# A Decoy Planet made of expanded polystyrene. How big can I make it?

Note: This is different from the previous question Building a full-sized Lego Earth - what would it look like at various levels?. In that case it was about making a solid full-size Earth from Lego and whether it would have a liquid core. This question is asking How big can a (non-Earth) planet realistically be whilst remaining relatively non-compressed. It does not have to be solid or even have a core at all if it would be stable without one. Answers already given show that the result is completely different.

I wish to make a fake planet to use as a decoy for a real planet. It will be painted to look the same from a distance but will be made from expanded polystyrene.

Question

How big can I make an expanded polystyrene (EPS) planet before it starts to collapse under its own weight?

The planet is not Earth but, for the sake of this question, it is to be placed in an Earth-like orbit around a Sol-like sun. It will not have a moon however.

Possible cross-sections - it can be solid or hollow, whatever gives the greatest stable size. However it must be self-supporting and not make use of any other material than EPS.

Note

If a full-fledged planet is impossible then I'd still like to know the maximum achievable size for an EPS spheroid in an Earth-like orbit around a Sol-like star.

How big can I make an expanded polystyrene (EPS) planet before it starts to collapse under its own weight?

To answer this we need to know how strong is the gravity generated by the prop. The weight will then be a consequence of this gravity.

Let's say the propbuilder who is making this prop goes for a shell of EPS, with vacuum inside.

A shell of radius $$R$$ and thickness $$T$$, much smaller than the radius, has a total volume $$V \approx 4\pi R^2T$$ and a mass $$M= 4\pi R^2T \rho$$.

The gravitation pull exerted by the shell on itself will be $$g_\mathrm s=G\frac{M_\text{shell}}{R^2}=G\frac{4\pi R^2T \rho}{R^2}=4\pi GT \rho$$

As you can see the force that a thin shell exerts on itself doesn't depend on its radius but only on its thickness. Keep it thin, and you can make it as large as you want.

EDIT

As HenningMakolm pointed out in their comment

Draw an equator on your spherical shell. The sum of all the northward components of gravity in the south hemisphere must be canceled out by southward gravity in the north -- but all that force is transmitted between north and south by compression across the equator. The total northbound gravity comes out as $$W_\mathrm s=\frac12 g_\mathrm s T\rho \cdot \pi R^2$$ whereas the area of the equatorial cross section this force needs to be distributed over is only $$A=2\pi R\cdot T$$. So even if you keep $$T, \rho$$ constant, the lateral pressure the shell needs to withstand to keep itself up grows in proportion to $$R$$.

The pressure will then be $$P=\frac{W}A =\frac14 g_\mathrm s \rho R = \pi \rho ^2 GTR$$

Based on the values reported on the table available at this link, density of EPS varies between $$16$$ and $$640\ \mathrm{kg/m^3}$$, while varies between $$16$$ and $$40\ \mathrm{MPa}$$. $$G$$ is $$6.6\cdot 10^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}}$$.

Putting the numbers in we get that, for the densest and more resistant EPS we get $$TR < \frac{40\cdot10^6\ \mathrm{Pa}}{\pi \cdot (640\ \mathrm{kg/m^3})^2 \cdot 6.6\cdot 10^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}}}=4.71\cdot 10^{11} \ \mathrm{m^2}$$

Coincidentally, $$10^{11} \ \mathrm m$$ is the order of magnitude of the Earth-Sun distance.

