The equation for the distance to the visible horizon can be derived from a simple triangle. If R = radius of the earth, H = height of viewing and D = distance to visible horizon is:
R^2 + D^2 = (H+R)^2.
Solving for D:
D = sqrt(2RH+H^2)
R=6378 km or 6378000 m
A person, 1.6 m in height would be able to see 4518 m
If you were standing on the highest floor of Taipei 101 in Tiawan (438 m) then the visible horizon would be 74750 m or just under 75 kilometers.
Standing on top of Mount Everest (8844 m) would allow you to see 336.0 km.
Standing in the International Space Station (~360 km) would allow you to see 2173 km.
Standing on the moon (384,403 km) would allow you to see 390,700 km
As can be seen, the distance to the visible horizon becomes less valuable a measure when extreme altitudes are used.
Computing the distance seen along the surface (S) can be found by using
cos(S/R) = R/(R+H) or S = R * arcos(R/(R+H))
This gives the following distances along the surface:
Person (1.6 m) = 4.518 km
Taipei 101 (438 m) = 74.74 km
Mount Everest (8844 m) = 335.7 km
ISS (~360 km) = 2,094 km
Moon (384,403 km) = 9,914 km
Computing the total surface area (SA) visible can be determined by SA=2piRh where h is the height of the spherical segment (from mathworld.wolfram.com)
From similar triangles:
(h + H)/D = D/(R+H) or h = D^2/(R + H) – H
Substituting . . .
SA = 2piR(D2/(R + H) – H)
Using the total surface area of the earth as
SA = 4piR^2 = 5.112*10^8 sq km, then:
Person (1.6 m) = 64.12 sq km
Taipei 101 (438 m) = 17550 sq km
Mount Everest (8844 m) = 35300 sq km (0.07% of total)
ISS (~360 km) = 1.37 * 10^7 sq km (2.7% or total)
Moon (384,403 km) = 2.514*10^8 sq km (49.2% of total)
Edited on January 29, 2006, 4:20 pm
Edited on January 29, 2006, 4:24 pm
Posted by Leming
on 2006-01-29 16:12:33