Assuming that the earth is a perfect sphere, in units of the earth's radius, how high must one be to see exactly
one half of the earth's surface?
Okay... okay... how about exactly one third of the earth's surface?
If one must ignore the effect of gravitation and atmospheric refraction of light, one can still see 1/2 the surface area if one may assume that the point one may observe the Earth from a height above its surface is heliocentric and that the Earth is rotatating upon its axis as it revolves around the sun. One only need be a point above the equator such that he sees the surface area of the sphere between the two spherical caps (actic- and antarctic-wise) that would have a height equal to half the Earth's radius.
As the Earth completes one rotation in reference to the observer, 1/2 of the Earth's surface would be observed. (As the Earth is given as a perfect sphere, one should not consider the effect of the Chandler wobble as it would be absent.) This height would be (SQRT(3) - 1)*R (approximately 0.73205*R), such that R is the radius of the Earth.
For observation of 1/3 of the Earth's surface, a much smaller height of (3/(2*SQRT(2)) - 1)*R approximately 0.06066*R) above the Earth's surface would need be maintained.
Edited on September 26, 2008, 1:53 pm
Posted by Dej Mar
on 2008-09-26 13:52:14