I'm trying to calculate how much the Space Station can see of the Earth from upthere using functions and calculus.
Here's the information needed:
Earth's radius: $r = 6400$
ISS altitude from Earth's surface: $a = 400$
$b = a + r$
I've plotted the upper surface of the Earth: $ y = \sqrt{r^2 - x^2}$
The horizon lines would be tangential to the Earth, therefore the slope would be equal to the derivative of Earth's surface.
$y = mx + b$, where $m = \frac{-x}{\sqrt{(r^2-x^2)}}$
combining both equations, I get that
$x = c = ±\frac{\sqrt{(r^2(a^2-r^2))}}{a}$
Then, using an integral, I calculated the arc length.
$s = \int_{0}^{c}{\sqrt{1+(\frac{-x}{\sqrt{r^2-x^2}})^2}}\text{ d}x$
$s = r \text{ arcsin(}c/r)$
Using $ r \theta = s$, we can get $\theta = \text{ arcsin(}c/r) $
and to find the percentage, I divided by $2 \pi$, which gives about 5%
The supposed answer is ~4%, so are there any error in my solution or is this just a problem from assuming stuff?
