Einstein felt that gravity should be treated as any other acceleration which is a change in velocity. The only stable acceleration not requiring a continuous energy supply would be a rotation. So
g = G m / r2 = vg2/ r where vg is the velocity associated with gravity and
( equation 1 ) vg2 = Gm / r
Rotation produces a centrifugal force ; whereas, gravity gives rise to a centripetal force. Gravity must then be a deceleration. That implies that the universe contains a rotational component which matter lags behind by an amount represented by vg, a loss of some universal spin.
Absolute Velocity Equation ( 2 )
Newton showed through F = ma + v dm/dt that the first part of that equation holds when mass is constant. Relativity explains that the second part is real suggesting that velocity is constant. I have read that Einstein had said that the absolute velocity of an object is constant but F= ma type motion is possible.
Absolute velocity is not a measurable property but we can be aware of some of the components. The earth spins on its axis, revolves around the sun, is a component of the Milky Way's rotation and revolving around the " Great Attractor ", and likely included in other celestial motions. The universe appears to be expanding at relativistic speeds.* And it also might be spinning if, perhaps, the primordial universal particle possessed spin or the “Big Bang" imparted spin as well as expansion. The lighter components of the universe would be moving faster while the denser ones would experience a larger lag related to vg.
This suggests that, while massless particles travel at the speed of light ( c ), objects with mass will travel at an absolute velocity equal to c vectorially minus universal velocity loss ( vg ). That is
( equation 2 ) c2 = vi2 + vg2
where the vi's are the individual velocities of the body. vg is linked to gravity and to motion.
Relation to Relativity
From Relativity: m = mo / ( 1  V2 / c2 )1/2 = mo / ( c2  V2 / c2 )1/2 = mo c / ( c2  V2 )1/2
equation ( 2 ) c2 = V2 + vg2 and vg2 = c2  V2
therefore m = mo c /( vg2 )1/2 so
equation ( 3 ) m vg = mo c and a similar calculation yields
equation ( 4 ) ro vg = r c
note that m r = ( mo c / vg ) ( ro vg / c ) or
equation ( 5 ) m r = mo ro
When v = v0, m and r are the values at the object’s absolute velocity.
Cosmic Rays
Cosmic rays are mostly atomic nuclei generated by celestial events. Their measured energy depends on their velocity relative to us. Those originating from the sun, sharing our solar system's celestial velocity, would show the lowest energy range. Those up to energy levels of 1014 ev or 1015 ev are believed to originate in our galaxy.
According to equation ( 2 ), the maximum velocity for any nuclei would be the square root of ( c2  vg2 ). That value is close to c and so would determine the maximum mass of the nucleus and its maximum energy.
Derivation of the equation for maximum mass
from equation ( 3 ) moc = mvg,0 = m ( Gm0 / r0 )1/2
so, mo2c2 = m2 ( Gm0/ r0 )
equation ( 6 ) m2 = mo ro c2 / G
Example calculation: maximum energy of a proton
using 1.67x 1024 g as mo and 3.2x 1013 cm as ro,
m2 = ( 1.67 x 1024g ) ( 3.2 x 1013 cm ) ( 3.00 x 1010 cm / sec )2 / 6.66 x 108 cm3 / g sec2
So, m = 9.0 x 105 g
The velocity is close enough to c to use that value in the calculation of energy.
E = m v2 = ( 9.0 x 105 g ) ( 3.00 x 1010 cm/sec )2 / 1.60 x 1012 erg/ev = 5.1 x 1028 ev The same calculation on other particles gives these projected energies:
particle  mo ( grams )  ro ( cm )  calculated m ( grams )  energy ( eV ) 
electron  9.1 x 1034  1.0 x 1016  3.5 x 1011  2.0 x 1022 
Fe56  9.37 x 1023  8.8 x 1013  9.95 x 1013  5.6 x 1020 
U235  3.90 x 1022  7.0 x 1011  2.74 x 1012  1.5 x 1021 
The maximum energies predicted are dependent on the value of ro selected ,with the heavier nuclei giving the larger energies.These values are far above observed cosmic rays values. These maximum values should also apply in particle accelerators although that may not be of practical concern as their current range is well below these values.
Rotation of Condensing Protostars
Gaseous clouds destined to become stars upon gravitational collapse have rotation which increases as their radius becomes smaller. An unresolved problem is that the Law of Conservation of Momentum predicts that the forming stars should be rotating faster than observed.
Equation ( 2 ) limits the amount of speed an object can locally acquire. The vg increase of a shrinking protostar would have to come from a reduction in some celestial motion. This same restriction would also apply to the increase in spinning speed. It appears that the spinning speed increases at the same rate as vg, so that:
