See Table 1 . The data of second and third column have been taken from [8] R. Ragusa et al. 2022 and using the formula $M_{200}=\frac{100H^2{R}_{200}^3}{G}$ it is calculated its mass associated for each radius. In the last column shows the relative difference for masses, which is always under 10%.

As the Virgo cluster is the nearest between the big clusters it is crucial to check the approximation for virial mass and radius with its data.

See in Table 2 the data of Virgo cluster, according [9] Olga Kashibadze, I. Kara-chentsev. 2020.

Using formula (2.3) with R₂₀₀ =1.7 Mpc it is got M₂₀₀ = 5.59 × 10¹⁴ $M_{\Theta }$ which matches with the mass published if it is considered the range of errors.

In conclusion: R₂₀₀ and M₂₀₀ are a very good estimation for the virial radius and the virial mass for a group of galaxies and a cluster of galaxies, when they are in dynamical equilibrium.

Due to the fact that the Direct mass formula has one parameter only, is enough to know the mass associated to a specific radius to be able to calculate parameter a². That is the situation when it is known the virial mass and the virial radius for a cluster of galaxies.

If it is considered that the virial radius is the border of halo cluster where galaxies are in dynamical equilibrium and at the same time is negligible the amount of Baryonic matter outside the sphere with such radius, then according DMbQG theory is possible to do an equation between M_VIRIAL (<R_VIRIAL) = M_DIRECT (<R_VIRIAL) (3.2.1) i.e.

$M_{VIRIAL}=M_{TOTAL}\left(<R_{VIRIAL}\right)=\frac{a^2\cdot \sqrt{R_{VIRIAL}}}{G}$ (3.2)

and clearing up

$a^2=\frac{G\cdot M_{VIRIAL}}{\sqrt{R_{VIRIAL}}}$ (3.3)

this formula is called parameter a² (M_VIR, R_VIR) because depend on both measures.

In chapter 2 was got this formula $R_{VIR}^3=\frac{G\cdot M_{VIR}}{100\cdot H^2}$ (2.2) as a good approximation between virial mass and virial radius. So using that formula and by substitution of virial radius in $a^2=\frac{G\cdot M_{VIRIAL}}{\sqrt{R_{VIRIAL}}}$ (3.3) it is right to get parameter a² depending on M_VIR only

$a^2={\left(G\cdot M_{VIR}\right)}^{5/6}\cdot {\left(10\cdot H\right)}^{1/3}$ (3.4)

This formula will be called parameter a² (M_VIR) as depend on M_VIR only.

With the virial data for Virgo cluster, see Table 2 , will be calculated its parameter a² with the formula (3.3) i.e. a² (M_VIR, R_VIR) and with the formula (3.4) i.e. a² (M_VIR).

The last formula is an approximation of the previous formula as it is supposed that R_VIR ≈ R₂₀₀. In Table 3 are calculated both formulas and fortunately its relative difference is negligible.

According [5] Chernin, A. D. in the current cosmologic model $\Lambda CDM$, dark energy has an effect equivalent to antigravity i.e. the mass associated to dark energy is negative and the dark energy have a constant density for all the Universe equal to ${\rho }_{DE}={\rho }_C\cdot {\Omega }_{DE}=6.444\times {10}^{-27}{\text{kg/m}}^3$ being ${\Omega }_{DE}=0.7$ and

${\rho }_C=\frac{3H^2}{8\pi G}=9.205\times {10}^{-27}{\text{kg/m}}^3$ the critic density of the Universe.

As DE density is constant, the total DE mass is proportional to Radius with power 3, whereas DM mass grows with radius power 0.5 so it is right to get a radius where DM is counter balanced by DE.

According to [5] Chernin, A. D.

$M_{DE}\left(<R\right)=-\frac{{\rho }_{DE}8\pi R^3}{3}$, (4.1)

Is the mass associated to DE or equivalently

$M_{DE}\left(<R\right)=-{\rho }_{DE}\frac{8\pi R^3}{3}=\frac{-H^2\cdot {\Omega }_{DE}}{G}\cdot R^3$ (4.2)

Notice that the author multiplies by two the volume of a sphere by reasons explained in his work.

