Register to reply

Pre-flood firmament was in orbit

by kjvonly
Tags: firmament, orbit, preflood
Share this thread:
kjvonly
#1
Sep5-04, 10:46 AM
P: 13
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//
Phys.Org News Partner Physics news on Phys.org
Vibrational motion of a single molecule measured in real time
Researchers demonstrate ultra low-field nuclear magnetic resonance using Earth's magnetic field
Bubbling down: Discovery suggests surprising uses for common bubbles
enigma
#2
Sep5-04, 11:48 AM
Emeritus
Sci Advisor
PF Gold
enigma's Avatar
P: 1,817
According to the Bible, there was a whole lot more rain than just 13 feet. Consider that everything was covered, including the mountains.

If all of that water did fall, where did it go?

If all of that water was in orbit (and water dissociates in orbit) how did it slow down enough to fall? Were talking delta-V's on the order of km/sec.

If you're going to resort to magic and superstition to explain your universe, kjvonly, keep the science out of it. There is so much evidence that the Bible's account of creation is false that creationist scientists (there wasn't anything else back then) who went out looking for proof in the mid 1800s discounted it as false. You have been lied to.
Nereid
#3
Sep5-04, 12:10 PM
Emeritus
Sci Advisor
PF Gold
P: 4,014
Welcome to Physics Forums kjvonly!

That's an amusing little set of calculations you posted, so I'm going to ask you to do some more, based on your assumptions:
1) what tidal effects would there be in the 15m water shell?
2) other than at the equator, write some equations giving the motion of the water, with specific reference to circular orbits
3) compare the mass of water in your hypothetical shell with the mass of water in the Earth's oceans, and that locked up in ice in Antartica and Greenland
4) how long after local sunset would a sphere '1720 miles above the Earth' remain sunlit? what would such a sphere look like to those on the ground, say 30 minutes after sunset on 21 March?
5) how would light from the Sun, Moon, and stars be altered by passage through a 15 m layer of water?
6) what would be the effect of a meteor hitting this water sphere?
7) by how much would this hypothetical sphere reduce the ground incidence of cosmic rays?
8) how would this sphere interact with the Earth's magnetic field and the solar wind? Specifically, how could aurora form?

tony873004
#4
Sep5-04, 01:13 PM
Sci Advisor
PF Gold
P: 1,542
Pre-flood firmament was in orbit

G*Mf*Me/Rf^2 = G*Mf*Ms/Rs^2
The R in this formula means Radius, or distance from Earth's center. So wouldn't the 1720 miles that you come up with using this formula mean 1720 miles from Earth's center, which is well below the surface, rather than 1720 miles of altitude?

If this water orbited as a miniature moon, it should have been clearly visible from the Earth as a bright satellite, and there should be pleanty of historical references to this moon. But it couldn't survive there anyway since 1720 miles is well within the Roche limit.

1/6 inch of rain per hour is more than a lite drizzle. We get flooding around here when rainfall exceeds 2 inches per day.
mathman
#5
Sep5-04, 07:06 PM
Sci Advisor
P: 6,057
There is, according to some scientists, a basis for the flood theory. Specifically, before around 5500 bc (I am not sure I got the date right), the Black Sea was a small fresh water lake, with the surface below ocean level. The Bosporus was closed. At about this time, the Bosporus opened up and water from the Mediterranean rushed in flooding settlements around the lake shore. Some people miraculously escaped, and the accounts, orally transmitted for several thousand years, ended up in various flood stories.
veryyoung
#6
Sep5-04, 11:48 PM
P: 6
the great flood was not so great. we now assume that the story of the great flood was derived from a story about a merchant stuck in the middle of the mediterranean sea with the animals he was to sell at market the next day. however the sea flooded a little and the merchant lost his bearings and so was lost at sea for forty days and forty nights. rhe writers of the bible saw the morales in this story and changed it to fit a more godly version including the impossible flooding of the entire earth.
kjvonly
#7
Sep7-04, 02:52 PM
P: 13
Quote Quote by enigma
According to the Bible, there was a whole lot more rain than just 13 feet. Consider that everything was covered, including the mountains.

