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Lego representations of Earth have been made at various scales, for example this one:

enter image description here

Suppose Slartibartfast made a full-size Earth from Lego (including the oceans being represented as solid blocks. It is made inside his huge cavern inside Magarathea.

Magrathea is an ancient planet located in orbit around the twin suns Soulianis and Rahm in the heart of the Horsehead Nebula.

It was the home of a new form of specialist industry: custom-made, luxury planet building. Hyperspatial engineers sucked matter through white holes in space to form dream planets - gold planets, square planets, glass planets, platinum planets, soft rubber planets with lots of earthquakes, planets covered with fish - all lovingly made to meet the exacting standards... https://hitchhikers.fandom.com/wiki/Magrathea

Clearly the gravitational force of the artificial Earth would affect it in some way. Presumably the core would melt but the surface would still consist of individual blocks.

Question

What would a cross-section of this planet look like in terms of layers? At what depth would Lego blocks be welded together by pressure and at what depth would they melt?

Notes

The average Lego density is 0.565 g/cm3 - https://www.wired.com/2016/12/heres-much-lego-brick-stepped-worth/

ABS maximum temperature is 80°C (176°F) and melt at 105°C (221°F) Polycarbonate plastic used for transparent bricks melt at 267°C (512.6°F) https://bricks.stackexchange.com/questions/547/how-much-heat-can-lego-bricks-withstand

Diameter of the Earth = 12.742 million meters - https://www.space.com/17638-how-big-is-earth.html

EDIT For the sake of calculations, let us use the classic brick with 8 knobs, 2×4. It can be assumed to weigh 2.22g

Units in diagram are all in millimetres. enter image description here


Additional Lego and ABS information

ABS Plastic in LEGO - https://legoways.com/abs-plastic-in-lego/

Acrylonitrile Butadiene Styrene (ABS) - https://www.makeitfrom.com/material-properties/Acrylonitrile-Butadiene-Styrene-ABS

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

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    $\begingroup$ Is that the density of just the plastic, or of the rectangular volume including empty spaces? Do we assume it's all one brick type? Different brick types have different volume proportions of empty space. Is there air or is this in a vacuum? Do you have any data on a Lego brick's compressional strength? $\endgroup$ – Spencer Jan 19 at 12:56
  • $\begingroup$ bbc.com/news/magazine-20578627 $\endgroup$ – Loong Jan 19 at 12:58
  • $\begingroup$ @Loong that has a crucial measurement for this; however, most of the article is about what happens under Earth gravity. It does bring up an interesting question to ask chasly: Are these idealized Lego bricks, or will we have to take manufacturing flaws into account? $\endgroup$ – Spencer Jan 19 at 13:05
  • $\begingroup$ Also, sphere or ellipsoid (or geoid!) might affect the number of bricks. $\endgroup$ – Spencer Jan 19 at 13:09
  • $\begingroup$ @Spencer - For simplicity the bricks can all be considered identical. I have added new information. The final shape of the Lego Earth should accurately model the real Earth. However if that makes the calculations difficult I don't mind it being regarded as a sphere. After all it is currently stationary inside Magarathea. $\endgroup$ – chasly from UK Jan 19 at 15:19
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Let's get out the calculator.

$M_E=5.972 \times 10^{24} \text{ kg}$

$\rho_E=5.51 \frac{\text g}{\text{cm³}}=551 \frac{\text {kg}}{\text{m³}}$

$V_E=\frac{M_E}{\rho_E}=1.083\times10^{21}\text{ m³}$

$=> M_L=V_E\times56.5\frac{\text {kg}}{\text{m³}}=6.12\times10^{23}\text{ kg}$

$=> g_L=-G\frac{M_L}{r^2}=1.006\frac{\text m}{\text{s²}}$

Result 1: Gravity on the surface

On the surface of the planet, the sheer mass of Lego-Earth creates about 1/9th of eath gravity, and as we assume equal density for all of the planet, we get this gravity graph, Y axis in m/s², X in meters: enter image description here

Result 2: The core does not melt.

As we don't have a density gradient to the core made from metal but the same plastic, the core does not melt.

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To the actual thing! Based on comments (of my other answer), I'll assume:

  • Cubic construction: Construct it like a cube, but stop building when you reach a spherical boundary. (It turns out that, because the so high Young's Modulus as compared to UTS value, it doesn't matter if you do a cubic construction, or spherical shell, or whatever).

