Assume Earth is a perfectly smooth sphere. Suppose also that you are
standing on the surface, but have zero height. As you sweep your gaze
from, say, the East horizon to the Westward one, how many degrees is
that angle between opposing extremes? For zero height it comes to 180
degrees. The sky occupies 1/2 of your spherical field of view, Earth
the other 1/2.
Now let's put you on a rocket, which lifts you straight up, away
from Earth. As you gain altitude, Earth of course gets farther away.
Naturally the angular size of the planetary sphere gets smaller &
smaller, occupying less than 1/2 of your whole, spherical field of
view. In fact, as your distance from Earth grows toward infinity, its
angular size tends toward zero. At infinity the sweep between
diametrically opposed horizons will be zero.
Given that you have some height, h, above Earth's surface, you also
have a distance between your position & Earth's geometric center, R. R
- h is, of course, Earth's radius, r.
Through your position at height h and any point on the distant
horizon is one line. Since Earth in this hypothetical arrangement is a
sphere, the line between you & the horizon, t, is tangent to the
horizon, which also means that it's at a right angle to an Earth
radius intersecting the surface at the tangent point. You can?t see
sky behind the horizon, so to determine the angular sweep through
clear sky between extremes of the horizon amounts to 360 degrees minus
twice the angle between the two lines R & t, given height h above the
surface. The trigonometric sine of this angle is r/R. Since we know
this number from plugging values into the two variables, the angle
then is the inverse-sine, or arcsin, of the ratio r/R. Since R is
really just r + h, this becomes arcsine(r/[r + h]). The answer to your
overall question then is 360 ?2arcsin(r/[r + h])
So as an example, let?s say you?re 10 miles above Earth?s
surface, and Earth has a radius of 3,800 miles. This gives you a
visual sweep of clear sky of
360 ? 2arcsin(3,800/[3,800 + 10])
~ 188.3 degrees |