If all of that water did fall, where did it go?

If all of that water was in orbit (and water dissociates in orbit) how did it slow down enough to fall? Were talking delta-V's on the order of km/sec.

If you're going to resort to magic and superstition to explain your universe, kjvonly, keep the science out of it. There is so much evidence that the Bible's account of creation is false that creationist scientists (there wasn't anything else back then) who went out looking for proof in the mid 1800s discounted it as false. You have been lied to.
Greetings enigma,

Your are a mean person.

The reason for my posting was to get constructive criticism. You did not address whether or not my calcuations were theoretically correct.

The flood waters did not just come from rain. They also came from 'the fountains of the deep'.

There is much archeological evidence, that mountains did not exist prior to the flood. For example, there is an archeological dig, in a mountain in Peru, of a pre-flood city which was actually a shipping port... The only reasonable explanation, is that the city, at one time had been much lower in elevation, but became much higher when the mountain was formed.

I have read, that if all the non-water, land mass were spread out equally, around the whole earth, so that there were no mountains or valleys, the whole earth would be covered with water approiximately 1 mile deep.

This shows, that there theoretically there was enough water to cover the whole earth.

Toby
kjvonly
#8
Sep7-04, 02:57 PM
P: 13
Quote Quote by Nereid
Welcome to Physics Forums kjvonly!
Thank you.

That's an amusing little set of calculations you posted, so I'm going to ask you to do some more, based on your assumptions:
?
I don't think I am able to answer all your questions. However, I will try to answer some of them ---if--- you honor me with first discussing whether or not my calculations were theoretically correct and if not, where is my mistake(s)?

Toby

Rest of quote:

1) what tidal effects would there be in the 15m water shell?
2) other than at the equator, write some equations giving the motion of the water, with specific reference to circular orbits
3) compare the mass of water in your hypothetical shell with the mass of water in the Earth's oceans, and that locked up in ice in Antartica and Greenland
4) how long after local sunset would a sphere '1720 miles above the Earth' remain sunlit? what would such a sphere look like to those on the ground, say 30 minutes after sunset on 21 March?
5) how would light from the Sun, Moon, and stars be altered by passage through a 15 m layer of water?
6) what would be the effect of a meteor hitting this water sphere?
7) by how much would this hypothetical sphere reduce the ground incidence of cosmic rays?
8) how would this sphere interact with the Earth's magnetic field and the solar wind? Specifically, how could aurora form?
kjvonly
#9
Sep7-04, 03:46 PM
P: 13
Quote Quote by tony873004
The R in this formula means Radius, or distance from Earth's center. So wouldn't the 1720 miles that you come up with using this formula mean 1720 miles from Earth's center, which is well below the surface, rather than 1720 miles of altitude?

If this water orbited as a miniature moon, it should have been clearly visible from the Earth as a bright satellite, and there should be pleanty of historical references to this moon. But it couldn't survive there anyway since 1720 miles is well within the Roche limit.

1/6 inch of rain per hour is more than a lite drizzle. We get flooding around here when rainfall exceeds 2 inches per day.
Thanks for the constructive criticism.

I agree with you, that I should have used a radius 'R' which should have included the earth's radius. That is:

Fge = G*Mf*Me/Rf^2

should have been:

Fge = G*Mf*Me/(Rf-Re)^2

where:
Re is the earth's radius = 3960 miles = 6.37 * 10^6

Hence,

Rf - Re = 1720miles
Rf = 1720 + Re = 5680miles

However, it is customary to specify the 'orbit' as the distance above the earth's surface (i.e. 1720 miles)... for instance, I have an article which speaks of a LEO (Low Earth Orbit) satellite as being 500 miles above the earth, not 4460 miles from the center of the earth.

In studying this topic, the articles, I looked at, were looking at orbits whose radius was way larger than the radius of the body they orbit around, hence, the radius of the body was negligible and the body was considered 'a point'. For instance, the moon is 400,000 miles from the earth, while the earth's radius is a pittance of 4000 miles. Geostationary satellites are roughly 25,000 miles in orbit which is 6 times that of the earth's radius.

