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.

(In reply to re: discussion by Ady TZIDON)

a) right, I didn't explicitly state that r=rho, just that r was the radius of the earth, and that the web site to which I referred used rho as the radius of the sphere.

b) The only places where I put more than two places after the decimal was in the relative area/area of earth, where I guess I just went overboard in making up the USING clause of the PRINT USING, and in ".249 km, or .1509 miles or 797 feet" as the height from which to observe a curved horizon, under the assumptions given in the problem, where I didn't go into an analysis of what the errors might be, but did restrict to about 3 significant figures (4 for the one that begins with digit 1).

As an aside: it's interesting that the radius of the earth is least at the poles and greatest at the equator, but that makes the radius of curvature greatest at the poles and least at the equator.

Posted by Charlie on 2006-01-30 09:25:45