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Restating Newtons universal law of gravity in terms of volume

Usually Newtons universal law of gravitation is stated in the following way:
  1.         F = Mm/r2
M and m are two point masses, at rest; r is the distance between the two. G is the gravitational constant, needed to make the dimensions match, as well as needed to make observations match with predictions. G is an empirical constant, the logical consequence of the fact that once we have chosen units for distance, time and mass..
Newton was interested in the force F. [In fact there are two equal but opposite forces: the force from M towards m, and the force from m towards M.]
In the context of galaxys, we're more interested in accelerations than in Forces. So we use:
  1.         F = ma
And then we substitute:
  1.         a = MG/r2
This is just the acceleration the lesser mass m experiences as a result of the larger mass M. This tells us this rate of acceleration is independent of the value of m: a grain of sand in the orbit of the earth around the sun experiences exactly the same acceleration towards the sun as the earth itself. [Since G is a constant, and M does not vary within my expressions, this comes down to: a ~ 1/r2.]

All of this is valid for a point mass M at rest, and - as Newton showed - also for a homegenous shell around that point mass (with the same center of gravity). So this works really well within our solar system, where 99% of all mass is concentrated in a small sphere, called 'the sun'.
But our galaxy is not a sphere. It is a disk with exponentially diminishing density, with about only a fifth of its mass in the centre (the bulge). It is not obvious how to use the above formula for acceleration, when a galaxy consists of lots of different masses, with a lot of different distances between them, in a certain distribution.
To work around this, I will reformulate Newtons law into Volume-change as a function of time (-interval).

Think a spherical volume around a point mass, with radius r0. The surface area of the sphere is:
  1.         area = 4πr02
I'll take a time-interval of Δt, for Δt → 0, so that the accelaration a during the interval may be taken to be constant. In that case, starting from rest:
  1.         Δs = -1/2 a Δt2
In the interval Δt the Volume of the sphere shrinks with (area x Δs), so
  1.         ΔVol = -2πar02Δt2
Now I substitute the acceleration a from (3) into (6).
Since Δt → 0, also Δs → 0 , and r0 → r
  1.          ΔVol = -2πMGΔt2
This is an interesting statement.
It says that the decrease in volume is determined only by time (squared) and by Mass.
The amount of change in volume does not depend on the radius of the sphere.
That means that
-it is independent of the size on the volume I put around my point mass M
-it is independent on the shape of the volume I put around my point mass M,
-it is independent on the distribution of the mass within the shape.

In the case of our own Milkyway (total Mass = 16,6 * 1040 kg) this means that, no matter the distribution of its mass, no matter the surrounding shape of volume I choose, as long as all mass is within that shape, in 1 second that volume will decrease by 69,554 * 1030 (m3). This will be the crux of my line of reasoning.

Please take note that I did not change Newtonian gravity in any way by formulating it in terms of volume and time.