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Tetrahedral Planets


Nazalassa
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I got bored in class yesterday, and let my imagination tell me stories about physics, rockets, wizards and magic, exploration, discoveries... And then I somehow imaginated a tetrahedral planet. A big tetrahedron in space, orbiting a star.
My math brain immediately wanted to know how gravity (and that sort of things) would work on such an hypothetical planet.

Spoiler

Yes I know, such a planet won't naturally occur. But all this is theory. And maths.
Mostly maths.

So I took a sheet of paper, and started doing math, searching for triangles in the tetrahedron, applying the Pythagorean theorem... And I got these results.
 

But first, let's define some things...

Let's call center [of the tetrahedron] the tetrahedron's center of mass (center of gravity too, which is equidistant to all of the tetrahedron's vertices, and also equidistant to the centers of all of the tetrahedron's faces, and also equidistant to the centers of all of the tetrahedron's edges, you see what I mean).
By "edge" I mean the middle of the edge, by "face" I mean the center of the face, and by "adjacent vertex" I mean vertex that is linked to the current vertex by an edge, vertex that belongs to the current edge, or vertex that belongs to the current face (depending wether I'm talking of a vertex, an edge, or a face).

If one unit = the side of the tetrahedron
The distance between a vertex the center the center the center an edge a face a face
and an adjacent vertex a vertex an edge a face an adjacent vertex an adjacent vertex one of its edges
is 1 √(6)  /  4 √(2)  /  4 √(6)  /  12 √(3)  /  2 √(3)  /  3 √(3)  /  6

 

If one unit = the distance between the center of the tetrahedron and one of its faces
The distance between a vertex the center the center the center an edge a face a face
and an adjacent vertex a vertex an edge a face an adjacent vertex an adjacent vertex one of its edges
is 2 * √(6) 3 √(3) 1 3 * √(2) 2 * √(2) √(2)

 

If one unit = the distance between the center of the tetrahedron and one of its vertices
The distance between a vertex the center the center the center an edge a face a face
and an adjacent vertex a vertex an edge a face an adjacent vertex an adjacent vertex one of its edges
is 2 * √(6)  /  3 1 √(3)  /  3 1 / 3 √(2) 2 * √(2)  /  3 √(2)  /  3

(Note that there may be errors, and that you can go from one table to another with a simple multiplication.)

 

So, as gravity is inversely proportionnal to the distance to the center of mass, that would mean (if we look at the second table) that the gravity at the center of a face is nine times stronger than the gravity at a vertex?
Cool...

Now you can "walk" into space, and also to a place were gravity is waaay weaker.


However, as you progress towards the edges (and vertices) of a face, the apparent "slope" will increase and increase, right? Because the angle between the plane orthogonal to the direction of gravity and the actual surface increases.
I have calculated some slopes:

Sloppy slopes, yey
The slope at a vertex an edge a face
is (roughly) arccos(1 / 3) ≈ 70.53° arccos(√(3) / 3) ≈ 54.74° arccos(1) = 0°

So a vertex looks like a mountain now. That's impractical for rocket launches.
Fortunately, the gravity there is a ninth of the gravity at the center of a face.

Edited by Nazalassa
Added "disclaimer"
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2 minutes ago, Nazalassa said:

I got bored in class yesterday, and let my imagination tell me stories about physics, rockets, wizards and magic, exploration, discoveries... And then I somehow imaginated a tetrahedral planet. A big tetrahedron in space, orbiting a star.
My math brain immediately wanted to know how gravity (and that sort of things) would work on such an hypothetical planet.

So I took a sheet of paper, and started doing math, searching for triangles in the tetrahedron, applying the Pythagorean theorem... And I got these results.
 

But first, let's define some things...

Let's call center [of the tetrahedron] the tetrahedron's center of mass (center of gravity too, which is equidistant to all of the tetrahedron's vertices, and also equidistant to the centers of all of the tetrahedron's faces, and also equidistant to the centers of all of the tetrahedron's edges, you see what I mean).
By "edge" I mean the middle of the edge, by "face" I mean the center of the face, and by "adjacent vertex" I mean vertex that is linked to the current vertex by an edge, vertex that belongs to the current edge, or vertex that belongs to the current face (depending wether I'm talking of a vertex, an edge, or a face).

