The Spring of the Earth , the Sun and the Universe

A spring term is added into Newton’s law of gravitation. The spring k of the earth is found to be 1.21 × 10/sec. The PPN gamma is a dependence of distance r from the sun. The expanding universe is due to the cosmological constant. The Hubble constant is found to be the square root of the cosmological constant. The query of the missing dark matter in the galaxies is clarified.


Introduction
The true nature of the cosmological constant is still unknown, whether it is physical or purely a mathematical conjecture.Secondly, its value is so small that the value can only be significant in large-scale space.We previously suggested that each source had its own cosmological constant, which hereinafter, to be renamed as the spring constant of the source as we did previously [1,2].
where the first term is the usual Newton's inverse square law while the second term refers to the spring constant k of the source, for instance, the earth.The spring constant k, which can either be extended or compressed, is in fact the cosmological constant if we look at Einstein's exterior field equations of the form is called the spring constant because it resembles an harmonic oscillator.Obviously, the Kottler solution (1) is the result of (2).Throughout the entire paper, only the 3dimensional case is to be considered

The Spring of the Earth
To be more rigorous when applying to the earth, the centrifugal acceleration 2  0 cos r   is taken into account and placed onto the left of (1).For θ = 0, the centrifugal ac-celeration at the equator will be maximum.Taking: the earth rotation ω = 7.3 × 10 −5 /sec; the earth radius r 0 = 6.4 × 10 6 m; the earth mass = 6 × 10 24 kg; the gravity of earth = The third term is usually referred to the fifth force as suggested by Fischbach et al [3][4][5] but we have pointed out previously that the Yukawa-like fifth force will yield an unreasonable value of r at the acceleration a = 0.For an escaping object, the spring will have the same direction as gravity The first two terms of (3) and (3a) give the value of 9.7516 m/s 2 .Hence, L. M. TSANG 1206 smaller than 9.7516 m/s 2 in (3) and 9.854 m/s 2 larger than 9.7516 m/s 2 in (3a).Both equations yield the result of k = 1.21 × 10 −8 /s 2 .
Of course, one can argue that the percent level of accuracy can also disprove (3) and (3a).Upon substituting the above value of k into (1) and setting a = 0, we have r = R = 32,000 km where such a spherical shell is gravity-free.This R is the maximum extension, and beyond this critical extension, only Newton's inverse law remains effective.To express mathematically: 2 where 32, 000 km where 32, 000 km where 32, 000 km The spring force is the type of intermediate range.Springs are the aether that fill up the entire space and attach onto the source.Galaev found aether to be a compressible viscous gas having the kinematic viscosity of [6].However, we are not in favor of such a aether wind concept.Axion is not spring since it can be produced in the core of the sun via the Primakoff effect.Then it travels to the earth where by interacting with a transverse magnetic field and detected by an X-ray detector.Similarly, the Poher's universon particle travelling at the speed of light [7,8] is not our proposed spring.However, the above three particles namely, the Galaev gas, axions and the universons travel through the spring.Our earth carries the spring aether while revolving around the sun, causing the null result of the Michelson-Morley experiments.

The Spring of the Sun
We had included the spring term in the Binet's equation but found that the spring constant of the sun varied from 10 −16 to 10 −21 /sec 2 (See Table 1).Furthermore, by comparing with some other authors, our obtained spring constant within the inner planets seems to be more reasonable.Later in this paper Section 8, we explain how the spring of the sun being obtained.We need to point out that by solving the Binet's equation, the spring of the sun does not affect the perihelion shift of a planet but this spring will affect the light deflection in the form of [1]: Although the exact value of γ cannot be determined, most authors agree that it must be a constant near to unity.Froesche [14] pointed out that 1/2(1 + γ) = 1 within 0.1%; Bertotti [15] pointed out that [16] pointed out that |1 − γ| lies between 10 −5 to 10 −7 ; and Shapiro [17] gave the value γ = 0.9998 ± 0.0004.Nearly all authors concluded that the value of γ is a constant close to unity within the solar system.But from (7), γ depends on the distance r, even though it is not easily to be observed.
Outside the solar system the sun's spring becomes weaker and finally breaks at r = 10 13 m.From (7), the value γ is close to unity since the second term is very small.The nominal value gamma = 1 applies for pure General Relativity.Modified gravitation theories, also including additional interactions, are expressed in the Parametrised Post-Newtonian formulation with parameter values, which may be different from 1.
The amplitude of the discrepancy from unity depends on the theory.The current experimental verifications (as produced by Gai and Vecchiato's group) are at the precision level 10 −5 with Cassini, and may reach in the near future the 10 −6 level with Gaia.Better measurements will constrain the theory.

