**1. Abstract**
On the Basis of tabular values of the gravitational constant. The calculated
mass of the nucleus of the milky way galaxy. The numerical value of the
gravitational constant is determined by the mass of the nucleus of the
milky way galaxy Abstract: Based on the tabular value of the gravitational
constant, the mass of the Milky Way galaxy is calculated. Based on the
physical meaning the gravitational constant, and the coincidence with the
calculated on the surface Of the Sun, the mass of the Milky Way galaxy
is calculated. gravitational constant on the surface of the Sun, the mass of
the galaxy The Milky Way.

**2. Keywords:
**
Gravitational constant, The core of the galaxy, The core mass of the
Galaxy, Physical meaning of the gravitational constant, Gravitational
constant on the surface of the Sun, The mass of the galaxy The Milky Way.

**3. Gravity in the center and on the surface of the Sun**
Let the Sun be in the empty space of the universe (on the “edge“The
universe). In the center of the Sun we introduce a spherical coordinate
system At the beginning of the coordinate system, we place the proton p.
Wave fronts of a high-frequency gravitational field re-emitted by matter
The suns fall on the proton p within the solid angle of 4π.When the
wave front interacts with a proton p, the energy of the proton p becomes
indeterminate for some time, according to the principle Heisenberg
uncertainties Δε*Δt ≥ h/( 2π) ; where h is Planck’s constant, Δε is the
uncertainty of the
proton energy over time Δt . When a wave front passes through a proton,
it re-emits excess energy into the surrounding space in the form of a
spherically symmetric gravitational wave. Calculate the gravitational energy incident on a proton p within a solid angle of 4π in one second.
The number of protons in one cubic meter of the Sun’s matter is equal
to n = p/m = M/(V*m) where p is the density of the Sun’s matter, M is
the mass of the Sun, V is the volume of the Sun, m is the rest mass of
the proton. Let us isolate an elementary volume dV inside the surface of
the Sun at a distance R from the proton p . With d V = R2 d θ d φ d
R Fig. 1). The number of protons in the volume d V is equal to d N = n
d V = R2 d θ d φ d R. In one second, the volume d V re-emits
gravitational energy in the amount of Δε*d N = R2 d θ d φ d
R ; where π = 3.14. The energy incident on the proton p from the entire
volume of the Sun in one second εр = ;
where 0 ≤ R ≤ R0 ; 0 ≤ θ ≤ π ; 0 ≤ φ ≤ 2π ; R0 - is the radius of the Sun.
The volume of the Sun V = π R0
3 . r- is the radius of the proton. or εр=
After solving the integral , we
get εр = ; Select a small region in the vicinity of the proton p
space free from the matter of the sun. Let’s put two protons p1
and p2
in
this cavity at a distance R from each other. Each of the protons p1
and p2 in
one second, it re-emits of gravitational energy within a solid angle of 4π.
Energy gets from proton p1
to proton p2 ε = = ; m
is the rest mass of the proton. In the last
expression R is the distance between the test protons p1
and p2
. The pulse
received by the proton p1
in one second
Δр = = ; where c is the speed of light in vacuum. The
force of attraction between protons p1
and p2
is defined as F = Δр . ( Δt
=0 ). Substitute = γ ; where γ is the gravitational
constant at the center the Sun. Hence γ = ; Substitute
the values of the quantities, we get the coefficient of gravity in the center
of the Sun γ = 2.9256* 10 – 10 н*м2
*кг – 2 Let’s calculate the coefficient
of gravity on the surface of the Sun. Let the center of the Sun S be located
on the OZ axis of the spherical coordinate system at a distance R0
from
its origin (Fig.2). R0
is the radius the Sun, M is the mass of the Sun. Let’s
put a test proton at the beginning a spherical coordinate system on the
surface of the Sun . Select inside The elementary volume of the Sun dV
= R2 d θ d φ d R The number of protons in one cubic meter of the
Sun n = = ; where p is the density of the substance the Sun, M
is the mass of the Sun, V is the volume of the Sun, m is the rest mass of
the proton. The number of protons in the dV volume is equal to d N = n
d V = R2 d θ d φ d R In one second, the volume d V re-emits
gravitational energy in the amount of Δε*d N = R2 d θ d φ d
R ; where π = 3.14. Where Δε is taken from the Heisenberg uncertainty
ratio Δε*Δt ≥ h/( 2π) ; where h is Planck’s constant, Δε is the uncertainty
of the proton energy over time Δt . At Δt = 1 sec. we take Δε = h/( 2π).
The energy incident on the test proton in one second from the volume d V
is equal to The energy incident on the proton p from the entire
volume of the Sun in one second εр =
; where 0 ≤ R ≤ 2R0 ; 0 ≤ θ ≤ ; 0 ≤ φ ≤ 2π ; After calculating the
integral, we get εр = (1 - ) Now let there be two test protons on the surface of the Sun with the re-emission energies of εр
. The
energy incident on proton p2
from the side of proton p1 will be defined as
ε = = (1 - ) ; In the last expression , R is the
distance between the trial protons p1
and p2
Pulse received by proton p2
in one second Δр = = ; where c is the
speed of light in vacuum. The force of attraction between protons p1
and
p2
is defined as F = Δр. Or = γ ; where
γ is the gravitational constant on the surface of the Sun. Hence
γ = ; (1)
Substitute the values of the quantities, we get the gravity coefficient
(gravitational constant) on the surface of the Sun (r- is the radius of the
proton). γ = 8.572* 10 – 11 н*м2
*кг – 2