[5] Chernin defines gravitating mass

$M_G\left(<R\right)=M_{DE}\left(<R\right)+M_{TOTAL}\left(<R\right)$ (4.3)

where M_TOTAL is baryonic plus dark matter mass, and defines R_ZG, Radius at zero Gravity as the radius where $M_{DE}\left(<R_{ZG}\right)+M_{TOTAL}\left(<R_{ZG}\right)=0$. i.e. where the gravitating mass is zero.

According to the previous equation it is got

$M_{TOTAL}\left(<R_{ZG}^{}\right)={\rho }_{DE}\frac{8\pi R_{ZG}^3}{3}$ (4.4)

Using (3.1) formula $M_{TOTAL}\left(<R_{ZG}\right)=\frac{a^2\cdot \sqrt{R_{ZG}}}{G}$ the Equation (4.4) leads to

$\frac{a^2\cdot \sqrt{R_{ZG}}}{G}={\rho }_{DE}\cdot \frac{8\pi \cdot R_{ZG}^3}{3}$, (4.5)

where it is possible to clear up

$R_{ZG}={\left[\frac{3a^2}{8\pi G{\rho }_{DE}}\right]}^{2/5}$ (4.6)

and as

${\rho }_{DE}=\frac{3\cdot H^2}{8\pi G}{\Omega }_{DE}$ (4.7)

then by substitution

$R_{ZG}={\left[\frac{a^2}{H^2\cdot {\Omega }_{DE}}\right]}^{2/5}$ (4.8)

This formula will be called R_ZG (parameter a²).

As the radius R_ZG is the distance to cluster centre where is zero the gravitating mass, it is right to consider R_ZG as the halo radius and its sphere defined as the halo cluster.

In previous chapter was got the value for $a^2={\left(G\cdot M_{VIR}\right)}^{5/6}\cdot {\left(10\cdot H\right)}^{1/3}$ (3.4), depending on M_VIR, so by substitution in (4.8) it is right to get

$R_{ZG}=\frac{{\left(G\cdot M_{VIR}\right)}^{1/3}\cdot \sqrt[15]{100}}{H^{2/3}{\Omega }_{DE}^{2/5}}$ (4.9)

In Table 4 are calculated R_ZG by two ways: Formulas (4.8) and (4.9). Both calculi are mathematically equivalents. See in Table 4 how the both values match perfectly. See in Table 2 the values for the Virial mass of Virgo.

With this important cluster of galaxies, it has been illustrated how the total mass,

calculated by $M_{TOTAL}\left(<r\right)=\frac{a^2\cdot \sqrt{r}}{G}$, is counter balanced by dark energy at mega

parsecs scale, and precisely this Radius at zero gravity determines the region size where the cluster has gravitational influence.

From (2.2) $R_{VIR}^3\approx R_{200}^3=\frac{G\cdot M_{200}}{100\cdot H^2}\approx \frac{G\cdot M_{VIR}}{100\cdot H^2}$ it is right to get $R_{VIR}={\left(\frac{G\cdot M_{VIR}}{100\cdot H^2}\right)}^{1/3}$. In previous epigraph was got $R_{ZG}=\frac{{\left(G\cdot M_{VIR}\right)}^{1/3}\cdot \sqrt[15]{100}}{H^{2/3}{\Omega }_{DE}^{2/5}}$ (4.9). So it is right to get the ratio

$\frac{R_{ZG}}{R_{VIR}}={\left(\frac{100}{{\Omega }_{DE}}\right)}^{2/5}\approx 7.277$ (4.10)

which is universal as it does not depend on virial mass associated to a specific cluster.

Using the data from Table 1 , first, second and third columns, the fourth column of Table 5 is calculated the zero gravity radius formula (4.9). In the last column is calculated the ratio.