The Spring of the Universe
The spring k of the universe in such a large scale structure can be easily explained without using higher dimensions nor vacuum as some authors suggested [18][19][20][21].The expanding mechanism can be in the form of L. M. TSANG 1207 where Ω is the negative pressure to separate matter when the spring force is acting against it.Strong evidences show that the accelerating rate is increasing [23][24][25].
Since none of the observable matter can produce an expanding pressure , the expansion can be due to the cosmological constant .Ignoring the first term of (8) and all the numerical factors, from (1) we immediately yield the result of We are aware that some authors also applied the cosmological constant in the expanding universe similar to our (9) [27].Differently, they inserted the Hubble law into the equation rather than deduced it consequently.From ( 9), we immediately obtain the approximate relationship of: 10 ~10 s  .The value of H agrees with observations [see also [28][29][30][31][32].We consider only the case of 3-dimensions.
Since expansion is mainly based on the red-shift and the present radius of the universe r = R ~ 10 billion light years ~ 10 26 m (from P. Butterworth, NASA), an accelerating rate of R can be very useful in the study of cosmology.Equation ( 9) shows a non-stop universe until the spring breaks.However, this may not be the case.By choosing the present radius and density of the universe, the mass M is found to be 3 51 Substituting the above into (1) and set a = 0, we get which is more or less the radius at our present time.The velocity 8 ~10 m s v Hr  is less than the speed of light.The universe is expected to cease its acceleration at our present time.

The Missing Mass in the Rotation Curve of Galaxies
Our initial plan is to evaluate the mass and spring constant of the Coma Cluster without the aids of rotation curve but based on the assumptions that:  the outer edge radius R of the cluster and the velocity ge in velocity outside r > R and the r [33,34] po must be known  there is no chan spring is assumed to be constant outside R. Authors including Kraft and van Leeuwe inted out some aberations in the observed velocity v ~ 0(10 6 ) m/s and the distance r.Zwicky [35] and Rubin et al [36] provided the first pieces of evidence that large amounts of dark matter do exist outside the visible region of most galaxies.Our present purpose is to clarify the so-called missing mass using spring theory.As a rough estimation,the core radius here can be treated as the outer edge radius of the cluster while the velocities in the following 4 papers are to be used in our calculation.The simplest way is to apply the viral theorem Outside the edge radius of the Cluster, the rotation curve is more or less flat, or dv/dr = 0 for r > R. The spring constant can be treated as a constant up to r = 4R even though it must be function of r.Hence we get: We admit our calculation above is too crude without a ro R = 2.7 kpc. which are ob 664: v = 200 km/s, R = 7.5 kp tation curve.In spite of this, the existence of a spring term can explain the missing mass.Next, we look at some rotation curves.Figure 1 shows the rotation curves of 4 NGC's [40].We select a suitable point before the Keplerian motion on each curve: For NGC 4594: v = 230 km/s, tained from the curve, whereas Zwicky gives v = 4 × 10 5 m/s, R = 4 × 10 29 m [35].
For NGC 2590, 1620 and 7 c where r = R is the outer edge radius of these NGC's.All 4 NGC's are found to have the same mass M ~ 10 41 kg using (11).Next, we select 3 points on each curve Each mass can only have one unique as 6.On the Rotating Universe g universe in 1949, the discrepancies among these authors, the lim spring constant signed to it.Comparing (11), ( 12), ( 13) and ( 14), we can deduce that for M = 10 41 kg, k = 10 −31 /s 2 .Since Goedel's proposal of a rotatin numerous authors gave different values of the angular velocity of the universe (ω rad/yr).These values were based on the cosmic microwave background, or on Einstein's field equations.Some of their values are selected as Table 2.
In spite of it of the angular velocity can still be estimated.Along the equator of the universe, where the centrifugal acceleration is maximum, we have: The third term of the above equation can be ignored.U Since the Hubble is t uare root of the co pon integration of the above, we get: Since s ~10 26 m, the speed at the outer rim will be <10 8 m/s, or less than the speed of light.This appears to be reasonable but problem arises: The centrifugal force along the axis of rotation will be zero.(15) can then be reduced to: Same as (9) which differs from (15) and should be detected.There is a problem to locate the axis of rotation.One finds no difficulty to produce an angular speed.For example,  can be postulated from the total derivative of the posi n r in the vector form of: tio which is, of course, meaningless.We suggest an isotropic,

The Electric Field
ensity surrounding a charge q 6) homogeneous and non-rotating universe which agrees with the Big Bang and Causality.
The electric field energy d is proportional to the square of the field intensity E, or L. M. TSANG 1209

Since a charge is always a m c mass δm
The above two masses on the right side of (17 no (18) where is a constant to be determ ccompanied by its electroagnetic mass δm, the total mass becomes M = mechanical mass + electromagneti (17) ) are n-separable from each other.The surrounding electric field energy, according to Gauss law, should be 2 mc  .Equation ( 16) can be re-written as where A, B are the constants of integration.By setting A be the charge q, (16) becomes The total electric field energy over the whole space is which is the Gauss law.Thus, The field intensity becomes The potential can be written as In short range, A, B need to be determined eventhough (2 el of an atom to find the va 5) has the form as (31) of [47].Obviously, for δm → 0, E → 0, indicating that the electromagnetic mass always accompanies with the charge.
We can use the Bohr mod lue of m tions from ( 19) and ( 30) among which we roughly estimate w n is a composite particle.

