## If the Earth Were a Cube

What if the Earth were a cube instead of (approximately) a sphere?  I saw this same strange question twice in the last couple of weeks, first in a Reddit post linking to a good Ask a Mathematician article, then again in a recent Straight Dope column, with the usual entertaining-while-informing reply from Cecil.  Both are very interesting reading, and I do not intend to rehash all of their observations here.  The purpose of this post is to provide some additional details that I found interesting, as well as to point out a couple of possible errors in the earlier write-ups.

I like “what if” questions like this, mostly because they are fun, but also because they are good exercise.  Thought experiments like this one often lead to additional or clearer insight into a more general problem.  In this case, what I found most interesting about this problem is how strange the effects of gravity would be on a cube-shaped planet; I learned that I did not understand gravity quite as well as I thought I did.  There is a reason why physicists prefer their chickens to be spherical.

But first, let’s define the problem.  For the most direct comparison with our experience, it seems reasonable to assume that our cube planet has the same mass and volume (and thus mean density) as the Earth, as in the figure below.

The Earth and a cube with the same volume.

An immediately noticeable difference is the enormous range of altitudes.  Using the center of each face of the cube as a reference, each edge is approximately 1,300 miles (2,100 km) higher (i.e., farther from the center of the cube), and each corner is 2,300 miles (3,800 km) higher.  Compare this with the Earth, where altitudes vary over just tens of kilometers.

(The Ask a Mathematician article gives larger altitudes that suggest an assumption of a larger cube that contains the Earth, with a side length equal to the Earth’s diameter.  This would imply that the cube has either a larger mass or a smaller density than the Earth.)

Beyond just these geometric differences, the physical effects of gravity are even weirder.  First, of minimal weirdness is the observation that gravity is much weaker near the edges and corners than at the center of a cube face.  This makes sense, since the edges and corners are “farther away” from the center of mass of the cube.  The figure below shows the magnitude of the force of gravity over the surface of each cube face, normalized by 1 “Earth g“:

The force of gravity on the cube surface, in Earth g's.

At the center of each cube face, the force of gravity is almost exactly 1 g; at each corner, however, it is just 0.646 g, meaning that a person weighing 200 lbs. here on Earth would weigh only 129 lbs.

(Using this same example, the Straight Dope article suggests that this weight is only 103 lbs.  This value assumes that the cube and the person at the corner are point masses, which is a safe assumption when the bodies in question are spherical.  But when they are not, things get complicated, even for relatively simple shapes like a cube, as we will see shortly.)

A slightly weirder effect is that, standing on the flat surface of a cube-shaped planet, the force of gravity is not always “down.”  That is, as you walk in a straight line from the center of a face toward a corner, gravity causes the flat face of the cube to seem to get steeper and steeper, so that you are eventually climbing instead of walking.  This also makes sense, since the force of gravity is directed approximately toward the center of the cube, which is only “straight down” at the center of each face:

The "steepness" of the perceived hill, or the angle in degrees between the gravity vector and the cube surface normal.

Finally, I think the most interesting part of this problem, and what caught my attention in the first place, is the following innocuous statement in the Ask a Mathematician article:

“… Gravity on the surface wouldn’t generally point toward the exact center of the [cube] Earth anymore.”

In other words, when calculating the force of gravity exerted by the cube, even on a point mass, the direction of that force is not always toward the center of (mass of) the cube.  This was a surprise to me; I had to think about it for a while to realize that, even with all of the nice symmetry, constant density, etc., of the cube, the correspondingly “nice” Shell Theorem, or Gauss’ flux law, etc., do not help us here.  We essentially have to resort to the triple integral to work out exactly how gravity behaves on our cube-shaped planet.  The details of the derivation are at the end of this post.

And it is not a small effect.  I was surprised by just how much the direction of the force of gravity deviates from the center of the cube, nearly 14 degrees in some places, as shown in the figure below.  The overall effect is essentially to reduce the “steepness” effect described above, so that the force of gravity is directed more nearly straight down than directly toward the center of the cube.  As expected, the deviation is zero at the center of each face, at the center of each edge, and at the corners.

The angle in degrees between the gravity vector and the vector to the center of the cube.

The Gravitational Potential for a Cuboid

I initially tried the brute-force numeric integration approach, but particularly for points near the surface of the cube where we are interested, the integrand is not very well-behaved.  At the other extreme, the Werner-Scheeres paper referenced below describes an interesting algorithm for computing the exact gravitational field for arbitrary polyhedra.  Fortunately, the cube is sufficiently simple that we can work it out by hand, with a little help from Mathematica.

