Angle of Depression: satellite in circular orbit 9800 mi up

[joanne1102]

A satellite is traveling in a circular orbit 9800 miles above the surface of the earth. FInd the angle of depression from the satellite to the horizon. Assume that the radius of the earth is 4000 miles.

The diagram provided shows a horizontal line from center of earth to edge of orbit where satallite is located length of 13,800 miles.
Line from center of earth to the horizon length of 4,000 miles.
Line connecting these two lines (from satellite to horizon) that forms a right triangle.

I think a right triangle needs to be formed that includes the angle of depression. I know what the final answer is, but I can't seem to come up with it.

 

[soroban]

Re: Angle of Depression

Hello, joanne1102!

I don't like their diagram . . .

A satellite is traveling in a circular orbit 9800 miles above the surface of the earth.
FInd the angle of depression from the satellite to the horizon. Assume that the radius of the earth is 4000 miles.

              S o - - - - E
                |\ T
                |A\
            9800|  \
                |   \
                |    \
              * * *   \
          *     |     *\  H
        *       |       o
       *    4000|     /  *
                |   /4000
      *         | /       *
      * - - - O o - - - - *

The satellite is at \(\displaystyle S.\)
\(\displaystyle O\) is the center of the Earth: .\(\displaystyle OH = 4000\)
\(\displaystyle A \:=\: \angle HSO,\;\;T \:=\: \theta \:=\: \angle ESH\:=\: 90^o - A\)
And we have: .\(\displaystyle SO = 13,\!800\)
. . We want \(\displaystyle \theta.\)

\(\displaystyle \text{Since }\angle SHO = 90^o\!:\;\;\sin A \:=\:\frac{4,\!000}{13,\!800} \:=\:0.289855072 \quad\Rightarrow\quad A \:\approx\:16.8^o\)


\(\displaystyle \text{Therefore: }\:\theta \:=\:90^o - 16.8^o \;=\;\boxed{73.2^o}\)