Mixing
A Decoy Planet made of expanded polystyrene. How big can I make it?
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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.
https://www.epsindustry.org/packaging/physical-properties
planets hard-science
asked Feb 2 at 15:36
chasly from UKchasly from UK
20k983174
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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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show 9 more comments
$begingroup$
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.
https://www.epsindustry.org/packaging/physical-properties
planets hard-science
asked Feb 2 at 15:36
chasly from UKchasly from UK
20k983174
$endgroup$
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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Into which environment is this prop placed?
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– L.Dutch♦
Feb 2 at 16:25
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@ L.Dutch - I mentioned that in the last sentence of my notes.
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– chasly from UK
Feb 2 at 16:33
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That's unclear. Are there other bodies? How big and how far are they?
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– L.Dutch♦
Feb 2 at 16:34
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There is no Moon. For the sake of argument let us say that everything else is the same as in our solar system.
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– chasly from UK
Feb 2 at 16:35
1
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Possible duplicate of Building a full-sized Lego Earth - what would it look like at various levels?. While there is quite a lot of difference between ABS and EPS, at this scale it should even out.
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– o.m.
Feb 2 at 18:28
|
show 9 more comments
$begingroup$
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.
https://www.epsindustry.org/packaging/physical-properties
planets hard-science
asked Feb 2 at 15:36
chasly from UKchasly from UK
20k983174
$endgroup$
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.
https://www.epsindustry.org/packaging/physical-properties
planets hard-science
planets hard-science
asked Feb 2 at 15:36
chasly from UKchasly from UK
20k983174
asked Feb 2 at 15:36
chasly from UKchasly from UK
20k983174
edited Feb 3 at 13:23
chasly from UK
asked Feb 2 at 15:36
chasly from UKchasly from UK
20k983174
asked Feb 2 at 15:36
chasly from UKchasly from UK
20k983174
asked Feb 2 at 15:36
chasly from UKchasly from UK
20k983174
20k983174
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.
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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Into which environment is this prop placed?
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– L.Dutch♦
Feb 2 at 16:25
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@ L.Dutch - I mentioned that in the last sentence of my notes.
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– chasly from UK
Feb 2 at 16:33
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That's unclear. Are there other bodies? How big and how far are they?
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– L.Dutch♦
Feb 2 at 16:34
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There is no Moon. For the sake of argument let us say that everything else is the same as in our solar system.
$endgroup$
– chasly from UK
Feb 2 at 16:35
1
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Possible duplicate of Building a full-sized Lego Earth - what would it look like at various levels?. While there is quite a lot of difference between ABS and EPS, at this scale it should even out.
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– o.m.
Feb 2 at 18:28
|
show 9 more comments
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Into which environment is this prop placed?
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– L.Dutch♦
Feb 2 at 16:25
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@ L.Dutch - I mentioned that in the last sentence of my notes.
$endgroup$
– chasly from UK
Feb 2 at 16:33
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That's unclear. Are there other bodies? How big and how far are they?
$endgroup$
– L.Dutch♦
Feb 2 at 16:34
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There is no Moon. For the sake of argument let us say that everything else is the same as in our solar system.
$endgroup$
– chasly from UK
Feb 2 at 16:35
1
$begingroup$
Possible duplicate of Building a full-sized Lego Earth - what would it look like at various levels?. While there is quite a lot of difference between ABS and EPS, at this scale it should even out.
$endgroup$
– o.m.
Feb 2 at 18:28
$begingroup$
Into which environment is this prop placed?
$endgroup$
– L.Dutch♦
Feb 2 at 16:25
$begingroup$
Into which environment is this prop placed?
$endgroup$
– L.Dutch♦
Feb 2 at 16:25
$begingroup$
@ L.Dutch - I mentioned that in the last sentence of my notes.
$endgroup$
– chasly from UK
Feb 2 at 16:33
$begingroup$
@ L.Dutch - I mentioned that in the last sentence of my notes.
$endgroup$
– chasly from UK
Feb 2 at 16:33
$begingroup$
That's unclear. Are there other bodies? How big and how far are they?
$endgroup$
– L.Dutch♦
Feb 2 at 16:34
$begingroup$
That's unclear. Are there other bodies? How big and how far are they?
$endgroup$
– L.Dutch♦
Feb 2 at 16:34
$begingroup$
There is no Moon. For the sake of argument let us say that everything else is the same as in our solar system.
$endgroup$
– chasly from UK
Feb 2 at 16:35
$begingroup$
There is no Moon. For the sake of argument let us say that everything else is the same as in our solar system.
$endgroup$
– chasly from UK
Feb 2 at 16:35
1
1
$begingroup$
Possible duplicate of Building a full-sized Lego Earth - what would it look like at various levels?. While there is quite a lot of difference between ABS and EPS, at this scale it should even out.
$endgroup$
– o.m.
Feb 2 at 18:28
$begingroup$
Possible duplicate of Building a full-sized Lego Earth - what would it look like at various levels?. While there is quite a lot of difference between ABS and EPS, at this scale it should even out.
$endgroup$
– o.m.
Feb 2 at 18:28
|
show 9 more comments
3 Answers
3
active
oldest
votes
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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 $V approx 4pi R^2T$ and a mass $M= 4pi R^2T rho $.