However, when looking at the firmament orbitting 1720 miles above the earth (of radius 4000), it cannot be said that 1720 >> 4000

Hence, the earth cannot be treated as a 'point'.

I do agree that my use of Rf in the appendix (for calculating the mass of the water in the firmament) was wrong. In calculating the mass of the firmament, Mf, I should have used 1720 + Re = 5680miles = 9.14 * 10^6

But keep in mind, the '1720 miles' was calculated independent of Mf.

Let's try it:

I had written: Rf = 2.769 x 10^6 meters

I should have used:

Rf = 9.14 * 10^6

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

Mf = 120 * [45* (9.14 x 10^6)^2 + 675*(9.14 x 10^6) + 3375 ]

Mf = 45.1 x 10^16 kg

It was: Mf = 4.14 x 10^16 kg

BTW, this mass calculation was done for a firmament 15m thick --- which is 50ft thick, not 13ft thick. Also, it is at an orbit of 1720 miles above the surface of the earth. 50ft thick at 1720 miles would translate to something much thicker if all this mass were to be located on the surface of the earth, namely 100 ft thick as follows:

45.1 x 10^16 = ~120 * [3*T*Re^2]

45.1 x 10^16 = 120 * [3*T*(6.37 * 10^6)^2]
T = 30meters = 100 ft
Nereid
#10
Sep7-04, 04:23 PM
Emeritus
Sci Advisor
PF Gold
P: 4,014
Quote Quote by kjvonly
Greetings enigma,

Your are a mean person.

The reason for my posting was to get constructive criticism. You did not address whether or not my calcuations were theoretically correct.

The flood waters did not just come from rain. They also came from 'the fountains of the deep'.

There is much archeological evidence, that mountains did not exist prior to the flood.
Really? How about geological evidence? Or evidence from trees (e.g. those which are known to be >10,000 years old, and show no evidence of a 'flood', and which couldn't possibly grow except at their present altitudes)? Or archaelogical sites which show evidence of continuous human habitation?
For example, there is an archeological dig, in a mountain in Peru, of a pre-flood city which was actually a shipping port... The only reasonable explanation, is that the city, at one time had been much lower in elevation, but became much higher when the mountain was formed.
And which site is that?
I have read, that if all the non-water, land mass were spread out equally, around the whole earth, so that there were no mountains or valleys, the whole earth would be covered with water approiximately 1 mile deep.
Source please!
Nereid
#11
Sep7-04, 04:41 PM
Emeritus
Sci Advisor
PF Gold
P: 4,014
Quote Quote by kjvonly
I don't think I am able to answer all your questions. However, I will try to answer some of them ---if--- you honor me with first discussing whether or not my calculations were theoretically correct and if not, where is my mistake(s)?
enigma already pointed out a serious flaw in your calculations - if the water was 'in orbit' at ~>1500 miles above the surface of the Earth, how did it lose its delta v? To take his question one step further, if that mass of water moving at orbital speeds (wrt the surface of the Earth) 'fell' to Earth, and the kinetic energy were converted to heat (where else would it go?), don't you think it would sterilize all living things on the surface of the Earth?

I also hinted at a far more serious flaw - "other than at the equator, write some equations giving the motion of the water, with specific reference to circular orbits". From your first post
Tf = period of the firmament's rotation around the earth
so you clearly assume that all this water is in orbit and rotating about an axis (which?). Now all orbits around the Earth must have the Earth's centre as a focus - so any orbit other than at the equator will be at an angle to the equator, which means that a blob of water will pass through the equatorial band of water twice every orbit ... if you know of a set of orbits for a liquid hollow sphere that are stable (around the Earth), do let us know ... otherwise your idea contains a fatal flaw (if you don't understand this, please read a book on elementary celestial mechanics and then come back and we can have a proper discussion).
mapper
#12
Sep7-04, 04:58 PM
P: 120
If all of that water did fall, where did it go?
Back then the world was flat so all the water simply fell off the sides silly. :)
enigma
#13
Sep7-04, 05:29 PM
Emeritus
Sci Advisor
PF Gold
enigma's Avatar
P: 1,817
Quote Quote by kjvonly
Greetings enigma,

Your are a mean person.
My is a "mean" person? Why? For telling the truth? You want me to get really "mean" and start pointing out mistakes?

