Question

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

Answer 1

See explanation for a nice approximation, using altitude H km and visible horizon ( nautical ) distance D km :
Circumference = `2pi(1/2D^2/H+H)` 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, `angleBSC = 90^o`..

Now,

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

Also, `cos alpha =cos(D/R)=(SC)/(BC)=R/(R+H)`

As D/R is small, `cos (D/R) = 1-1/2(D/R)^2`, nearly

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

`R/(R+H)=(1+H/R)^(-1)=1-H/R+H^2/R^2`, nearly. So,
.
`1-H/R+H^2/R^2=1-1/2(D^2/R^2)`, giving

`R = H +1/2 D^2/H` km, nearly.

Circumference = `2pi(1/2D^2/H+H)` 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