We can generalize slightly by considering a cuboid with side lengths $(2a, 2b, 2c)$.  The acceleration due to gravity is the gradient of the potential function $U$ defined at a point $(x_0,y_0,z_0)$ by

$U(x_0,y_0,z_0) = G\rho\int_{-a-x_0}^{a-x_0}\int_{-b-y_0}^{b-y_0}\int_{-c-z_0}^{c-z_0}\frac{1}{\sqrt{x^2+y^2+z^2}}dx\,dy\,dz$

where $G$ is the gravitational constant and $\rho$ is the density of the cube.  Note the change of variables to shift the origin to $(x_0,y_0,z_0)$, which has the convenient effect that the integrand does not involve any of the limits of integration.  So after each of the three integrations, we can eliminate any summation term that does not involve all three variables x, y, and z, since the term will evaluate to zero in the final result.

Mathematica does most of the heavy lifting, with some nudging simplification, yielding the following expression for the potential function:

$U(x_0,y_0,z_0) = G\rho(w(x,y,z)+w(y,z,x)+w(z,x,y))]_{x=-a-x_0}^{a-x_0}]_{y=-b-y_0}^{b-y_0}]_{z=-c-z_0}^{c-z_0}$

$w(x,y,z) = x y \ln(z+\sqrt{x^2+y^2+z^2}) - \frac{1}{2}x^2 \arctan{\frac{y z}{x\sqrt{x^2+y^2+z^2}}}$

References:

1. R. Werner and D. Scheeres, Exterior Gravitation of a Polyhedron Derived and Compared With Harmonic and Mascon Gravitation Representations of Asteroid 4769 Castalia.  Celestial Mechanics and Dynamical Astronomy, 65 (1997):313-44. [PDF]

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### 8 Responses to If the Earth Were a Cube

1. spenczar says:

Interesting. The gravitational potential formula you ended up with is not symmetric when you interchange x, y and z. That seems worrying – why should the gravitational potential be so clearly coordinate dependent?

• Hmmm. I double-checked my LaTeX transcription, and I did miss the parentheses around the sum, which I corrected. But perhaps you are referring to the intermediate function $w$? You are correct that this function is not symmetric in the coordinates, but $U$ is, being a sum over three “rotated” valuations of $w$.

2. What if you add the effect of the cube Earth’s rotation? Then you would have less gravity on the edges at the equator and the gravity wouldn’t pull straight down at the corners (vertices). But I think the effect would be minimal, like on round Earth, where the gravity at the poles is not noticeably different then at the equator.

• Right on all counts, assuming that the axis of rotation is through the centers of two opposite faces of the cube. The effect is indeed minimal; the magnitudes of accelerations at corresponding points on a rotating vs. non-rotating planet differ by less than half of one percent, even at the corners of the cube.

• Kudos for this analysis of gravity on the flat surfaces of a cube.

Twenty years ago I presented papers and published on this topic (see references below).
I did not use calculus but presented a geometric proof that gravity vectors changed direction along the “flat” surface of a cube. This disproved the contention from Aristotle to Asimov that gravity pulls perpendicular to a flat surface. They and contemporary textbooks neglect the “mass” below “flat” surfaces and continue to misunderstand the simple positions of “up and down”.

Perhaps textbooks will eventually recognize these flaws in (theoretical) flat earth models. Though mostly theoretical, there may be a practical application when vehicles land on large, irregular space objects like asteroids with large “flat surfaces.

http://gravityresearchfoundation.org/pdf/awarded/1991/thorndill.pdf

Published abstract of a poster paper presented at:
American Association for the Advancement of Science
Annual Meeting in Washington, D.C.
February 18, 1991
ancient flat earth gravity vectors corrected gravity on flat cube

A Reexamination and Correction of the Ancient Flat Earth-Round Earth Controversy
DAVID THORNDILL, (Essex Community College, Baltimore).

Aristotle, Ptolemy, and Copernicus gave several proofs for a spherical earth. They were so convincing that almost every educated person since the time of Aristotle has accepted these proofs. Yet most of these proofs for a spherical earth also fit many flat earth models. This paper will show how the following ancient observations which have been described as spherical earth “proofs” also fit cuboid flat earth models:

1. The pole star and the zenith appear to move as one travels north or south.
2. Stars disappear from the southern horizon as one travels north.
3. Ships sailing away appear to sink over the horizon.
4. Lunar eclipses have been recorded at earlier local time in western locations.

The earlier flat earth models assume the zenith, “up”, is perpendicular to the flat surface of the earth. Analysis of gravity vectors shows how “up” changes as one moves from the center to an edge of a flat earth (planetoid). A corrected flat earth model is presented.

P.S, See http://faculty.ccbcmd.edu/~dthornd1/EGSflatEarth.doc for graphics

3. This was an interesting read, thanks! I think you would find the referenced Werner/Scheeres paper interesting, too; their algorithm would have utility in exactly the type of situation you describe, modeling the gravity of an irregularly shaped object such as an asteroid. The idea is to approximate an arbitrary shape as a polyhedron, for which their algorithm can efficiently compute the gravity potential *exactly*.