• Technically, you've calculated the local value of g (gs as you've noted), which is acceleration, rather than force. – WhatRoughBeast Feb 2 at 19:38
• @WhatRoughBeast, correct, but the "own weight" is a consequence of the "own gravity" – L.Dutch - Reinstate Monica Feb 2 at 20:03
• While the force the thin shell exerts is independent of size it's ability to resist that force isn't. The bigger your sphere the flatter the surface and thus the lower it's ability to resist buckling. – Loren Pechtel Feb 2 at 22:58
• Draw an equator on your spherical shell. The sum of all the northward components of gravity in the south hemisphere must be canceled out by southward gravity in the north -- but all that force is transmitted between north and south by compression across the equator. The total northbound gravity comes out as $\frac12 g_s T\rho \cdot \pi R^2$ whereas the area of the equatorial cross section this force needs to be dirstributed over is only $2\pi R\cdot T$. So even if you keep $T, \rho$ constant, the lateral pressure the shell needs to withstand to keep itself up grows in proportion to $R$. – Henning Makholm Feb 3 at 2:54
• @HenningMakholm, good point, I will expand my answer using that. Thanks – L.Dutch - Reinstate Monica Feb 3 at 5:17

The surface gravity of the planet cannot be such that any object standing on the surface will begin to crush the polystyrene. With the exception of a meaningless variance in thickness (meters compared to millions of kilometers) that I'll mention momentarily, this represents the maximum radius of the planet.

And some constants...

• G = 6.674×10−11 m3/s2kg

• Density = 1 lb/ft3 = 16.0185 Kg/m3 (calculations for all other densities are left as an exercise for the reader.)

• Compressive Strength (max) = 17 psi = 703.068836 kg/m2

So, using a test mass of 1kg, our surface gravity must equal the compressive strength or 703.07 = G * (4/3𝜋r3) * Density / r thus...

• r = 1.57x1011 m = 1.57x108 Km (compared to Earth's 6.37x103 Km)

Notes...

• Higher densities means a smaller radius, so it's not work calculating against the other two densities of EPS.

• A hollow in the sphere means, roughly, that the outer radius is equal to r, above, plus the hollow inner radius. This isn't exactly true because gravity weakens as the inner radius expands. But for the purposes of this answer, it's close enough to true for government work (regardless the tag).

• I'm ignoring the fact that a radius so large means it's possible for weird and wonderful things to happen inside the planet, like rotational forces exceeding the compressive strength causing liquefaction which will cause all kinds of compromising issues, like friction leading to heat leading to other kinds of breakdown. Planetary science is much more complicated than I've presented here, leading me to the conclusion that it would be simpler to simply coat the planet with enough EPS to smooth out the surface and begin spray painting. At least then all you have to worry about is the effect of solar heating, tidal damage, and the fact that you just pushed all your atmosphere into space.

HOWEVER...

What this answer does not contemplate is that the higher density of a hidden planet (higher G at the surface of said planet than for EPS of the same radius), will seriously reduce the maximum radius of the EPS shell around it. If your goal is to actually wrap this around something, that something's gravity must be considered. This answer demonstrates that you could high a pretty big planet (bigger than Earth) inside the EPS shell, but the maximum radius indicated is for a hidden planet of G=0. As the hidden planet's gravity increases, the max radius of the EPS shell must decrease.

Note that if you're tempted to ask, "how big a planet can I hide with an EPS shell before that shell breaks down?" you can't simply assume Earth's density and start increasing the radius. As radius increases, density changes because internal pressures are increasing, too. It's a pretty complicated calculation, I suspect.

• Ah, I hadn't considered disguising a planet - That is a fascinating idea though. Rather I literally wanted to build a duplicate of the original that would deceive an enemy into attacking it instead of the real thing. It could be great to combine the two ideas – chasly from UK Feb 2 at 19:29
• Ah-hah. I apologize that I jumped to a conclusion, but I hope and believe my answer addresses your original intent. – JBH Feb 2 at 19:30
• 1.57x10⁸ km, you say? That's within 5% of the distance between earth and sun. So you have a polystyrene Dyson sphere, nice. – Fabian Röling Feb 2 at 20:59
• @FabianRöling Holy Cow! I hadn't even stopped to consider how close the ball would get to the sun. Hah! This design has bigger problems (along the lines of, "this is what happens when you put marshmallows on a stick near a campfire" problems). Gratefully, the OP doesn't need anything like that distance to achieve the stated goals. – JBH Feb 2 at 21:29
• Are you sure the constraint you start out with is the only relevant one? Are you assuming that each successive shell of polystyrene will support its own gravity by lateral compression? Otherwise you would have increasing pressure as you move inwards in your Dyson-sized polystyrene sphere. – Henning Makholm Feb 3 at 2:40

