## 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.

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“:

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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### 74 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?

• possiblywrong says:

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. Benjamin Lefebvre says:

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.

• possiblywrong says:

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.

• DAvid Thorndill says:

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.

Click to access 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. possiblywrong says:

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*.

4. welok says:

Why humans (including myself most of the time) think of a cube when think about the other side of a sphere (or square when thinking of the other side of circle)?
The other side of it would be a piramid.
Said that, make a article with the same question, but with a piramid instead of a square.

• possiblywrong says:

I think I understand your point– roughly, if a sphere (or circle in two dimensions) has the “most” number of faces (edges), in a limiting sense, then the object with the *fewest* number would be a tetrahedron (or triangle), not a cube (or square)?

This is a good point. The resulting “weirdness” of gravity is qualitatively not that different, though; that is, there are still extreme high/low altitudes, there is still symmetric variation in the magnitude of the gravitational force, and the direction of the force is still not always toward the center of mass of the tetrahedral planet. The actual quantitative differences would simply require a similarly tedious triple integral :).

5. Andrew MacDowell says:

I am sure this is a foolish question but if one travels into the Earth does one become lighter?
As you move towards the centre, does the mass above you increase therefore attract you more and more and reduce your weight. But then increasingly, does the mass to either side attract you as well so at the centre you are pulled in all directions?

I know that the gravitational attraction of mountain ranges can be measured.

• possiblywrong says:

Not foolish at all, this is a great question, with two answers: a “textbook” answer and a more realistic one. First, the Shell Theorem referenced in the original post helps us here, since it tells us that as we move toward the center of the earth, we experience *no* net force from the “shell” of the earth that is farther from the center than we are. (Note that this shell isn’t just “above” us.) So the only mass that we have to consider is the ever-shrinking sphere of earth “below” us.

So there are two competing factors: as we move inward, the mass of the sphere exerting a net force on us is shrinking… but so is the distance from the center. If we assume that the earth has uniform density, then the former effect wins out, so that gravity decreases exactly linearly with distance, reaching zero at the center.

However, the actual earth is *not* of uniform density, but instead gets denser as we move inward, so much so that gravity actually *increases*… but only up to a point, beyond which it necessarily decreases to zero at the center.

This Wikipedia page has a good plot showing curves of gravitational acceleration vs. distance for these various assumptions.

• Andrew MacDowell says:

Thank you. I’ve been reading about Einstein’s theories relating to gravity. I’ve been struggling with the shell theorem – I haven’t really gone into it in depth so perhaps I misunderstand. I do not see why we have to account only for the mass that is below us. I just find it hard to appreciate the idea that at the centre there is no gravity – why isn’t the mass all around you exerting a gravitational attraction? I can understand that the opposing attractions sum to no overall acceleration. Is it because gravity is a field effect like magnetism – there are lines of ‘flux’ that flow? In other words a massive object creates its own field, the structure/pattern of which means that the ‘individual’ gravitational fields created by masses within the total mass are lost – subsumed into the structure of the total field. But then we are not dealing with an electrostatic field that focuses around poles and also, we can measure the the different gravitational fields of different parts of the Earth, so that seems to refute that idea.

I see that gravity wells are always shown surrounding the surface of a mass – the bottom of the well is not at the centre of the mass. If the gravity at the centre of the Earth is zero then it cannot be influencing space time so there is no gravity well. So do we have a very strange idea that as you travel towards the centre of the Earth, you climb out of the gravity well in the same way as when moving away from the surface This does not feel right intuitively – it is too chaotic. I believe that at the centre is the point of maximum gravitational effect – it is the bottom of the gravity well: the opposing gravitational forces create the point of maximum distortion of space time: although weightless we experience the maximum time dilation.

