You’re standing on the shore watching a ship sail away. When it reaches the horizon it starts disappearing due to the curvature of the Earth. Assuming your eyes are $1.5\,\mathrm{m}$ above sea-level and the Earth is a sphere of radius $6,\!400\,\mathrm{km}$, how far away is the horizon (along the straight line between your eyes and the base of the ship)? If you were instead on a cliff, $50\,\mathrm{m}$ above sea-level, how far away would the horizon be then? If you don’t have a calculator, estimate the distance. Show hint

The line is tangent to the Earth, and therefore perpendicular to the radius.

It’ll be easier to understand if we draw a very small planet, such as Le Petit Prince lived on. When the ship is at position $1$, it’s completely visible. At $3$ you might be able to see the mast, but the base of the ship has disappeared below the horizon – the planet gets in the way of seeing it. Position $2$ is the critical point, beyond which the ship starts to disappear. So, what’s special about this point? If we draw a line of sight passing through position $2$, it will be tangent to the planet, and the important thing about tangents to circles is that they make a right angle with a line from the centre of the circle.

It’ll be neater to use letters for the distances, so let $r$ be the Earth’s radius, $d$ be the distance to the horizon, and $h$ be the height of the eyes above sea-level. \begin{align*} (r+h)^2 &= r^2 + d^2\\ d &= \sqrt{(r+h)^2 - r^2} \end{align*} If you have a calculator, you can put those values in now. Otherwise, it’ll help to expand and simplify: \begin{align*} d &= \sqrt{r^2 + 2rh + h^2 - r^2}\\ &= \sqrt{2rh + h^2} \end{align*} or use the difference of two squares to simplify: \begin{align*} d &= \sqrt{(r+h)^2 - r^2}\\ &= \sqrt{(r+h - r)(r+h + r)}\\ &= \sqrt{h(2r + h)} \end{align*} If $h = 1.5 = 3/2$ and $r = 6,\!400,\!000$ (remembering to convert it to metres), then $2rh + h^2$ is approximately $19$ million. Since $4000^2 = 16,\!000,\!000$ and $5000^2 = 25,\!000,\!000$, the distance is somewhere between $4$ and $5\,\mathrm{km}$. (Those with a calculator will say it’s about $4.382\,\mathrm{km}$).

In the tower case, $h = 50$, so $2rh + h^2$ is $640,\!002,\!500$. Noting that $25,\!000^2 = 625,\!000,\!000$ and $26,\!000^2 = 676,\!000,\!000$, the distance is somewhere between $25$ and $26\,\mathrm{km}$ ($25.298\,\mathrm{km}$ on the calculator). Or if you’re lazier, $20,\!000^2 = 400,\!000,\!000$ and $30,\!000^2 = 900,\!000,\!000$, so it’s somewhere between $20$ and $30\,\mathrm{km}$.

For most purposes, the $h^2$ term can be ignored because it’s much smaller than $2rh$, so a good rule of thumb for distance to the horizon is $d = \sqrt{2rh}$ or $d = 8\sqrt{h/5}$ if $d$ is measured in km and $h$ in m, since $r = \frac{8000^2}{10}\,\mathrm{m}$.