Formulation of the problem

Let us consider the following problem:

We have at our disposal a certain amount of matter with specific density η and volume V. This amounts to the mass M = ηV capable of generating gravitational attraction. Now suppose that we wish to use this matter to construct an object which, at one point of its surface, attracts small material objects with a maximum possible force.

Figure 1. Getting maximum gravity from a fixed amount of matter.

A given amount of matter of constant density is to be arranged so as to maximize the gravity at a barypole B. The matter will have to be located all on one side of a baryplane passing through B which we imagine perpendicular to the image (the solid horizontal line is its footprint). The horizontal disk (a) and vertical column (b) arrangements are not optimal since a transfer of a small amount of mass indicated by the arrows increases the gravity at B. The optimal solution is an axially symmetric body like the one shown in (c) whose optimal contour curve r(θ) needs to be determined.

One feels that all the mass should be located on one side of a plain passing through the point of maximum attraction because otherwise the gravitational forces due to the masses located on different sides would partially cancel each other. We will call the point of maximum attraction the barypole and the plain passing through it the baryplane. Now, if the matter were arranged in a flat disc adherent to the baryplane as shown in Fig.1a, much of it would be located along the rim of the disc, too remote from the barypole to be effective. Equally unsatisfactory would be an arrangement along a long rod perpendicular to the baryplane (Fig.1b).

In both cases, in fact, one just needs to take a piece of the material lying far away from the barypole and move it somewhere closer in order to increase the gravity at the barypole and thus prove that the arrangement is not optimal. For an arrangement to be optimal, it must be impossible to increase the barypole gravity it by any transfer of material from surface location to another.

It therefore appears that the solution should be some kind of a compact blob of mass arranged more or less as shown in Fig.1c. The optimal body is likely to be convex and, due to lack of other indications, possess a symmetry axis passing through the barypole and perpendicular to the baryplane. As such, it can be described by means of its 2D cross-section contour line r(θ) with θ being the polar angle and r the distance of the corresponding surface points from the barypole. The problem is to find out the contour function r(θ) for which (a) the volume of the body equals V and (b) the gravity at the barypole is maximum. One feels that there should exist a well-defined solution and one hopes that it is unique.

The solution: gravitoids and gravidomes

In order to find the solution we will
   (a) assume a generic contour curve,
   (b) calculate the volume V of the corresponding body,
   (c) calculate the gravity G at the barypole B, and
   (d) find the equation for r(θ) which makes G stationary under the constraint of constant V.

Detailed workout of each point:
   (a) Let r(θ) be continuous in [0,π/2] and possess a derivative anywhere in (0,π/2).
   (b) In the polar coordinates {ρ,θ,φ} shown in Figure 2, an infinitesimal volume element is

(1)   .

For the volume V we therefore have

(2)   .

   (c) The calculation of the gravity vector Q at point B is simplified by the symmetry which tells us that Q is parallel to the axis. We therefore need to sum up just the parallel components due to all the masses present in the volume elements dV. Since the density η is constant, the mass present in dV is ηdV which, according to Newton's law, contributes to Q by an element dQ whose magnitude is |dQ| = G(ηdV)/ρ2, where G is the gravitation constant. Since the projection of dQ onto the axis is dQ = |dQ| cosθ, the formula for Q = |Q| becomes

(3)   .

   (d) Applying the tools of variational calculus, assume that r(θ) is subject to an infinitesimal functional variation δ(θ).
This leads to the following variations of V and Q:

(4)   .

The first of these equations exploits the calculus of derivatives according to which the variation in r³(θ) equals 3r²(θ)δ(θ).

In practice, we need to consider only those variations δ(θ) which leave the volume constant, i.e., for which δV = 0. Under this constraint, we would like Q to be extreme which means that we should have also δQ = 0 for any admissible δ(θ).

One of the basic approaches of variational calculus is to form the linear combination (δV + λ δQ) and see whether there exists a constant value of the Lagrange multiplier λ for which the combination is null regardless of δ(θ). If we are lucky and such a constant value exists then satisfying the constraint δV = 0 implies that the desired condition δQ = 0 is valid as well. In our case

(5)   .

which is zero regardless of δ(θ) if and only if

(6)   .

Consequently, λ is constant if and only if r²(θ) is proportional to cos(θ), which is the condition defining the optimal contour curve. We will write

(7)   .

with D being the value of r(0), i.e. the distance between the barypole B and the opposite contour point A which we will call antipole. Figure 3 shows the normalized optimal contour curve for D = 1. In what follows, we will call gravitoid the optimal contour curve and gravidome the body obtained by its rotation around the axis.

Once the constancy of λ is guaranteed, its actual value becomes irrelevant. What interests us more is the relationship between D and the gravidome volume V, as well as its barypole gravity Q. Substituting (7) into (2) and (3) and evaluating the elementary integrals, one obtains

(8)   .

The shapes of a gravitoid curve and of a gravidome body are therefore fully defined by a single parameter (such as the depth D or the volume V). All such shapes are similar and can be obtained by simple scaling with respect to the barypole of the normalized shape curve. In this sense, Figure 3 represents all of them.

Figure 3. A gravitoid (cross section of a gravidome).