If one unit = the side of the tetrahedron
The distance between a vertex the center the center the center an edge a face a face
and an adjacent vertex a vertex an edge a face an adjacent vertex an adjacent vertex one of its edges
is 1 √(6)  /  4 √(2)  /  4 √(6)  /  12 √(3)  /  2 √(3)  /  3 √(3)  /  6

 

If one unit = the distance between the center of the tetrahedron and one of its faces
The distance between a vertex the center the center the center an edge a face a face
and an adjacent vertex a vertex an edge a face an adjacent vertex an adjacent vertex one of its edges
is 2 * √(6) 3 √(3) 1 3 * √(2) 2 * √(2) √(2)

 

If one unit = the distance between the center of the tetrahedron and one of its vertices
The distance between a vertex the center the center the center an edge a face a face
and an adjacent vertex a vertex an edge a face an adjacent vertex an adjacent vertex one of its edges
is 2 * √(6)  /  3 1 √(2)  /  6 1 / 3 √(2) 2 * √(2)  /  3 √(2)  /  3

(Note that there may be errors, and that you can go from one table to another with a simple multiplication.)

 

Strong gravity (moon strength or higher) tends to compress solids into spheres.

So unless a planet is hollow and very low gravity... it will always be a sphere.

 

It's the same reason you are unlikely to find a donut-shaped planer occurring naturally... because gravity would cause the open hole to shrink until the planet collapsed in on itself and filled it.

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5 minutes ago, Spacescifi said:

It's the same reason you are unlikely to find a donut-shaped planer occurring naturally... because gravity would cause the open hole to shrink until the planet collapsed in on itself and filled it.

"Occuring naturally" being the key word here. In theory, if a donut-shaped planet would spin fast enough, the spin could allow it to maintain its shape.

As for the theorhetical tetrahedron planet, the stress would cause the corners to pull inwards, forcing it into a rounder shape. Although, if made out of a strong enough material/structure, this deformation would halt, so you'd have a sort of 'inflated tetrahedron' shape.

Edited by intelliCom
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2 minutes ago, intelliCom said:

"Occuring naturally" being the key word here. In theory, if a donut-shaped planet would spin fast enough, the spin could allow it to maintain its shape.

 

Well yes but.... why would you want a donut shaped planet beyond the rule of cool?

 

Besides, the only quick and efficient way of making one involves using a deathstar in a far more surgical manner.

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I just did that for the maths behind it :)

Spoiler

Also, considering that I was thinking of a high-magic fantasy world when I got the the idea of a tetrahedral planet, I guess some wizard created an orbital tetrahedron for fun and used magic to keep it in tetrahedral shape. That's the only explaination.

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There was a whole article with pictures on topic of polyhedral planets in Tekhnika MolodezhiTechnology for the Youth magazine in late 1970s, but can't find it online.

After having created the planet, make it ocean and watch its shape.

As the mass of any such planet will be anyway concentrated around its geometrical center, the ocean surface will have a nearly-spherical shape.

This demonatrates that the normal vector of gravity will be pointing at the center of the planet on the water surface and at the coastline, and on top of the vertices and edges.

The vertices will look like mountain peaks, connected with edge mountain ridges.

The ocean will consist of lakes in the centers of the faces, separated with these ridges.

Walking away from the lake towards the edge or vertex, you will quickly notice that the gravity normal vector (pointing at the center of the planet) makes you hardly climbing up.

Some of the peaks and ridges can stick out of the atmosphere, but if they don't, then you will have several separated local weather system at every face.

P.S.
If you make a solid model of the planet (say, out of clay) and attach a thread to an arbitrary point, the planet will be hanging down more or less showing you a local gravity vector at the point of attachment.
Say, stick a needle with thread into that point.