It is clear that the ratio R_ZG/R_VIR got in this sample of celestial bodies match very well with the theoretical ratio formula (4.10)

Similarly in Table 6 , it is done the same ratio for the Virgo cluster with an optimal result, for the Virial radius see Table 3 , for the Zero gravity radius see Table 4 .

In epigraph 4.1 was shown how this equation $M_{TOTAL}(<R_{ZG})={\rho }_{DE}\cdot \frac{8\pi \cdot R_{ZG}^3}{3}$ see (4.4) is used to define R_ZG. By simplification it is got

$M_{TOTAL}\left(<R_{ZG}\right)=\frac{H^2\cdot {\Omega }_{DE}}{G}\cdot R_{ZG}^3=M\cdot R_{ZG}^3$ (4.11)

being M = 5.3984·10⁻²⁶ (I.S.)

So that is the total mass formula, using the equation between total mass and dark energy at zero gravity radius.

In Table 7 is shown calculus of total mass for Virgo.

Using rightly the direct mass formula see (3.1) at R_ZG it is got

$M_{TOTAL}\left(<R_{ZG}\right)=\frac{a^2\cdot \sqrt{R_{ZG}}}{G}$ (4.12)

By substitution of (3.4) $a^2={\left(G\cdot M_{VIR}\right)}^{5/6}\cdot {\left(10\cdot H\right)}^{1/3}$ and (4.9) $R_{ZG}=\frac{{\left(G\cdot M_{VIR}\right)}^{1/3}\cdot \sqrt[15]{100}}{H^{2/3}{\Omega }_{DE}^{2/5}}$ in (4.12) it is got

$M_{TOTAL}\left(<R_{ZG}\right)=\frac{\sqrt[5]{100}}{{\Omega }_{DE}^{1/5}}\cdot M_{VIR}$ (4.13)

and calling

$U=\frac{\sqrt[5]{100}}{{\Omega }_{DE}^{1/5}}$ (4.14)

U ≈ 2.7 then

$M_{TOTAL}\left(<R_{ZG}\right)=U\cdot M_{VIR}$ (4.15)

So may be stated that according Dark matter by gravitation theory, the total mass (baryonic plus DM) enclosed by the sphere with radius R_ZG is equivalent to 2.7 times the Virial Mass.

In Table 8 are compared the masses for Virgo using the formula (4.15) and the previous one (4.11). It is clear that both are mathematically equivalents.

By the direct formula for total mass, may be calculated the total mass associated to a radius, which is a fraction of R_ZG, with the following property: if $\left(R=f\cdot R_{ZG}\right)$ then

$M_{TOTAL}\left(<R\right)=M_{TOTAL}\left(<f\cdot R_{ZG}\right)=\sqrt{f\cdot R_{ZG}}\cdot \frac{a^2}{G}=\sqrt{f}\cdot U\cdot M_{VIR}$ (4.16)

because $M_{TOTAL}\left(<R_{ZG}\right)=U\cdot M_{VIR}$. See (4.15)

e.g., using the data for Virgo cluster R_ZG = 12.87 and R_VIR = R₂₀₀ = 1.7687 Mpc, then R_VIR/R_ZG = f = 0.137428 and by the formula (4.16)

$M_{TOTAL}\left(<R_{VIR}\right)=M_{TOTAL}\left(<f·R_{ZG}\right)=\sqrt{f}·U·M_{VIR}\approx 0.99999M_{VIR}$.

The formula (4.2) at R_ZG becomes

$M_{DE}\left(<R_{ZG}\right)=\frac{-H^2\cdot {\Omega }_{DE}}{G}\cdot R_{ZG}^3$ (4.17)

that it is just the opposite value to $M_{TOTAL}\left(<R_{ZG}\right)=\frac{\sqrt[5]{100}}{{\Omega }_{DE}^{1/5}}\cdot M_{VIR}$ see (4.13) because the total gravitating mass enclosed into the sphere of zero gravity radius is zero by definition.

