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How to calculate the latitude

  1. Jul 6, 2012 #1
    Hello everyone!
    First of all,this is not a homework question :D It's just a problem that I want to solve but I don't know how(it was given at the astronomy olympiad 2011)
    Observations were done by the naked eye on June 16, 2008, Universal time was
    used. An observer has registered that a star passed zenith at 0 h18m, and at 8h17m its height above the horizon was 87°12'. Find the latitude of the observations.
    (you can find it here http://www.issp.ac.ru/iao/2011/ )
    I know that if the star passes zenith that means that the zenith distance z = 0 because its height would be 90 degrees.This implies that the latitude is equal to the star's declination.
    Then I calculated after the formula: z(zenith distance) = latitude - declination.
    z = 90 degrees - 87°12' = 2 degrees and 48 minutes
    So,the latitude would be 2 degrees and 48 minutes + declination.But what do I do with that time given ?How do I use it?
    I would greatly appreciate your help!
    Thanks in advance!
     
  2. jcsd
  3. Jul 6, 2012 #2

    TSny

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    I'll give it a shot. I will take into account only the rotation of the earth on its axis and not the orbital motion of the earth around the sun.

    Introduce a Cartesian coordinate system that is fixed relative to the stars. The origin is at the center of the earth and the z-axis is along the axis of rotation of the earth. At the instant of the first observation, we may assume the observer is located at a point in the x-z plane. If θ is the latitude of the observer, then we may construct a unit vector r pointing from the center of the earth to the observer as r = cosθ i + sinθ k. This vector also points to the star since the star is directly overhead at this instant.

    For the second observation, the earth will have rotated through an angle [itex]\phi[/itex] that is easy to determine from the time between observations. I find [itex]\phi[/itex] = 119o 45'. The observer will have rotated to a new position relative to the fixed Cartesian frame. Using standard formulas for rotation about the z axis, the new unit vector pointing toward the observer will be r' = cosθcos[itex]\phi[/itex] i + cosθsin[itex]\phi[/itex] j + sinθ k

    The dot product of r and r' equals the cosine of the angle between the position of the star and the zenith at the final observation. Call this angle β. (β is given to be 2o 48').

    Thus, cosβ = r[itex]\cdot[/itex]r' = cos2θcos[itex]\phi[/itex] + sin2θ.

    Using sin2θ = 1-cos2θ and solving, I find cosθ=[itex]\sqrt{(1-cos\beta)/(1-cos\phi)}[/itex] from which I find θ = 88o 23' = latitude of observer.

    [Added note: Using half-angle trig identity, the result may be written as cosθ = [itex]\frac{sin(β/2)}{sin(\phi/2)}[/itex] ]
     
    Last edited: Jul 6, 2012
  4. Jul 7, 2012 #3
    Thank you for your answer but I am afraid I don't understand this method(using unit vectors I mean).I only understand this part
    Can you explain it to me more simply or is there another way,simplier than this one?
     
  5. Jul 7, 2012 #4

    TSny

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    I've attached some figures that show how to get the answer just using trig. Fig. 1 shows the setup. O is the center of the earth.

    When the observer is at A the star is overhead. As the earth rotates through the angle [itex]\phi[/itex], the observer moves to point B on a circle with center O'. Fig. 2 shows how to express the radius of this circle in terms of the co-latitude angle [itex]\alpha[/itex] and the radius of the earth. The figure also shows the angle [itex]\beta[/itex] between the zenith at B and the star.

    Fig. 3 shows looking down on triangle ABO'. C is the midpoint of AB. The angle AO'B is the angle [itex]\phi[/itex] that the earth rotates through. Thus you can express CB in terms of R, [itex]\alpha[/itex], and [itex]\phi[/itex].

    Fig. 4 shows the triangle ABO. Angle AOB = [itex]\beta[/itex]. From this triangle you can get an expression for sin(β/2). You can solve this expression for sin[itex]\alpha[/itex]. Finally, sin[itex]\alpha[/itex] = cosθ because [itex]\alpha[/itex] and θ are complimentary.
     

    Attached Files:

  6. Jul 9, 2012 #5
    Thank you very much for the solving the problem.
    Can you explain, please, why angle AOB = β??
     
  7. Jul 9, 2012 #6

    TSny

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    The star is directly overhead when the observer is a point A. So, the line OA points toward the star. When the observer is at point B, the line OB points toward directly overhead (zenith). So, the angle between OA and OB is the angle between the direction of the star and directly overhead at point B. When the observer is at B, the star is 87o12' above the horizon. So, the angle β between the direction to the star and directly overhead is the complement of this angle, or 2o48'.
     
  8. Jul 9, 2012 #7
    oh, i'm grateful...thank You very much

    p.s. I was searching in the net solutions for the IAO 2011, 2010 and earlier years a lot. There are not any answers or solutions for problems of that olympiad. So can You help me please(cause in Your messages I see professionalism and kindness)? Best regards.
     
    Last edited: Jul 9, 2012
  9. Jul 10, 2012 #8
    So AOB triangle is a spherical triangle,right?What about that time given in the problem (0 h 18 m)?Is it the hour angle or something?
    And one more question:why is BOO' the colatitude?
     
    Last edited: Jul 10, 2012
  10. Jul 10, 2012 #9

    TSny

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    Triangle AOB is just a regular plane triangle whose 3 sides are straight lines all lying in a plane. (A spherical triangle would be a triangle drawn on the surface of a sphere with sides formed by arcs of "great circles".)

    I interpreted the times as just ordinary time in hours and minutes. I interpret 0 hour as midnight. So, 0 h 18 min would be 18 minutes after midnight. The problem states that the times are "Universal Time (UT)" in which a day is the mean solar day. That is, 24 hours is the time for the sun to return to the same point in the sky. The time for the earth to rotate once on its axis relative to the stars is about 4 minutes less than 24 hours of UT time. I did not worry about this small effect in the calculations.

    To see why BOO' is equal to [itex]\alpha[/itex], look at Fig. 1 of my attachment. You can see that the observer rides around on his circular latitude line. Think of this circle as the top of a cone with the vertex of the cone at the center of the earth O. Hopefully you can see the cone drawn in the figure. The axis of rotation of the earth is also the axis of the cone. So, if you pick any point on the circle, say point B, and draw a line from O to that point, then that line will form the angle [itex]\alpha[/itex] with the axis of rotation of the earth. It will be the same angle [itex]\alpha[/itex] as line OA makes with the axis of rotation.
     
  11. Jul 10, 2012 #10
    Ok,got it now.Thank you very much :D
     
  12. Jul 10, 2012 #11
    Thank you TSny for your answers. They really helped me to understand the solutuon of this problem.
    Can you, please, try to answer for this life problem too?
    https://www.physicsforums.com/blog.php?b=4091 [Broken]
     
    Last edited by a moderator: May 6, 2017
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