San José State University

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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 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/r2and 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 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 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∫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 ((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[(ζ−cos(θ))/(1+ζ²−2ζcos(θ))3/2]dθ or, equivalently F = −(GM/R²)H(ζ) where H(ζ) = (1/(2π))∫02π[(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.

 ζ H(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.

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 R² so this has to be taken into account as well.

#### F = −∫0Rmax(GM(R)/r²)H(ζ)dR = −∫0Rmax(GρπR²/r²)H(ζ)dR = −Gρπ∫0Rmax(1/ζ²)H(ζ)dR

In the last formula the variable of integration can be changed from R to ζ=r/R, which means R=r/ζ and hence that dR=−(r/ζ²)dζ. Thus

#### F(r,Z) = −Gρπr∫0Z(−H(ζ)/ζ4)dζ and hence F(r,Z) = Gρπr∫0Z(H(ζ)/ζ4)dζ

where Z=r/Rmax. The expression Q(Z) =∫0Z(H(ζ)/ζ4)dζ takes the form of

The force on a unit mass at a distance r from the center of a disk of radius Rmax then takes the form

#### F(r,r/Rmax) = GρπrQ(r/Rmax)

Since the mass of the disk Mdisk is equal to ρπRmax² the above formula can be expressed as

#### F = G(Mdiskr/Rmax²)Q(r/Rmax) or, equivalently F = (GMdisk/Rmax)(Q(Z)/Z³)

The function Q(Z)/Z³ takes the form of

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

(To be continued.)