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kjvonly
kjvonly is offline
#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moon, it should have been clearly visible from the Earth as a bright satellite, and there should be pleanty of historical references to this moon.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