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Connection between right ascension and time

  1. Sep 27, 2013 #1
    I'm a physics major currently taking my first astro class. We're covering the basics at the moment but I am having trouble visualizing this question from our textbook. To preface this, I understand that declination is to the celestial sphere what latitude is to the Earth and RA is to the celestial sphere what longitude is to the earth. I also know that RA is measured from the vernal equinox. The problem references Hemingway's "The Old Man and the Sea" and describes a man in Cuba lay in his boat shortly after the sun set one September night and saw Rigel rising. I'm supposed to find what is incorrect about this. I'm fairly certain that Rigel wouldn't appear in the night sky until much later than the sun sets. Rigel's RA is 05h 14m. The longitude of Cuba (in a very general sense) is approximately 80 degrees west. But where do I go from here? Do I calculate how many hours away Cuba's longitude is from Rigel's RA? How do I factor in the time?
  2. jcsd
  3. Sep 27, 2013 #2


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    You need to determine what 'clock' is used to define RA, and the clock used by the 'Old Man'. The rest is easy.
  4. Sep 27, 2013 #3
    So the 'clock' used to define RA is that 0h is at the vernal equinox, and goes up east from there. The clock used by the Old Man allow him to read the time at approximately sunset in Cuba in late September?
  5. Oct 7, 2013 #4
    There is time in the Hour Angle:
    [tex]LMST = GMST + time + Longitude/15[/tex][tex]HA = LMST - RA[/tex]where LMST is Local Mean Sidereal Time in Hours. GMST is Greenwich Mean Sidereal time. The Sidereal Time above London, England. And time is in hours also.
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