36939

Assumptions:
Radius of the Earth - 6371km
Vertical height of Felix's jump - 39km
Radius of Earth is constant.

Solution:

Given that Felix could see the curvature of the Earth from his maximum displacement above the surface, two tangents can be drawn (length: a, name: A) from a vertex 39km above the Earth's surface. If one of these tangents is extended by length a parallel to vector A, a right angled triangle can be drawn from the center of the earth to Felix's maximum. Hence, by Pythagoras, (2a)^2 = 2(6371+39)^2, from which a can be resolved as 4533km (nearest km).

By basic trigonometry, Sin(Θ)=Opp/Adj, therefore Sin(Θ)=(6371+39)/(2a) (where Θ is the angle between the radius and A at the point of descent). When solved, Θ is found to be 45 degrees. Hence, 2Θ, the angle between Felix and the respective horizons in the northern and southern hemispheres, is 90 degrees.

Therefore, as angle A (for the sake of ease) is 90 degrees, and and the angles forming a kite with sides of length r,a,a,r, are also 90 degrees (tangent-radii angles = 90 degrees). Hence the angle of the sector, given by O, must be (360-(90*3)) 90 degrees.

This shows, as the angle of the sector is 90 degrees, approximately 1/4 of the Earth's surface was visible by Felix.

This can be proven as the length of the sector is given by rO (O in radians):
- O = 90 degrees = pi/2 radians
- Circumference of Earth = 2*pi*r
- Proportion visible = rO/2*pi*r (cancel r) = O/2pi
- O/2pi = (pi/2)/(2pi) = pi/4pi (cancel pi) = 1/4

SOLUTION: FELIX COULD SEE APPROXIMATELY 1/4 OF THE EARTHS SURFACE FROM HIS CAPSULE.