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Calculating longitude and latitude given the duration of sunset

  1. Apr 13, 2013 #1
    Hello!
    I've been trying to solve this problem for some time but I can't seem to find the answer.
    Can you help me?
    1. The problem statement, all variables and given/known data
    On the night of 5th april an observer watches the sunset(η = 3.5 minutes,η = correction of time). The solar disk touched the horizon at 20h 02 m 02 s and then the Sun's disk disappears below the horizon at 20h 05m 10s.
    Given the distance Earth-Sun= 149,6 * 10 ^6 km and the solar radius = 696 000 km find the latitude and longitude of the place.

    3. The attempt at a solution
    What I have tried so far was to write the formula for angular diameter θ = 206265d/D =>θ = 0.53 degrees.
    I have thought about finding the hour angle so after that I can use this formula:
    sin φ = tan(90 - H)*tan h,where φ=latitude of the observer,H = hour angle,h = altitude of the Sun which is 0 at sunset
    Now my question is:how do I calculate the hour angle using that angular diameter?
    Thanks in advance!
     
  2. jcsd
  3. Apr 13, 2013 #2
    What is the kind of time given in the problem? Solar, mean solar, civil, etc?
     
  4. Apr 13, 2013 #3
    It doesn't mention that in the problem but I am assuming it is civil time...
     
  5. Apr 13, 2013 #4

    haruspex

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    Draw a diagram showing the sun touching the horizon at point A (the top of the sun's disc being at point B here), and the last remnant disappearing below it at point C. ABC is a right angled triangle. Can you relate the hypotenuse to the elapsed time and determine its ratio to AB?
    Since April 5th is 16 days after the equinox, you might have to make an adjustment for that.
     
  6. Apr 16, 2013 #5
    So in that ABC triangle wouldn't AB be the diameter of the sun ? and AB/AC= sine C
    The time elapsed between the moment when the Sun touched the horizon and the moment when it disappeared below it would be 20 h 05 m 10 s - 20 h 02 m 02 s = 03m 08 s.
     
  7. Apr 16, 2013 #6

    haruspex

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    where 'diameter' means angular diameter, the angle subtended at the observer, yes.
    The way I defined the points, the hypotenuse is BC.
    Yes, but how does that relate to the angular 'distance' BC?
     
  8. Apr 16, 2013 #7
    Well,then that distance BC would be the distance travelled in that 3m 8s. So should I divide that theta angle(angular diameter)by some rate of change in the altitude of the sun so I can get the time elapsed?Wouldn't that be angle C ? I thought as well about multiplying the duration of the sunset with the speed of light so I can get the side BC of the triangle.Is this a bad idea?
     
  9. Apr 16, 2013 #8

    haruspex

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    Yes, but compare that with the angular distance you know it would travel in 24 hours.
     
  10. Jun 28, 2013 #9
    But how does this help me to find the latitude of the place?
     
  11. Jun 28, 2013 #10

    haruspex

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    You should now have the length of AB (as an angle subtended at the eye) and the length of BC (likewise). That tells you the angle ACB. How do you think that would be related to latitude at the equinox?
     
  12. Jun 28, 2013 #11
    I guess that at equinox the side BC would be smaller but I don't know how to relate the latitude to that.What I know is that the declination is 0 and that it sets in the west. For angle ACB I've found 42,94degrees where AB=1919,25 arcseconds and BC=2817"Then what about using some cosine formula for spherical triangles with latitude and declination?
     
  13. Jun 28, 2013 #12

    haruspex

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    The triangle is sufficiently small that you don't need to worry about spherical effects - treat it as a planar triangle.
    If at the equator, what angle would the apparent arc of the setting sun make with the horizon? What angle if at one of the poles?
     
  14. Jun 29, 2013 #13
    At equinox , right? If at the equator then doesn't this mean that the arc/path of the Sun is perpendicular to the horizon? Or are you asking about the altitude of the Sun above the horizon? I am not sure if I understand correctly.
    What about angle ACB in the right triangle ABC? Does this angle actually has to do with the rate of change in altitude of the Sun? Am I getting it wrong?
     
  15. Jun 29, 2013 #14

    haruspex

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    Yes, so ACB in this case is a right angle, yes?
    It's the angle the apparent path of the sun makes to the horizon as it sets. Can you see how that relates to latitude? If not, consider how the plane of the equator relates to the apparent path of the sun.
     
  16. Jun 29, 2013 #15
    At equinox,yes. The angle ACB which the Sun makes with the horizon would be 90 but that angle is actually 90-latitude so sin ACB=AB/BC becomes: cos latitude=AB/BC. But this happens at equinox. Shouldn't there be a general relation between declination,latitude and this angle ACB?
     
  17. Jun 29, 2013 #16

    haruspex

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    As I wrote in post #4, April 5th is 16 days after the equinox. That is not much, so it should be possible to make a relatively simple adjustment.
    What angle will the setting sun's trajectory make at the equator at a solstice? What sort of function do you think will express how that angle varies over the year?
    Suppose that at a particular time of year T, the setting sun makes an angle ##\phi## to the vertical at the equator. At latitude ##\theta## it makes an angle ##\theta## to the vertical at the equinox. What angle do you think it will make at latitude ##\theta## at time T?
     
  18. Jun 30, 2013 #17
    After using some spherical trigonometry although you told me there's no need for that I've found this relationship between declination,latitude and the angle that the setting sun makes with the horizon: cos angle = sin latitude/cos declination. So at the equator at a solstice the angle is 90,right?
    The Sun moves on the ecliptic 1 degree per day so I should multiply this angular velocity by the time(16 days) to find out how the declination angle changed from the equinox(0) to now (at time T) Is this correct?
     
  19. Jun 30, 2013 #18

    haruspex

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    Maybe - I'm not exactly sure what you're proposing. Can you put all that into actual equations?
     
  20. Jul 1, 2013 #19
    Ok,so my idea was:ω = Δx/ΔT where ω = 0,98 degrees/per day and ΔT=16 days. From here I get the arc length made by the Sun in 16 days from the equinox. Then I calculated what fraction represents this number from the total 360) =>f = 4,35% Taking into consideration the fact that the maximum declination is 90 degrees I wrote:
    f/100*90=3,92 degrees so I can deduce that from the equinox the declination changed from 0 to 3,92 degrees. Am I wrong? Please correct me if I am mistaken.
     
  21. Jul 3, 2013 #20

    haruspex

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    All that is saying is that 16 days is 4,35% of a year. The declination is not a linear function of that. It'll be a sine function, no? But maybe that small enough not to matter. So what do you now get for the answer?
     
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