San José State University

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Thayer Watkins
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Gravitational Force in a Thin Disk of Matter

For mass distributed in a spherical arrangement there are some marvelous theorems that simplify the analysis. These include:

Such theorems do not hold for matter not distributed spherically. In particular they do not hold for matter distributed in a thin disk. To see why the internal statics of matter distributed in a disk is completely different from that of matter distributed in a sphere consider first the case of a spherical shell.

Take any point P within the spherical shell and construct a circular cross section cone with its apex at P. Let dΩ be the solid angle for the cone and σ be the areal mass density on the spherical shell. Let r1 and r2 be the distance from the point P to the shell along the axis of the cone. The cone subtends areas proportional to dΩr1² in one direction and to dΩr2² in the other direction. The constants of porportionality are equal to the cosines of the angles between the direction of the cone axis and the normals to the spherical surface. These normals are just the directions of the unit radial vectors from the center of the sphere. The angles are exactly the same at both intersections. This is shown in On the Angles of Intersection of Lines and n-Spheres. Let β be the constant of proportionality.

The mass acting on the point P along the axis of the cone is then approximately σdΩr1² in one direction and σdΩr2² in the other. The net gravitational along the axis of the cone is

dF = Gσ(βdΩr1²)/r1² − Gσ(βdΩr2²)/r2²
which reduces to
dF = GσβdΩ − GσβdΩ = 0

Thus for any direction the net force is zero to the first order of approximation.

For a ring let ρ be the linear mass density. For an arbitrary P within the ring pass a line through P and construct two lines which have an angle with the line of ½dθ. Let the midline between these two lines be denoted as L.

The lengths of the ring which are subtended by dθ are proportional to r1dθ and r2dθ. The constant of proportionality it the cosine of the angle between the line L and the radial vector from the center of the circle. As noted previously the proof that the two angles are equal is at On the Angles of Intersection of Lines and n-Spheres. Let β be the constant of proportionality. The net force on a unit mass at P is

dF = G(ρ(βr1dθ))/r1² − G(ρ(βr2dθ))/r2²
which reduces to
dF = Gρβdθ/r1 − Gρβdθ/r2
and further
dF = Gρβdθ(1/r1 − 1/r2)

Thus there is a net attraction toward the point on the ring which is closest to P.

To make the analogy with the spherical shell closer consider a cylindrical band of width b. The areas subtended by two planes separated by an angle dθ are (r1dθ)b and (r2dθ)b. The net force is then

dF = Gβbr1dθ/r1² − Gβbr2dθ/r2²
which reduces to
dF = Gβbdθ(1/r1 − 1/r2)

Therefore there is a net attraction to the closer point on the cylindrical band.

For mass outside of a ring there is a net attraction toward the ring. Thus for mass near the edge of a disk there is a net radial force toward the center of the disk. For mass near the center of the disk, the attraction of the outer part of the disk might outweigh the attraction toward small mass near the center. Thus there might be a net radial attraction away from the center of the disk and over time the disk could evolve toward a ring shape. Matter exactly at the center would be held in balance.

There are some galaxies having a ring-shape such as the one below.

However this form is not typical and later it will be shown that there is no net attraction away from the center of a disk. First consider the case of the gravitational attraction of a ring.

The Gravitational Attraction of a Ring

Consider mass distributed throughout a thin ring of radius R at a uniform linear density of ρ. The polar coordinate system has its origin at the center of the ring. An element at radius R and angle θ has xy coordinates of (Rcos(θ), Rsin(θ)) has a distance s from the point at r and angle 0 (xy coordinates (r, 0)) given by

s² = (r−Rcos(θ))² + (Rsin(θ))²
which reduces to
s² = R² + r² − 2rRcos(θ)

The gravitational force at (r, 0) due to an infinitesimal element of length Rdθ located at (R, θ) has a magnitude of G(ρRdθ)/s². Now it is important to adopt the convention that any force directed toward the center is negative and any force directed away from the center is positive.