$M_G\left(<R_{ZG}\right)=M_{TOTAL}\left(<R_{ZG}\right)+M_{DE}\left(<R_{ZG}\right)=0$ (4.18)

See epigraph 4.1

Therefore $M_{DE}\left(<R_{ZG}\right)=-\frac{\sqrt[5]{100}}{{\Omega }_{DE}^{1/5}}\cdot M_{VIR}=-U\cdot M_{VIR}$ see (4.15), being U ≈ 2.7, so joining (4.17) and (4.18) it is got

$M_{DE}\left(<R_{ZG}\right)=\frac{-H^2\cdot {\Omega }_{DE}}{G}\cdot R_{ZG}^3=-U\cdot M_{VIR}$ (4.19)

The formula (4.19) may be used to calculate the mass associated to DE at a specific radius R, writing such radius as a fraction of R_ZG i.e. $\left(R=f\cdot R_{ZG}\right)$ then (4.2) is written as

$M_{DE}\left(<R\right)=M_{DE}\left(<f\cdot R_{ZG}\right)=\frac{-f^3\cdot H^2\cdot {\Omega }_{DE}}{G}\cdot R_{ZG}^3=-f^3\cdot U\cdot M_{VIR}$ (4.20)

For example using the Virgo data R_ZG =12.871 Mpc and R_VIR = R₂₀₀ = 1.7687 Mpc, the ratio R_VIR/R_ZG = f = 0.137428 and $M_{DE}\left(<R_{VIR}\right)=M_{DE}\left(<f\cdot R_{ZG}\right)=-f^3\cdot U\cdot M_{VIR}=-0.007\cdot M_{VIR}$ i.e. M_DE (<R_VIR) is negligible, whereas $M_{DE}\left(<R_{ZG}\right)=-U\cdot M_{VIR}$.

In the epigraph 4.1 was defined the gravitating mass M_G = M_DE + M_TOTAL, where M_TOTAL is baryonic plus dark matter mass and M_DE is the negative mass associated to DE.

The best way to calculate the gravitating mass is using the formulas got in epigraph 4.4 and 4.5 where are calculated the both types of masses associated to a radius, which is a fraction of R_ZG, i.e. $R=f\cdot R_{ZG}$, these formulas are (4.16) and (4.20).

Joining both ones it is got

$M_G\left(<R\right)=M_G\left(<f\cdot R_{ZG}\right)=\left[\sqrt{f}-f^3\right]\cdot U\cdot M_{VIR}$ (4.21)

where $U\approx 2.7$. This way the gravitating mass depends on the dimensionless factor f.

As the gravitating mass is defined into the halo cluster, its dominion begins at R_VIR so the gravitating mass depending on f begins at R_VIR/R_ZG. By (4.10) formula R_ZG/R_VIR =7.277 then R_VIR/R_ZG = 1/7.277 therefore the dominion of the gravitating mass function depending on $f=R/R_{ZG}$ begins at 0.13732.

In this epigraph this function will be studied up to f = 1 i.e. when the radius reaches the zero gravity radius.

In Figure 1 is represented the ratio gravitating mass/virial mass versus the ratio f = R/R_ZG in its dominion and close to f = 0.5 the function reach the maximum.

It is clear that such function will have a maximum, that it is found easily by derivation, $f_M=\frac{1}{\sqrt[5]{36}}\approx 0.488$, so the radius at maximum is

$R_M=\frac{1}{\sqrt[5]{36}}\cdot R_{ZG}\approx 0.488\cdot R_{ZG}$. (4.22)

By substitution of this value f_M into the mass gravitating formula (5.21) it is got

$M_G^{MAXIMUM}\left(<R_M\right)=1.57\cdot M_{VIR}$ (4.23)

So may be stated the following important result:

For any cluster of galaxies at a half of R_ZG, it is reached the maximum of gravitating mass which is 1.57·M_VIR.

