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