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Minimum Altitude

  1. Jan 10, 2016 #1
    From my astrophysics class, we were given that the minimum altitude of a star relative to your latitude on Earth ([itex]\phi[/itex]) is given by:
    $$h_{min} = \delta + \phi - 90°$$ (where [itex]\delta[/itex] is the declination angle of the star)

    Supposing you were trying to view as many stars as possible in the night sky (assuming perfect weather conditions / no light pollution everywhere on the Earth), where would be the best place to go?

    Using the formula, you want [itex]h_{min}[/itex] to be as positive as possible for any given star declination [itex]\delta[/itex]. This means you should make you [itex]\phi[/itex] as large as possible, which would be 90 degrees (i.e. North pole). Is my reasoning correct, from this formula, that the best place to view as many stars as possible (again assuming perfect weather) would be at the north pole?
     
  2. jcsd
  3. Jan 10, 2016 #2

    Bystander

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    How many stars in one hemisphere? (North pole)
    Is there any place on earth you can view two hemispheres?
     
  4. Jan 10, 2016 #3
    That would be at the equator, correct? That makes sense (e.g. a lot of observatories in South America), but why does the formula say otherwise?
    Is this because the formula only works for Northern hermisphere latitudes?
     
  5. Jan 10, 2016 #4

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    I suspect it's a matter of the interpretation of the question.
     
  6. Jan 17, 2016 #5

    Tom.G

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    For naked eye viewing, wouldn't the best spot be where the Milkyway was at zenith at local midnite?
     
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