Obsolete... We take the unit of mass (electron m_e) surrounded of a sphere with radius R and guess that the flux of gravity is changing around this point:

G = -Km_e/R²

The initial gravity, where K is gravity constant, the flux is:

Ф_G= = Km/R²ds = Km4πR² /R²= 4πKm

Where area integral is equal with area of sphere (4πR²) because G is perpendicular on ds:

dФ_G = d(Km4π)/dt = 4Kπd/dt(m) (1)

E = mc² -> d/dt (m) = 1/c²dE

According to Planck formula a photon that will yield the energy dE to the unit of mass:

So: d/dt (m) = 1/c²dE = hν/c²

Were c is the light speed:

1. -> dФ_G = 4Kπhν /c²

ν = dФ_Gc²/(4Kπh) (2)

d/dt(m) =d/dt m_{0 (} )= m₀ d/dt(₎

4Kπhν /c² = 4Kπd/dt (m) = 4Kπ m₀ d/dt()

Were V = dR/dt and m₀ is the rest of the unit mass so:

hν /c² = m₀ d/dt()

hν /m₀c² = -(V/(c²())*dV/dt

hν /m₀ = -(V/()*dV/dt

dV/dt = -(hν/m₀V)( )

With the approximation ()~1+ 3/2(V²/c²)

V(1+ 3/2(V²/c²)) dv = -(hν/m₀ )dt

With V =2.19*10⁶m/s and R=2.12*10¹⁰m

V(1+ 3/2(V²/c²))dv = -(hν/m₀ )dt

From the moment t₀ to t₁ where T= t₁- t₀ and from the variation of speed V = V₀-V_1.

The signification of the minus sign is that the speed of the electron is decreasing:

V² /2 + 3/8 (V⁴/c²)=hνT/m₀

With V =2.19*10⁶m/s (2.19*10⁶m/s is the speed of the electron in the Bohr model of the first orbit, principal quantum number = 1) and T=3.38*10^-2s.

The time T is the latency time and I expect that from the apparition of the laser beam anti-gravity will appear in ~ 4 hundredth of seconds.

3) ν =m₀(V²+3/4(V⁴/c² ))/(2hT); ν = 9.74*10¹⁶Hz

And the wave length λ ~ 3nm;

In this approximation we take the unit of the participating electron mass (m_e) in the gravity process and interact with the photon, as a result latency time of ~ 4 hundredth of seconds will appear. I expect that the nucleus of the atoms (protons and neutrons) will interact with the photon at the wavelength 380.4nm and also a latency of time of ~ 4 hundredth of seconds will appear.

According to the Quarks Informational Theory and the Relativity Theory the gravity propagation is with the speed of light. At the resonance of the graviton and photon will produce antigravity in a period of time equal with 2R/c were R=6370Km the radius of the earth and the latency is 4.24*10^-2s.; the result is in conformity with the above calculation.

The interaction between the photon and the electron or the nucleus (protons and neutrons) is generic because its are tied with strong electrostatic forces and its behave like a whole. The only one that differs is the attached wave of the electron on the orbital and the attached wave of the nucleus in its vibratory movement.

The attached wave of the electron on the orbital in conformity with L. de Broglie (λ = h/p) is: λ ~7.2 10^-10m = 7.2Å so there is a photon that can interact with the graviton and the frequency is 30.8Å on witch we can realize anti-gravity.