Greetings,



I would like to present some calculations which show that the Bible's 'pre-flood' firmament, which I believe was a 13 to 50ft layer of water, could have orbited 1720 miles above the earth.



I would also like to ask for your constructive criticism --- for your review.



Critics of the flood ask 'where did all the rainwater come from?'. They ridicule CRI's 'vapor canopy theory', and for good reason --- in order for the air to hold 40 days worth of rain, you would need a cloud 80 miles tall and this would have blocked out the light. Further, if the firmament were a cloud, then the situation would be no different than today. But we know it is different because now we have rainbows. There are good indications in the bible, that it did not would rain before the flood.



If the firmament was just vapor, the situation would be similar to what it is today, and it would rain. The reason has to do with temperature. Today, air near the earth's surface gets heated from the sun, it expands taking with it evaporated water, it rises, where it hits cold air, which condenses out the vapor as rain. If, however, the firmament was a warm layer of water above the earth, warm air would sink, and the vapor would condense on the earth as dew.



A 13ft thick layer of liquid water is a 'minimum' amount for the firmament's thickness, because if it rains only 4 inches per day (1/6 inch per hour - a lite drizzle) for 40 days/nites, you get:

4"/day * 40 day * 1ft/12inch = ~13ft



I am told, that the Hebrew word used for the Noah's flood 'deluge' indicated a heavy rain; so it's likely that it rained more than 4 inches per day.



50ft thick is like a maximum thickness because using Lambert's law, 60% of the

Green light (along with the rest of the visible light wavelengths)

will pass though 50ft of water (See calculations in Appendix 2). Now,

as Green light passes through the water into the air/space towards the

earth, it is refracted and it's wavelength changes, so it becomes red.



16"/day * 40 day * 1ft/12inch = ~52ft



There are satellites in a LEO (Low Earth Orbit), but these are

relatively close to the earth (400 to 1200 miles above it). They are

low in order to get a clearer view of the earth; however, they don't last

long because atmospheric drag slows them down. 1720 miles would

be out of reach of the present day atmosphere. It is possible that

before the flood, the atmosphere extended right up to the water layer.



We want to find out how high the firmament was orbitting in the sky:



Rf = radius from the earth's surface to the firmament



Now, there are 4 forces which affect the firmament's orbit around

earth.

1. Fge - earth's gravity

2. Fgs - sun's gravity

3. Fae - acceleration around the earth

4. Fas - acceleration around the sun



Fge + Fas = Fgs + Fae --- equation 1



The standard physics textbook 'Geostationary satellite orbit' simply

equates:

Fge = Fae



So, substituting this into equation 1, we get:



Fge + Fas = Fgs + Fge



and we are left with:



Fas = Fgs



When this is done, the orbit is like 20,000 to 35,000 miles in space

--- depending on the speed (and thus the period). However, what if,

the Firmament's orbit was designed so that the Earth's gravity was way

larger than the firmament's acceleration around the earth? -- that is:



Fge >> Fae



Well, consider equation 1:



Fge + Fas = Fgs + Fae



rewritting it:



Fge - Fae = Fgs - Fas



Since, Fge >> Fae, this reduces to:



-------------------------------

Fge = Fgs - Fas --- equation 2

-------------------------------



So, let's find Fgs and Fas:



----------------

Some statistics:

----------------



Te = period of the earth's rotation = 24 hrs * 60 minutes/hr * 60

seconds/minute = 86,400 seconds

Tf = period of the firmament's rotation around the earth = Te (this is

an assumption, however, even if Tf is shorter or longer, Rf is not

affected that much)

Ts = period of sun = 365 days * 24hrs/day * 60 minutes/hr * 60

seconds/minute = 3.15 * 10^7 seconds

G = Universal Gravitational constant = 6.67259 x 10 ^-11 Newtons* meter^2/Kg^2

Me = Mass of earth = 5.98 x 10^24 Kg

Ms = Mass of the sun = 2.0 x 10^30

Rs = Distance from the earth (and firmament) to the sun = 1.6 X 10^9 m

Re = The Radius of the earth = 3960 miles



----------------

Some Formula's:

----------------



1. Newton's law for Fgs and Fge

Fgs = G*Mf*Ms/Rs^2

Fge = G*Mf*Me/Rf^2



2. info from websites for Fas,

Fas = Mf * Rs * (2 * pi / Ts)^2



-----------------

Solve for Rf

using equation 2 from above

-----------------



Assume for now (we'll go back and prove it later), that: Fgs >> Fas



Fge = Fgs - Fas --- equation 2



G*Mf*Me/Rf^2 = G*Mf*Ms/Rs^2 - Mf * Rs * (2 * pi / Ts)^2



G*Me/Rf^2 = 6.67259 x 10^-11 * 2.0 x 10^30 /(1.6 X 10^9)^2- 1.6 X 10^9 * (2 * pi / 3.15 *



10^7)^2



MF* G*Me/Rf^2 = Mf*52 - Mf*0.0000635



Notice, that Mf*52 >> Mf*0.0000635

Notice, that Mf*52 is really Fgs, and Mf*0.0000635 is really Fas



So Fgs >> Fas, equation 2 can thus be reduced from:

Fge = Fgs - Fas

to:

Fge = Fgs



and substituting in Fge and Fgs, we get:



G*Mf*Me/Rf^2 = G*Mf*Ms/Rs^2



the G & 'Mf' cancel out, and we get:



Me/Rf^2 = Ms/Rs^2



Rf = sqrt[Me*Rs^2/Ms]



Rf = sqrt[5.98 x 10^24*(1.6 X 10^9)^2/2.0 x 10^30]

Rf = sqrt[7.68 X 10^12] = 2.765x 10^6 meters



Rf = 1720 miles



-----------------------

Solve for Mf, Fas, Fgs, and Fge

-----------------------



Now, Appendix 1 uses this value of Rf, and solves for Mf



Mf = 4.14 x 10^16 (see appendix 1)



Fas = Mf * Rs * (2 * pi / Ts)^2

Fas = 4.14 x 10^16 * 1.6 X 10^9 * (2 * pi / 3.15 * 10^7)^2

Fas = 2.63E+12



Notice: Fas/Mf = 2.63E+12/4.14 x 10^16 = 0.635 x 10^-4



Fgs = G*Mf*Ms/Rs^2

Fgs = 6.67259 x 10^-11 * 4.14 x 10^16 * 2.0 x 10^30 /(1.6 X 10^9)^2

Fgs = 2.16E+18



Notice: Fgs/Mf = 2.16E+18/4.14 x 10^16 = 52



Fge = G*Mf*Me/Rf^2

Fge = 6.67259 x 10^-11 * 4.14 x 10^16 * 5.98 x 10^24/(2.765 x 10^6)^2

Fge = 2.16E+18





Notice, that Fgs >> Fas



Double check this --- recall:

G*Mf*Ms/Rs^2 = 2.16E+18 --- equation 3



Rf = sqrt [G*Mf*Me/2.16E+18]

Rf = sqrt [6.67259 x 10^-11 * 4.14 x 10^16 * 5.98 x 10^24/2.16E+18]

Rf = sqrt [76.48 * 10^11] = 2.765E+6 meters = 1720 miles







Toby



Appendix 1 - finding the Mass of the firmament.



Assuming that the firmament was 1720 miles (2.769 x 10^6 meters) from

the earth's surface, and that it was 50 foot thick (15meter), let's

figure out what the mass of the firmament would be.



Rf+15 = radius from surface of the earth to outside part of firmament.



We want to take the bigger sphere (of radius Rf + 15) and subtract the

smaller sphere (of radius Rf)



sphere volume = 4/3 * PI * Radius^3



Mf = mass of firmament

Mf = [4/3 * PI * (Rf+15)^3 - 4/3 * PI * Rf^3] cuft * 64lb/cuft *0.45kg/lb

Mf = 120 * [(Rf+15)^3 -Rf^3 ]



Notice, the following Term:

(Rf+15)^3



Using Binomial Expansion, we get:

(Rf+15)^3

= Rf^3 + 3Rf^2*15 + 3*Rf*15^2 + 15^3

= Rf^3 + 45*Rf^2 + 675Rf + 3375



Now substitute this back into the equation:



Mf = 120 * [Rf^3 + 45*Rf^2 + 675Rf+ 3375 - Rf^3 ]

Mf = 120 * [45*Rf^2 + 675Rf+ 3375 ]



Recall, Rf = 2.769 x 10^6 meters



Mf = 120 * [45* (2.769 x 10^6)^2 + 675*(2.769 x 10^6) + 3375 ]



Mf = 4.14 x 10^16 kg//