  • Isothermal (or slow) construction. There's plenty of time for heat to dissipate to (or from) the environment, thus the temperature of the entire structure is the temperature of the environment (in the vacuum nothingness of space, that would probably be around 3K -- but then we're back to insufficient data. So, our environment will magically be held at 300K). For an estimate of how slow the construction rate has to be in order to be isothermal, one would need to know the thermal diffusivity of ABS (or, equivalently, its heat capacity and thermal conductivity).

That said, about ABS, as given by the OP in the question:

  • Young's Modulus at roughly 2~5 GPa. That means, it needs these kinds of pressure to cause a significant deformation in percentage compared with the original size within the linear hookean regime.

  • Bending Modulus (or flexural Modulus) is also about 2~5 GPa. This means that these are the pressures that will cause significant curvature change (at least in the context of linear hookean elastic plate theory).

  • Ultimate Tensile Strength (UTS) is about 40 MPa. That is "the equivalent" of the Young's Modulus for plastic deformation regimes. This means, at these stresses, the deformation will be plastic (instead of elastic), and will lead to tensile rupture.

  • Flexural Strength is about 72 MPa. That is "the equivalent" of the bending modulus for the plastic regime. At these pressures, it will lead to rupture.

Basically that means, that the lego blocks basically don't deform much, but they soon break: it doesn't stay too much in the linear regime.


Calculating the all materials' data

For a thin plate, the bending modulus is given by the equation below: $$ B = \frac{1}{12}\frac{h^3 E }{1 - \nu^2} $$

Where $h$ is the thickness of the plate, and $\nu$ is the Poisson ratio of the material, which is related to shear modulus by the following equation: $$ E = 2G(1+\nu) = 3K(1-2\nu) $$

Where $G$ is the shear modulus (describes what happens when one applies shear stresses), and $K$ is the bulk modulus of the material (proportional to the change in volume of the object under hydrostatic pressures) (we won't use $K$).

Plugging the given values for $E$ and $B$, together with the provided $h$ value from the shape of the block, we find: $$ \nu\approx 1, \quad\quad G\approx\frac{E}{4}, \quad\quad K\approx \frac{E}{3} $$

$\nu\approx 1$ means the brick will tend to volumetric expand/compress under pressure, instead of giving preference to the direction we're applying the stresses. Also, $G$ and $K$ are close enough to $E$, meaning, it resists shear stresses, tensile stresses and hydrostatic pressures around the same way (or, strain responses are in the same order of magnitude).

All of this data is much less than UTS and Bending strength, meaning, linear analysis is meaningless for pressures in the megapascal range.


Critical Radius

All of this, suggests the existence of a critical radius, call it $R_c$, which will aid the transition between an ABS blob, and the bricks itself. When pressures go above 40 MPa, rupture will happen, and the structure will collapse, as the material won't be able to support it, and, the lego blocks will disappear, and all we have is granular ABS under hydro-static pressures.


Below the critical radius

Below the critical radius, we have an ABS blob subject to huge hydrostatic pressures. Our goal is to calculate the density, and that can be done by the very definition of the bulk modulus: $$ K = -V\frac{dp}{dV} = \rho\frac{dp}{d\rho} $$

If we assume constant $K$, we have a differential first order linear equation to solve. Fortunately, this one is rather easy: $$ K = \rho\frac{dp}{d\rho} = \frac{d}{d\rho}\left[\frac{p^2}{2}\right] \quad\implies\quad \frac{p^2}{2} = K\rho $$

Thus, we find our density is dependent on the pressure: $$ \rho(p) = \frac{p^2}{2K} $$

By the radial hydrostatic equilibrium equation for a planet (or stellar object), we have: $$ \frac{dp}{dr} = \frac{GM}{r^2}\rho = \frac{4}{3}\pi G r\rho^2 = \frac{4}{3}\pi G r\frac{p^2}{4K^2} $$

Yet another differential equation to solve. Fortunately, this one is easy too. $$ \frac{dp}{dr} = \frac{4}{3}\pi G r\frac{p^2}{4K^2} \quad\implies\quad \frac{1}{p^2}\frac{dp}{dr} = \frac{\pi G r}{3K^2} $$

