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Math Help - Satellite Distance Calculation from observer

  1. #1
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    Satellite Distance Calculation from observer

    A satellite is in orbit 800km above the surface of the earth.
    The satellite is viewed and imaged at an angle of 40 degrees above the earths surface.

    calculate the distance between you and the satellite.

    So far I have:

    triangle
    point A (you)
    point B (centre of the Earth)
    point C (satellite)

    length A-B is the diameter of Earth approx 6370km
    length C-B is diameter of Earth + 800km (altitude of satellite)
    angle B-A-C is 90+40 degrees

    however this triangle doesn't allow me to find the distance between A-C

    any suggestions would be great.

    Regards,
    David
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  2. #2
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    Quote Originally Posted by Moon1975 View Post
    A satellite is in orbit 800km above the surface of the earth.
    The satellite is viewed and imaged at an angle of 40 degrees above the earths surface.

    calculate the distance between you and the satellite.

    So far I have:

    triangle
    point A (you)
    point B (centre of the Earth)
    point C (satellite)

    length A-B is the diameter of Earth approx 6370km
    length C-B is diameter of Earth + 800km (altitude of satellite)
    angle B-A-C is 90+40 degrees

    however this triangle doesn't allow me to find the distance between A-C

    any suggestions would be great.

    Regards,
    David
    Hmmm...I'm not seeing how to do it either. I think we need more information.

    Two small corrections: AB is the radius of the Earth, not the diameter, and similarly CB is the radius of the Earth + 800 km.

    -Dan
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    Draw an arc with center C and radius (6370 + 800) km. Let ABD be the radius of the new arc. Drop CE perpendicular to BD = x. Let DE = h.

    Now AE= d = x*tan(40)

    Using the properties of circle, you can write

    h*( 2r - h)) = x^2.

    Here h = 800 - x*tan(40).

    So [ 800 - x*tan40][2*(6370) + 800 + x*tan40] = x^2

    Solve for x. Once you know x, you can find AC.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Draw an arc with center C and radius (6370 + 800) km. Let ABD be the radius of the new arc. Drop CE perpendicular to BD = x. Let DE = h.

    Now AE= d = x*tan(40)

    Using the properties of circle, you can write

    h*( 2r - h)) = x^2.

    Here h = 800 - x*tan(40).

    So [ 800 - x*tan40][2*(6370) + 800 + x*tan40] = x^2

    Solve for x. Once you know x, you can find AC.
    Hi! Sorry it took so long for me to get back to this. You may well be right, but I am confused on your diagram? The OP had B as the center of the Earth, C as the position of the satellite, and A the observer's point. You are drawing an arc at C of radius 6370 + 800 and defining D such that the radius of the new arc is ABD. I can't figure out where D is?

    -Dan
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    Sorry. It should be BAD = BC is the radius of the new arc concentric to B.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Sorry. It should be BAD = BC is the radius of the new arc concentric to B.
    Okay, I think I have point D, but then you have d = x tan(40) and I don't see where that's coming from at all.

    However you have inspired me. I'm guessing we are going to be saying the same things, just that I'm not clear on your diagram. Let me sketch out what I saw.

    Let B be the center of the Earth. Let C be 6370 km + 800 km from B, ie. 800 km above the Earth's surface. Let A be a point on the Earth's surface such that C is at a 40 degree elevation from A. I am going to extend the radius BA. Now drop a line from C to the extended line BA meeting at a point CE such that CE is perpendicular to the extended BA. Simply put, CEB is a right triangle. Call d = AE and c = CE. Now, CEA is also a right triangle and the angle ECA is 40 degrees. Thus we know that d/c = tan(40). Using the Pythagorean theorem on triangle CEB:
    c^2 + (d + 6370)^2 = (6370 + 800)^2 \implies c^2 + (c~tan(40) + 6370)^2 = (6370 + 800)^2

    Solve for c. This also gives d = c*tan(40). We can again use the Pythagorean theorem on triangle CEA giving c^2 + d^2 = AC^2.

    Is this more or less what you had been looking at?

    -Dan
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  7. #7
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    Hello, Moon1975!

    A satellite is in orbit 800 km above the surface of the earth.
    The satellite is viewed at an angle of 40 degrees above the earth's surface.
    Calculate the distance between you and the satellite.


    So far I have:

    Triangle ABC:
    . . point A: you
    . . point B: centre of the Earth
    . . point C: satellite

    Length AB is the radius of Earth, approx. 6370 km
    Length CB is the radius of Earth + 800km (altitude of satellite)
    \angle BAC is 90 + 40 = 130 degrees

    However this triangle doesn't allow me to find the distance AC.
    Yes, it does.
    Code:
                    C
                    o
                   *|
                  * |
                 *  |
                *   | 800
               *    |
              *     |
             *      *
            *       |
           *        |
        A o 130d    |
            *       | 6370
              *     |
           6370 *   |
                  * |
                    o
                    B

    Law of Sines: . \frac{\sin C}{AB} \:=\:\frac{\sin A}{BC}

    . . \sin C \:=\:\frac{6370\sin130^o}{7170} \:=\:0.685351559

    . . Hence: . \angle C \:\approx\:43.263^o

    Then: . \angle B \;=\;180^o - 130^o - 43.263^o \;=\;6.737^o


    Law of Sines: . \frac{AC}{\sin6.737^o} \:=\:\frac{71670}{\sin130^o}

    . . AC \:=\:\frac{7170\sin6.737^o}{\sin130^o} \:=\:1098.014065


    Therefore: . AC \;\approx\;1098.0\text{ km.}

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