The radial component of the force is the important factor. Let φ be the angle between the radial line to (R, 0) and the point (R, θ). This is the angle the force makes with the radial line to (r, 0). The cosine of the angle φ of the force is equal to (r−Rcos θ)/s. Thus the radial component of the force due to the infinitesimal element is

dFr = −Gρ[R(r−Rcos(θ))/s³]dθ

The tangential component of the total force, by symmetry, vanishes. The total force is then simply the integral of this expression from 0 to 2π radians; i.e.,

Fr = −Gρ∫0[R(r−Rcos(θ))/s³]dθ
or, equivalently
Fr = −GρR∫0[(r−Rcos(θ))/s³]dθ

The mass M of the ring is 2πRρ so the above formula may be put into the form

Fr = −GM(1/2π)∫0[(r−Rcos(θ))/s³]dθ

This expression may be simplified by dividing the numerator and denominator of the integrand by R. The numerator then becomes ((r/R)−cos(θ)). The denominator, which is s³/R, is better expressed as R²(s³/R³). (The factor R² will later be brought outside of the integration.) Since

s³/r³ = (s/r)³
and
s = (r² + R² − 2rRcos(θ))½
s/R reduces to
((1 + (r/R)² − 2(r/R)cos(θ))½

Letting the ratio (r/R) be denoted as ζ the force can be expressed as

F = −((GM/R²)(1/2π)∫0[(ζ−cos(θ))/(1+ζ²−2ζcos(θ))3/2]dθ
or, equivalently
F = −(GM/(2πR²))H(ζ)
where
H(ζ) = ∫0[(ζ−cos(θ))/(1+ζ²−2ζcos(θ))3/2]dθ

There might be an analytical evaluation of H(ζ) but for now a numerical evaluation will suffice. However it is easily seen that as ζ → 0 (which corresponds to r<<R) H → 0 and that as ζ → ∞ (which corresponds to r>>R) H → 1/ζ² → 0.

ζH(ζ)
0.1 -0.318
0.2 -0.658
0.3 -1.048
0.4 -1.528
0.5 -2.167
0.6 -3.100
0.7 -4.644
0.8 -7.855
0.9 -22.139
1.0 0.000
1.1 27.529
1.2 11.542
1.3 7.568
1.4 5.607
1.5 4.410
1.6 3.600
1.7 3.015
1.8 2.574
1.9 2.231
2 1.957
2.1 1.733
2.2 1.548
2.3 1.393
2.4 1.261
2.5 1.147

Because of the negative sign in the formula for force, a negative value for H(ζ) corresponds to a positive force; i.e., a force directed away from the center. The singularity is at ζ=1.0.

This force is for a single thin ring. For a disk of radius Rmax there has to be an integration over R from 0 to Rmax. The mass of an infinitesimal ring is proportional to its area of 2πRdR so this has to be taken into account as well. Let σ be the areal mass density. (The density ρ used before was the linear mass density.)

Starting with an infinitesimal element of RdθdR, its mass is σRdθdR and the gravitational force due to the element is −G(σRdθdR)/s². The radial component of the gravitational force due to the element is then

dFr = −[GσR(r − Rcos(θ))/s³]dθdR

Thus the total force on a unit mass at (r, 0) is

F = −∫0Rmax(GσR(r−Rcos(θ))/s³]dθdR
and after the numerator and denominator
of the integrand are divided by R³
F = −Gσ∫0Rmax[(ζ−cos(θ)/(R(1+ζ²−2ζcos(θ))3/2)]dθdR
which can be expressed as
F = −Gσ∫0Rmax(1/R)H(ζ)dR

where, as before,

H(ζ) = ∫0[(ζ−cos(θ)/(1+ζ²−2ζcos(θ))3/2]dθ

Thus,

F = = −Gσ∫0Rmax(1/R)H(r/R)dR

Below is the relationship between the force F and the distance from the center of the disk. There is a singularity at the edge of the disk because the mass within the disk is assumed to be continuous. Within the disk the attraction of the infinitesimally close material toward the center is counterbalance by the infinitesimaly close material beyond the point. At the edge of the disk the counterbalancing force from the material beyond the point is not there.

The computation shows that even within the disk the net force is directed toward the center.

For large values of r relative to Rmax the force is roughly proportional to 1/r².

(To be continued.)


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