Henceforth: $$ -\frac{d}{dr}\left[\frac{1}{p}\right] = \frac{\pi G r}{3K^2} \quad\implies\quad \frac{1}{p} = \frac{\pi G r^2}{3K^2} + C $$

Where $C$ is a constant to be determined by the actual pressure at the critical radius: $$ \frac{1}{p_C} = \frac{\pi G R_c^2}{3K^2} + C \quad\implies\quad C = \frac{1}{p_C} - \frac{\pi G R_c^2}{3K^2} $$

Therefore, for $r < R_c$, we have a pressure profile of: $$ p(r) = \frac{1}{\displaystyle\frac{\pi G (r^2 - R_c^2)}{3K^2} + \frac{1}{p_c}} = \frac{3K^2p_c}{\pi G p_c (r^2 - R_c^2) + 3K^2} $$

Where $p_c$ is the critical pressure (that is, the pressure at the critical radius).


Transition range

The critical radius itself is not a region where the thing suddenly change its behaviour. It is a smooth transition. To calculate the exact radial size of the transition region, we would need the Yield Strength Point of the ABS, which is a data that we do not have.

But, it is probably safe to say, under a good approximation, that the size of the transition region will be much less than the radius of the entire planet. With that in mind, we approximate saying that the critical radius is a point of sudden change.

Thus, under the critical radius, we have ABS blob, and below critical radius, we have completely intact lego bricks. I know, not realistic, but again, we do not have the Yield point, and, transition region is probably too small.


Above critical radius

That can be done my simply doing the planet's hydrostatic radial equation once again. Radial hydrostatic equilibrium would give: $$ \frac{dp}{dr} = \frac{GM}{r^2}\rho(r) = \frac{4}{3}G\pi r\rho^2 $$

This time, given lego blocks are intact, $\rho$ is assumed to be constant, and the density of the lego bricks. We know from Young's Modulus, the lego won't deform significantly at all, thus, no change is density is good enough approximation.

Thus, to find the pressure profile is a direct integration: $$ p(r) = \frac{4}{3}\pi G(R^2 - r^2)\rho^2 $$

Where $R$ is the radius of the lego planet.


Finally, calculation of the Critical Radius!

Finally, the thing that we are aiming to calculate!

Thus, the critical pressure, that is, at $r=R_c$, is simply: $$ p(r=R_c) = \frac{4}{3}\pi G R_c^2 \rho^2 $$

Where we used the formula for $r\ge R_c$. However, for $r\le R_c$, we calculated: $$ p(r=R_c) = p_c $$

Because we require continuity at the critical radius, we'll declare these two to be equal. Therefore: $$ p(r=R_c) = p_c = \frac{4}{3}\pi G R_c^2 \rho^2 $$

Thus, the critical radius is: $$ R_c = \sqrt{\frac{3}{4}\frac{p_c}{\pi G \rho^2}} $$

Wow.

If we plug the critical pressure to be 40MPa (the UTS value), we get $R_c \approx 0.10 R$, where $R$ is the total radius of the planet.


Radial profile of the pressure

We calculate the pressure everywhere in the planet: At the critical radius, above critical radius, and below critical radius. Thus we can build a full radial profile of the pressure.

I am too lazy at the moment to plot it. But if anyone want to do that, go ahead and edit my answer. I might do it later (or I might never do).

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I think that this answer should have been a comment, but it would be too long, thus, an answer. Feel free to remove this answer.

Basically, you want to know three things:

What would a cross-section of this planet look like in terms of layers?

Depends on the construction. Isn't that obvious? If you construct it like a cube, having a cubic world in the end, and remove the legos until you get a sphere.. the cross section will be..... well..... you got it. If you construct a spherical shell at each time.. you will have a different cross section.

At what depth would Lego blocks be welded together by pressure?

Depends on the construction. Resistance of materials is highly dependent on its geometry, and where the forces are being applied. The same amount of material can resist much more or much less, if it is made in a certain shape or geometry.

Each type of construction will yield different types of forces applied under the geometrical form of the lego block. The material response will be different at each of them. For instance, a cube like world, the forces will come at all different shapes and and angles across the lego block, for different positions in the interior (that will create significant shear stresses in almost all blocks).

