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Basic Astronomy problem

  1. Jan 20, 2008 #1
    1. The problem statement, all variables and given/known data
    It's a very simple (I guess) astronomy problem but I have no idea how to start and what to do, so please show me something.

    One star Σ has (right ascension) α=1h25m18s. What is the (sidereal time) ST at Thessaloniki (φ=40.5°, λ=-23°) when the star is at its upper culmination?

    Thank you,
    Panos
     
  2. jcsd
  3. Jan 20, 2008 #2
    How much of this do you understand so far?

    Do you know what is meant by right ascension? Sidereal time? Latitude and longitude? Upper culmination? If you don't understand any one of these terms, you'll have trouble solving the problem; if you do understand them, then you should show us how far you've gotten with them so far.
     
  4. Jan 21, 2008 #3
    I have a little knowledge about all that, but I don't think I have fully understood them.
    For example, when we say that a star has α=1h25m18s we mean that this star is at it's upper culmination on the meridian that is about 20° east of Greenwich or something?

    Maybe what I'm saying is completely wrong, but I haven't had anyone to explain all that new stuff to me and I'm missing a basic point here.
     
  5. Jan 21, 2008 #4
    No, that doesn't sound right to me. The rt. asc. of the star tells you where it is on the celestial sphere. It will be at its upper culmination for you when your local meridian passes that angle of rt. asc. on the celestial sphere.

    I think the thing you need to use here is the Greenwich Hour Angle (GHA), which is defined such that GHA = (local sideral time) - (star's rt. asc.), i.e. it tells you how much siderial time has passed since the star was on the local meridian (upper culmination).
     
  6. Jan 21, 2008 #5
    OH! I think I just figured out what you are saying, and yes, now I think you are correct. You just need to add that this star is at upper culmination for that meridian at a sidereal time of zero.
     
  7. Jan 22, 2008 #6

    dynamicsolo

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    If the question is asking for the local sidereal time at Thessaloniki, you need to use the local hour angle. GHA will give you the sidereal time at Greenwich.

    Sidereal time is a local measure which tells you about the orientation of your location on Earth relative to the direction of the vernal equinox. Another way of putting this is that the local sidereal time tells you the right ascension of stars on your meridian at that moment.
     
    Last edited: Jan 22, 2008
  8. Jan 22, 2008 #7

    andrevdh

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    The hour angle of the star will be zero when it is in culmination at Thessaloniki.

    http://en.wikipedia.org/wiki/Hour_angle

    or if you look in Wikipedia at the definition of ST it mentions that ..."and when sidereal time is equal to an object's right ascension, the object will be at its highest point in the sky, or culmination," ....

    which means that ST is the RA of the objects on the local meridian.
     
    Last edited: Jan 22, 2008
  9. Jan 22, 2008 #8
    So the answer is

    ST(Thessaloniki)= [tex]\alpha[/tex] + H
    ST= [tex]\alpha[/tex] + 0
    ST= 1h25m18s ?


    -edit-
    I changed the alpha letter with a latex one, because I typed it before (i'm greek so I have an eng/gr keyboard) without thinking that you won't be able to see it. Sorry if it caused any mess.
     
    Last edited: Jan 22, 2008
  10. Jan 22, 2008 #9

    dynamicsolo

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    That should be correct for the local sidereal time. I believe that belliott4488 may be thinking of a different situation where Greenwich Hour Angle is called for, such as needing to make the calculations starting from GMT. Nowadays, at observatories, this is all handled with atomic time standards, precise geographical location of observatory positions, and computer software. (It's still good to know how to calculate sidereal time if it's something you work with much...)
     
  11. Jan 22, 2008 #10
    Thank you all for your help :)

    Yes it says so in the book, that this is no more of use, but we still have to learn it.
    Is this not being taught anymore at the us universities in the introductory astronomy lesson?
     
    Last edited: Jan 22, 2008
  12. Jan 22, 2008 #11

    dynamicsolo

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    Well, it's still always a good idea to know how something is calculated and how to at least estimate a result, even if computers are available to do the detailed work.

    It occurred to me that the other place the Greenwich Hour Angle is used is in traditional navigation, where you have star positions and a GMT clock, but use the difference between GHA and the hour angle you measure for a star at your location to work out your longitude. (Again, nowadays, we use timing signals from satellites in the GPS network and a portable, dedicated microprocessor to work out positions on the Earth. But mariners and aviators still learn traditional methods of reckoning because you can't always be sure your GPS "box" will be functioning...)

    I think astronomy students are still shown this topic here, but it's probably gotten reduced in emphasis of late...
     
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