So when a collapsing body reaches a point whereby the density and radius are such that it creates a singularity – is it that the centre of that body has a such an intense gravitational field pulling away from the centre in all directions that it opens a hole in space time? Why then would all the matter disappear into that hole if the gravitational forces are pulling away from it? Is it because the singularity is an utterly different phenomenon – as soon as space time is annihilated and it appears, there is no longer an opposing gravitational field created by the mass pulling away from the centre because there is no point in space time for this to act upon and the whole gravitational field then becomes directed inward into the singularity- hence the massive increase in the gravitational attraction? Or more likely, because in that region of space time, we have entered a situation where our laws of physics do not apply, we cannot even talk of gravity – it is completely beyond our ken.

The more I think about gravity, the more fascinating and perplexing it becomes.

6. U can never no it says:

But how will it affect day and night?

• possiblywrong says:

If the Earth were a cube, there would effectively be only six “time zones,” one for each cube face, since everyone living anywhere on a particular face would see the sun rise and set at exactly the same times. (Compare this with our actual time zones, which are a more artificial discretization; people in the eastern part of a time zone see the sun rise and set earlier than people in the western part of the same time zone.)

People on the northern “polar” face would experience alternating six-month periods of constant daylight (late March to late September) and constant darkness. For the southern polar face these six-month periods would be reversed. The remaining four faces would have a more typical experience, with alternating 12-hour periods of daylight and darkness.

• David Thorndill says:

Not so. Time is determined by the position of the overhead sun–at noon–on a level surface (not perpendicular to the side of a hill). One does not count sunrise as the first ray of sun above a building or hill or mountain. The edges of the cube will appear to be mountains except in the center of the face of a cube.
Each face will have several time zones.
The problem is that on a completely flat face of a cube only the center is level. Since our planet has rolling hills and valleys, any similar cube surface will only have level places where the gravity vector (the line from the surface to the center of the cube) is perpendicular to the surface.
See http://faculty.ccbcmd.edu/~dthornd1/EGSflatEarth.doc for diagrams and a more complete analysis.

• interval2016 says:

We can’t be sure about how inhabitants of a cubic earth would set their clocks, but I don’t expect them to make their time zones according to David Thorndill’s definition of time. For practical purposes, time zones should be set according to time of sunrise and sunset, and those are uniform in each cube face. In a cubic earth sun would rise at the same time in all points of a time zone, while in our earth time is overhead at (about) the same time in the whole time zone.

An interesting point is that if the axis of the cubic earth goes through the centres of two faces, no time zone will have daily saving time, since all four non polar faces will have the same day/night distribution than our earth in the equator – that is, equal length of day and night all year round.

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8. jimi says:

Andrew – to reply to your comments about gravity “wells” these usually represent the gravitational potential well around a mass with zero being at infinity. As already stated, for a uniform density Earth the the gravitational field is proportional to radius when inside. Hence, the potential well is not shaped in the traditional hyperbolic curve as outside but will be a parabola ending in a turning point at the centre representing zero field. It will still be at a lower potential than zero and will require energy to get “out” as it were. I estimate this to be about -94MJ/Kg as opposed to -63MJ/Kg at the surface (assuming Earth density 5500 Kg/cubic metre). What is really interesting is what would happen if you fell right through the Earth in an evacuated tunnel !? I wonder how that would change for cube-Earth, hmmm…
cheers

J

• Randy Meyer says:

The scenario of a straight cylindrical hole passing through the center of our Earth interests me. What would be the diameter of that cylinder in order to totally consume the surface atmosphere? Also I have always believed that after jumping into the hole inertia would propel you part way back up the hole on the other side, until gravity stopped you and pulled you back to the center again and then inertia propelled you back up again in the opposite direction you came down, in a kind of damped sinusoidal oscillation, until you finally came to rest in the center. But if the gravitational force is decreasing at the same time you are falling do you then slow down as you fall deeper, cancelling the inertia so that you just feather down to a final rest in the center at zero g? One other interesting observation. And think about this from both ends. A straight line staircase from the surface down to the center. Standing up at the surface you have to think “No way will this ever get down to the center, it’s not going straight down. I will miss the center and resurface again somewhere else.” Which begs the question, at which point in the descent did it become an ascent? But now place yourself starting at the center and building a staircase up to the surface. You have to hit the surface eventually, it’s the only destination there is. And when you finally get up to the last final steps near the top you look behind you at the steps going down and think “No way can those steps lead down to the center, it’s not aiming at it.” Maddening.

