In Thunderf00t's Why do people laugh at creationists? (part 33), in an attempt to debunk NephilimFree's crazy hypothesis about how water from Noah's flood caused craters on the moon; Thunderf00t used rudimentary physics to calculate the maximum possible height of the water if NephilimFree's scenario were even remotely possible.

It occurred to me that it's really a fluid dynamics question, and not simply a question of mass falling from a height, as Thunderf00t attempted it, and so I thought it would be interesting to put it to the test and see how accurate Thunderf00t's method is using equations of fluid dynamics.

NephilimFree's basic claim is that water existed 6 miles (~10km) beneath the Earth's crust, surrounding the entire earth. He claims the force exerted by the entire crust on that much water was enough for water to be ejected after a worldwide earthquake, out through geysers with enough velocity to escape Earth's gravity and hit the moon, leaving the craters we see today. I believe his claims are based on what is known among creationists as the "hydroplate theory".

The scenario, as set up in Thunderf00t's video, based on NephilimFree's claims, is that a mass of rock sits on top of a reservoir of water 10km below the surface. The average density of the earth's crust is 3.3 g/mL. (This is illustrated from about 6:10 through the video).

Thunderf00t made two mistakes in his calculations. First, he rounded up and said that the rock was about 4 times more dense than the water, even though he gave the densities as 3.3 g/mL and 1 g/mL, respectively. Performing his same calculations using the more correct values actually results in a height of 33 km instead of 40 km.

His second mistake was that this height is calculated relative to the top of the water, 10 km below the surface. However, he forgot to take this into account when he concluded that the water would reach a maximum altitude of 40 km. Taking this into account, and using the more correct value of 33 km, gives an altitude of 23 km above the surface. (Altitude above sea level depends on where this hypothetical geyser is located and its elevation).

In the following equations, this is what the following symbols mean: A = Area, V = Volume, ρ = Density (Greek letter rho), m = Mass, g = Gravity

To keep things simple, we can calculate the water pressure per square meter, under the weight of 10km of rock. The volume of this unit cross section of rock is thus:

A = 1 m²
h = 10 km
  = 10,000 m

V = Ah
  = 1 m² · 10,000 m
  = 10,000 m³

The average density of the earth's crust:

ρ = 3.3 g/mL
  = 3,300 kg/m³

m = Vρ
  = 10,000 m³ · 3,300 kg/m³
  = 33,000,000 kg

This means that there would be a mass of 33 kilotonnes per square metre pushing down on the water. Using Newtonian physics to calculate the force per m², and thus the water pressure.

g = 9.8 m/s² (acceleration)

F = mg
  = 33,000,000 kg/m² · 9.8 m/s²
  = 323,400,000 (kg·m/s²)/m² = N/m² = Pa
  = 323,400 kPa

To put this into perspective, atmospheric pressure is just over 101 kPa; a car tyre is about 220 kPa; a bicycle tyre is about 400 kPa. Now that we have the water pressure, I believe we should be able to calculate the height of a geyser that could be produced from this.

Using Bernoulli's principle, we can find the velocity of the water exiting the hole. I only learned about this while trying to solve this problem, but I tried hard not to make any mistakes. Bernoulli's equation for this is:

½ρv₁²   ρgh₁   P₁ = ½ρv₂²   ρgh₂   P₂

Let v₁ = 0 m/s (initial velocity at the bottom of the shaft)
Let P₁ = 323,400,000 Pa
Let P₂ = 101,325 Pa (1 atmosphere, pressure at the earth's surface)
Let h₁ = 0
Let h₂ = 10,000 m
Let ρ  = 1,000 kg/m³ (approximate density of water)
Let g  = 9.8 m/s²

½ρv₂² = ½ρv₁²   ρgh₁   P₁ - ρgh₂ - P₂

v₂² = 2(½ρv₁²   ρgh₁   P₁ - ρgh₂ - P₂) / ρ

Since h₁ and v₁ are both 0, this simplifies to:

v₂² = 2(P₁ - ρgh₂ - P₂) / ρ


v₂² = 2 · (323,400,000 Pa - (1,000 kg/m³ · 9.8 m/s² · 10,000 m) - 101,325 Pa) / 1,000 kg/m³
    = 2 · (323,400,000 Pa - 98,000,000 Pa - 101,325 Pa) / 1,000 kg/m³
    = 225,298,675 / 500
    = 450,597.35
v₂  = 671.265 m/s

Earth escape velocity = 11.2 km/s
                      = 11,200 m/s

So, the water velocity is well below the escape velocity of earth, meaning that there is no chance that the water would make it to the moon, and, as expected, creationists are defeated by science. But we can determine how high the geyser will reach.

It should be noted that this velocity ignores the effects of any friction from the shaft as the water passes through. The diameter of the shaft, or nozzle size, was also not accounted for because it makes no difference in this case. Thuderf00t demonstrated this in a recent video. With a narrower opening, the flow rate in this case would be decreased, while the pressure and velocity would not change. For the velocity to increase with a narrower shaft, then the pressure at the top of the shaft would have to be reduced to keep the same flow rate.

Ignoring air resistance, mostly because I don't know how to apply the correct physics for that, we can calculate the maximum possible height that the geyser can reach. Using the equation:

v²  = u²   2as
2as = v² - u²
s   = (v² - u²) / 2a

Let u = 671.265 m/s (Initial velocity, was v in equations above)
Let a = -g = -9.8 m/s²
Let v = 0 (Velocity at highest point)

Since v is 0, we can remove it from the equation:

s = -u² / 2a
s = -u² / 2 · -9.8
  = -u² / -19.6
  = u² / 19.6

And since we know from before that that u² is 450,597.35, we can use that value directly.

s = 450,597.35 / 19.6
  = 22,989.66 m
  ≅ 75,423 feet

So, the water would thus reach a maximum altitude of nearly 23 km , which matches up pretty well with the corrected version of Thunderf00t's equations. In fact, if atmospheric pressure were 0, rather than 101 kPa, then these above equations would have worked out to be exactly 23 km. That's about twice the height of a commercial aeroplane flying at an altitude of about 11.5 km (38,000 feet).

If drag were to be taken into account for this (if anyone here knows the equations for that), we would find that maximum height to be significantly reduced.