I have a Geodesic planet roughly the same diameter of Earth but hollow. It has a surface crust approximately 20 miles deep with landmasses and oceans on its surface. It has giant reserves of "unobtanium" to give the mass required for a gravity of 1g.

My question is would this planet suffer any oddities from its shape, Would there be odd areas with gravity higher or lower than other areas even with the crust and geodesic shell having the same width and density throughout?

Edit: The planet is a truncated polyhedron with 12 decagonal faces and 20 triangular phases. The unobtanium is distributed in the shell of the polyhedron which is about 50 miles thick and located beneath the 20 miles of crust.

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    $\begingroup$ The gravitational accelerarion on Earth's surface varies from place to place, from 9.825 m/s² in Oslo, Norway, down to 9.776 m/s² in Mexico City. That's a variation of 0.5%. (And you didn't tell us how many faces does the geodesic polyhedron have -- it will approximate a sphere better and better as the number of faces increases.) $\endgroup$
    – AlexP
    Jun 8, 2020 at 22:49
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    $\begingroup$ Would you mind clarifying what you mean by a "geodesic planet"? $\endgroup$
    – HDE 226868
    Jun 9, 2020 at 0:09
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    $\begingroup$ Where is the reserve of unubtanium kept? Equally distributed through the crust, in the center, or offset in some sector of the geodesic shell? $\endgroup$ Jun 9, 2020 at 0:35
  • $\begingroup$ So I understand your planet to be a truncated polyhedron made out of some material 50 miles thick, made partially of unobtanium, and on top of this layer is a crust, 20 miles thick, on which are the oceans and land masses, mountains, valleys, and such? And inside of this is basically hollow? $\endgroup$ Jun 9, 2020 at 11:45
  • $\begingroup$ @justin thyme the second.....yes! $\endgroup$
    – Thalassan
    Jun 9, 2020 at 15:27

2 Answers 2


There are a few little oddities comming in my mind here:

At first: gravity As long as your unobtanium is equaly distributed, your center of gravity still lies in the center of the planet. The faces of your planet are rather large, so there happen two oddities to gravity. Gravity will vary between center and endge of a face more then gravity varies on earth (the rotational differences from earth will happen also if your planet rotates), I'm not shure how much, I have to admitt. Also, your gravity will be 'downwards' just in the middle of a face. Near the edges gravities pull will be directed mostly downwards and a bit sideways, directed to the center of the face. By that, crossing a face will not feel like walking on a flat surface but more like crossing a concave area.

Second: magnetism As your planet lacks a rotating metal core, there will be no magnetic field protecting it from cosmic rays. If you want this planet to be habitable, there has to be some device fullfilling that part.

Third: tectonic Your planet lacks any tectonics, therefore if you implement an atmosphere with water cycle and so on, over thousands and millions of years you will loose your edges to weather (at least until it reaches a surface that hard that it couldn't be eroded.

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    $\begingroup$ Magnetosphere can happen in different ways. 3 examples: Jupiter magnetosphere happen due metalic liquid hydrogen. Venus because induction in the high atmosphere and Ganimede for interactions of the different layers of ice and water, with different levels of salts inside. Actually we have a kind of human-made magnetosphere too. One 4th problem is keep all chemical elements cycles (water, carbon, phosphorus, nitrogen, etc) working in good levels in the litosphere $\endgroup$ Jun 9, 2020 at 16:41
  • $\begingroup$ If the unobtanium is distributed in just the right way (more of it near the edges and vertices than in the middle of the faces), it might actually be possible to have gravity be perpendicular to the ground over most of the surface of the planet. Maybe. I'm not sure. $\endgroup$ Jun 10, 2020 at 22:23

It seems to me, and I am sure someone will correct me if I am wrong, that gravity inside the hollow sphere would be very interesting. All of the mass creating the gravity would be surrounding you on the outside, not the inside.

If you were just on the inside of the shell structure, there would be much more mass on the 'down' side of you from the other side (towards the center), so you would be pulled towards the equilibrium point - the center of the sphere. Would your acceleration due to gravity increase, or decrease, on the trip towards the center, as the competing forces balance out and approach equilibrium? Would you speed up in your decent towards the center, or slow down? It seems to me that, at the center, you would have no further acceleration due to gravity. All of the gravitational forces would be equalized in every direction. Would you feel them - 'weightless'?

Another point - if your unobtanium produces gravitational effects normally, through mass, your shell would have to be extremely 'massive' and therefore 'dense' in order to produce 1 g. The stresses on the shell structure would be primarily compressive. The structure, therefore, would have to be able to withstand tremendous compressive forces. This begs the question be asked, 'How compressive is the material forming the shell?', and how are these compressive forces distributed through the structure? The denser the material, the more plausible this is. However, if something very, very massive, for instance hit the planet in the center of one of your decagonal faces, would this be enough to deform the panel? The main support structures of a geodesic dome are in the connecting 'folds' between the panels. The panels themselves transfer any forces to these fold lines. Any force perpendicular to the center of the panel (towards the center of the sphere) would become a tension force. That is, the strength of a geodesic dome is in the structure of the boundaries between the panels (compressive), the folds. The actual surface of the panels themselves (tension) is the weak point. You might find that the centers of your panels deform towards the sphere center, forming concave plates, and the folds, or joints, between the plates forming ridges, or peaks. Water would flow 'downhill', from the high point of the folds/joints to the low point center of the plates. The more facets your geodesic dome has, the less this distortion would be. Perhaps 12 would be on the very low side of providing structural integrity. The closer to an infinite number of plates you get, the more it approximates a sphere, with pure compressive forces and no tension forces with a perfect distribution of material.

Now, if your 'unobtanium' creates gravity through some other mechanism than 'mass', we have a different scenario. Since 'unobtabium' is speculative in nature, could the gravitational effect be somehow related to something other than 'mass', therefore involving a completely different calculation of the stresses on the structure? This scenario makes the entire solution totally speculative in nature. In such a case anything goes. You are free to assume a perfectly rigid structure that transfers tension and compression forces perfectly, without distortion.

  • $\begingroup$ Gravity on the interior is really not all that interesting. In most of the interior, the mass distribution can be well-approximated as spherical, and per the shell theorem the net force will be zero--not towards the center. Near the polyhedral edges and vertices, gravity would pull weakly outwards. $\endgroup$ Jun 11, 2020 at 3:17
  • $\begingroup$ @Logan R. Kearsley The possibility that the shell theorem would apply occurred to me, but if it did, then the entire structural principle of the geodesic sphere would be invalidated. The integrity of the sphere depends on everything being in compression, due to gravity 'pulling' it in. Thus, if the shell theorem applied, then things would be even more interesting. In order for the unobtanium to produce gravity of 1g universally at the surface of the shell (not outside), would it not have to be assumed that the shell was somehow not hollow gravitationally, by some quirk of unobtanium? , $\endgroup$ Jun 11, 2020 at 4:33
  • $\begingroup$ Nope. Every bit of matter that makes up the shell is, by definition, not in the hollow interior of the shell, so the forces do not cancel out. All of the material of the shell near its exterior surface feels the gravity of the material of the shell that is nearer the interior surface (and thus provides force to keep the structure in compression)--and all material external to the shell feels the full gravity of the entire shell. $\endgroup$ Jun 11, 2020 at 5:45
  • $\begingroup$ @Logan R. Kearsley It's that 'infinitely thin' part of shell theory that is the glitch. How can anything be part of an infinitely thin shell? In this case, the 'shell' is actually 70 miles thick. So, yes the outer surface 'shell' layer would not encase a 'hollow' sphere, but a sphere with some substance and mass. Only the inner 'surface' layer would fit the definition of a sphere enclosing a hollow area, and with only 12 decagonal faces and 20 triangular phases on something with the radius of earth, that would not be a sphere.The symmetry would not be uniform. $\endgroup$ Jun 11, 2020 at 14:41
  • $\begingroup$ As a curiosity, would time dilation due to gravity be absolutely uniform within the hollow sphere? If the universe is expanding, does that mean there is a 'hollow sphere' forming in the center of the expanding universe? $\endgroup$ Jun 11, 2020 at 14:44

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