Creationism was discounted as a viable scientific solution over 150 years ago. That's the facts of the world. If you think otherwise, you are merely believing the lies you've been brainwashed with.

You did not address whether or not my calcuations were theoretically correct.
Sorry. You're right. I didn't address that. Let me correct my error:

Your calculations are not theoretically correct.

Your 'hypothesis', like every other creationist 'hypothesis', merely suspends those areas of physics which don't agree with it. Whether it's because of misunderstandings (like I'm sure yours are) or through deliberate deception (like most published creationist works), none of them fit all the data we know to exist.

The flood waters did not just come from rain. They also came from 'the fountains of the deep'.
Ah yes. "Fountains of the deep". Must have come welling up from those non-existant holes in the Earth...

There is much archeological evidence, that mountains did not exist prior to the flood.
That's funny. I'd think that if the archeological evidence was there, then the archeologists would be throwing up red flags.

HINT: They aren't.

For example, there is an archeological dig, in a mountain in Peru, of a pre-flood city which was actually a shipping port...
No kidding. What archeological dig was that?

I have read, that if all the non-water, land mass were spread out equally, around the whole earth, so that there were no mountains or valleys, the whole earth would be covered with water approiximately 1 mile deep.

This shows, that there theoretically there was enough water to cover the whole earth.
Do you have any conception of the amount of heat that would be released by re-arranging the surface of the Earth in such a short period of time? You wouldn't have oceans... you'd have a giant sauna.

Now, regarding your previous calculations:
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 acceleration of the Earth around the sun would have nothing to do with a stable orbit. The entire system has the same specific angular momentum about the sun, therefore it can be considered an inertial space.

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
An acceleration is a force? What happened to the mass? What exactly are you referring to here? (There is a reason I didn't spend the time refuting this earlier...)

When this is done, the orbit is like 20,000 to 35,000 miles in space
--- depending on the speed (and thus the period).
What? How? I see nothing which leads to that conclusion. Speed has nothing to do with the period. Welcome to the big leagues. Orbits aren't circular.

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
Force is not an acceleration. Therefore this statement is meaningless. Similar statements in subsequent <snipped> section

<snip>

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
The Earth doesn't take 24 hours to rotate 360 degrees. It takes ~23hrs 56 min. The additional 4 minutes are spent re-aligning your zero meridian with the sun due to the Earth's orbit.


<snip>

Ts = period of sun = 365 days * 24hrs/day * 60 minutes/hr * 60
seconds/minute = 3.15 * 10^7 seconds
Earth doesn't take 365 days to orbit the Sun. It's closer to 365.2425. You forgot leap year. Regardless, the Earth's orbit about the Sun affects an object in Earth orbit barely a lick.

<snip>

----------------
Some Formula's:
----------------

<snip>

2. info from websites for Fas,
Fas = Mf * Rs * (2 * pi / Ts)^2
I don't know where you got this, but any gravitational force is dependant on both masses. The acceleration would be nothing more than the force divided by the mass of the body you're looking at. If this is anything different than the Earth's acceleration, then the firmament won't stay in orbit for long, that's for sure. (Not to mention the fact that it's irrelevant, as I mentioned above).

Assume for now (we'll go back and prove it later), that: Fgs >> Fas
The two terms don't have the same units, therefore this statement is gibberish. A force has units of Newtons, or kg*m*s^-2. Accelerations have units of, well, acceleration m*s^-2. You can't simply equate the two.

Look... I'd love to go on with this, but I think I've already proven that your calculations aren't right. They don't even make sense.