• possiblywrong says:

I can’t answer your question about “consuming the surface atmosphere.” Regarding the idea of jumping into a hole through the earth, this Wikipedia page should answer most of your questions. (Note, however, that most of those “nice” results depend on the assumption that the earth’s density is uniform throughout, which it isn’t.)

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10. Randy Meyer says:

Thanks for the quick reply. Could you give me the http: of that wiki article, for some reason it didn’t get embedded in your reply link. I’ll just copy/paste it in my browser. As for the answer to the hole diameter to displace atmospheric volume, I’ll just get some numbers and figure it out. My suspicions are that it is not a very wide hole that will suffice. Thanks again.

• possiblywrong says:

Sorry– I edited the comment to correct the link, it should work now. (At any rate, it’s the “gravity train” entry.)

11. Sander van der Horst says:

We had this discussion at lunch in my office 😀
We couldn’t really figure out what would happen with aircrafts and spaceships. Of course the orbit wouldn’t be round anymore, but neither square. Could you elaborate on this?

• possiblywrong says:

This is generally a hard problem. That is, given a spacecraft’s (or a ballistic missile’s, or whatever) initial location and velocity, it is relatively *easy* to compute its trajectory, or where it will travel through the changing gravity field over time. The result might be that it will travel in a periodic orbit around the planet (as with a spacecraft), or that it will impact the planet at some point (as with a missile).

It is much harder in general, however, to work the problem backward, so to speak: that is, to figure out what initial location and velocity will *result* in a periodic orbital trajectory around a planet that has an irregular shape such as a cube.

Some work has been done on this, though. Check out this paper, in particular Figure 8a, for an example of how complex the shapes of such periodic orbits can be.

12. Benjamin Molko says:

what would day and night be like in a situation like this?

• possiblywrong says:

See my earlier response to a similar question (search for “time zone” in this page of comments).

13. Benjamin Molko says:

thanks

14. Benjamin Molko says:

Also what would the edge look like and could you jump off it? Would gravity pull you back onto that side instantly?

15. David Thorndill says:

The edges would be like tops of mountains or big hills… Imagine a symmetrical situation. You climb up the mountain to the peak (edge) then down the other side, The same force down as the force up the hill. See: . http://faculty.ccbcmd.edu/~dthorndill/EGSflatEarth.doc Astronomical observations which have been used to confirm a spherical geoid also fit a cuboidal model. A presentation and poster paper at the European Geophysical Society meeting in Edinburgh, Scotland, April 9, 1992. Abstract publication in Annales Geophysicae.:.

16. Benjamin Molko says:

thank you so much. You saved my science project

17. Amaka says:

So on this cubed planet, would there still be northern and southern hemispheres? If so, will they be on EACH face of the cube, so that there will be a total of 12 hemispheres (6 faces x 2 hemispheres, north & south)? How will the weather and temperatures differ on each face/hemisphere? And when the cube orbits, will it still have the 23.5 degree tilt as our spherical Earth has currently? On which part will the axis go through, the top and bottom faces or the diagonal opposite corners? I’m sorry for my myriad of questions!

• possiblywrong says:

I think it depends– that is, since this is a made-up scenario, we are free to imagine that the cube rotates about whatever axis we want. (Rotational inertia would play a part, but in this case the two natural axes, (1) through the centers of two opposite faces or (2) through diagonally opposite corners, have the *same* rotational inertia.)

