Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple astronomy question.

  1. Feb 5, 2010 #1
    Your observatory lies at a latitude of 45 N and longitude 90 W. Which object or objects can you observe from this location? An object at +87 dec, an object at -40 dec, and an object at -67 dec. I drew up a little celestial sphere for this and decided that 87 dec would be the only object I can observe. Is this correct? We couldnt observe any object with a negative dec from that location could we? thanks.
  2. jcsd
  3. Feb 5, 2010 #2


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Ask yourself what you would see at the equator. Or at 1 degree north or south. Doesn't it make sense that you would be able to see more than one hemispere? What do you think you would see there?
  4. Feb 5, 2010 #3


    User Avatar

    Staff: Mentor

    Is this for schoolwork, Freeman?
  5. Feb 5, 2010 #4
    Check World Wide Telescope, I think it has an option of showing night sky at a certain location.
  6. Feb 5, 2010 #5
    Yes. Does this make me persona non grata in this forum?

    I know there are online resources that will do the work for me. But I want to know how to do it on my own. So I drew up a celestial sphere on a piece of paper and figured out the horizon. This stuff is new to me.

    Is there a shortcut without drawing a celestial sphere diagram? I think our professor expects us to figure this out by drawing a celestial sphere.
  7. Feb 5, 2010 #6
    But would it be possible to see anything in negative dec from that 45 n and 90w lat/long is the question. That isnt part of the question but something I just wanted to know anyway. Because that is what my celestial sphere diagram was showing me. Though I'm not sure I did it right so here I am.
  8. Feb 5, 2010 #7


    User Avatar
    Gold Member

    In fact, this is first and foremost a homework forum. It's just a matter of how we answer it.
  9. Feb 5, 2010 #8
    So am I right or not?

  10. Feb 5, 2010 #9


    User Avatar
    Homework Helper

    Here's the celestial navigator's (greatly helpful) way of visualizing positions on the celestial sphere. Imagine a line going from the center of the Earth to a star. The latitude on Earth's surface that it crosses is the star's declination. At the specific spot that this line intersects Earth's surface, the star would be at the zenith for a local observer.

    Move away from this intersection point by a great-circle angle of one degree in whichever direction and the star will appear to be at 89 degrees. Move by 5 degrees and it'll be at 85 degrees. So somebody 5 degrees away in latitude will see the star at a maximum angle of 85 degrees (as the Earth rotates, this angle decreases).

    So if I'm standing at 45 degrees North, how far can the intersection point be before I can no longer see a star?
  11. Feb 6, 2010 #10
    At the equator?
  12. Feb 6, 2010 #11


    User Avatar
    Homework Helper

    The equator is 45 degrees away from 45 degrees North, so the star will be 45 degrees away from the zenith at its highest point.
  13. Feb 6, 2010 #12
    so the longitude is irrelevant here?
  14. Feb 7, 2010 #13
    So you're saying I can see till 45 S.
  15. Feb 7, 2010 #14


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Let's start from the beginning. If you've drawn a picture, this should all become clear. If you're at 45 deg. N latitude, then the point on the celestial sphere that is directly overhead (i.e. at the zenith) is at 45 deg. N declination, agreed?

    Therefore, what angle from it is the N. celestial pole?

    Based on that, what is altitude of the N. celestial pole (i.e. what is its angle measured upwards from the N. horizon)? Therefore, what declination are you going down to at that horizon?

    Therefore, what is its angle from the opposite (S) horizon? Or, in other words, what declination are you going down to on that side?

    Hint: what is the altitude (or elevation angle of the zenith)?
  16. Feb 7, 2010 #15
    The NCP is at 90 dec. We are at 45. So the angle to the NCP is 45. Tha altitude of the NCP is = observers latitude. Correct? So NCP altitude is 45.

    The declination of the southern most visible star is observer latitude minus 90. Correct?

    I found a simlar way to do it. if 45 (co latitude=angle from NCP right?) plus the star's dec add up to more than zero, then the star can be seen sometime during the year. So 45 + 87 can been seen. 45-40 = 5. So that can still be seen. 45-67 is the only one that cant be seen. Correct?
    Last edited: Feb 7, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook