__The
Total Force on the Sun as it Orbits Around the Center of the Milky Way __

As the Sun orbits in the
Milky Way, Newton 2 indicates that the total force acting on it must point
toward the galaxy’s center and have the value

__F___{total} = -|M_{Sun}|w^{2}** r**,
(1)

where w is
the angular rate of galactic spin, and ** r** is the Sun’s
displacement from the galaxy’s center. The value of w is
approximately 9.955106388286159e-16 radians
per second , and r is approximately 2.55447e+20 meters. M

F_{total x}=
-5.0353237720337573e+20 nt.
(2)

Since all forces are Real,
it is noteworthy that |M_{Sun}| in Eq.1 is inertial mass, which is Real.
In this article we shall assume that

F_{total x}=F_{grav
x}+F_{gravmag x},
(3)

where F_{grav x} is
the gravitational force exerted upon the Sun in the galaxy's gravitational field
, and F_{gravmag x} is the
gravitomagnetic force. Let us begin by determining
F_{gravmag x}. According to GMT the rotating Milky Way engenders
a gravitomagnetic field which, at *internal*
points, has the *same* direction as the
galaxy’s spin vector.

We can model the galaxy as
a spinning disc of matter of uniform mass density s=m_{MW}/pR^{2} where m_{MW} is the mass of the Milky Way and
R is its radius. Analogous to a rotating disc of positive electric charge’s B
field, at internal points there should be a gravitomagnetic field that points
in the same direction as the galaxy’s spin vector. We may readily derive the
gravitomagnetic field for the Milky Way by applying the GMT transformation rules
to the well-known ** B** field of a
rotating disk of positive

B = swR/2e_{o}c^{2}
( 4)

= qw/2pe_{o}Rc^{2}.

It is noteworthy that B is
not a function of r. ** B** is of course normal to the disk surface for all r<=R..

The gravitomagnetic
equivalent of Eq. 4 would then be

O = 2pGswr_{q}/c^{2}
(5)

=2Gi|m_{MW}|w/R_{MW}c^{2}

=5.8814292458671056e-21 i.

Viewed from the inertial
frame in which the center of the Milky Way is at rest, the Sun should experience
an *inward*-pointing gravitomagnetic
force of magnitude

F_{gravmag }= m_{Sun}(v)(O)
(6)

= i|m_{Sun}|(wr)(2Gi|m_{MW}|w/R_{MW}c^{2}

Eq. 6 solves to

F_{gravmag }=
-2974845204074973.5 nt. (7)

And, since F_{grav}
and F_{gravmag} both point toward the galaxy’s center,

F_{grav} =
F_{total }- F_{gravmag }= -5.035294023581716e+20 nt.
(8)_{ }

Thus F_{gravmag} is
less than F_{grav}:

F_{gravmag}/F_{grav}
= 5.907987081077939e-06.
(9)

Nevertheless under the
influence of gravity alone, the Sun (and all of the galaxy’s stars) would not
remain at fixed distances from the galactic center. The
addition of the gravitomagnetic force is just what is required in order
to hold the galaxy together.