The drawing shows the gravitoid curve (bold red) with its barypole B and antipole A. The corresponding gravidome is formed by rotating the gravitoid around the axis AB. Also shown are the depth D and the width W of the two objects and the gravidome's insribed sphere (gray, smallest), equivalent volume sphere (blue, intermediate) and equivalent gravity sphere (green, largest).

Gravidome numerology

Equations (8) link together the gravidome volume V, its barypole gravity Q and its depth D, leaving independent just one of them. Figure 3 shows also the gravidome width W whose numeric value can be determined as follows.

Using Figure 2 and Eq.(7), we see that the width w(θ) corresponding to polar angle θ equals

(9)   .

Finding the polar angle θ which maximizes the latter ratio and the corresponding maximum value of W/D is elementary.
It turns out that

(10)    (see OEIS A208745).

making the gravidome about 24% wider than it is deep.

From Eqs.(8) it is easy to see that the volume of a gravidome is exactly 8/5 that of its inscribed sphere with diameter D. Vice versa, the diameter of a sphere with the same volume as the gravidome (the V-equivalent sphere) equals D multiplied by the cubic root of 8/5 (about 1.1696).

From the space engineering point of view, a more interesting question is the diameter and volume of the G-equivalent sphere - one which has the same surface gravity as that at the barypole of a gravidome. We know that the gravity outside a sphere (including its surface) is the same as though all the mass were concentrated in its center. The surface gravity of a sphere of diameter d filled with a material of density η therefore equals to

(11)   .

Comparing this with Eqs.(8), the diameter d of the G-equivalent sphere is (6/5)D and the ratio of its volume to that of the gravidome equals (27/25) = 1.08. In other words, using a gravidome instead of a sphere to produce a certain gravity, the required mass is 8% less. One can also reverse the question: Given a fixed amount of material, how much greater will be the gravity if I shape it as a gravidome rather than as a sphere? The answer is 2.6 % (the cubic root of 1.08).

From Eqs.(8) and (11) it follows that, regardless of the shape of an object, the gravity Q on its surface is directly proportional to the density η and to the object's linear dimension (D or d). Now assume that somebody offers you a material whose density is twice as high as the density of the stuff you are currently using. This will enable you to produce the same gravity as before with gravidomes (or spheres, if you prefer them) scaled down by a factor of 2 and thus reduce their volume by a factor of 8. Consequently, their mass will be reduced by a factor of 4 (2 times denser material in 8 times smaller volume).

Considering this, plus the fact that raw matter is always heterogeneous, it is likely that sorting its components according to their density might help. For example, consider a comet with mass M half of which is ice (density η_ice = 1) and half rock (η_rock = 5). That amounts to an ice volume of V_ice = M/2, a rock volume of V_rock = M/10, and a total volume of V_total = 6M/10 so that, when unsorted, the comet has an average density of η_total = 10/6. Suppose now that we sort the material and construct a gravidome only from the recovered rock Then, according to the last of Eqs.(8), the ratio between the barypole gravity of the rock-only gravidome and that of the gravidome build from the unsorted stuff is

(12)   .

This is a huge success, since we are getting 65% more gravity using only half of the comet !!! Vice versa, if we insisted on getting the same boost in gravity just by adding more unsorted comet matter, we would need a total of 4.5 comets ! The trick clearly consists in the fact that the denser the matter, the more mass can be located close to the barypole.

So what if we used all the matter we have, but sorted it first into classes with different densities. The question is: how do we arrange the various lots within the novel, heterogeneous gravitational attractor? Surprisingly, this is an easy problem which can be solved by plain inspection with no math at all.

Consider first just the two outer components A and B in Figure 4 (ignoring C and D). The two nested bodies A and B are gravidomes which share a common barypole and a common rotation axis. The inner barydome is made of the higher density material while the lower density material fills the space between the barydomes.

Figure 4. A four-component gravinion.

A gravinion consists of a number of gravidomes nested as shown. The density of the materials increases from the outermost gravidome towards the innermost one.When raw mass (debris, asteroid rock and ice, etc.) is sorted according to its density, melted and arranged into a gravinion, the resulting gravity at the barypole is much higher than using unsorted mass.

To understand that this is really the best arrangement, consider first a single barydome A having a homogeneous density η_A (the smaller of the two). We know that, for A by itself, a barydome is the optimal solution with the highest possible barypole gravity. Now imagine that we introduce inside barydome A a body with the higher density η_B. Rather than replacing part of the barydome A volume, however, we will think about it as if we were superposing a body whose density equals the difference η_B-η_A (since gravity forces are additive, this does not alter them in any way). Now all we need to do is to optimize the location and the shape of a body with the difference density located inside A but otherwise quite independent of it. The solution to this second problem is of course again a gravidome with the same barypole. To complete the solution, one just needs to adjust the sizes of the gravidomes so as to make sure that the volume of the inner one matches that of the available denser material, and the difference of the two gravidome volumes equals the volume of the lighter material.

Using recursion, the solution is easily extended to multi-component situations. Suppose that we have n components, each with a density ηk and volume Vk, and assume that, without lack of generality, η₁ > η₂ > ... > η_n. The solution is a series of nested gravidomes with depths D₁ < D₂ < ... < D_n < defined by

(13)   .

The gravity at the barypole is then

(14)   .

and it is always greater than the maximum barypole gravity achievable when using a homogeneous mixture of the components.