Okay. Let's just do some very
rough calculations just for the fun of it. I'm excited to hear about your response.
Let's assume the earth is a perfect sphere with a radius of 6378 km (equatorial radius). The volume of water on earth not already in the oceans, which we shall call Vw, is about 1.606 * 10^7 mi^3 , which equals 6.694 * 10^16 m^3 (we're using SI units here).
That's all we need, really. We know how to calculate the volume of a sphere from its radius, which is what we'll be doing now:
After some easy calculations, we arrive at the following cubic equation:
Wolfram Alpha wouldn't be good for anything if not calculating the solution. If you want to try it yourself, here's the input:
w^3 + 3q w^2 + 3 w q^2 - (3 V)/(4 pi) = 0, q = 6378*10^3, V = 6.694*10^16
w is what we're looking for (rW), q is rE and V is Vw.Wolfram Alpha
gives us, by virtue of mathematics, the real solution of w being about 131. We ommitted dimensions from our input values, but they're all in meters. The calculation seems to be correct, but what does it tell us?
Well ... assume we'd take all the water currently not in the oceans already, extract it from the air, melt the ice etc. and dump it in the ocean--how much would the water levels rise? The answer is: About 131 meters.
Nice global flood, isn't it?
(More on radioisotope dating later)
(About water quantities, #7). These figures look okay to me, Encarta's value for water content in oceans is slightly higher, but the order of magnitude seems to be right: http://encarta.msn.com/media_461547746/The_World's_Oceans_and_Seas.html