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Q: field of view from earth ( No Answer,   2 Comments )
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 Subject: field of view from earth Category: Science > Earth Sciences Asked by: eemfilbob-ga List Price: \$10.00 Posted: 03 Apr 2006 09:55 PDT Expires: 03 May 2006 09:55 PDT Question ID: 714977
 ```If I am on a flat suface on earth what is my field of view from horizon to horizon of the universe?```
 `How tall are you? It makes a difference.`
 ```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```