(Because irl the thread axis will be passing to the Earth center through the CoM of the planet model).

Thus, you can picture the gravity and draw it on paper.

Edited by kerbiloid
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I did some more calculations.
If you draw a line from one of the tetrahedron's vertex to the center of the opposite face, the tetrahedron's center of mass (and of gravity) will be at 1/4 of the line from the center of the face, and 3/4 of the line from the vertex.

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Don't shoot the messenger, but...

https://solarsystem.nasa.gov/planets/in-depth/

Quote

What is a Planet?

The most recent definition of a planet was adopted by the International Astronomical Union in 2006. It says a planet must do three things:

  1. It must orbit a star (in our cosmic neighborhood, the Sun).
  2. It must be big enough to have enough gravity to force it into a spherical shape.
  3. It must be big enough that its gravity cleared away any other objects of a similar size near its orbit around the Sun.

So regardless of how it came to be, or if it once started out as a natural planet, once you make/magic this thing, it is not technically a planet anymore... :D

It would have to be fairly light to stay in its 'sculpted' shape, and any sort of atmosphere or surface water would wear down the sharp edges within no time (geologically speaking) and fill up the 'valleys' with sediment quickly (valley being the center of the polygon).

If you want a KSP example look no further than Gilly, although not nearly as regularly shaped as a Pythagorean solid you can imagine it being more regularly shaped and standing on the surface of it wouldn't be all that different. You'd effectively have 2 'biomes', the low lands and the high lands, with an arbitrary cut-off point somewhere in between. As we all know, when it comes to launching a rocket from Gilly it really isn't worth it to stumble/bounce yourself to a high point first, the gravity is so low on such non-spherical bodies that you'd just pretty much jump off into space.

For the heck of it let's say we have a planet sized body made out of nanobots that actively kept themselves in such a shape (and we disregard the enormous amounts of waste heat of all the work needed to fight gravity, which I have no doubt would melt the entire thing). I think the interesting thing about it would just be to walk from the center of a polygon to its edge. You'd be walking on a perfectly flat surface, as far as the eye could see, yet with every step you took away from the center of the polygon it would feel like you're walking up a steeper and steeper incline. I think it would be very disorienting, possibly nausea inducing... but only if the body was small enough to actually notice a difference in the direction of gravity after a short distance.

I think that's kinda the problem with the idea. If the body is so big it has enough gravity for normal surface travel, then it is probably also big enough for local differences in gravity to be unnoticeable at typical surface traveling speeds. If OTOH the body is small enough to notice the gravity shift over short distances, then its gravity would be so low that surface movement was just a series of ballistic trajectories (like 'driving' on Gilly or even walking on Minmus) and it probably wouldn't bother your sense of balance and direction any more than being in free fall would.

Note on Donut-shaped planets, Sixty Symbols did an episode about that:

Spoiler

 

 

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19 minutes ago, Beamer said:

It would have to be fairly light to stay in its 'sculpted' shape, and any sort of atmosphere or surface water would wear down the sharp edges within no time (geologically speaking) and fill up the 'valleys' with sediment quickly (valley being the center of the polygon).

Technically, it's possible to have an atmosphere near the center of the faces, but not around the edges/vertices, as they are three times further to the center of gravity than the center of the faces.

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7 minutes ago, Nazalassa said:

Technically, it's possible to have an atmosphere near the center of the faces, but not around the edges/vertices, as they are three times further to the center of gravity than the center of the faces.

But erosion doesn't limit itself to the top of mountains, quite the opposite in fact. The further you go down a slope the more water and debris from higher up will stream over it. It would just eat away at the slopes and the tops would collapse as the slopes became to steep to support their integrity. You would need some active process to 'top off' the mountains and slopes to combat this. On earth mountains form due to plate tectonics and volcanic activity, those processes require a molten core which would cause your body to become spherical. A totally rigid body would just wear down, it would have no active geological processes to keep its shape.

 

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2 hours ago, kerbiloid said:

A cubic planet plate tectonics.

