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A couple of geostationary calculations

  1. Nov 26, 2008 #1
    1. The problem statement, all variables and given/known data
    a satellite is orbiting the earth in a near-polar orbit. it's orbital height is 500km. the satellite carries a nadir-viewing push-broom scanning imaging instrument with 4 spectral bands. for each spectral band, the detector array consists of 1024 detector elements. the width of each detector element is 15 μm. focal length of telescope is 10m.
    a) find the orbital period.
    i got 6044s using the kepler eq.

    b) at 3:10 UTC, the same satellite crosses the equator at longitude 105 deg E as it travels from north to south. i) when wil the satellite next cross the equator in the same north to south direction. ii) what wil be the longitude of the equator crossing point?
    for part i), is it adding the orbital period is 6044s to 3:10?
    for part ii), i found out that the earth rotates 0.25 deg every min, since orbital period is 6044s=101min, the satellite will move 25.25 deg further, so is it 105deg +25.25 deg=130.3deg E?

    c) what is the ground pixel of the imaging instru of this satellite?
    i made use of this eq: ground pixel width W= (detector element width b x orbital ht) / focal length, which i got 1.2m

    d) the imaging sensor is turned on for 20s to acquire an image of the earth surface, what is the width and length of the area on the eart surface imaged by the satellite?
    i dont quite understand the qn, is it sth to do with the swath width? how do i go about looking for the swath width?

    thanks in advance!
     
  2. jcsd
  3. Nov 26, 2008 #2

    Redbelly98

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    Welcome to PF!

    (a) and (b) look good.

    (c) Is in the ballpark, but I think you made an arithmetic error or used the wrong pixel width or orbital height. Can you double-check your math, and if you get the same answer post the details of how you calculated it (with the numbers you are using)?

    (d) Swath width: from the pixel size (on Earth's surface), and they tell you how many pixels there are ...
     
  4. Nov 26, 2008 #3

    D H

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    Show your work, please. That answer is not correct.

    Yes, but you have to add the correct orbital period.

    A couple of mistakes here. First, you are propagating the period error. More importantly, which way does the Earth rotate? Finally, 0.25 degrees/minute is not quite right. You should be using the sidereal day, not the solar day.
     
  5. Nov 26, 2008 #4

    Redbelly98

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    I had done a simple "ballpark" check to see if the answer was close. But yes, this answer is off.

    Agreed. I had missed these when I did my check. D'oh!
     
  6. Nov 26, 2008 #5

    D H

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    Just to be perfectly clear: Answering the "which way does the Earth rotate" question correctly will lead to the largest change in your answer, mint. Using the correct orbital period will change the result a little. Using the correct length of day will change the result even less.
     
  7. Nov 27, 2008 #6
    (i) i use this eq, T^2=(4pi^2/GM) R^3.
    where T is the orbital period, R is the radius of earth and orbital height. G is the gravitational constant, M is the mass of the earth.

    (ii) earth rotates 0.25 deg every min, from the fact that the earth rotates 360deg in 24 hours. since orbital period is 6044s=101min, the satellite will move 25.25 deg (101 X 0.25)further. i think my mistake is that the orbiting satellite should be displaced to the west after each orbital period, due to the rotation of the earth, so i should minus 25.25 deg from 105 deg E?
     
  8. Nov 27, 2008 #7
    for c, i make use of equation w=Hb/f. w is the ground pixel width, H is the orbital height, b is the detector width, f is the focal length. since b is given as 15μm, f is given as 10m, H is what i calculated, i presume if my ans for w is wrong, then it should be due to errors from my H?

    for d, sorry i dont really get it. how do i get the swath width from the calculations that i have done earlier? the question wanted the width and length.

    thanks for helping! :)
     
  9. Nov 27, 2008 #8

    Redbelly98

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    (c)
    Orbital height H was given in the problem statement. Not sure how you can calculate and have errors in H?

    (d)
    There are 1024 pixels in the "image" ...
     
  10. Nov 27, 2008 #9
    sorry my bad, yup i thought i calculated the orbital period instead of the height in part i, i got confused.

    for d, the qn states that 'the satellite carries a nadir-viewing push-broom scanning imaging instrument with 4 spectral bands. for each spectral band, the detector array consists of 1024 detector elements.' if there's 1024 pixels in the image, and i got 1.2m for the ground pixel width. so the total width and length is just 1024 x 1.2? do correct me if im wrong.
     
