# How can altitudes be used to calculate the circumference of the earth?

Jan 23, 2017

See explanation for a nice approximation, using altitude H km and visible horizon ( nautical ) distance D km :
Circumference = $2 \pi \left(\frac{1}{2} {D}^{2} / H + H\right)$ km

#### Explanation:

Let B be the beacon from the top of a Light House (LH), at a

height H meters, from sea level.

Let D be the nautical distance traveled by a ship S,

sailing away from the LH, and

$\alpha$ radian = the angle subtended by the arc of length D, at the

center C of [the Earth](https://socratic.org/astronomy/our-solar-

system/the-earth).

When the beacon B just disappears beneath the horizon, from

the sea-level Telescope of the ship, then

BS touches the Earth at S, and so, $\angle B S C = {90}^{o}$..

Now,

D = R $\alpha$ km, where R km is the radius of the Earth.

Also, $\cos \alpha = \cos \left(\frac{D}{R}\right) = \frac{S C}{B C} = \frac{R}{R + H}$

As D/R is small, $\cos \left(\frac{D}{R}\right) = 1 - \frac{1}{2} {\left(\frac{D}{R}\right)}^{2}$, nearly

Now, H/R is small, and so, the RHS

$\frac{R}{R + H} = {\left(1 + \frac{H}{R}\right)}^{- 1} = 1 - \frac{H}{R} + {H}^{2} / {R}^{2}$, nearly. So,
.
$1 - \frac{H}{R} + {H}^{2} / {R}^{2} = 1 - \frac{1}{2} \left({D}^{2} / {R}^{2}\right)$, giving

$R = H + \frac{1}{2} {D}^{2} / H$ km, nearly.

Circumference = $2 \pi \left(\frac{1}{2} {D}^{2} / H + H\right)$ km, nearly

For sample data H = 100 meters = 0.1 km, D = 11.3 km.

circumference = 40, 116 km, nearly.

Here, the assumed visible-horizon ( nautical ) distance D = 11.3 km,

against the altitude 100 meters,

For this formula and R = 6371 km, the visible-horizon (nautical )

distance, for the Statue

of Liberty of height H = 41 meters is

D = 22.9 km, nearly