  Reveal hidden contents

61e6316d104849a0287894b2b5b05763.gif


Tetrahedral planet plate tectonisc.

  Reveal hidden contents

giphy.gif


The real world, spherical planet plate tectonix.

  Hide contents

 

people who like these things should never watch any of the hellraiser movies. 

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On 11/26/2022 at 3:02 PM, Nazalassa said:

So, as gravity is inversely proportionnal to the distance to the center of mass, that would mean (if we look at the second table) that the gravity at the center of a face is nine times stronger than the gravity at a vertex?
Cool...

Did you consider that at the center of a face a sizable portion of the tetrahedron's volume is above you? Now the shell theorem is specifically about spherical objects but I suspect the three points of that face do end up to canceling somewhat. How much that effects the local gravity is beyond my mathing abilities though.

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10 hours ago, monophonic said:

Did you consider that at the center of a face a sizable portion of the tetrahedron's volume is above you? Now the shell theorem is specifically about spherical objects but I suspect the three points of that face do end up to canceling somewhat. How much that effects the local gravity is beyond my mathing abilities though.

The vertices are above you in absolute altitude but they are "below" you in the sense that they are beneath the plane tangent to the surface at your feet.

On 11/27/2022 at 12:18 PM, Beamer said:

Don't shoot the messenger, but...

https://solarsystem.nasa.gov/planets/in-depth/

Quote

What is a Planet?

The most recent definition of a planet was adopted by the International Astronomical Union in 2006. It says a planet must do three things:

  1. It must orbit a star (in our cosmic neighborhood, the Sun).
  2. It must be big enough to have enough gravity to force it into a spherical shape.
  3. It must be big enough that its gravity cleared away any other objects of a similar size near its orbit around the Sun.

So regardless of how it came to be, or if it once started out as a natural planet, once you make/magic this thing, it is not technically a planet anymore... :D

Well, technically, this definition merely says "big enough" to have enough gravity to force it into a spherical shape. It doesn't have to be spherical.

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Quote

A planet is a celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit.

Both cube and tetrahedron are nearly round. Nearly.

P.S.
Actually, if we have a cube axaxa A x A x A, the tangential sphere of A diameter will have a volume pi * A3 / 6 ~= 1/2 of the total cube volume.

So, a spherical ocean (six lakes at the face centers) will occupy most part of the cube voulme, and we can treat it as quasispherical.

Edited by kerbiloid
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21 minutes ago, kerbiloid said:

So, a spherical ocean (six lakes at the face centers) will occupy most part of the cube voulme, and we can treat it as quasispherical.

What if there is no ocean?

--

OK, technically, a tetrahedron IS topologically a sphere. Until you dig a tunnel that has two ends, at which point it topologically becomes a torus... Or a mug. Or both. Your choice.
Topology solves anything.

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10 minutes ago, Nazalassa said:

What if there is no ocean?

It will be a desert plain, geometrically flat, but concave by feeling.

  

10 minutes ago, Nazalassa said:

Until you dig a tunnel that has two ends, at which point it topologically becomes a torus... Or a mug. Or both. Your choice.

If keep digging again and again, it will be a sponge.

Then we have a plenty of little homes for creatures of any size.

Edited by kerbiloid
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Just now, kerbiloid said:

People live on the mountain slopes.

Yes, but they see the world as flat, wherever they are -- they will see the world as flat if they live at -- or near -- the center of a face, and as flat, but sloppy, if they live further from the center of the face. It will always seem flat. If people are travelling at random on a face, they will experience different gravity orientation, even if they see the world as flat, and I guess they won't really understand what's happening.

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9 minutes ago, Nazalassa said:

Yes, but they see the world as flat, wherever they are -- they will see the world as flat if they live at -- or near -- the center of a face, and as flat, but sloppy, if they live further from the center of the face. It will always seem flat. If people are travelling at random on a face, they will experience different gravity orientation, even if they see the world as flat, and I guess they won't really understand what's happening.

You see the spherical Earth flat.

They'll see the flat face spherical.

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