# How is the circumference of the earth calculated?

Nov 16, 2016

From visible horizon.

#### Explanation:

From visible horizon:

Let h be the height of the flag mast of a ship, from sea level. If d is

the maximum sea-level distance from which the top could be seen.

the circumference of the Earth is nearly

$2 \pi \left({d}^{2} / h + h\right)$.

Explication:

If h = 20 meters = 0.02 km and d = 11288 meters,

$= {d}^{2} / h + h =$

$= {\left(11.288\right)}^{2} / 0.02 + 0.02$

$= 6370.967$ km

The circumference

=$2 \pi$ x (radius)

$= 40029.966$ km, nearly

Proof:

d=(radius)(angle subtended at the center, in radian)=$R \alpha$

Here the proportion $\frac{\alpha}{R}$ is very small.

If T is the Top of flag mast, O is the observer, C is the center of the

Earth,

$\cos \alpha = 1 - {\alpha}^{2} / 2 = 1 - {\left(\frac{d}{R}\right)}^{2}$, nearly

$= \frac{C O}{C T} = \frac{R}{R + h} = {\left(1 + \frac{h}{R}\right)}^{- 1} = 1 - \frac{h}{R} + {\left(\frac{h}{R}\right)}^{2}$, nearly,

as h/R is very small.

So, the approximation formula is

${d}^{2} / {R}^{2} = h \frac{R - h}{R} ^ 2 \to R = {d}^{2} / h + h$, nearly.