It is impossible for a sphere to be hovering in orbit around a planet. The individual droplets would simply bump into each other every time they pass a node. It just wouldn't be stable. If you're considering a ring like Saturn's, it wouldn't be possible due to interactions with the Moon. This also doesn't consider the mechanism which causes the droplets to slow down ~3 km/sec so they would fall to Earth. It also doesn't consider that you can't have water in a stable state in a vacuum. It would dissociate into monatomic hydrogen (which would escape the Earth/Moon system) and monatomic oxygen (which would eventually decay and end up in the atmosphere).

Like I said above (and I wasn't being mean... unless being realistic and obeying the laws of the universe is "mean"), if you want to resort to magic or superstition to explain the world, go ahaid but leave the science out of it. Science has already analyzed ALL the data, and our current best bet of what was, is, and will happen is what you read in your standard biology, geology, paleontology, physics, etc. textbooks. They all agree. Creationism is patently false.
enigma
#14
Sep7-04, 05:46 PM
Emeritus
Sci Advisor
PF Gold
enigma's Avatar
P: 1,817
If I come off as harsh, Toby, I'm sorry.

It isn't you I'm angry with.

It's the intellectually dishonest luddites who perpetrate the belief that creationism has any basis in fact or reality that I'm mad at. I'm furious that they pressure school districts into actually debating whether it should be taught along with actual science in science classes. I'm furious that they are destroying any chance of hundreds of thousands of kids to actually have any chance of succeeding in sciences.

Unfortunately, you've already bought into it, so unseating the garbage may be difficult to do. If you actually took the time and read about the world (take a biology or geology or astronomy class), you'd see how rediculous creationism actually is.
mapper
#15
Sep7-04, 06:46 PM
P: 120
Also, just because most able minds can see that creationism is a load of BS does not mean that they also believe in no god. You can still believe in something greater then yourself, like many of us here.

From reading written text from a few thousand years ago can you honestly say to yourself that there is no room for error in either translation over the centuries or even just simple minded people trying to make an explanation of something they simply couldnt understand.

If you showed a caveman from a few thousand years ago a lighter from today you think when he goes back to his tribe to tell others he could tell them exactly how it worked or what it even was? It would sound like magic/paranormal to those that have never seen such a thing, even to him and he was witness to it? No?

Think about asteroids, meteors, comets or anything that isnt common in everyday experiences. The stories from the past are simply a way of trying to make sense of something they dont understand.

Religion had its moment in the sun. Back then we didnt have science as we know it today so it was the science of the past. Unfortunately if you wanted to do real science and say anything against what religion believes to be true it was heresy and you were burned at the stake for such.

Religion was just a set of rules placed with some scary stories so people could follow those rules. Look at the basic 10 commandments. What were the rules before those commandments? If we didnt live in a society without those basic 10 commandments wouldnt we be living in a terrible place? But I can tell you, we once were living like that. Someone with some great ideas seen a way to create law and order to the lands. Sick of chaos/rape/murder/thieves, dog eat dog if ya will. There was someone that laid out a good set of rules that would make a better way of life if everyone could participate.

Unfortunately over the centuries religion has corrupt. It was power, (Eventually may just be the demise of mankind.) and people did everything to maintain that power. What better way to deal with a power threat then to simply call someone a witch and have them hanged, burned or stoned for whoever said the earth revolves around the sun rather then the sun around the earth. If you cant see the literally hundreds of examples of power abuse throughout the centuries because of religion im very sorry for you. Open your mind if not for but a moment to the ideas.

The most terrible thing are these people that refuse to let go, becoming fanatics, blowing themselves up for their cause. They are getting smarter though, rather then blow themselves up they can send out their children to blow themselves up for them cause children are more easily brainwashed before the age of true reasoning and doubt.

I myself believe in something, but I don’t believe an ounce of crap ive read in any bible. There just may be something so outrageous, so great our simple minds cant grasp or understand. That doesn’t scare me, the people that think they know everything on how and why were here are the ones that scare me.
kjvonly
#16
Sep7-04, 07:08 PM
P: 13
Quote Quote by tony873004
The R in this formula means Radius, or distance from Earth's center. So wouldn't the 1720 miles that you come up with using this formula mean 1720 miles from Earth's center, which is well below the surface, rather than 1720 miles of altitude?

