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Math Help - Aeronautics problem

  1. #1
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    Aeronautics problem

    A satellite S is in a geosynchronous orbit; that is, it stays over the same point T on Earth as Earth rotates on its axis. From the satellite, a spherical cap of Earth is visible. The circle bounding this cap is called the horizon circle. The line of sight from the satellite is tangent to a point Q on the surface of Earth. If the radius of Earth, CQ, is 3963 miles and the satellite is 23300 miles form the surface of the Earth at Q, what is the radius of the horizon circle, PQ, in miles?

    I've been staring at this problem for hours. I'm completely stumped.
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  2. #2
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    Re: Aeronautics problem

    Quote Originally Posted by explodingtoenails View Post
    A satellite S is in a geosynchronous orbit; that is, it stays over the same point T on Earth as Earth rotates on its axis. From the satellite, a spherical cap of Earth is visible. The circle bounding this cap is called the horizon circle. The line of sight from the satellite is tangent to a point Q on the surface of Earth. If the radius of Earth, CQ, is 3963 miles and the satellite is 23300 miles form the surface of the Earth at Q, what is the radius of the horizon circle, PQ, in miles?

    I've been staring at this problem for hours. I'm completely stumped.
    Actually you are dealing with a right triangle (see attachment)

    1. Calculate the length l.

    2. then r (it's the height in the right triangle) is calculated by:

    r = \dfrac{R \cdot l}{R+23300}
    Attached Thumbnails Attached Thumbnails Aeronautics problem-satell_horizont.png  
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  3. #3
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    Re: Aeronautics problem

    Start by drawing a picture: draw a circle, representing the earth, and a point, representing the satellite. Draw lines from the point tangent to the circle, at Q and Q', representing the lines of sight. Finally, draw a line from the satellite to the center of the earth, and lines from the center of the earth to the points at which the tangent lines touch the circle. You now have two right triangles and you know the length of one leg, the radius of the earth, and the hypotenuse, the radius of the earth plus the height of the satellite, so you can calculate the distance from the satellite to Q and Q' and the angle, \theta, in either of those triangles, at the satellite ( cos(\theta)= "near side/hypotenuse".

    Draw the line QQ', crossing the line from the center of the earth to the satellite at point P. That gives another pair of right triangles having angle \theta and hypotenuse equal to the distance from the satellite to Q or Q'.

    (earboth, whose post I did not see before writing this, has a very nice picture with his solution.)
    Last edited by HallsofIvy; August 13th 2011 at 05:21 AM.
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    Re: Aeronautics problem

    earboth, how did you get the formula for r?
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    Re: Aeronautics problem

    Quote Originally Posted by explodingtoenails View Post
    earboth, how did you get the formula for r?
    In a right triangle with the legs a and b and the hypotenuse c the area is calculated by:

    A = \frac12 \cdot a \cdot b and A = \frac12 \cdot c \cdot h

    where h is the height in the right triangle.

    That means you have:

    \frac12 \cdot a \cdot b = \frac12 \cdot c \cdot h ~\implies~h = \frac{a \cdot b}c

    As I've mentioned before the h correspond to the radius r.
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    Re: Aeronautics problem

    Why is the bottom part of the fraction R + 23300 when the hypotenuse is only 23300?
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    Re: Aeronautics problem

    Quote Originally Posted by explodingtoenails View Post
    Why is the bottom part of the fraction R + 23300 when the hypotenuse is only 23300?
    The hypotenuse of the right triangle is the line segment CS. According to the wording of the question

    |\overline{CS}|= \underbrace{|\overline{CT}|}_{=R} + \underbrace{|\overline{TS}|}_{= 23300}
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