Weather, temperature, etc., is beyond my knowledge of what might happen. But note that the Ask a Mathematician post provides some interesting ideas about how the weirdness of gravity would cause oceans to “collect” near the centers of each face.

• Amaka says:

Thanks that helped me 🙂 one more thing, will there still be hemispheres?

18. possiblywrong says:

@Amaka: It depends on what you mean by “will there be hemispheres.” If you mean can we arbitrarily draw an “equator” in a plane perpendicular to the rotational axis, and call the upper half the “Northern Half” and the lower half the “Southern Half” (better names since “hemisphere” refers specifically to a spherical shape), then yes, sure, we can.

But if you mean would the weather or other environmental phenomena be fundamentally different in those northern and southern halves, then the answer is more complicated. First, keep in mind how it works here on Earth: it’s colder at the poles than near the equator, and this would be true even if our rotational axis were *not* tilted relative to the plane of revolution about the Sun. But the weather is “opposite” in the northern and southern hemispheres (when it’s winter in one, it’s summer in the other, and vice versa)… and this is *only* true *because* of the axis tilt.

On a cube Earth, the north and south faces (i.e., the opposite faces that the rotational axis passes through) would be similarly colder. However, the other four faces would have essentially the same conditions over their entire faces (i.e., both the northern and southern halves)… and this would be true whether the axis were tilted or not, since the Sun’s rays strike the cube face surface at the same angle over the entire face (where here on Earth, that angle depends on latitude).

• Michael Mendelsohn says:

If the axis of rotation is a cube diagonal (or close), and if it’s still tilted, then we’d have two opposite “hemispheres” (hemicubes?) consisting of 3 cube faces each; but it would not be colder at the poles. The seasons would cycle yearly as they do on spherical earth.

For the “opposite face centers” diagonal, we’d have the polar faces and the equatorial faces. While the equatorial faces would be perpendicular to the sun twice a year, making for two season cycles per year, the weathor wouldn’t really be seasonal, much as it isn’t on earth, unless the opposite conditions on the polar faces generate seasonal weather patterns – but that would be a yearly cycle.

19. Amaka says:

Thank you so much definitely makes sense

20. Stephane Jaspert says:

Rolling earth cube rock / Art Paris
http://jaspert.free.fr/earth_cube_art.html

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24. Michael Mendelsohn says:

What would the surface of an ocean look like on cubeworld?

• possiblywrong says:

The “Ask a Mathematician” article linked in this post provides at least a qualitative answer here; gravity would cause ocean water on each face to “collect” in a sort of bubble centered on the face, so that “shores” would be near the edges/corners of each face.

• Michael Mendelsohn says:

Yes, but the drawing in that article shows the ocean surface as spherical, but from the data you posted, the gravity vector near the center of the faces is more perpendicular to the surface than it points to the center, so the ocean should be flatter than in that illustration because the water surface normal should be that gravity vector. But due to water pressure constraints, gravity should also be nearly equal at every point of the water surface.

The earth has ca. 1350 million km³ ocean water, so one face of cube earth might have a central ocean of 225 million km³ water. If the ocean sat on top of a perfectly flat cube face, what would its diameter be, and its maximum depth?

25. Michael Mendelsohn says:

The video game “The Inner World” posits a universe made of earth, with the “world” being a hollow sphere in this universe. What would the gravity be on the surface of this inner sphere, assuming Earth diameter?

The center of the sphere has no gravity, obviously. What would the distribution of air density be from the center to the surface be like, assuming Earth standard air pressure at the surface?

• possiblywrong says:

If I understand you correctly, then this is a situation where the answer *is* really nice, due to the shell theorem: there would be zero gravity not only at the exact center of the sphere, but *everywhere* inside the the hollowed volume.

• Michael Mendelsohn says:

The shell theorem is a nice find, but It looks to me as if it would only hold for finite outer objects: it basically says, as you move from the center, you move away from one side, but as the side gets “bigger”, the decline in gravity is compensated.
But in The Inner World, the outside is infinite, so the masses on each side of the observer are never uneven.

