Hello everyone my name is Glenn, this is my first post and I apologize in advance for this being somewhat lengthy. I run a local church group in Front Royal Virginia and has been active in speaking and teaching for nearly 20 years. Normally the topics that I speak on has to do with basic Christianity and theological arguments. But for the past few years I've wanted to put together something substantial about Noah's flood.
I don't hold a degree in any of the sciences that would surround this topic so as you can imagine this has been slow going. What I have been studying is the Hydro plate theory by Dr. Walt Brown. I think he is a very intelligent man and puts forth a very good argument that is well-founded in the Scriptures; this is the area that I know, so when someone brings forth something that matches up to God's word I take notice.
While I was studying his theory I came across one area that might be problematic with his theory which is pointed out in the quotation which is attached that was established by someone else. This has to do with the assumed returning material which could potentially overheat the atmosphere and the oceans. If you're familiar with Dr. Brown's theory he talks about a large amount of material that was launched into space. http://www.creations...calNotes16.html
For the record I do not have a problem with the water that escaped from inside of the earth as being too hot. I feel that Dr. Brown covers this sufficiently here:
The biggest problem that I see with the argument that was advanced was the assumptions that were made but I'm hoping that some of the people on this forum can see things and raise arguments that I am not capable of.
One last note is that I have been planning to speak to Dr. Brown directly about this but have not yet found the opportunity, but if and when the opportunity arises any answers that I get from him I will definitely add to this post.
Thank you in advance.
I did an analysis of only one part of Brown's model ... the launch of all the meteroids and comets in the solar system and found if his model were 90% efficient at getting the material out of Earth's gravity, the return of the remaining 10% material would contain enough energy to boil every drop of water on the planet. In short, a flood would have been the least of Noah's problems.
Surface area of Earth 5.14e14 square meters
Mass of water on Earth 1.36e21 kg
Mass of material sent to space according to Brown 2.74e21 kg - http://www.creations...calNotes16.html
Specific heat of ice 2030 j/kg
Specific heat of water 4184 j/kg
Specific heat of steam 1860 j/kg
Latent heat of melting/freezing 334,000 j/kg
Latent heat of condensation/evaporation 2.5e6 j/kg
Velocity of material that did not escape Earth's gravity (average) 5.6 km/sec
Boltzmann constant = 5.67e-8
The energy of re-entry will boil all the water on Earth 10 times.....
Amount of energy to raise all the water on Earth from freezing and boil it.....
Multiply the mass of the oceans times the specific heat of water times the number of degrees (C or K) temperature change. That would be 1.36e21 * 4184 *
100 = 5.39e26 joules to bring all the water on Earth from freezing to boiling. Then multiply the mass of the water times the latent heat of evaporation. This is 1.36e21 * 2.5e6 = 3.4e27 joules.
Adding the two gives us 3.97e27 joules of energy needed to bring all the water on Earth from freezing and boil it.
Amount of energy from material that did not reach escape velocity......
Kinetic energy = one half the mass times the velocity squared. Using Brown's model and an efficiency of 50% we multiply the mass of material sent to space times the velocity squared then divide by two. That gives us 2.74e21 * 5600 * 5600 / 2 = 4.29e28 joules of kinetic energy in material that did not reach escape velocity.
The number of times the energy can evaporate all the water on Earth.....
Divide the total amount of energy available by the amount of energy needed to boil all the water on Earth. This gives us 4.29e28 / 3.97e27 = 10.82 times all the water on Earth can be boiled by the energy of re-entry.
The amount of material that will provide enough heating to boil every drop of water on Earth would need to have a kinetic energy of 3.97e27 joules. With a launch velocity of 5.6 km/sec, the mass of material needed is: m = 3.97e27 * 2 / 5600 / 5600 = 2.53e20 kg will have enough kinetic energy to boil every drop of water on the planet.
Taking the total material launched in Brown's model (2.74e21 kg) and adding 2.53e20 kg that will need to return to boil all the water we find that the minimum total mass launched that will still boil all the water is 2.99e21kg. Calculating the efficiency by 2.74e21 / 2.99e21 = 91.54% efficiency will till boil all the water on the planet.
At least 75% of the returning material will be rock......
Brown says his model sends 2.74e21 kg of material out of Earth's gravity. At 50% efficiency, an equal amount will be sent to space only to return to the surface. He also says about half of the present water in the oceans is the result of his flood model. The mass of Earths surface water is 1.36e21 kg. This means his model uses 6.8e20 kg of water. While Brown says his model continues to send water out of underground storage for some time after the launch phase, that would only reduce the amount of water available to return to Earth. If we take all of the water Brown says ended up in the oceans and compare that to the 2.74e21 kg returning to Earth in a 50% efficient model, we get 6.8e20 / 2.74e21 = 24.82% of the material returning to Earth is water. The rest (75.18%) will be rock. As stated, according to Brown's model, a higher percentage would need to be rock since he doesn't send all his water to space.
If the material takes 40 days to fall, atmospheric temperatures will be over 7700F....
Using the Stefan-Boltzmann equation, we can calculate the average temperature needed to radiate the heat of reentry to space. The equation is w = kat^4 where w= the watts; k= the Boltzmann constant (5.67e-8); a= the radiating surface area and t = the temperature in Kelvin. The energy of the returning material is 4.29e28 joules and 40 days equals 3.46e6 seconds. The number of watts equal the number of joules divided by the number of seconds, so we get 1.24e22 watts. Solving for t, we get t= (w/ka)^.25 or t = (1.24e22 / 5.67e-8 / 5.14e14)^.25 = 4544.58K. Converting 4544K to F, we find 7721F is the average temperature of the atmosphere to radiate that heat to space in 40 days.
If the material takes 150 days to fall, atmospheric temperatures will be over 6000F.....
All we need do for this one is change the time period from 40 days to 150 days. Doing this, we find the number of watts changes from 1.24e22 to 3.32e21.
Using the same equation, we get a temperature of 3266K or about 5400F.
If falling material cools to absolute zero, friction with the atmosphere will still heat it to over 10,000F.....
In order to be as favorable as possible to Brown, we will consider water which requires much more energy to heat than granite. At 5.6 km/sec, each kilogram of falling material will have kinetic energy = 0.5 mass times velocity squared. This is then =.5 * 1 * 5600 *5600 = 1.57e7 joules. The energy to raise the temperature to freezing, melt it, raise it to boiling, and boil it can be calculated by multiplying the specific heat of ice by 273, adding the latent heat of melting, add the specific heat of water times 100 and add the latent heat of evaporation.
This is then (273 * 2030) + 334000 + (100 * 4184) + 2.5e6 = 3.81e6 joules to raise a kilogram of water from absolute zero and boil it.
We can now take the remaining 1.19e7 joules and heat the steam. Dividing 1.19e7 joules by the specific heat of steam (1860 j/kg), we find we can heat the water another 5936C. The final temperature of the water will be 6036C which converts to 10898F.