If this water orbited as a miniature moon, it should have been clearly visible from the Earth as a bright satellite, and there should be pleanty of historical references to this moon. But it couldn't survive there anyway since 1720 miles is well within the Roche limit.

1/6 inch of rain per hour is more than a lite drizzle. We get flooding around here when rainfall exceeds 2 inches per day.
Consider the following website:

http://scienceworld.wolfram.com/physics/RocheLimit.html

=================================
The Roche limit is the orbital distance at which a satellite with no tensile strength (a "liquid" satellite) will begin to be tidally torn apart by the body it is orbiting. A real satellite can pass well within its Roche limit before being torn apart. Consider a rigid body of mass M and radius R with a satellite with mass m and radius r orbiting at a distance d. The Roche limit is reached when a loose chunk of material. (of arbitrary mass, u ) is tidally attracted to M more than it is gravitationally attracted to m. This occurs when:

Ftidal = Fbinding

Plugging in,

2GMur/d^3 = Gmu/r^2

d = r * (2M/m)^(1/3)
=================================

That is, if the ‘liquid’ satellite is closer than 'd', it will be torn apart, and chunks of liquid, the size of ‘u’ will be pulled toward the earth.

I’m not exactly sure how we could apply this to our 15m thick, firmament situation. I guess the firmament could be considered as being composed of millions of small satellites each of radius r (= 15/2 = 7.5); in which case, the Roche limit would be as follows:

The mass of each firmament sphere, Mfs, would be:

Sphere Volume cubicmeter * 64lb/cuft * 1 cuft/0.0283 cubicmeter *0.45kg/lb

m = [4/3 * PI * 64 * 0.45/ 0.0283] Radius^3
m = [4260] 7.5 ^3 = 1.80 x10^6

d = r(2M/m)^(1/3)

d = 7.5 (2*5.98 x 10^24/1.8 x 10^6)^(1/3) = 13.9 x 10^6m = 8645miles

As you pointed out earlier, this is the distance from the center of the earth. This liquid spheroid satellite is only 4685 miles (8645-3960) above the earth’s surface.

In the case of the firmament, however, the firmament is NOT a speroid satellite of radius R. It's like a complete eggshell all around the earth, whose 'yolk' is the earth, and the 'white' is the atmosphere. Hence, since we are considering forces which would pull this arbitrary chunk of mass, u, apart; we must consider the other gravitational forces from the mass on each side of this so-called spheroid. That is, not only does the water in the ‘chunk’ itself attrack u; so too, the mass on each side attact it as well.

In other words, the ‘m’ becomes like ‘3m’, and thus the old ‘d’ becomes d’ as follows:

d’ = r(2M/3m)^(1/3)

d = 7.5 (2*5.98 x 10^24/ [3*1.8 x 10^6])^(1/3) = 9.65 x 10^6m = 6000 miles

6000 miles is not too far off from the 5680 miles I had calculated as being how far the firmament was from the center of the earth, i.e. 3960 (earth’s radius) + 1720 miles (firmament above the surface) = 5680.

I want to also point out, that the above Roche Limit was derived ignoring the affect of a 'third body', that is, 'the sun', Fgs upon the chunk: ‘u’.

Fgs = G*u*Ms/Rs^2

Here is the new diravation:

2GMur/d^3 = Gmu/r^2 + G*u*Ms/Rs^2
2Mr/d^3 = m/r^2 + Ms/Rs^2

d^3 = 2Mr/[ m/r^2 + Ms/Rs^2 ]
d = {2Mr/[ m/r^2 + Ms/Rs^2 ]}^(1/3)

d = {2*5.98 x 10^24*7.5/[ 1.80 x10^6/7.5^2 + 2.0 x 10^30/1.6 X 10^9^2 ]}^(1/3)
d = {9.0 x 10^25/[ 3.2 x10^4 + 7.8125 x10^11]}^(1/3)
d = {9.0 x 10^25/[ 3.2 x10^4 + 7.8125 x10^11]}^(1/3) = 48160meters = 30 miles
kjvonly
#17
Sep7-04, 07:38 PM
P: 13
Toby wrote:

