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Rotation speed for a satellite

  1. Mar 3, 2016 #1
    1. The problem statement, all variables and given/known data
    A geostationary satellite is located at 0'N 0'E (degrees), 36000 km above a spherical Earth with radius R(earth) = 6370 km. To scan the fieldof view, the satellite rotates around its own axis(oriented parallel to the Earth's rotation axis). It records one (constant latitude) scanline per rotation with the scan mirror steering the scanline in latitude with every rotation. A scan of the complete disc from North to South is allowed to take 25 min. With additional 5 min being required for returning the scan mirror back to its start position, one complete scan can be provided every 30 min.

    When the horizontal resolution of a single scan pixel at the subsatellite point is 7.5 x 7.5 km, what is the rotational speed required to archieve this scan cycle ?



    2. Relevant equations

    αlat,lon = 2tan-1(Resolution/R(geo))

    v = wr

    v = 2H/t

    3. The attempt at a solution

    v = 2H/t = (2*7.5)/(30 min) = 0.5 km/min

    w = v/r = (0.5 km/min)/(36000 km) = 1.38*10^-5 rad/s

    Which is wrong. What is it that i dont understand here ?
     
  2. jcsd
  3. Mar 3, 2016 #2

    haruspex

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    I don't understand your attempt. What is the logic behind your first equation there?

    I would start by finding how many rotations are needed to cover the North-South distance.

    Edit: I suspect that either you have not understood the question or I haven't. It might help if you were to describe your interpretation.
     
  4. Mar 10, 2016 #3
    I dont know what to do, if i have the e.q, αlat,lon = 2tan-1(Resolution/R(geo)) and
    f9a62419f9da2c54cf180349edcb356c.png
    And if one rotation takes 1 hour. What should i do ?
     
  5. Mar 10, 2016 #4

    gneill

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    Start by making a sketch showing the angle that the vertical scan must cover in order to "see" the entire image of the Earth from pole to pole. Draw the scan profile at the distance that you know the resolution of the scan. How many scan lines are required to cover that profile?
     
  6. Mar 10, 2016 #5
  7. Mar 10, 2016 #6

    gneill

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    Notice how the lines that you drew from the satellite to the poles cut through the circle of the Earth? That means there's some small portion that's lying outside of the scan. So those lines really should meet the Earth surface at a tangent, and not simply connect to the poles.

    Sketch in the scan profile: That would consist of a circular arc centered at the satellite that touches the surface of the Earth at the sub-satellite position (where you also happen to know the resolution). What's the value of the angle ##\alpha##? What's the length of that arc bounded by the tangent lines?
     
  8. Mar 10, 2016 #7

    haruspex

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    The pixel area is quoted in sq km, not sq seconds of arc, so I assume that for the purposes of this problem the satellite altitude is considered great enough that we can treat the earth as a flat disc.
    @Firben, consider a rotation of the satellite during which it scans at some latitude theta. Sketch the region of the earth scanned.
     
  9. Mar 10, 2016 #8

    gneill

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    @haruspex, I'm assuming that the satellite distance r is sufficient that the profile arc can be well approximated by a concatenation of short straight line segments of length r⋅dα = 7.5 km. This because the resolution at the center of the arc is 7.5 x 7.5 km, and by following an arc of the same radial distance from the satellite the resolution will be uniform over the whole arc. I think this will be simpler to manage mathematically than the flat disk.
     
  10. Mar 10, 2016 #9

    haruspex

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    Ok, so you are suggesting converting the 7.5km to an angle? Your r is the satellite altitude I take it.
    That's a bit more complicated than my method (which only needed earth diameter/7.5km), but it is certainly more accurate.
     
  11. Mar 10, 2016 #10

    gneill

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    Yes, that r is the 36000km "height" of the satellite. But all I need is the length of the arc, which is trivial once the subtending angle is known.
     
  12. Mar 10, 2016 #11
  13. Mar 10, 2016 #12

    haruspex

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    The drawing is a little inaccurate. Where the tangent touches, what is the angle between the tangent and the radius? What is the distance from the satellite to the centre of the earth? What does that give you for alpha?
     
  14. Mar 10, 2016 #13

    gneill

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    They want to know how fast the satellite needs to rotate in order to completely map the observed surface area in 25 minutes. The mapping is done in horizontal strips, the width of which is dictated by the resolution. You should try to determine how many such strips will be required to complete the image.
     
  15. Mar 11, 2016 #14
  16. Mar 11, 2016 #15

    gneill

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    Here's my sketch:
    upload_2016-3-11_8-39-30.png

    Find the value of ##\alpha##. Find the length of the arc ab. How many "slices" of width 7.5 km does it take to sum up to the length of the arc? That's how many rotations are required to scan the arc. You need to complete that many rotations in the specified time.
     
  17. Mar 12, 2016 #16
    If i use the following eq.

    α = 2*tan-1(3.75/36000) ≈ 0.0119°
    and
    s = rα = 36000*0.0119 ≈ 428.4 km

    It takes 57 slices to sum up the length of the arc (428.4/7.5) = 57.12

    then v = 57.12km/30 min = 1.9 rad/s

    Is this right ?
     
  18. Mar 12, 2016 #17

    gneill

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    No, that's the angular size of one "slice", not the angular extent of the whole arc (from a to b on the diagram that I posted above).
    You've used an angle in degrees. That won't work in the formula arclength = radius x angle. You must use radians. The calculation you've specified would return the width of the slice in kilometers, not the length of the entire arc.
    Nope. Find the angle subtending the whole arc from a to b, then the arc length from that. Then find out how many "slices" fit the whole arc. Note that 30 minutes is not the scan time; it includes the 5 minutes required to re-position the apparatus to begin again.
     
  19. Mar 13, 2016 #18
    How can i find the angular size of the entire arc a and b ? I dont know the angle α, if im using s = rα . I dont have much information to after
     
    Last edited: Mar 13, 2016
  20. Mar 13, 2016 #19

    gneill

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    Look at the diagram. Are there any right-angled triangles that include α for which you know some side lengths?
     
  21. Mar 13, 2016 #20
    If im using α = tan-1(Re/(Re+RGeo) is that what you mean ?
     
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