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Jmartin
2003-Aug-17, 08:47 PM
My prior posting of 8/15/03 explained a possibly new method of determining the Hubble Constant, Ho, using the universal gravity constant, G. This posting uses the premise of the prior posting and new assumptions to yield possibly more accurate results. That premise states that the universal gravity constant of 6.67E-11 cubic meters / kg X sec^2 means that the universe is growing by 6.67E-11 cubic meters per second per second for each kg of mass in the universe.

If one assumes that a maximum velocity of recession of almost the speed of light has always existed in the universe and that the universe is 14 billion years old, the prior formulas can be easily used to yield a present Hubble constant of 70 km/s/Mpc and a mass of the universe of 1.12E+54 kg. If one assumes that that mass has remained almost constant since the universe was 1 billion years old, the formulas will show that the value of Ho has fallen from 981 km/s/Mpc and the value of G has risen from 4.76E-12 m^3/kg X s^2 during the last 13 billion years.

Jmartin
2003-Aug-18, 01:14 PM
My prior posting of 8/17/03 gave the mass of the universe as 1.12E54 assuming that a maximum velocity of recession of almost the speed of light has always existed in the universe. Without that assumption and using the original formula adapted to obtain mass (mass in kg = 8 pi X c^3 X the age of the universe in seconds / 3 G), one obtains a mass of 1.50E+54 kg.

Jmartin
2003-Aug-18, 05:34 PM
My original posting of 8/15/03 states that it might be possible to obtain Ho by using the formula of 2 G / 3 X age of the universe in seconds. However, that formula predicts a maximum velocity of recession of 2 / 3 c. In order predict a maximum velocity of recession of almost c, one can postulate that an additional factor, possibly a gravitational constant for cold dark matter exists. In addition, that constant, like the universal gravitational constant for ordinary matter would mean that the universe is growing by an additional 3.33E-11 cubic meters per second per second according to the mass of cold dark matter in the universe. For example if the mass of cold dark matter equals four times the mass of ordinary matter, then the numerical part of that constant will be 8.33E-12.

snowflakeuniverse
2003-Sep-03, 01:56 AM
Jmartin

O.K. Im intrigued. I like dimensional analysis and your interpretation of g to actually represent something dimensionally real, appeals to me. g does have the dimensional unites of volume/(kg x sec^2). This potentially means that dimensionally there is a possibility that there is a relationship which does yield something meaningful.

m/sec^2 means a linear acceleration m/ss = a
s = cumulative distance transversed while accelerated = 1/2 a t^2

volume/sec^2 would imply a volumetric acceleration? v/ss = g?
would this mean that total volume is 1/2g T^2 ?

Could you further explain The density of the universe figures in determining the Hubble constant. To determine that density, one does not need to know the mass of the universe since mass cancels out in the formula of 2 X mass / mass X age of universe^2 X G. The determined density is then multiplied by the volume of a sphere having a radius of one Mpc to yield the mass contained within that sphere. Consecutively multiplying that mass by G and the age of the universe yields the volume of expansion of the sphere per time. By dividing that volume by the surface area of the sphere, one obtains the Hubble Constant. However, cancellations allow the Hubble constant to be obtained by using the very simple formula of 2 X 3.09E22 meters / 3 X age in seconds. (Note that 3.09E22 = 1Mpc). 

Thanks

snowflake