**4. Calculate The Mass Of The Milky Way Galaxy**
Formula (1) is good with a tabular value γ = 6.672*10 – 11 нм2
кг – 2. From
here, mentally, the Milky Way galaxy can be represented as a homogeneous
huge star with a radius of Н. The density of which is equal to the density
of the Sun. Let About the center of the core of the Milky Way galaxy. R0
is the radius of the globular region of the galaxy centered at point O and
mass M. The solar system is located at a distance of H + R from the center
of the galaxy (Fig. 1). We introduce the system coordinates so that the
center of the core O is on the z axis at a distance H from the origin of the
coordinate system. Let’s put a test proton at the beginning of the system
coordinates to the point o. The galactic core determines the distribution of
the gravitational constant within the galaxy. The gravitational constant is
no longer constant inside the galaxy. The farther away from the center of
the galaxy, the smaller it is. The mass of the Galactic core is determined
in terms of the gravitational constant inside the Solar system (table value).

**5. The Basic Part**
Let the center of the nucleus of the galaxy О. The radius of the ball
region of the galaxy is R0
. The mass of a galaxy M. The Solar system is
at a distance H from the center of the galaxy (fig.1.1.). We introduce
a coordinate system. To the center of the kernel was on the z axis at a
distance H from the beginning of the coordinate system. Place a test
proton in the origin of the coordinate system at point p. In inside the ball
volume. dV = R2 d θ d φ d R . The number of protons in one cubic
meter inside the ball n = = ; where V – the volume of a sphere
of radius R0
, m is the mass of a proton, ρ is the density substance inside the
ball, M – mass of the substance inside the bowl. The number of protons
inside the volume dV is equal to dN = R2 d θ d φ d R The
volume dV in one second, emits gravitational energy in the amount of Δε
* dN = R2 d θ d φ d R ; where Δε taken from Heisenberg’s
uncertainty principle: Δε * Δt = ; Take Δt = 1 second, then Δε = .The energy of the incident proton p per second from the Volume dV is
equal to Δε * dN ; where r – the radius of the proton ( r = 1.5 *10 –
15 m); π = 3.14. R – the distance from the volume dV before the beginning
of the coordinate system. The energy of the incident proton p. 1 second of
the entire volume of a sphere
εp = ; where 0 ≤ θ ≤ arctg θ , 0 ≤ ≤ 2π ;
H – R0 ≤ R ≤ H + R0 ; tg θ = . In view of the smallness of the angle
θ we write arctg θ = θ = ; After calculating the integral get εp =
Let on the edge of the Solar system. Far away from massive objects.
There are two protons at a distance R from each other. Energy falling on
the p2 proton from the proton p1 is equal to ε = εp =
The impulse received by the proton p2
in one second is equal to Δp =
The force of attraction between the protons p1
and p2
is equal to
F = Δp = γ = ; where с – the speed of light in vacuum.
Hence the weight of the Central region of the milky way galaxy M =
γ ; where γ the table value of the gravitational constant (γ =
6.672*10 – 11 н m2
кг – 2). Take Н = 2.65*1020 м. then М = 1.3*1053 кг

**6. Conclusion**
The galactic core in the weight specifies a numeric value the gravitational
constant inside the Solar system.

Yanbikov Vil. The mass of the Milky Way galaxy. Annals of Clinical and Medical Case Reports 2023