Assuming the hypothesis that R_vir = R₂₀₀ and M_vir = M₂₀₀, i.e. the virial sphere $Densit{y}_{VIR}=\frac{M_{vir}}{Vo{l}_{VIR}}=200{\rho }_C$ being ${\rho }_C=\frac{3H^2}{8\pi G}$ the critical density of Universe.

Then into the halo cluster sphere the

$Densit{y}_{TOTALMASS}\left(<R_{ZG}\right)=1.4\cdot {\rho }_C$ (4.24)

Being R_ZG the halo radius.

Proof:

$\begin{array}{c}Densit{y}_{TOTALMASS}^{CLUSTER}\left(<R_{ZG}\right)=\frac{M_{TOTAL}\left(<R_{ZG}\right)}{Vo{l}_{CL}\left(<R_{ZG}\right)}=\frac{M_{VIR}\cdot U}{Vo{l}_{vir}\cdot {\left(\frac{R_{ZG}}{R_{VIR}}\right)}^3}=\frac{M_{VIR}\cdot U}{Vo{l}_{vir}\cdot U^6}\\ =\frac{200\cdot {\rho }_C}{U^5}=\frac{200\cdot {\rho }_C\cdot {\Omega }_{DE}}{100}=1.4\cdot {\rho }_C\end{array}$

In that chained equalities has been used two properties before got: (4.15) and (4.10).

As into the current $\Lambda CDM$ model the DE density has a constant value, ${\rho }_{DE}=0.7{\rho }_C$in every place, in particular

$Densit{y}_{DARKENERGY}^{CLUSTER}\left(<R_{ZG}\right)=0.7\cdot {\rho }_c$ (4.25)

Corolarius

The ratio density of total mass versus density of DE into the halo cluster is 2.

At the present, it is accepted that $\frac{{\Omega }_M}{{\Omega }_{DE}}=3/7$ as a global average in the current Universe, so it is worth to calculate an extension of halo cluster, R_E, where it is reached such ratio. Now it is calculated the radius of a sphere that verify

$\frac{Densit{y}_{TM}^{SPHERE}\left(<R_E\right)}{{\rho }_{DE}}=3/7$ (4.26)

$As{M}_{TM}\left(<R\right)=Densit{y}_{TM}\cdot Volume$ and by (4.2) $M_{DE}\left(<R\right)=2\cdot {\rho }_{DE}\cdot Volume$

then by substitution into (4.26) it is got

$\frac{Densit{y}_{TM}^{SPHERE}}{{\rho }_{DE}}=\frac{2\cdot M_{TM}\left(<R\right)}{M_{DE}\left(<R\right)}=3/7$ (4.27)

Using (4.16) and (4.20) it is right to get the second one equality below.

$\frac{Densit{y}_{TM}^{SPHERE}\left(<R\right)}{{\rho }_{DE}}=2\cdot \frac{M_T\left(<R\right)}{M_{DE}\left(<R\right)}=\frac{2\cdot \sqrt{f}}{f^3}$ (4.28)

By equation (4.27) and (4.28) it is got $\frac{2\cdot \sqrt{f}}{f^3}=\frac{3}{7}$ whose solution is

$f={\left(\frac{14}{3}\right)}^{2/5}\approx 1.85$ (4.29)

Therefore the radius of extended halo searched is

$R_E\approx 1.85R_{ZG}$ (4.30)

and by (4.16)

$M_{TOTAL}\left(R_E\right)\approx 3.67M_{VIR}$ (4.31)

In order to validate such calculus is enough to check that the Density of total mass at R_E radius is $0.3\cdot {\rho }_C$

$D_{TM}\left(<R_E\right)=\frac{M_{VIR}\cdot U\cdot \sqrt{1.85}}{Vo{l}_{vir}\cdot {\left(\frac{R_{ZG}}{R_{VIR}}\right)}^3\cdot {1.85}^3}=\frac{200\cdot {\rho }_C\cdot {\Omega }_{DE}}{100\cdot {1.85}^{5/2}}=\frac{1.4}{{1.85}^{5/2}}\cdot {\rho }_C=0.3\cdot {\rho }_C$(4.32)

as it was expected.

