Chapter 5, Problem 69RE

Solution

a.

To write: The height $h$ as a function of $\theta$ .

The height $h$ as the function of $\theta$ can be expressed as $h=4000\left(\sec \frac{\theta }{2}-1\right)\text{miles}$ .

Given information: Earth is spherical with a radius of $4000\text{mi}$ . The given figure is shown below.

  

Figure (1)

Calculation:

Draw a line that bisects angle $\theta$ and form two similar right triangles.

  

Figure (2)

It is given that Earth is spherical with a radius of $4000\text{mi}$ . So, $AC=4000\text{mi}$ . The angle $CAS$ is $\frac{\theta }{2}$ .

Calculate the length $AS$ .

  $AS=\left(4000+h\right)\text{mi}$

In $\Delta ACS$, apply the cosine triangle property.

  $\begin{array}{c}\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \left(\frac{\theta }{2}\right)=\frac{AC}{AS}\\ \cos \left(\frac{\theta }{2}\right)=\frac{4000}{4000+h}\\ 4000+h=\frac{4000}{\cos \frac{\theta }{2}}\end{array}$

Further simplify the above equation.

  $\begin{array}{l}h=\frac{4000}{\cos \frac{\theta }{2}}-4000\\ h=4000\cdot \sec \frac{\theta }{2}-4000\\ h=4000\left(\sec \frac{\theta }{2}-1\right)\end{array}$

Thus, the height $h$ as the function of $\theta$ can be expressed as $h=4000\left(\sec \frac{\theta }{2}-1\right)\text{miles}$ .

b.

To calculate: The approximate value of $\theta$ for a satellite $200$ miles above the surface of Earth.

The approximate value of $\theta$ for a satellite $200$ miles above the surface of Earth is $35.51°$ .

Given information: The satellite is $200$ miles above the surface of Earth.

Calculation:

From part (a), the function for the height $h$ as the function of $\theta$ is $h=4000\left(\sec \frac{\theta }{2}-1\right)\text{miles}$ .

Substitute $200$ for $h$ in the above function.

  $\begin{array}{c}200=4000\left(\sec \frac{\theta }{2}-1\right)\\ \frac{200}{4000}=\sec \frac{\theta }{2}-1\\ 0.05+1=\sec \frac{\theta }{2}\\ 1.05=\sec \frac{\theta }{2}\end{array}$

Further express the function as cosine to find the value of $\theta$ .

  $\begin{array}{c}1.05=\frac{1}{\cos \frac{\theta }{2}}\\ \cos \frac{\theta }{2}=\frac{1}{1.05}\\ \frac{\theta }{2}={\cos }^{-1}\left(\frac{1}{1.05}\right)\\ \theta =2\left(17.75°\right)\end{array}$

Further simplify.

  $\theta =35.51°$

Thus, the approximate value of $\theta$ for a satellite $200$ miles above the surface of Earth is $35.51°$ .