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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

Ftotal = -|MSun|w2r,          (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. MSun=1.989e30 kg. Substituting in Eq. 1 we find that

Ftotal x= -5.0353237720337573e+20 nt.         (2)

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

Ftotal x=Fgrav x+Fgravmag x,         (3)

where Fgrav x is the gravitational force exerted upon the Sun in the galaxy's gravitational field , and Fgravmag x is the gravitomagnetic force. Let us begin by determining  Fgravmag 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=mMW/pR2 where mMW 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 charge. At the center of the disk of charge the B field has a magnitude of

B = swR/2eoc2   ( 4)

          = qw/2peoRc2.

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 = 2pGswrq/c2 (5)

          =2Gi|mMW|w/RMWc2

          =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

Fgravmag = mSun(v)(O)   (6)

                = i|mSun|(wr)(2Gi|mMW|w/RMWc2 ).

Eq. 6 solves to

Fgravmag = -2974845204074973.5 nt.      (7)

And, since Fgrav and Fgravmag both point toward the galaxy’s center,

Fgrav =  Ftotal - Fgravmag =  -5.035294023581716e+20 nt.     (8)

Thus Fgravmag is less than Fgrav:

Fgravmag/Fgrav = 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.