For example for the Virgo cluster R_E = 1.85·12.9 Mpc = 23.9 Mpc.

The gravitating mass is zero at R_ZG the line integral for potential goes up to R_ZG, so the formula is

$V=\underset{R}{\overset{R_{ZG}}{{\displaystyle \int }}}\left[\frac{-G{M}_G\left(<r\right)}{r^2}\right]\cdot \text{d}r$ (5.4)

The gravitating mass $M_G\left(<r\right)=M_G\left(<f\cdot R_{ZG}\right)=\left[\sqrt{f}-f^3\right]\cdot U\cdot M_{VIR}$ (4.21) will be used to do the integral, doing some little changes.

As $f=r/R_{ZG}$ and calling $K=U\cdot M_{VIR}$, by substitution in (4.21) it is got

$M_G\left(<r\right)=K\cdot \left[\frac{\sqrt{r}}{\sqrt{R_{ZG}}}-\frac{r^3}{R_{ZG}^3}\right]$ (5.5)

that by substitution in (5.4) results

$V\left(R\right)=-GK\underset{R}{\overset{R_{ZG}}{{\displaystyle \int }}}\left[\frac{r^{-3/2}}{\sqrt{R_{ZG}}}-\frac{r}{R_{ZG}^3}\right]\cdot \text{d}r$ (5.6)

whose result is

$V\left(r\right)=\frac{5GK}{2\cdot R_{ZG}}-GK\cdot \left[\frac{2}{\sqrt{r}\cdot \sqrt{R_{ZG}}}+\frac{r^2}{2\cdot R_{ZG}^3}\right]$ (5.7)

where its dominion ranges from R_VIR up to R_ZG.

Notice that $V\left(R_{ZG}\right)=0$ and $V\left(r\right)$ is negative inside its dominion.

In this epigraph will be developed the equation $V_E\left(r\right)=V_{HF}\left(r\right)$ (5.1) to calculate the $R_{ZV}$

From (5.2) $V_E^2=-2\cdot V\left(R\right)$ So

$-2\cdot V\left(R\right)=V_{HF}^2=H^2\cdot R^2$ (5.8)

and by substitution of potential formula (5.7) it is got

$\frac{-5GK}{R_{ZG}}+2GK\cdot \left[\frac{2}{\sqrt{r}\cdot \sqrt{R_{ZG}}}+\frac{r^2}{2\cdot R_{ZG}^3}\right]=H^2\cdot r^2$ (5.9)

reorganising that equation it is got:

$\frac{4GK}{\sqrt{R_{ZG}}}\cdot \frac{1}{\sqrt{r}}+\left[\frac{GK}{R_{ZG}^3}-H^2\right]\cdot r^2=\frac{5GK}{R_{ZG}}$ (5.10)

and multiplying that equation by the factor $\frac{R_{ZG}}{GK}$ it is got a better expression

$4\sqrt{R_{ZG}}\cdot \frac{1}{\sqrt{r}}+\left[\frac{1}{R_{ZG}^2}-\frac{H^2\cdot R_{ZG}}{GK}\right]\cdot r^2=5$ (5.11)

Which is the equation for zero velocity radius.

That equation is not possible to solve with algebraic methods, but is quite easy to solve numerically for specific data.

In Table 9 there are the Virgo cluster data. Virial radius and mass data come from [9] O. Kashi-badze.

Using such data it is possible to calculate the coefficient for the equation (5.11)

$4\sqrt{R_{ZG}}\cdot \frac{1}{\sqrt{r}}+\left[\frac{1}{R_{ZG}^2}-\frac{H^2\cdot R_{ZG}}{GK}\right]\cdot r^2=5$ As U ≈ 2.7 and K = U·M_VIR then

GK = 2.259E+35 (I.S. units), $\frac{H^2\cdot R_{ZG}}{GK}=9.047\text{E}-48{\text{m}}^{-2}$, $\frac{1}{R_{ZG}^2}=6.3397\text{E}-48m^{-2}$ and $4\sqrt{R_{ZG}}=2.5208\text{E}+12m^{1/2}$

The equation (5.11) is easy to be solved numerically.