vg2 = Gmo / r = ( v spin )2 and, ( v spin )2 r = Gmo = a constant
If there were enough matter, theoretically, this process could continue to form a black hole where
vg2 = 0.5c2 and spin would also be 0.5c2, according to equation (2.)
Example calculation: What would the spinning speed of a cloud of one solar mass and radius of 3.39 x 1013 km with a rotational velocity of 3.375 x 102 km/sec upon collapsing to the radius of the sun, 6.95 x 105 km
(v1 spin)2 r1 = ( v2 spin )2 r2
v22 = ( 3.375 x 102 km/sec )2 ( 3.39 x 1013 km ) / 6.95 x 105 km and v2 = 236 km/sec instead of the value exceeding c as predicted by the Law of Conservation of Momentum.
Solar System
The Solar System presents a similar, although more complex, situation as the stellar rotation case. It consists of objects sharing the same celestial velocity. The planets, rather than collapsing along with other debris to form the sun, collapse independently to acquire their own spin while revolving around the sun at the same time. The latter two velocities are lumped together here; rotation being viewed as a special case of revolution where the revolving distance is zero. This means that v2g should be equal to the sum of v2spin + v2revolving. The chart below shows the results for the planets:
Rotation  Revolution 
( v2s + v2r)1/ 2 cm/sec 
(Gm/r)1/2  
Mercury  3.01x102  4.79x106  4.79x106  3.00x105 
Venus  1.80x102  3.50x106  3.50x106  7.30x105 
Earth  4.65x104  2.98x106  3.01x106  7.90x105 
Mars  2.41x104  2.41x106  2.42x106  3.55x105 
Jupiter  1.26x106  1.31x106  1.82x106  4.20x106 
Saturn  1.00x106  9.64x105  9.72x105  2.50x106 
Uranus  4.06x105  6.80x105  4.75x104  
Neptune  2.68x105  5.43x105  6.10x105  16.4x105 
Pluto  2.85x102  4.74x105  4.74x105  5.33x104 
Moon  1.02x105  1.02x105  1.68x105  
Sun  2x105 
~2.20x107* 
4.355x107 
*Galactic Velocity cm/sec
Other than Mercury, the agreement was within a factor of ten for the planets. Uranus’ horizontal axis spin does not allow addition of its two velocities. Inclusion of a planet’s moons should probably be included in the calculation for a more precise result. The sun’s high temperature should also be included as an energyvelocity equivalent.
Beta Pictoris b was recently found to be the fastest rotating planet known to date. Its 56,000 mph is equal to 2.5 x 106 cm / sec.
Taking the orbit to be 9.0AU and its period to be 20.5 years means its orbiting velocity is 1.3 x 107 cm / sec. Combining the velocities vectorially would result in an overall velocity of 1.3 x 107 cm / sec.
If this total velocity squared is expressed as G x mass / radius , then using the mass as being 3J and the radius 1.65 rJ gives:
v2 = ( 6.67 x 108 cm3 / g sec2 ) ( 1.33 x 1031g ) / 1.18 x 1010 cm
and v = 2.7 x 107 cm / sec
Reuters: The headspinning speed at which Beta Pictoris b whirls, the scientists said, lends support to the notion that a planet’s rotational velocity is closely related to its size: the bigger, the faster.
“Yes, the relation between mass and spin velocity was already known in our solar system,” said University of Leiden astronomy professor Ignas Snellen, another of the researchers.
“We now extend it to a more massive planet to see that the relation still holds. We need to observe more planets to confirm this is really a universal law,” Snellen added.
This paper suggests that, while mass and radius are factors in determining rotational velocity, also so is the orbital velocity which is determined by the distance of the planet from its sun.
Our moon shares the same velocities as the earth but,in addition, has its own orbital velocity. That speed should be amenable to the same type calculation. The result is shown in the above table. Below is a table showing the results for seven other larger moons. The calculated values are within a factor of ten of the average actual values for all of these.
m (grams) x 1025  r ( cm ) x 108  v (calc) cm/ s x105  v (ave obs) cm/s x105  
Callisto  10.76  2.41  1.73  8.20 
Titan  13.45  2.576  6.00  1.87 
Triton  2.14  1.35  4.38  1.03 
Io  8.93  1.82  1.81  17.3 
Ganymede  14.82  2.634  1.94  10.9 
Europa  4.80  1.56  1.43  13.7 
Iapetus  0.1805  0.7345  0.405  3.24 
For smaller moons ( ~1019 grams ),the discrepancy appears to be several hundred to a thousand fold. These moons’ speeds would be affected greater by other gravitational forces.
The calculated velocities of stars are higher than the observed rotational velocities, ranging from within a factor of ten to as much as 200. The difference could be the galaxy revolving, as is the case of the sun.
Solar Mass  Solar Radius  Oberved Rotation cm/s x 107 
Calculated Rotation cm/s x 107 

