First, let us be specific about the superposition of collinear forces, at least in one case. Others are similar. Assuming the form GM_AM_B/(R_eff)² and referring to Figure 2, panel 2A, we find that

1 [ R e f f ( c f ; S 2 ) ] 2 = 2 [ R e f f ( p f ; S 2 ) ] 2 − 1 [ R e f f ( f f ; S 2 + B ) ] 2 (1)

where, in order of their appearance, “cf” refers to curved-flat; “pf” refers to point-flat (because the sphere acts as a point source), and “ff” refers to flat-flat, as also shown in the first panel. The separation for each R_eff, S₂ and S₂ + B, are also shown. The factor of 2 exists because the mass of the sphere is twice that of the hemisphere.

Next, we present the formulas for the magnitude of the force between a sphere and a hemisphere required in Figure 2(A), Figure 2(B), and 3 of Figure 2. In panels 2A and 2B, the flat surface of the hemisphere faces the sphere, but with A and B reversed. In panel 3, the curved side of the hemisphere faces the sphere. A simple integration yields the following for the force, F_pf, in Figure 2(A):

F p f = G M A ( 2 M B ) [ 3 A 3 ] [ ( 1 3 S 2 ) { ( 2 S 2 − A 2 ) A 2 + S 2 + 3 A S 2 + ( A ( A − S ) − 2 S 2 ) ( A + S ) } ] (2)

where we identify the multiplier of [GM_A(2M_B)] with (1/R_eff)², S = S₂ + B, and M_A and M_B are the two hemispherical masses. In the case of panel 2B, A and B are reversed in the above formula. The equation for the force between the curved side facing the sphere and the sphere can be obtained by a similar integration, resulting in a similar formula. More simply, it can be obtained by subtracting from the force between two spheres the force between the sphere and the hemisphere (flat side facing the sphere), as shown in Figure 4. The force between the two spheres is simply proportional to 4 / ( S 3 + A + B ) 2 .

As was stated earlier, this situation is the only one requiring a nontrivial calculation. It was performed in two ways. The first involves a direct integration of the force between all of the points in one hemisphere and all of the points in the

other. Because integration over two volumes was required, it started out as a six-fold integration. However, azimuthal symmetry reduced it by one integral and another integral could be done analytically, thus reducing the problem to a still-unpleasant four-fold integration. The second way involved expansions of the potential created by one hemisphere along its axis for positions z ≤ A and z ≥ A. The off-axis values were then expressed in terms of the original expansion multiplied by Legendre polynomials, as in electrostatics.

Since the gravitational force between the two halves of a solid sphere, such as the earth, is of some interest to inquiring minds [2] , we shall consider two applications of the result immediately above. The first concerns how the gravitational force between two contacting spheres, each of the same mass as the individual hemispheres, compares to the force between the two hemispheres. Since the ratio of the radii, sphere to hemisphere, is 1/2^1/3, the ratio of the two forces, spheres to hemispheres, can easily be deduced as (2^2/3/3) ≈ 0.53. Thus, the force between the two hemispheres is almost twice that of the two spheres. The second application is concerned with the rotation rate of the earth required to separate its two halves under only the combined action of the fictitious centrifugal force and gravity. The calculation is done with reference to Figure 7, in which is shown a sphere of radius A split along a plane perpendicular to the plane of the paper and passing through the central axis of rotation. The split sphere is rotating with angular velocity ω. The coordinates (r, φ, z) define the position of a rotating point. The total “centrifugal force”, F_c, is obtained by integrating the component of the force perpendicular to the interface plane over the hemisphere.

F c = 2 ω 2 ρ ∫ 0 A d z ∫ 0 ( A 2 − z 2 ) 1 / 2 r 2 d r ∫ − π 2 π 2 cos φ d φ = π 4 ω 2 ρ A 4 (6)

Equating F_c to [G(πρA²)²]/3, solving for ω, and expressing ω as 2πN_r, where N_r is the number of rotations per second, we obtain N_r = [Gρ/(3π)]^1/2. This results in ≈ 20 × 10⁻⁵ rotations/sec. required to just separate the two halves of the sphere. The earth’s rotational speed ≈ 1.16 × 10⁻⁵ rotations/sec., which ≈ (1/17) of the required speed. Of course, the separation speed would be much greater if the cohesion within the earth, which serves a kind of geological corpus callosum, were taken into account. Furthermore, tidal friction is causing the earth’s rotational speed to slowly diminish. Thus, we have nothing to worry about.

In this section, we present a graphical array of results, in addition to the one already presented in Figure 6. The graphical array will be of (1/R_eff,A)² vs. S_A for the various configurations in Figure 1 and of Alpha (the spelled out version of the Greek letters in Figure 2) for those same configurations. Figure 8 is a more detailed view of the behavior of (1/R_eff,A)² for S_A ≤ 3, b = 1, and for its three distinguishable configurations. The dotted line is added to show the linear behavior in the immediate vicinity of S_A = 0. The ordering in size is consistent with the amount of mass in one hemisphere that is close to the other hemisphere. That order is flat-to-flat, flat-to-curved, curved-to-curved. Figure 9 is similar to Figure 8, but with b = 0.25. For these graphs, (1/R_eff,A)² is larger than in Figure 8 because one of the hemispheres is so much smaller. The ordering in size in Figure 9, in general, is also consistent with the amount of mass in one hemisphere that is close to the other. Curiously, however, there is a cross over between con-

figurations 2A and 2B at S_A = 0.25, which is maintained down to S_A = 0. Figure 10 and Figure 11 are the same as Figure 8 and Figure 9, except that the entire range of values are shown on a log-log scale. The dashed line in Figure 10 shows how the behavior at large values of S_A varies as 1 / S A 2 . Figure 12 and Figure 13 illustrate the behavior of Alpha for various configurations and the two values of b (1 and 1/4). Alpha 1 (or Alpha 1/4) is the value of Alpha for a hemispherical radius = 1 (or 1/4). As expected from the comment in the introduction, this parameter approaches (3/8) or (5/8) for large values of S_A, depending on whether

the flat or curved side faces the other hemisphere. That condition is denoted by “flat” or “curved” in Figure 12. The curves in Figure 12 and Figure 13 can be interpreted as indicating that the force between the two hemispheres decays away more slowly than 1/S² and only approaches that behavior at very large vales of S (as shown in Figure 10 and Figure 11). As Alpha diminishes, the increasing separation between the two fictitious point masses with S is partially offset by their move toward each other within their respective hemisphere.

