RE: Physics of a Mesopotamian Flood--miracles

Glenn Morton (GRMorton@gnn.com)
Mon, 13 May 1996 21:20:35

Hi Dennis,

You wrote of Sodom and Gomorrah:

>I thought the area was volcanically active,hence a catastophy waiting to
>happen. Could Lot's wife have been miraculously hit by a big splatter of
>molten salt (partly because she didn't run for cover fast enough)?
>

There is some volcanism in the Jordan area from around 4000 years ago
(2000 B.C.) see friedrich Bender, _Geology of Jordan_, Berlin: Gebruder
Borntraeger, 1974 p. 108.

Due to the fact that I apparently am missing a set of photos and
illustrations with my copy of this book I am unable to exactly say where
those eruptions were. As near as I can tell, they were the same age as
the Jebel el Drouz basalts in Syria.

>
>Under Glenn's theory, wouldn't only Noah's fore knowledge of the flood
> need to be miraculous. (?)
>

That is all that would be miraculous. But frankly that is not what I
consider a good point. It was God's judgment.

>Question for Glenn: Why didn't God just have Noah & co. climb out of
> the Med. basin? Too steep? Too many animals? ??? That's a question
>for Dick too: Why didn't Noah just leave the flood plain, driving the
>animals before him?
>

The devastated area would have been much, much larger than merely the
basin. I did a little atmospheric physics in an attempt to estimate how
far out the rainfall would occur. My conclusion is that it would be very
likly to rain as far as 400-800 kilometers away from the Mediterranean. I
had never calculated this aspect before. Thanks for prodding me.

Math below.

glenn

Consider the basin:

<pre>
Basin
North new air coming out of the basin South
replacing old air
<---| |--->
<---| |--->
<---| |--->Air being
<---| |--->replaced
<---| |--->
<---| |--->
----------- ---------------- -
----------- ---------------- .
----------- ---------------- .
----------- --------------- 5 km
----------- ---------------- .
----------- ---------------- .
----------- ---------------- .
----------- ---------------- .
--------------------------------------------------------------- -

All of what I am doing can be found in R. M. Goody and James C. G. Walker,
_Atmospheres_ Prentice Hall, 1972. The scale height of the is 8.4 km.
This means that the pressure increases according to the rule

P(z) = P(0) e^-z/H

where P(0) is the pressure at the surface
P(z) is the pressure at elevation z
H is the scale height

to find the pressure at the bottom of the empty mediterranean we re
arrange the above equation and realizing that z is now defined as sealevel
with P(z)=1.013 x 10^6 dynes/cm^2

P(0)=P(z)/(e^-5/8.4)

P(0)= 1.013 x 10^6/.55

P(0)= 1.837 x 10^6 dynes/cm

To find the mass of the air in the empty Mediterranean basin we can use
the pressure to find the mass of the air above a square centimeter. We
merely divide by gravitational acceleration. (Goody and Walker p. 12)

Mass/cm^2 = 1.837 x 10^6/ 981 gm/sec^2= 1872 grams

Goody and Walker were kind enough to calculate the atmospheric mass above
sealevel. It is 1032 grams. Subtracting we have an extra 840 grams/cm^2
over the entire basin which must be accomodated. The Mediterranean has
2.5 million cubic kilometer area. The mass of air in the basin is

840 g/cm^2 x 2.5 x 10^16 cm^2= 2.1 x 10^19 grams.

This is fascinating.

The total mass of the atmosphere is 5.29 x 10^21 grams (Goody and Walker
p. 12)

This extra air represents 4 thousandths of the atmosphere! This means that
the air below sealevel in the basin will eventually cover .004 of the
surface area of the earth.
The surface area of the earth is 5.1 x 10^18 centimeters

Thus

5.1 x 10^18 * .004 = 2.02 x 10^16 centimeters

Remembering that as air rises the moisture condenses to form rain. The
rain will extend at least as far as the replacement zone probably much
further because a lot of air is being moved and disturbed.

The area in which direct replacement occurs is

2.5 x 10^16 + 2.02 x 10^16 =4.52 x 10^16 cm^2

Representing the Mediterranean by a rectangle 4150 x 600 km we can
superscribe another rectangle of similar ratio which equals the total
surface area of the replacement zone. If I did my math correctly this is
a 400 km band around the Mediterranean where there would be a high
likelihood of devastating rain during the fill-up of the basin.

If the area was twice as large, it would require an 800 km hike for Noah
and co.

Foundation,Fall and Flood
http://members.gnn.com/GRMorton/dmd.htm