San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 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:

• The gravitational force exerted by a spherical shell or a spherical ball on an outside point is the same as if the entire mass were concentrated at the center of the sphere.
• For a point within a spherical shell there is no net force.

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 an area of dΩr1² in one direction and dΩr2² in the other direction.

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

#### dF = GσdΩr1²/r1² − GσdΩr2²/r2² which reduces to dF = GσdΩ − GσdΩ = 0

Thus for any direction the net force is zero.

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.

The net force on a unit mass at P is

#### dF = G(ρr1dθ)/r1² − G(ρr2dθ)/r2² which reduces to dF = Gρdθ/r1 − Gρdθ/r2and further dF = Gρdθ(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 = Gbr1dθ/r1² − Gbr2dθ/r2² which reduces to dF = Gbdθ(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 outweighs the attraction toward small mass near the center. Thus there is a net radial attraction away from the center of the disk and over time the disk would evolve toward a ring shape. Matter exactly at the center would be held in balance.

## 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 Rd&heta; located at (R, θ) has a magnitude of G(ρRdθ)/s². 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∫02π[R(r−Rcos(θ))/s³]dθ or, equivalently Fr = GρR∫02π[(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)∫02π[(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 (1−(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)³and s = (r² + R² − 2rRcos(θ))½s/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)∫02π[(1−ζcos(θ))/(1+ζ²−2ζcos(θ))3/2]dθ or, equivalently F = (GM/r²)H(ζ) where H(ζ) = (1/(2π))∫02π[(1−ζcos(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 → ∞ (which corresponds to r<<R) H → 0 and that as ζ → 0 (which corresponds to r>>R) H → 1.

 ζ H(ζ) 0.1 1.007570999 0.3 1.073742106 0.5 1.245620571 0.7 1.692441706 0.8 2.270761711 0.9 4.642991428 1.0 0.0 1.1 -3.449766463 1.2 -1.106285155 1.3 -0.615289403 1.4 -0.402346904 1.5 -0.28477633 1.6 -0.211925335 1.7 -0.163423773 1.8 -0.12945636 1.9 -0.104743953 2.0 -0.08621929

This is for a thin ring. For a disk there has to be an integration from 0 to Rmax. The mass of an infinitesimal ring is proportional to R² so this has to be taken into account as well. In terms of the above, this would involve integrating ζH(ζ) from 0 to Ζ.

#### F = ∫0Rmax(GM(R)/r²)H(ζ)dR = ∫0Rmax(GπρR²/r²)H(ζ)dR = Gρ∫0Rmaxζ²H(ζ)dR

In the last formula the variable of integration can be changed from R to ζ=R/r, which means that dR=rdζ. Thus

#### F(Ζmax) = Gρ∫0Ζmax(ζ²H(ζ))dζ

(To be continued.)