Altair  1.79  1.63 to 2.03  2.40  3 
Betelgeuse  7.7 to 20  950 to 1200  0.05  0.25 to 0.45 
Epsilong Eridani  0.82  0.735  0.024  3.3 
Formalheut  1.92  1.84  0.93  3.17 
Pollux  2.0  8.8  0.028  1.48 
Rigel  18  74  0.40  1.5 
Sirus  0.978  0.0084  0.16  33 
Spica  10  7.4  1.99  3.6 
Tau Ceti  0.78  0.79  0.024  3.1 
Cosmology
Perhaps, by the reverse engineering of collapsing bodies, we can get insight of the dynamics of our expanding universe. A subatomic size particle with the mass of the universe, upon creation, might expand according to some yet unknown Quantum behavior to a size where non Quantum laws take over. At that point, Equation 2 would be violated, as vg2 >>>c2, prompting an instantaneous expansion to bring vg 2= ½ c2. This would account for the hypothesized Inflation at 1033 seconds after inflation.
At that point, c2= v2g + v2spin + v2expansion. Assuming vspin to be zero, the expansion velocity2 would then also be ½ c2. Upon expanding, v expansion would increase and vg would be correspondingly reduced in accordance with Equation 2, which is what is observed.
Equations 1 and 2 may be written as: m ( c2 – V2 ) = Gm2/r , where V is the absolute velocity of the object. This equation is stating that the total positive energy of an object is equal in amount to its gravitational energy which is negative. The net total energy of objects in our universe, and the universe itself ,is zero, just as a planet revolving around a sun at a certain velocity while spinning at an equivalent opposite velocity could have no intrinsic net movement. Some physicists have speculated in the past that this is the case with our universe.
If this theory is correct our universe was created not from energy but from nothing!
Calculation of the Expansion Rate of the Radius of Universe
Selecting the mass of the universe to be 6x1054 grams and its radius 4.40x1028 cm, then
v2g = Gm/r = ( 6.66x108cm3/gsec2 ) ( 6x1054g ) / 4.40x1028cm = 9.08x1018cm2/sec2
v2expansion = c2  v2g =9.00x1020  0.18x1020 = 8.82x1020 cm2/sec2
vexpansion = 2.97x1010cm/sec depending on the actual value m and r
Calculation of the Age of the Universe
At Inflation: v2g = ½ c2 = G m / r
r = 2Gm/c2 = 2 ( 6.67x108 cm3/gsec2 ) ( 6x1054g ) / 9.00x1020cm = 8.9x1026cm
Estimating the average expansion velocity between the point of inflation ( ½ c2 ) and today ( 2.97x1010 cm/sec ) to be 2.6x1010 cm/sec then,
time required = increase in radius of the universe / average expansion velocity
t = ( 4.40x1028cm  0.09x1028cm ) / 2.6x1010cm/sec = 1.66x1018 sec
Today’s accepted age of the universe based upon observations is 13.7 billion years or 4.32x1017sec
An exercise
A particle with Planck’s mass ( hc / G )1/2 and radius of Planck’s length ( hG/ c3 )1/2 would have its
vg2 = Gm/r = G( hc/G x c3/hG )1/2 = c2
This describes a tiny motionless black hole.Its absolute mass determined by mA2= mPrPc2 /G and mA turns out to equal mP. That is consistent with the particle not violating equation ( 2 ) by having motion, v2 + vg> c2. The energy of this particle may be expressed as mPc2 or mP2G/rP.
Calculation of the radius of quarks
Below is a table of the six quarks. A midrange energy value is selected. New values should be substituted as their certainty increases for the forthcoming calculation. These values are converted to grams in the next column for use in equation 6.
When a quark and antiquark combine to form a meson the assumption is made here that they attain their absolute velocity which determines the energy of that meson. The Phi meson, made up of a Strange quark and Strange antiquark, has a rest energy of 1020 MeV/c2 so that each is assigned half of that amount and that value is entered in the next column. In a like manner, the J/PSI meson, consisting of a charm quark and charm antiquark, has a rest mass of 3096 MeV/c2 so each is assigned a maximum value of 1548 MeV/c2. The Upsilon meson ( b, b ) at 9460 MeV/c2 is used to get the value for the bottom quark. The Kaon meson ( u, s ) of 493.7 MeV/c2 was used to get the up quark value by subtracting the strange quark value. In the same way, the down quark value was gotten from the Pion meson ( u, d ) rest mass of 493.7 MeV/c2. Other mesons could also be selected for this purpose which would slightly change the results. It is the method being featured here rather than the precision of the results.
Quark 
m0 MeV/c2 
m0 x1027g 
m MeV/c2 
m x10^25g 

