotation? let me describe a demonstration for you. find a pair (2) of coins of the same denomination (US quarters work pretty good for this). Place one coin on the table, heads up, and firmly place your thumb on it so that it doesn't move from here on out. Put the second coin right next to it, so that the tip of the the second coin's head touches the tip of the first coin's head. Now rotate/revolve the second coin and let it move around the first coin such that the edges of the coins do not slide relative to one another. Things to notice:
  • Each location on the second coin's circumference touched the first coin once and only once. So, relative the first coin, the second coin had one rotation.
  • Relative to where you are sitting, the second coin was rotated around twice! If the second coin starts out upside down when atop the first coin, it is upside again at the bottom of the first coin, then flips once more before reaching the top. It rotated around 1 additional rotation with respect to the background objects (like yourself) than it did the object it was revolving around.
kr75
#3
Jul25-13, 09:59 AM
P: 16
Thanks a lot for the detailed explanation and taking time out for it. I have understood most of the concepts since I read and re-read about those to get it right. I still have some trouble when solving problems that relate to finding the position of stars in the sky at a given time, but I guess I need to keep solving more to get a better hang of those


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