Consider an Inner World W of diameter d, an observer on its surface, and a sphere S of radius d centered on the observer. The sphere S contains the Inner World W and some extra earth; everything outside the sphere S is full of earth, and since the observer is at the exact center of S, for symmetry reasons the net gravitational force of the earth outside S exerted on the observer is zero. This means that whatever gravitational force is exerted on the observer must be produced by the mass present in S; that is the volume of S minus the empty volume of W. Since that mass is not symmetrical, the shell theorem does not apply to it.

But this approach does yield the gravity: mirror W across the observer to yield W’, which is full of earth and also still within S; the rest of S that is not W or W’ is full of earth and symmetrical to the observer, so it cancels out to zero gravity. Thus, the gravity exerted on the observer is that exerted by W’, which means the gravity on the surface of the inner world is equal to the gravity on the outside of a sphere of equal diameter.
If the inner world is earth-sized, then gravity at its surface is 1g.

Nice! I didn’t know that when I asked the question.

26. possiblywrong says:

@Michael, ah, I see, I didn’t catch that the mass “outside” of the inner world was infinite, in which case I think you’re right, the gravity at a point inside the hollow would be equal in magnitude– but in the opposite direction– to the corresponding point inside our “normal” earth.

• David Thorndill says:

Thorndill, David 1992. Astronomical observations which have been used to confirm a spherical geoid also fit a cuboidal model. A presentation and poster paperat the annual EGS meeting in Edinburgh, Scotland, April 9, 1992. Abstract publication in Annales Geophysicae: General Assembly of the European Geophysical Society. and poster paperat

Thorndill, David 1991. A reexamination and Correction of the Ancient Flat Earth and Round Earth Controversy A poster paper presentation at the annual meeting of the American Association for the Advancement of Science in Washington, D.C., February 18, 1991. Abstract publication in AAAS Annual Meeting Abstracts of Papers.

• David Thorndill says:

Mr. Mendelsohn: . I would be greatly interested a calculation of a size and depth of the 225 million km³ ocean you proposed on Dec. 20th. The abstracts and paper above are found at: http://faculty.ccbcmd.edu/~dthorndill/publications.htm I had proposed oceans with curved surfaces on the faces of a cubic planet in 1991. My interest was to discount the popular view that a curved oceanic surface (with ships disappearing over the horizon) was proof that the earth is spherical.

27. Jeanette Rogge says:

Would living things exist if the earth were a cube

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29. Sundar Vedantham says:

Hi, Is the cube earth image used on this post is under creative commons license? I am writing a small book in the Indian language of Tamil discussing various thought experiments. I will not make any money out of this endeavor as I am serving as a volunteer trying to educate school children through this effort. While discussing the cube shaped earth, I’d like to use the image without violating any copyright laws. In case you have the copyright, can you please provide permission for reuse?
Regards.﻿

• possiblywrong says:

Sure, attribution is always appreciated, but no problem, you are welcome to use it.

30. Melisa Suldahan says:

What are the consequences of the rotation and tidal distortions if the earth is cube in shape? I need your help.

• possiblywrong says:

The centrifugal acceleration from rotation is relatively small compared to gravity, less than half a percent or so. The tidal effect is an interesting question, to which I don’t have a good answer; I think this is definitely at best an academic exercise, since a planet-sized cube would simply not be able to remain cube-shaped for long– gravity would tear it apart, so to speak.

31. Alanis Marie Ballans says:

If it became like this, how long would it then take to collapse itself back into a sphere-like shape, and what effect would that have on, say, life on the faces?

• possiblywrong says:

I don’t know. Part of the problem is that gravity would make it essentially impossible for a mass the size of the Earth to *get* into a cube shape to begin with. Assuming we could mash it into that shape, how quickly it would pull itself back into a sphere is a physics question beyond my expertise.

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