# Different distributions of mass

In the following I will consider 'volume', even 'empty volume' as an entity that can move and accelerate due to gravitational forces. This is because I can allways mark the surface of any volume by as many points as needed to determine its place and shape. Each of those points can have a mass of limit to zero, since the amount of acceleration does not depend on the mass of the accelerated object.

Previously, I derived:

7) | ΔVol = 2πMGΔt^{2} |

*shape*of this volume may change, but the total amount of volume stays the same.

This is a well known fact in the context of tidal forces, and it is easy to see why this is true: if you picture an amount of mass, and around it you place some shape of volume. Around this first volume you place a second volume. Now, if one second passes, both volumes shrink with exactly the same amount, so the volume in between has to remain unchanged. But the shape is different.

Another consequence is, that the amount of acceleration at any point

*does*vary with the distribution of mass. I shall illustrate this for three different cases: a point-mass, a line-mass and an area-mass.

## point-mass

The point where alle lines converge represents an attracting point mass.

Start with the outer square with an area of W (light-blue area). In Δt it moves towards the center. The darker blue area (W') depicts the place of this square after Δt seconds. The distance travelled is Δs. As al four corners move towards one point, the square has to shrink in two dimensions. (length and width)

Compare the outer square to the inner square. Suppose the latter is located at 90% between the center and the outer square. In that case, the area W' is only 0,9

^{2}of the area of the outer square, i.e. W' = 0,81 W.

In the same Δt the second square W' also moves towards the center, covering a distance Δs'.

Since the volume W * Δs = (0,81 W) * Δs', it follows that Δs = 0,81 Δs'. So s' is larger than s, and it is a relation of the second degree.

Because this is freely scalable, it implies that for any pointmass the relationship a ~1/r

^{2}holds.

This is hardly surprising, since this is where we started from in the first place.

## line-mass

The black line represents in infinitely long line with a fixed mass (M) per meter (l), so M/l = constant.

Start with the outer ractangle with an area of W. In Δt it moves towards the line (at a right angle, as depicted).

The distance travelled is Δs.

As al four corners move towards to the line at a right angle, it has to shorten in only one dimension.

Compare the outer rectangle to the inner rectangle. Suppose this one is located at 90% between the center and the outer rectangle. In that case, the area of the inner rectangel is only

*0,9*of the area of the outer rectangle, i.e. W' = 0,9 W. In the same Δt the second rectangle moves towards the line, covering a distance Δs'.

Since the volume W * Δs = (0,9 W) * Δs', it follows that Δs = 0,9Δs. Because this is freely scalable, it implies that for a line mass the relation a ~ 1/r.

I made a numerical simulation of a point attracted to a line, and the exact relationship appears to be

8) | a = 2G/r * M/l |

Excel sheet |

## area-mass

There is really not much to draw to illustrate what happens in the case of an infinite plane with a constant density (M/l^{2}= constant). If you take a square parallel to that infinite plane , alle corners of that square accelerate straight towards the plane. Since the area of the square does not change (unlike with a point- or a linemass), the amount of acceleration is not affected by the distance to the infinite plane. You can find my simulation in the Excelfile linked above. The relation appears to be:

9) | a = 2πG * M/l^{2} |

Now that I have covered the basic concepts I am going to use, I can go on and make a numerical approximation of the acceleration of different locations in our galaxy.