For a spherical-shell-by-shell world, most of the forces will be radial (thus less shear stresses on the block). And so on. Each of these cases, will have a different response from the lego blocks.

More, if the pressure is too high, it is possible the lego might break, and even granulate, so we will have to look at the responses of the ABS itself (young modulus, bulk modulus, shear modulus, etc).

The maximum I can do, is to calculate the pressure profile assuming the entire thing is a continuum body (no empty spaces), and search if there is an area which surpasses the material properties of ABS itself.

Radial hydrostatic equilibrium would give: $$ \frac{dp}{dr} = \frac{GM}{r^2}\rho(r) = \frac{4}{3}G\pi r\rho^2 $$

Thus, to find the pressure profile is a direct integration: $$ p(r) = \frac{4}{3}\pi G(R^2 - r^2)\rho^2 $$

Thus, pressure on the core, that is, at $r=0$, is simply: $$ p(r=0) = \frac{4}{3}\pi G R^2 \rho^2 $$

If $R = 6371 Km$ and $\rho = 565 Kg/m^3$, we get $p(0) = 3.6224 GPa$. Though, I couldn't find anywhere Young's Modulus and Shear Modulus of ABS, to make a comparisom and determine if it will collapse or not. But, as you can see, it is well within the Young's Modulus of several (or most) materials, thus, probably within the regime of linear deformation, and thus, no collapse.

Polymers and plastics can have very high, or very low Young's Modulus, thus, ABS's data is indeed required. And keep in mind our small analysis assumed a continuum: lego blocks has many empty spaces. This further increases the need of the ABS data, and how exactly the lego planet is built, for one to make proper calculations.

What depth would they melt?

Depends on the construction. You can, for instance, do an isothermal quasi-static construction: Put it in contact with a heat reservoir of any chosen temperature, to make sure your entire structure has the same desired temperature $T$, that could be any of your choice. Then construct it quasi-statically: Put the first lego block in place. Wait for thermalization. Slowly, put the second lego block in place, then wait for thermalization. In the end, the entire structure will have the same temperature, and no part of it will melt.

Or perhaps you can do a sudden adiabatic construction: If a cubic lego brick earth-sized of bigger, happen to magically appear in our universe (or, to be 'constructed' extremely quickly), they would suddenly feel gravity and deform (or worse: collapse) under their weight. That elastic energy would cause heating, and then the core probably would reach considerable temperatures in the process (further increasing collapse probability), creating a temperature gradient.

Your construction methods will likely be in between the two (or not), and thus, the core temperature will be in between the two. Thus, temperature is highly dependent on how you build it.

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  • $\begingroup$ You could construct it like a cube and when any part of that cube intersects with the surface of your desired sphere you stop building at that point. However that still doesn't say whether it's possible to build a cube that big. Wouldn't it simply collapse under its own gravity? $\endgroup$ – chasly from UK Jan 20 at 12:28
  • $\begingroup$ @chaslyfromUK That's what I tried to answer here: we have insufficient data to answer all your questions. For instance, plugging your data (density and planet's radius), and assuming the whole thing a continuum, I calculate the pressure on the core to be at 3.62 gigapascal. This is well within the range of several materials to resist the collapse. To answer if it will collapse or not, I need the young's modulus and shear modulus of ABS, and then to apply those to the specific shape of the lego block, under a specific construction. I have neither of these: all you provide is shape. $\endgroup$ – Physicist137 Jan 20 at 15:05
  • $\begingroup$ @chaslyfromUK I edited my answer to make it more clear. $\endgroup$ – Physicist137 Jan 20 at 15:22
  • $\begingroup$ @ Physicist137 - I didn't provide those quantities because I wasn't aware of their existence. If a complete answer is impossible then I'll forgo the air-space and projecting toggles of Lego and assume they are just rectangular solid blocks that stick together by some other force. I've done my best to provide the other information and included it at the bottom of the question. $\endgroup$ – chasly from UK Jan 20 at 15:22
  • $\begingroup$ I'm now assuming that you build as though making a cube but at every stage you build until you reach a spherical boundary. This boundary is updated every meter or so. Thus the planet always assumes a roughly spherical shape but has a 'crytalline' structure. $\endgroup$ – chasly from UK Jan 20 at 15:25

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