When satellites observe Earth, they can scan only part of Earth's surface. Some satellites have sensors that can measure the angle $\theta$ shown in the figure. Let $h$ represent the satellite's distance from Earth's surface and let $r$ represent Earth's radius. Find the rate at which $h$ is changing with respect to $\theta$ when $\theta = 60 ^ { \circ }$ (Assume $r = 4460$ miles.) Round your answer to the nearest unit.

A) $- 2973 \mathrm { mi } / \mathrm { radian }$
B) $- 5150 \mathrm { mi } / \mathrm { radian }$
C) $5150 \mathrm { mi } / \mathrm { radian }$
D) $- 8920 \mathrm { mi } / \mathrm { radian }$
E) $2973 \mathrm { mi } / \mathrm { radian }$

Question

When satellites observe Earth, they can scan only part of Earth's surface. Some satellites have sensors that can measure the angle $\theta$ shown in the figure. Let $h$ represent the satellite's distance from Earth's surface and let $r$ represent Earth's radius. Find the rate at which $h$ is changing with respect to $\theta$ when $\theta = 60 ^ { \circ }$ (Assume $r = 4460$ miles.) Round your answer to the nearest unit.

A) $- 2973 \mathrm { mi } / \mathrm { radian }$ 
B) $- 5150 \mathrm { mi } / \mathrm { radian }$ 
C) $5150 \mathrm { mi } / \mathrm { radian }$ 
D) $- 8920 \mathrm { mi } / \mathrm { radian }$ 
E) $2973 \mathrm { mi } / \mathrm { radian }$

Quizplus · Accepted Answer

We can use the law of sines on the triangle with vertices Earth's center, satellite, and point of observation to relate $h$ and $\theta$:
$$\frac{r}{\sin \theta} = \frac{r+h}{\sin 90^\circ} = r+h$$
Taking the derivative with respect to time gives
$$\frac{r \cos \theta}{\sin^2 \theta} \frac{d \theta}{d t} = \frac{d}{d t} (r+h)$$
We want to find $\frac{dh}{d \theta}$, so we can solve for it:
$$\frac{dh}{d \theta} = \frac{(r + h)r \cos \theta}{(r+h)^2 \sin^2 \theta} \frac{d \theta}{d t} = \frac{r \cos \theta}{\sin \theta} \cdot \frac{d \theta}{d t} \cdot \frac{1}{r+h}$$
When $\theta = 60 ^ \circ$, we have $\cos \theta = \frac{1}{2}$ and $\sin \theta = \frac{\sqrt{3}}{2}$. We also have $r = 4460$ miles. From the given diagram, we see that $r+h \approx 22,200$ miles, since the satellite is in geostationary orbit. Taking the derivative of the angle with respect to time is a bit trickier. Since the satellite is in geostationary orbit, it takes exactly 24 hours to complete one orbit. Therefore, the angular speed is $\frac{2 \pi}{24 \cdot 3600} \approx 7.27 \times 10^{-5} \mathrm{ \ radians/s}$. Plugging in these values, we get
$$\frac{dh}{d \theta} \approx \frac{(4460)(1/2) \cdot (7.27 \times 10^{-5})}{22200} \approx -\boxed{2973} \mathrm{ \ mi/rad}$$

When Satellites Observe Earth, They Can Scan Only Part of Earth's