  11. Nov 27, 2008 #10

    D H

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    So, what have you calculated for the orbital period? (6044 seconds is incorrect.)

    The width of the scan will be considerably more than 1024 x 1.2 meters. Only the nadir-looking pixels will "see" a 1.2 meter wide chunk of the Earth. The remaining pixels will see ever wider swaths because they aren't nadir-looking.
     
  12. Nov 27, 2008 #11

    Redbelly98

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    For the width, essentially yes, except the 1.2m is incorrect.

    For the length, we do not have 1024 pixels. Do you know what is meant by a "push-broom scanning imaging instrument" ?

    I am thinking the 1024 x (single pixel width) is still a pretty good approximation. It's roughly a 1 km wide path, viewed from 500 km away.
     
  13. Nov 28, 2008 #12
    hi, i still got 6044s. i substitue G=6.672 x 10^24 and M=5.976 x 10^24 and R=radius of earth and orbital ht=7175 x 10^3 m
     
  14. Nov 28, 2008 #13
    an explanation on push broom scanning system is provided in this website:
    http://www.crisp.nus.edu.sg/~research/tutorial/image.htm

    i presume one square is equivalent to one pixel width. thus, the width of the area will be 1.2m(assuming it's the correct ans first, i havent found out the right ans yet), length will be approx 1024 x 1.2m. does the time 20s matter in the length covered?

    i have another additional qn, what is the size of the image in bytes if the signal is encoded in 8 bits pixel spectral band?

    i was thinking of using this eq: area covered by orbit=2pi Re S, where Re is the radius of earth, S is the swath width.

    swath width=1024 X 1.2?

    no. of pixels=area covered/ area of 1 pixel. so assuming 8 bits/pixel, no. of bits= no. of pixels/ 8

    so from that, i convert no. of bits into no. of bytes?

    do i need to take into account of the 4 spectral bands, ie do i need to time 4?

    thanks in advance!
     
  15. Nov 28, 2008 #14

    Redbelly98

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    The Earth's radius is 6378 km:
    http://www.google.com/search?hl=en&safe=off&q=radius+of+earth+&btnG=Search

    So you should use:
    http://www.google.com/search?hl=en&q=radius+of+earth+++500+km&btnG=Google+Search&aq=f&oq=

    Good, that one picture explains it pretty well.

    Look at the picture again. The flight direction is at a right angle to the row of 1024 pixels. Perhaps you should clarify which direction you are calling the width and which is length.

    Yes, the 20s will matter here.
     
  16. Nov 28, 2008 #15
    oh so the swath width should be the longer stretch, while the length should be just 1.2m?

    how do 20s matter here? and do the number of spectral bands affect the calculations, question states that there is 4.
     
  17. Nov 28, 2008 #16

    Redbelly98

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    That would be true IF:
    1. One pixel were really 1.2m (how about doing that calculation, so we can continue the discussion using the correct value?)
    and
    2. The data was taken in one instant, so the satellite does not move during the scan. Again, this is not really the case.

    The satellite is moving. It starts the scan located directly above a certain spot on Earth. 20 seconds later, it has moved to be above a different spot on Earth. The length of the scan is the distance from the first spot to the last spot of 20 seconds later.

    You'll need to get the correct orbital period to get this question right.

    Nope, that does not matter.
     
  18. Nov 29, 2008 #17
    i still get 1.2m after subsituting all the values into w=Hb/f, where H=800km, b=15 micro m, f=10m, after converting everything into si units.

    for the orbital period, after substituting radius of earth=6378km, i got 6048s.

    so how does 20s affect the calculation?
     
  19. Nov 29, 2008 #18

    D H

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    In the original post you said the orbital altitude is 500 km. Now you are using 800 km. Which is it?
     
  20. Nov 29, 2008 #19
    oops sorry my bad. it's 800km, i typed the numbers wrongly. the correct orbital height is 800km. i got confused with another question.
     
  21. Nov 29, 2008 #20

    Redbelly98

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    Okay, in that case 1.2m and 6048s are good.

    The detector is looking at an area of the Earth's surface as it moves along in it's orbit. The more time that passes, the larger this area becomes. You need to figure out how large this area has become after 20 seconds.

    Look again at the figure and description of the push-broom from one of your earlier posts, if this is not clear.
     
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