Using f = Radius/R_ZG it is got that f = 0.602 is a very good approximation for the solution. Therefore the R_ZV = f·R_ZG = 7.75 Mpc.

Considering (5.11) as the equation for the R_ZV then according DMbQG theory the ratio R_ZV/R_ZG is universal and its value is R_ZV/R_ZG ≈ 0.602016.

Proof

The equation (5.11) is $4\sqrt{R_{ZG}}\cdot \frac{1}{\sqrt{r}}+\left[\frac{1}{R_{ZG}^2}-\frac{H^2\cdot R_{ZG}}{GK}\right]\cdot r^2=5$ where $K=U\cdot M_{VIR}$ being $U=\frac{\sqrt[5]{100}}{{\Omega }_{DE}^{1/5}}$ (4.14) and $R_{ZG}=\frac{{\left(G\cdot M_{VIR}\right)}^{1/3}\cdot \sqrt[15]{100}}{H^{2/3}{\Omega }_{DE}^{2/5}}$ (4.9) then using the values for K, U and R_ZG , doing some algebraic substitutions and transformations it is not difficult to get the following equation.

$4\sqrt{R_{ZG}}\cdot \frac{1}{\sqrt{r}}+\frac{1}{R_{ZG}^2}\left[\frac{{\Omega }_{DE}-1}{{\Omega }_{DE}}\right]\cdot r^2=5$ (5.12)

and defining $f=r/R_{ZG}$ the equation (5.12) becomes

$4\cdot \frac{1}{\sqrt{f}}+\left[\frac{{\Omega }_{DE}-1}{{\Omega }_{DE}}\right]\cdot f^2=5$ (5.13)

that considering ${\Omega }_{DE}=0.7$ becomes:

$4\cdot \frac{1}{\sqrt{f}}+\left[\frac{-3}{7}\right]\cdot f^2=5$ (5.14)

By elementary algebraic operations it is got this equivalent equation:

$9\cdot f^5+210\cdot f^3+1225\cdot f-784=0$

Thanks Wolfram alpha software, this is its only real solution

f ≈ 0.602016 (5.15)

By this new formulation of the equation for R_ZV calculus, it has been demonstrated that the ratio f = R_ZV/R_ZG is universal and depends on ${\Omega }_{DE}$ solely.

Notice how close are the value (5.15) with the one got for Virgo in epigraph 5.3.

Using the gravitating mass formula (4.21) being $R_{ZV}=f\cdot R_{ZG}$ and by substitution of $f\approx 0.602$ at formula (4.21) where $U=\frac{{100}^{1/5}}{{\Omega }_{DE}^{1/5}}\approx 2.6976$ it is right to get that

$M_G\left(<R_{ZV}\right)\approx 1.5\cdot M_{VIR}$ (5.16)

In the clipped text below, from [9] O. Kashibadze. 2020, the authors gives the interval for [7, 7.3] Mpc for the R_ZV.

rding DMbQG theory, the zero velocity radius R_ZV =7.75 Mpc, see epigraph 5.3, so the relative difference is only 6%.

This is a very good match between experimental results and theoretical results by DMbQG theory.

Also at the concluding remarks from [9] O. Kashibadze.2020, the authors give for the total mass M_T (<R₀) = (7.4 ± 0.9)·10¹⁴ $M_{\Theta }$. As this value is got by dynamical measures in fact this value must be considered as gravitational mass.

The theoretical value (5.12) calculated in epigraph 5.5 is M_G (<R_ZV) ≈1.5·M_VIR = 9.48E14 $M_{\Theta }$ which is 14 % bigger regarding the value given by the authors, if it is considered the upper value of the interval. So both results may be considered compatibles.

Table 10 summarized and compared the observational results and theoretical results.