For example, there is an archeological dig, in a mountain in Peru, of a pre-flood city which was actually a shipping port... The only reasonable explanation, is that the city, at one time had been much lower in elevation, but became much higher when the mountain was formed.
Nereid replied:
And which site is that?
I own a book written by David Fasold entitled ‘The Ark of Noah’. It's in Bolivia, not Peru. The Book has pictures. The following website references it:

http://www.biblesearchers.com/ancients/noah/noah5.shtml
------------------------------------------------------------
Fasold on the 300 day year - David Fasold gives evidence that the calendar data of Tiahuanaco, the ancient megalithic city which now sits high in the Andean highlands of Bolivia at about 17,000 feet, had a solar calendar engraved on the giant Gateway to the Gods which depicted a solar year of 291.2 days which included a ten-month solar year with the lunar cycles approximating a ten to one ratio. (Fasold, Ibid, p. 62-63)
------------------------------------------------------------
kjvonly
#18
Sep7-04, 07:55 PM
P: 13
Toby wrote:

Quote Quote by Nereid
I have read, that if all the non-water, land mass were spread out equally, around the whole earth, so that there were no mountains or valleys, the whole earth would be covered with water approiximately 1 mile deep.
Nereid wrote:
"Source please!"


First off, there are literally hundreds of websites which claim that 75% of the earth is covered with water. Here are 2 of them:

------------------------------------------------------
http://www.spa3.k12.sc.us/WebQuests/...est/earth.html

Earth: The Water Planet- Seventy percent of the Earth's surface is covered by water.

Although three-quarters of the earth is covered with water, 97.6 percent our water is salty and. 1.9 percent is frozen into the polar ice caps. This means that only about half a percent of our. planet’s water resources is fresh water.
www.nmapa.org/Policy/Water.pdf
------------------------------------

It seems obvious, at least to me, that if the remaining 25% of the earth, which is above sea level, was removed and spread out in the ocean, the whole earth would be under water.

But consider some facts


www.byhowell.com/HAIKU/FACTS.htm
--------------------------------
Ocean Facts
More than 70 percent of the Earth's surface is covered by water. The average depth of the ocean is 4 km.
www.byhowell.com/HAIKU/FACTS.htm
--------------------------------

That is, the average depth below sea level is 4km, i.e. 2.5 miles deep!

Now, in the United States, OREGON is one of the highest states. Consider:

--------------------------------
http://www.wssb.org/pnw/PNW%20Chapters/Chpter02.doc (MICROSOFT WORD)
Oregon’s average elevation above sea level is 3,300 feet ...
www.wssb.org/pnw/PNW Chapters/Chpter02.doc -
--------------------------------

This is roughly 2/3rds of a mile high.


Illinois is a typical state:

-------------------------------
http://www.1911encyclopedia.org/I/IL/ILLER.htm
The surface of lilinois is an inclined plane, whose general slope is toward the S. and S.\V. The average elevation above sea-level is about 600 ft.;
--------------------------------

This is roughly 1/9th of a mile.

I hope I made my point. Here is one last piece of evidence:


http://www.answersingenesis.org/Home...ok/flood12.asp
-----------------------------------
That is why the oceans are so deep, and why there are folded mountain ranges. Indeed, if the entire earth’s surface were leveled by smoothing out the topography of not only the land surface but also the rock surface on the ocean floor, the waters of the ocean would cover the earth’s surface to a depth of 2.7 kilometers (1.7 miles). We need to remember that about 70% of the earth’s surface is still covered by water. Quite clearly, then, the waters of Noah’s Flood are in today’s ocean basins.
-----------------------------------


Register to reply

Related Discussions
Orbit Transfer to Lower Apoapsis of a Mars Orbit Introductory Physics Homework 11
Historical accounts of a 'Great Flood' Social Sciences 24
Noah's flood, fact or fiction Introductory Physics Homework 44
I lost my calculator batteries in a flood.... Fun, Photos & Games 14