up 
2.4 
4.3 
419 
7.46 

down 
4.9 
8.7 
419 
7.46 

charm 
1290 
4077 
1548 
27.56 

strange 
100 
178 
510 
9.08 

top 
172,900 
307,380 

bottom 
4190 
7449 
4730 
84.2 
It is interesting to apply these results in the next chart. Here the actual energy of particles is compared to the value calculated from the quark chart. In all cases shown, the actual energies are lower than the calculated one, with one exception. The assumption will be made here that this is due to the quarks contained have fallen short of their absolute velocity. The exception, Z(4430), contains heavier quark charges which would raise its potential energy. The potential energy component here has been assumed to be small enough to ignore.
meson 
quark makeup 
actual MeV/c2 
calc MeV/c2 
% difference 

Rho 
u ( d ) 
775 
838 
9.9 

D,null 
c ( u ) 
1865 
1967 
5.2 

D,star 
c ( d ) 
1869 
1967 
5.0 

Kaon 
u ( s ) 
494 
929 
50 

Pion 
u ( s ) 
140 
838 
83 

baryon 

Lambda,null 
u d s 
1115 
1348 
17 

Lambda,+ 
u d c 
2281 
2386 
4.4 

Sigma,+ 
u u s 
1189 
1348 
12 

Sigma,null 
u d d 
1232 
1257 
2.0 

Delta,++ 
u u u 
1232 
1257 
2.0 

Cascade,null 
u s s 
1315 
1439 
8.6 

proton 
u u d 
1.67x1024g 
2.24x1024g 
25.4 

Z(4430) 
c (c) d (u) 
4430 
3935 
12.6 

Theta,+ 
u u d d (s) 
1540 
2186 
29.6 
Moving to the below chart, those values are then used in equation 6 to calculate the quark radius: r0 = Gm2 / m0c2
eg. Up quark: r0 = (6.67x108cm3 g1 sec2) ( 7.46x1025g)2 / (4.3x1027g) (3.00x1020cm sec1)2 =
9.6x1051 cm
The same calculation is made for the other quarks and is entered into the first column. If the assumptions made are valid, this figure would be the radius of each quark. If the quark mass falls short of its absolute value, this figure would represent the minimum of the uncertainty range of the quark radius
quark 
r0 x1052cm 
r x 1052cm 
up 
96 
0.55 
down 
47 
0.55 
charm 
1.4 
1.1 
strange 
3.4 
0.67 
bottom 
6.96 
6.2 
In the next column the radius at maximum speed is listed as calculated from equation 5:
eg, Up quark: r = r0 m0 / m = 2.4MeV X 96 X 1052cm / 419 MeV = 0.55 x 1052cm
While the r0 range is almost 70 fold, the r range has shrunk to almost 10 fold.

The author gratefully acknowledges the assistance of Dr. Andrew Nickel in the preparation of this paper.