Without the use of high-speed computers and sophisticated computational software, such as Wolfram Mathematica or, perhaps, MATLAB, it would be extremely difficult to perform the four-fold integration that makes up Equation (5). This is particularly true because of the coupling among the variables of integration. However, the use of a power expansion of the on-axis gravitational potential of hemisphere A (or B), followed by a term-by-term multiplication by Legendre polynomials to produce off-axis values of the potential simplifies the calculation. This is commonly done when dealing with electrostatics problems containing azimuthal symmetry, and it is discussed in many textbooks [3] . This method can be applied to the current problem because, as in electrostatics, it requires a solution of Laplace’s equation. The only possible drawback is that the result is an infinite series, which may require a largenumber of terms to produce the desired accuracy. Fortunately, this will not be true of the number of terms required for an accurate solution to the current problem.

This method of solution for configuration 1 is presented in conjunction with Figure 14, where a hemisphere of radius A, assumed to be the source of the potential, Φ(R, θ), is shown with the hemisphere of radius B, assumed to be in the gravitational field of hemisphere A.

Once the potential, Φ(r, θ), is determined, its negative z-derivative (the z- component of its gravitational field) is calculated and integrated over hemisphere B to determine the total force between the two hemispheres. In the case of Figure 14, when R ≤ A, the on-axis potential can be written as a power series in R^l, whereas for R ≥ A, it can be written as a power series in R^−(l+1). Thus, the potential at angle θ is of the form:

Φ ( R , θ ) = ∑ l = 0 ∞     C l R l P l ( cos θ ) ;     R ≤ A (7a)

Φ ( R , θ ) = ∑ l = 0 ∞     D l R − ( l + 1 ) P l ( cos θ ) ;     R ≥ A (7b)

C_l and D_l are parameters to be determined and P l ( cos θ ) is the Legendre Polynomial of order l, all of which are unity at θ = 0. In order to proceed, we must first determine the on-axis potential. After a simple integration it can be shown to be:

Φ ( R , 0 ) = − G ( 2 π ρ a ) [ ( A + R ) 3 3 R − ( A 2 + R 2 ) 3 2 3 R − A R − A 2 2 ] (8)

Ignoring constants in the above expression, which contribute nothing to the force, results in the following expansions for R ≤ A and R ≥ A, respectively.

Φ ( R , 0 ) = − G ( 2 π 3 ρ a A 3 ) ( 1 A ) [ − ( 3 2 ) ( R A ) + ( R A ) 2 − ( 3 8 ) ( R A ) 3 + ( 1 16 ) ( R A ) 5 − ( 3 128 ) ( R A ) 7 + ⋯ ] (9a)

and

Φ ( R , 0 ) = − G ( 2 π 3 ρ a A 3 ) ( 1 A ) [ ( A R ) + ( 3 8 ) ( A R ) 2 − ( 1 16 ) ( A R ) 4 + ( 3 128 ) ( A R ) 6 − ( 3 256 ) ( A R ) 8 + ⋯ ] (9b)

From these two equations, the constants C_l and D_l can be directly determined. To obtain Φ(R, θ), we multiply each term in the above two expansions by the appropriate Legendre polynomial, express the expansion in terms of the hemispheric coordinates z, r, φ and, to obtain the force, integrate its negative z-deriv- ative over z from s to ( s + ( B 2 – r 2 ) 1 / 2 ) and over r from 0 to B. The azimuthal integration over φ is simply replaced by 2π. In addition, we note that cos θ = z / ( z 2 + r 2 ) 1 / 2 . When all of this and a few other manipulations are completed, we identify (1/R_eff,A)² for R ≤ A as:

1 ( R e f f , A ) 2 = ( 3 b 3 ) ∫ 0 b r a d r a ∫ S a S a + ( b 2 − r a 2 ) 1 / 2 d z a ∂ ∂ z a [ 3 2 ( z a 2 + r a 2 ) 1 2 − ( z a 2 + r a 2 ) P 2 ( z a ( z a 2 + r a 2 ) 1 2 ) + ⋯ ] ;       R ≤ A (10a)

We note the use of scaled coordinates and the fact that the z_a-integral is particularly simple because it is the integral of a derivative. The integral over r_a was done numerically. For R ≥ A, the results are similar:

1 ( R e f f , A ) 2 = ( 3 b 3 ) ∫ 0 b r a d r a ∫ S a S a + ( b 2 − r a 2 ) 1 / 2 d z a ∂ ∂ z a [ − ( z a 2 + r a 2 ) − 1 2 − ( 3 8 ) ( z a 2 + r a 2 ) − 1 P 1 ( z a ( z a 2 + r a 2 ) 1 2 ) + ⋯ ] ;     R ≥ A (10b)

For simplicity, only the first two terms of each expansion are shown in Equations (10a) and (10b). Figure 15 and Figure 16 show the results of this method for b = 1 and b = 0.25, respectively, along with those using Wolfram Mathematica. With only a very few terms of the expansion, the two sets are virtually indistinguishable on the graphs.