1) As Felix stepped from the capsule at a height of 39km, approximately what proportion of the Earth's surface could he see?
-> First, we need to assume that the Earth is spherical so that each point on its surface is exactly the same distance away from it's centre. The radius of earth is 6,400,000 metres to 2 significant figure. The total surface area of a sphere can be found by using 4πr².
-> total surface area = 4πr²
-> total surface area = 4π*{(6.4*10^6)²}
-> total surface area = (1.64*10^14)π m³
That was the easy part. For the next part, please refer to figure 1. In Figure 1, The spherical cap shaded in lavender colour is the area that Felix (F) can see from 39,000m above the Earth's surface. From figure 1, we can deduce the following:
-> OX or OP (Radius of the Earth) = 6.4*10^6m (approx.)
-> OF (Distance between the centre of Earth and Felix) = 39,000m + 6.4*10^6m
= 6.439*10^6m
-> FL is a line that is tangent to the surface of earth at point X therefore, the angle FXO is a right angle therefore Triangle FXO is a right angle triangle.
->Angle FQX is a right angle therefore Triangle FXO is a right angle triangle.
Since we are interested in the curved surface area of the spherical cap, we can use the formula, A = 2Ï€rh, where h is the the height of the cap (PQ), r is the radius of the sphere (OX) and A is the curved surface area of the spherical cap. We don't know the height of the cap therefore to find it, we need to do some simple trigonometry.
Triangle FXO is a right angle triangle therefore FX can be found by using Pythagorean theorem.
-> a² = c² - b²
-> (FX)² = (OF)² - (OX)²
-> (FX)² = (6.439*10^6)² - (6.4*10^6)²
-> FX = 7.08*10^5m (approx.)
We also need the angle OFX. The need for this data will become apparent in a minute. We can find out angle OFX by using sine rule.
-> {Sin(OFX)}/(6.4*10^6) = {Sin(90)}/(6.439*10^6)
-> OFX = Sin^-1{(6.4*10^6)/(6.439*10^6)}
-> OFX = 83.69 degrees (approx.)
This information now allows us to find the distance QX which is the radius of the circle of the base of the spherical cap. To find QX, we use the Sine rule again.
-> QX/{Sin(OFX)} = FX/{Sin(90)}
-> QX = {Sin(OFX)}FX
-> QX = 7.03*10^5m (approx.)
The distance FQ can be found by using the Pythagorean theorem.
-> (FQ)² = (FX)²-(QX)²
-> FQ = 7.78*10^5m (approx.)
Since we know the distance between Felix and the surface of the Earth (FP) and we know the distance between Felix and the centre of the circle of the base of the spherical cap (FQ), we can now find out the height of the spherical cap (PQ) by doing the following:
-> PQ = FQ - FP
-> PQ = 3.88*10^5m (approx.)
We can now use the formula, A = 2Ï€rh to find out the curved surface area of the spherical cap.
-> A = 2Ï€*(6.4*10^6)*(3.88*10^5)
-> A = 4.96*10^11π m³ (approx.)
To find out how much this area is in terms of the total surface area of Earth, we can divide the curved surface area of the spherical cap by the total surface area of the Earth and multiply it by 100 to get the answer in percentage.
-> [{(4.96*10^11)Ï€}/{(1.64*10^14)Ï€]*100
-> 0.30% (approx.)
In conclusion, by my calculation (assuming that there is no mistake), Felix Baumgartner could only see approximately 0.3% of the Earth's total surface area from the height of 39km above Earth's surface.