Rise of Continents Due to Flooding
Alan Feuerbacher
Regarding the Flood, the Society has made the general claim that "under the added weight of the water, there was likely a great shifting in the crust. In time new mountains evidently were thrust upward, old mountains rose to new heights, shallow sea basins were deepened...." The Good News to Make You Happy book is a bit more specific, and says on page 82:^{247}
The tremendous pressures involved would bring about great changes on the surface of the earth, pushing up mountain ranges and forming depressions to receive the waters. Within the space of about one year these had receded into the oceans as we know them today.
It should be noted that under the weight of the water, which would produce a downward force but no horizontal force, there would be no reason to think that continents would be forced sideways. The only effect should be a tendency for the continents to rise. We shall now examine, using the physics of floating bodies, how the continents would actually rise under the influence of a sudden influx of enough water to flood them significantly deeper than their average height above sea level.
First we need some data to calculate the average density of continental crust. The average density p of both oceanic crust and the magma on which it sits is about 3.0 gm/cm^{3}, and the volume of continental crust is about 7 x 10^{9} km^{3}.^{248} The formula for the volume of a sphere is
V = 4/3 PI r^{3}.
The mean radius of the earth is 3982 miles, or 6,408 kilometers.
First let's calculate the average thickness of the continental crust using this data. Let the thickness of this crust in kilometers be T and the radius of the earth in kilometers be R. A spherical shell with the radius of the earth and the thickness of this crust has a volume of
V_{s} = 4/3 PI [ R^{3}  ( R  T )^{3}].
The continental area is approximately 29% of the earth's surface, so we have an approximate expression relating the volume of the crust, V_{c}, to the volume of the shell, V_{s}:
V_{c} = 0.29 V_{s}.
Using the above data for V_{c} = 7 x 10^{9} km^{3}, we solve and find that V_{s} = 2.4 x 10^{10} km^{3}. Then using this number for V_{s}, we solve for T in the above formula, and find T = 47 km. So the continental crust averages about 47 km thick.
Now according to the God's Word or Man's book,^{249} the average height of land above sea level is 0.84 km and the average depth of the oceans is 3.79 km. So the average height of the continents above the sea floor is
0.84 + 3.79 = 4.63 km.
Then the average depth of the continental crust below the ocean floor is
47.00  4.63 = 42.37 km.
To calculate the average density of the continental crust we use Archimedes' principle for a floating object, which essentially says that the weight of liquid displaced by the object is equal to the weight of the object. The continents are floating on magma, which is in turn covered by the oceans. The situation can be modelled in a simple manner by the following diagram, where I've modelled the continent as a rectangular box floating on a high density magma layer and surrounded by a lower density water layer. The exact dimensions of the box are unimportant, as they cancel out in the course of the calculations. Also note that I'm freely using the most convenient units for the various quantities without necessarily specifying what they are, as they also cancel out in the end.
p1 = 0
  
^   ^
    h1
   v
 water surface
   ^
p2     h2
   v
 seafloor surface
   ^
  pm M  
h     h3
  Area A  
v   v
  
p3
The magma in the lower layer has density p_{3} = 3.0 gm/cm^{3}, the water in the middle layer has density p_{2} = 1.0 gm/cm^{3}, and the upper layer is air with negligible density p_{1} = zero. The thickness of the box below the seafloor is h_{3}, the thickness in the water is h_{2}, and the height above the water is h_{1}. The total height of the box is h. We have already found these thicknesses above. The box has density pm, which is to be determined, mass M, and horizontal cross sectional area A.
An elementary principle of physics is that the pressure at a given depth in a fluid is equal to the acceleration due to gravity times the density of the fluid times the depth: P = gpd. The force on a floating object is zero in the horizontal direction, since all the forces due to pressure cancel out, but the upward force is equal to the pressure at the bottom of the object times the area of the bottom: F = PA. The net force on a floating object is zero, since it is in equilibrium, so the sum of all the individual forces acting on it must be zero. These include the force due to pressure and the force due to gravity (the weight of the object). The force due to gravity is the mass times the acceleration due to gravity:
F = Mg.
In equation form we have the net upward force
F_{net} = 0 = PA  Mg,
where the minus sign is used because the weight acts downward while the pressure acts upward. The pressure at the bottom of the box is due to the weight of the water and the weight of the magma. Putting all this in equation form we have the net upward force:
F_{net} = Agp_{2}h_{2} + Agp_{3}h_{3}  Mg.
Setting this equal to zero, and letting the area A be 1, we can solve for the mass M:
M = p_{2}h_{2} + p_{3}h_{3}
We already know the densities and the depths, so we find that
M = 1.0 x 4.0 + 3.0 x 42.37 = 131.1
Dividing by the volume (height of box h x area A) we get the average density of continental crust:
131.1 / 47 = 2.79 gm/cm^{3}.
Now we can proceed to find how much the box could rise if it became completely covered by water. The changed situation is shown in the following diagram.

^

d  p2
   
 ^   ^
     h2
v    v
seafloor surface
   ^
  pm M  
h     h3
  Area A  
v   v
  
p3

The symbols are as before, but now there is only the water and magma that surround the box. As before, the total upward force minus the total downward force is zero in equilibrium. There is an extra component of downward force due to the pressure of the water on top of the box.
This component is F = Agp_{2} x (d  h_{2}), since the top of the box is at depth d  h_{2}. The equation for net upward force becomes:
F_{net} = Agp_{2}d + Agp_{3}h_{3}  Agp_{2} x (d  h_{2})  Mg
Again setting area A equal to 1, using the facts that the mass of the box M is equal to its volume times its density (M = Ahpm) and h = h_{2} + h_{3}, and solving for h_{2} we find:
p_{3}  p_{m}
h_{2} = h 
p_{3}  p_{2}
All the numbers on the right have already been found, and plugging them into the equation we find the result is
h_{2} = 4.94 km.
Subtracting the original height above the sea floor, 4.63 km., we find the net increase in height:
4.94  4.63 = .31 km.
So the height could increase by 310 meters, or about 1000 feet.
Using reasonable assumptions I've shown that the Flood could have caused continents to rise by no more than about 1000 feet. But note that due to the response time of the earth's crust, which is measured in tens of thousands of years, as we've seen elsewhere, there would be only a tendency to rise. The flooding for one year could not have produced much rise at all, because of the slow response time. Since the water would have been distributed evenly over the earth, there is no location that should have received significantly more disturbance due to the water pressure than anyplace else. Therefore I must conclude that there was little tendency either for continents to rise significantly or for mountains to be pushed up in any particular place. If you disagree with this conclusion, then please show why, using quantifiable arguments.
Footnotes
^{247} Good News to Make You Happy, Watchtower Bible and Tract Society of New York, Inc., Brooklyn, New York, 1975.
^{248} H. W. Menard, Islands, p. 92, 193, Scientific American Books, Inc., New York, 1986.
^{249} The Bible  God's Word or Man's?, pp. 111112, Watchtower Bible and Tract Society of New York, Inc., Brooklyn, NY, 1989.
