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Satellite angle, remote sensing

  1. Jan 12, 2016 #1
    1. The problem statement, all variables and given/known data
    You have a geostationary satellite (location 0◦N 0◦E, 36000 km above geoid). Assume the Earth to be spherical with a radius of REarth = 6372km

    When observing a region centered at 60◦N 0◦E:

    What is the incidence angle of the satellite’s line of sight (at pixel center; surface parallel≡ 0 ◦ , nadir≡ 90◦ )?

    2. Relevant equations
    φ = 60◦
    r = 6370 km

    3. The attempt at a solution

    http://s716.photobucket.com/user/Pitoraq/media/Rs_zpstcj5qby5.png.html?sort=3&o=0

    Im not sure how to calculate the angle of incidence by that figure. This is from an exam and the solution is:

    b = r × sin φ = 5517 km
    a = r − r × cos φ = 3185 km
    scan-angle (off nadir) β: tan β = b/(h+a) = 5517/(36000+3185) = 0.1408, β = 8.014◦
    parallel surface: 90◦ − φ = 30◦
    incidence-angle: γin = 90◦ − (φ + β) = 21.99◦

    where do they get the relation 90◦ − φ = 30◦ from ? and where is β in my figure ?


     
    Last edited by a moderator: Jan 12, 2016
  2. jcsd
  3. Jan 12, 2016 #2

    NascentOxygen

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    Staff: Mentor

    β is the acute angle on the far right of your sketch.
     
  4. Jan 13, 2016 #3
    But where do they get the relation γin = 90◦ − (φ + β) = 21.99◦ from ?
     
  5. Jan 13, 2016 #4
    I got the answer by using the sinus law (v = 21.9). But what about the mean horizontal resolution in direction, respectively (in km)? You might assume the Earth to be locally flat.
    If i call the right side of the figure d,then d is = 41639 km
    The answer is:
    tan α(lon) × d × 2 = 2.89 km
    Why did they add those terms together ?
     
  6. Jan 13, 2016 #5

    NascentOxygen

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    Staff: Mentor

    From there being 180° in a ∆. Your diagram is too small; you need a large, neatly drawn figure that is so spacious that you can imagine yourself walking around in it----that's how I view geometry.
    γin needs to be shown measured against a tangent to the Earth's surface, so there's your 90° between a radius and that tangent.
     
  7. Jan 13, 2016 #6
    Yes, i saw it when i plotted a new figure. But where do the relation tan α(lon) × d × 2 = 2.89 km came from ?
     
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