For an observer at height h above the surface of the Earth,

i) What area, A, of the Earth's surface is visible?
ii) At what altitude, h, does the curvature of the Earth become apparent?

You may assume one can detect, with the human eye, an angle of one degree between the two ends of a line (i.e. two tangents at either end of the visible horizon, appear to intersect at an angle of one degree), that the average human field of view is 180°, also that the Earth is a sphere of radius 6378 km, or you may provide your own figures for the calculations.

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)

Inputing values:

  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