The horizon

You are standing on the beach, with your feet in the water, looking out at the ocean in the distance.

How far away from you is the horizon, that is, the farthest point your line of sight reaches on the surface of the water?

Given: radius of the Earth $R = 6371$ ~km.

I solved itMark it done Add to my listKeep it in your list

I like this problem4 likes

Liked by

Hints

Hint 1

Open this only if you want a small nudge before looking at the solutions.

Solutions

Reveal solutionsAre you sure? Give it a try first.

Solution by goldfinch · translated by Ancient Tree

Discussions0 useful votes

(NB: the figure below clearly has a scale problem, but it allows us to carry out the reasoning.)

Let BC be the height of the person looking at the horizon (more precisely, BC is the height between their eyes and the surface of the water), CH their line of sight, and AH the radius of the Earth. The line of sight is tangent to the Earth at point H, the horizon point beyond which the surface of the Earth is no longer visible to the observer at C because of the curvature of the Earth beyond H. The distance we want to determine is HC.

The triangle AHC is right-angled at H, so by the Pythagorean theorem,
$AC^2=AH^{2}+HC^{2}$ $(R+BC)^2=R^{2}+HC^{2}$ $R^{2}+BC^2+2\times R\times BC=R^{2}+HC^{2}$

We can neglect $BC^2$ compared with $2\times R\times BC$ and obtain the following expression for HC:
$HC=\sqrt{2 R\times BC}$

Numerical application: the value of BC is not given in the problem statement, but we can easily estimate its order of magnitude.

For a person who is $1m70$ tall standing in $10 cm$ of water, we have $BC=1.60 m$ .
Then $HC = 4.52 km$ .

For a person who is $1m90$ tall standing in $10 cm$ of water, we have $BC=1.80 m$ .
Then $HC = 4.79 km$ .

Report