L.Dutch calculates the acceleration of gravity just outside a thin hollow polystyrene shell as $$g_s = 4\pi GT\rho$$ where $$G$$ is the universal constant of gravity, $$T$$ is the thickness of the shell, and $$\rho$$ is the density of the polystyrene. He observes that this is independent of $$R$$, the radius of the sphere.

However this is not the entire story, because the shell has to withstand its own weight without crumbling, and this is not just a function of the local acceleration of gravity. The shell is basically a self-supporting $$360^\circ$$ arch, and in constant gravity larger archs (of the same cross section) can support smaller weights.

The force of gravity per unit area of the shell is $$\tfrac 12 g_s T \rho$$ (where the $$\tfrac12$$ is because the gravitational strength drops to $$0$$ just inside the hollow shell due to the shell theorem).

If we draw an equator around the shell, the integral of the southbound component of gravity everywhere on the north hemisphere works out nicely as $$\tfrac 12 g_s T \rho \cdot \pi R^2$$ (there are some nasty-looking trigonometric factors as we write down the integral, but they all happen to cancel out before we integrate, if we parameterize the north hemisphere by its perpendicular projection on the equatorial plane. Note that the new factor is just the area of the circle cut out of the equatorial plane by the shell!)

All this force needs to be transmitted between north and south by lateral compression of the shell across the equator. The area of this cross section is $$2\pi R \cdot T$$ so the average lateral pressure must be $$\frac{\tfrac12g_s T \rho \pi R^2}{2\pi R T } = \pi GTR\rho^2$$ Now there is an $$R$$ factor again, so we cannot make the shell arbitrarily large without it crushing itself from the sides. The exponents here make sense too:

• $$\rho^2$$ makes sense because gravity increases by the product of two masses, and making each segment of shell twice as massive increases the forces fourfold.
• $$R$$ makes sense because each meter of equator needs to support the entire wedge of shelf going from itself up to the north pole.
• $$T$$ makes sense because making the shell twice as thick makes the forces four times as large, but on the other hand also distribute the forces over twice as much cross section.

So according to this calculation we should use the light class of EPS (the compressive strength scales slower than the square of the density). We should also make the shell thin -- at least until it becomes so thin that it starts to buckle.

I think the above calculation actually takes care of making the shell not buckle under its own gravity [at least that's what I thought until I ran the numbers below] -- but if we make it too lightweight, it might start buckling under external forces, such as gusts of stellar wind or the exhaust from the camera crews' rocket engines.

To be on the safe side, let's dimension the shell such that the force of its self-gravity is about 1 mN/m². Then at being hit by the long-distance puff of a solar flare (which can reach ram pressures of several μPa, based on their measured effect on Earth's magnetosphere -- the ordinary solar wind is a few orders of magnitude weaker than that) should still be negligible compared to the self-gravity.

We can then solve $$2\pi GT^2\rho^2 = 1\;\rm mPa$$ for $$T$$ to find a thickness of about $$96\;\rm m$$. Let's make it an even 100 meters for back-of-the-envelope purposes.

Now we can solve $$\pi GTR\rho^2 = 12\;\rm p{.}s{.}i{.}$$ for $$R$$, getting about $$\large R \approx 15 \times 10^6 \;\rm km$$ which is just about a tenth of an astronomical unit, or 40 times the distance to the moon.

It looks like there is room enough for a cautious prop designer to increase $$T$$ by a safety factor of 10 and still build a 1:1 stand-in for Jupiter if he wants (assuming he can source enough polystyrene).

• Thanks. I was wondering about solar wind. – chasly from UK Feb 3 at 10:15