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

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

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

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