The Gravitational Field
There are several solu Again, in short range gravitational field, the value of A, B in (25) need to be determined.The Binet equation of a planet becomes where h, u are the conservation of angular momentum and the reciprocal of r respectively.By solving (33), we obtain the perihelion shit of which, of course, is much smaller than the result of general relativity.However, we are able to find the spring of the sun

Short Range Interactions
It is known that once the mass approaches or lesser than It is not reason set or able to A q GM  in very short electric or gravitational interacti there exists a big amount of energ on since ere are heavy p belongs to the so ergy of a bos y and th articles inside the nucleus.Equation (34) urce of a nucleon, hence the spring en along on oscillating where M is tentatively set to be the mass of the testing nucleon.For = 1.5 fm λ , mass of pi-meson m π = 132 MeV and where N is greater at longer distance and so the spring is weaker.Equation ( 36) only refers to a single line of force connecting two interacting bodies.We are not sure wh if a unit spring k 1 has the Planck length.
The β decay can be exp one of the springs is released from compression wi 5), the sp en ether lained 3 dimensionally as that th its one end attaching to the nucleus while the other free end pushing an electron outwards; a process similar to the expanding universe where the gravitational force cannot hold the matter from flying out.From (3 ring ergy on the left equals to the energy of a W boson.The spring kicked out an electron, exerting an energy of 80 GeV, which is much greater than that of an electron.This excited electron, upon the acceptance of a transfer of momentum from the spring, will release an anti-electron neutrino.Such a mechanism of kicking out an e   is performed by a spring with its one end attaching onto the electron while the other free end pushing the e   out of the electron.We have no idea why an electron neutrino turns out to be an anti-neutrino after leaving the elec-tron.The energy of such a spring is 90 GeV, which also known as the Z boson.Theorem In short range interaction.The spring

Conclusion
We admit that most of our numerical values in this paper ar is to in th .By comparing the spring of (10 −16 /s 2 ), galaxies of M = nd the universe (10 −35 /s 2 ), we show mass, the lesser is the value of k.

NCES
, Vol. 17, 2012, pp.18-21.doi:10.1016/j.newast.2011.05.004 e approximate since our main purpose troduce e spring term into the theory the earth, (10 −8 /s 2 ), the sun 10 41 kg (10 −31 /s 2 ) a that the larger the Conversely, the spring between 2 nucleons is found to be 1.1 × 10 46 /sec 2 .The myth of the fifth force, aether, the missing mass in the galaxies, dark matter, Hubble constant and the cosmological constant and all these problems can be solved by a simple 3 dimensional "spring".There are rooms for the spring theory to fit into particle physics.Our present work gives a start.
Since the first term of ( 37) is known, the second term will immediately yield the deceleration of 9.854 m 2 (same as Equation (3a)).

/s
Copyright © 2013 SciRes.JMP θ = 42˚, the latitude of Massachusetts where Pound and Rebka performed their experiments at Harvard.Equation (1), including the additional negative term of the centrifugal −27 kg/m 3 = mass density inside the universe.

1 2 H
 and the Hubble law v Hr  where H = Hubble constant   18  17.5 Different authors had different values of velocity v and co cky[35]: v = 0.78 × 10 m/s, R = 2 × 10 m om Shao et al[37]: velocity dispersion v = 0/s, core radius R = 5.2' = 0.936 × 10 19 m 3) from Omer and Wilson[38]: R = 100' = 1.20 4) f Here, we choose 1 kpc = 3 × 10 19 m throughout this per and take the average from the above to get v = 0.85 × 10 6 m/s and R = 1.3 × 10 20 m.The mass of the Coma Cluster of galaxies is found to be, neglecting the spring term, 2 42

Figure 2 . 41 ~10
Circular speed versus radius of our galaxy; curve Table 2. Different values of ω from various authors.Hawking[42] g M and the spring k.The results are rve of Figure2, we select two suitable points between the D and B curves[41]:


. The force equation, the energy equation and the co rvation of angular momentum can be written as, respectively, m is the rest mass of the orbiting electron.bing (26) and (28), we obtain or less the value of some electron neutrinos.If this is true, an electro In analogy to the electric field, the gravitational field energy (21) can be written as of 10 kg, gravity becomes strong interaction and the potential energy depends on the distance of the two interacting particles only.Consider the the energy level of proton-electron electric interaction, Bohr's atom −13.6 eV does not ag is the spring linking the 2 nucleons.o spring and smaller mass has stronger spring is that: if two ed spring constant will become f why larger mass has weaker particles are connected by N identical springs each of k 1 , the combin 1


is the length of the spring.
Same as(3) which is only true at Harvard, or likewise the State of Massachusetts.In 1965 Pound and Snider refined the appa tus so that the energy shifts on the upward and downward path