How to calculate curvature of Earth per surface kilometer I was watching a video regarding Flat Earthers giving curvatures of Earth that sounded way too big, so I decided to calculate how much the surface of Earth would drop with respect to a line perpendicular to the ground per a certain distance from this point.

I'll use variables as defined in my awfully drawn diagram.
Let $s = 1km$.
Given that the radius of Earth is 6371km, we can find 
$\theta = \frac sr = \frac {1km}{6371 km} = \frac {1}{6371} rad = 1.56*10^{-4} rad$
I think h could found by
$h = 6371\sin(\frac{\pi}{2}) - 6371\sin(\frac{\pi}{2}-1.56*10^{-4}) = 7.84*10^{-5} km$ 
which converted to meters would result in a curvature of $0.07\frac{m}{km}$.
EDIT Note that the formula is not linear, so in order to get the "fall" of Earth for 2 kilometres, multiplying the result by 2 is not enough.
The formula is $h = 6371-6371*\sin(\frac{\pi}{2} - s/6371)$ where $s$ is the distance to the object in kilometers. 
Did I make any mistake? This value seems quite small and I'd like to make sure. Thanks in advanced.
 A: Your answer is correct. 
See, e.g. the approximate formula given in the Wikipedia entry for horizon, which lists $d \approx 3.57 \sqrt{h}$. We see that a horizon of 1 kilometer is approximately corresponding to a height of 
$$ (1 / 3.57)^2 = 0.0785 \text{ meters} $$

Incidentally, I would've used $\cos$ instead of $\sin$ when writing the formula. Then you get 
$$ h = r * (1 - \cos \theta) \approx r \cdot \frac12 \cdot (\frac1r)^2 $$
where $r$ is the radius of the earth, and $\theta = 1/r$ is the angle in radians. The approximation uses Taylor expansion for $\cos\theta \approx 1 - \frac12 \theta^2 + \ldots$. So you get immediately 
$$ h \approx 1/r $$
when $h$ and $r$ are given in the same units.  

Let me give finally a remark concerning the discrepancy with this comment. The TL;DR is basically that the quantity $h$ is not linear as a function of the distance. Since the approximation is done by approximating the circle by a parabola, we actually have that the approximate height scales like the square of the distance to the horizon. For a quadratic relationship 
$$ 0.0785 \text{ meters } * \frac{1.6092^2}{1^2} \approx 0.20 \text{ meters} \approx 8 \text{ inches } $$
we see that the 8 centimeter per kilometer estimate is actually compatible with the 8 inches per mile estimate. 