As R_VIR = 1.7 Mpc its twice value is 3.4 Mpc. As R_ZG = 12.9 Mpc then f = 3.4/12.9 = 0.2635 and by (4.21) $M_G\left(<2\cdot R_{VIR}\right)=1.335\cdot M_{VIR}=8.4\times {10}^{14}{M}_{\Theta }$.

This value match perfectly with the interval of masses given below in the clipped text.

The total mass mentioned in Kashibadze paper it is considered in this paper as gravitating mass because in the whole paper of Kashibadze et al. they use always the concept of total mass as a result of dynamical measures so it is more suitable to interpret his total mass as gravitating mass.

Anyway, considering the total mass given by the formula (4.16) then

$M_{TOTAL}\left(<2R_{VIR}\right)=1.386\cdot M_{VIR}=8.73\times {10}^{14}{M}_{\Theta }$ that match with the upper value of mass range given by the authors.

Below is the clipped text of a paper published for a team of well known astrophysicist.

In page 3 of that paper, they state that at the nearby clusters the mean local density of matter is ${\Omega }_m^{local}=0.08$, whereas the global mass density in the Universe is ${\Omega }_m^{Global}=0.24$ (data year 2007).

Currently that data updated for scientific community is ${\Omega }_m^{Global}=0.3$.

The authors suggest that a possible solution for this tension would be that the total mass for cluster haloes must be three times the virial mass. That is justly what is found in this paper studying the DM at cluster scale as universal law in the framework of DMbQG theory.

In chapter 4, the formula (4.13) $M_{TOTAL}\left(<R_{ZG}\right)=\frac{\sqrt[5]{100}}{{\Omega }_{DE}^{1/5}}\cdot M_{VIR}=U\cdot M_{VIR}$,

shows that the total mass (baryonic and DM) enclosed into the halo cluster is U ≈ 2.7 times the virial mass equal to $M_{TOTAL}\left(<R_{VIR}\right)\approx M_{200}$.

However in the epigraph 4.9 has been calculated an extension of the halo cluster

up to radius R_E = 1.85·R_ZG to obtain a ratio $\frac{{\Omega }_{TM}^{LOCAL}}{{\Omega }_{DE}^{LOCAL}}=\frac{{\Omega }_{TM}^{GLOBAL}}{{\Omega }_{DE}^{GLOBAL}}=3/7$. Now using the formula (4.16) it is right to calculate the total mass enclosed by such sphere $M_{TOTAL}\left(<R_E\right)=M_{TOTAL}\left(<f\cdot R_{ZG}\right)=\sqrt{f}\cdot U\cdot M_{VIR}$ being f = 1.85 and the factor $\sqrt{f}\cdot U=3.67$

then

$M_{TOTAL}\left(<R_E\right)=3.67\cdot M_{VIR}$ (6.1)

Therefore with such factor the parameter ${\Omega }_m^{local}=0.08$ is increased up to a

${\Omega }_m^{local}=0.08\times 3.67=0.2936$ (6.2)

because according [6] Karachentsev et al. 2014, the coefficient ${\Omega }_m^{local}=0.08$ is calculated considering the virial masses of clusters into the Local Universe, and as according DMbQG such total mass is increased by the factor 3.67 if it is considered an extended halo with radius R_E = 1.85·R_ZG then the coefficient ${\Omega }_m^{local}=0.08\times 3.67=0.293$ match perfectly with ${\Omega }_m^{global}=0.3$.

This result enables an experimental test to validate these theoretical findings: If at the present Universe, the average distance between clusters is about its R_E associated to each one, then the DMbQG theory would explain the current ${\Omega }_m^{GLOBAL}=0.3$.

In other words, considering that almost the total baryonic matter at the current Universe is enclosed inside the virial radius of the clusters, as it is confirmed by multiples measures, the DMbQG is able to justify that ${\Omega }_m^{LOCAL}={\Omega }_m^{GLOBAL}$ on condition that the average distance between clusters is the extended radius R_E.