The gravitation pull exerted by the shell on itself will be $$g_mathrm s=Gfrac{M_text{shell}}{R^2}=Gfrac{4pi R^2T rho}{R^2}=4pi GT rho$$
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 $W_mathrm s=frac12 g_mathrm s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be distributed over is only $A=2pi Rcdot T$. 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 $6.6cdot 10^{-11} mathrm{m^3 kg^{-1} s^{-2}}$.
Putting the numbers in we get that, for the densest and more resistant EPS we get $$TR < frac{40cdot10^6 mathrm{Pa}}{pi cdot (640 mathrm{kg/m^3})^2 cdot 6.6cdot 10^{-11} mathrm{m^3 kg^{-1} s^{-2}}}=4.71cdot 10^{11} mathrm{m^2}$$
Coincidentally, $10^{11} mathrm m$ is the order of magnitude of the Earth-Sun distance.
answered Feb 2 at 18:56
L.Dutch♦L.Dutch
91.4k29211439
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Technically, you've calculated the local value of g (gs as you've noted), which is acceleration, rather than force.
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– WhatRoughBeast
Feb 2 at 19:38
1
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@WhatRoughBeast, correct, but the "own weight" is a consequence of the "own gravity"
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– L.Dutch♦
Feb 2 at 20:03
2
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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.
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– Loren Pechtel
Feb 2 at 22:58
4
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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 $frac12 g_s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be dirstributed over is only $2pi Rcdot T$. So even if you keep $T, rho$ constant, the lateral pressure the shell needs to withstand to keep itself up grows in proportion to $R$.
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– Henning Makholm
Feb 3 at 2:54
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@HenningMakholm, good point, I will expand my answer using that. Thanks
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– L.Dutch♦
Feb 3 at 5:17
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show 3 more comments
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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.
Gravity = G * Mass(M1) / radius2
F = mA or, Weight = Mass(M2) * Gravity
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 hard-science 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.
answered Feb 2 at 19:09
JBHJBH
48.3k699228
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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
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– chasly from UK
Feb 2 at 19:29
1
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Ah-hah. I apologize that I jumped to a conclusion, but I hope and believe my answer addresses your original intent.
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– JBH
Feb 2 at 19:30
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1.57x10⁸ km, you say? That's within 5% of the distance between earth and sun. So you have a polystyrene Dyson sphere, nice.
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– Fabian Röling
Feb 2 at 20:59
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@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.
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– JBH
Feb 2 at 21:29
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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.
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– Henning Makholm
Feb 3 at 2:40
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show 4 more comments
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L.Dutch calculates the acceleration of gravity just outside a thin hollow polystyrene shell as
$$ g_s = 4pi GTrho $$
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
$$ 2pi R cdot T $$
so the average lateral pressure must be
$$ frac{tfrac12g_s T rho pi R^2}{2pi R T } = pi GTRrho^2 $$
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
$$ 2pi GT^2rho^2 = 1;rm mPa$$
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
$$ pi GTRrho^2 = 12;rm p{.}s{.}i{.} $$
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).
answered Feb 3 at 4:04
Henning MakholmHenning Makholm
84869
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Thanks. I was wondering about solar wind.
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– chasly from UK
Feb 3 at 10:15
add a comment |
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3 Answers
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3 Answers
3
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$begingroup$
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 $V approx 4pi R^2T$ and a mass $M= 4pi R^2T rho $.
The gravitation pull exerted by the shell on itself will be $$g_mathrm s=Gfrac{M_text{shell}}{R^2}=Gfrac{4pi R^2T rho}{R^2}=4pi GT rho$$
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 $W_mathrm s=frac12 g_mathrm s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be distributed over is only $A=2pi Rcdot T$. 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 $6.6cdot 10^{-11} mathrm{m^3 kg^{-1} s^{-2}}$.
Putting the numbers in we get that, for the densest and more resistant EPS we get $$TR < frac{40cdot10^6 mathrm{Pa}}{pi cdot (640 mathrm{kg/m^3})^2 cdot 6.6cdot 10^{-11} mathrm{m^3 kg^{-1} s^{-2}}}=4.71cdot 10^{11} mathrm{m^2}$$
Coincidentally, $10^{11} mathrm m$ is the order of magnitude of the Earth-Sun distance.
answered Feb 2 at 18:56
L.Dutch♦L.Dutch
91.4k29211439
$endgroup$
$begingroup$
Technically, you've calculated the local value of g (gs as you've noted), which is acceleration, rather than force.
$endgroup$
– WhatRoughBeast
Feb 2 at 19:38
1
$begingroup$
@WhatRoughBeast, correct, but the "own weight" is a consequence of the "own gravity"
$endgroup$
– L.Dutch♦
Feb 2 at 20:03
2
$begingroup$
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.
$endgroup$
– Loren Pechtel
Feb 2 at 22:58
4
$begingroup$
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 $frac12 g_s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be dirstributed over is only $2pi Rcdot T$. So even if you keep $T, rho$ constant, the lateral pressure the shell needs to withstand to keep itself up grows in proportion to $R$.
$endgroup$
– Henning Makholm
Feb 3 at 2:54
$begingroup$
@HenningMakholm, good point, I will expand my answer using that. Thanks
$endgroup$
– L.Dutch♦
Feb 3 at 5:17
|
show 3 more comments
$begingroup$
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 $V approx 4pi R^2T$ and a mass $M= 4pi R^2T rho $.
The gravitation pull exerted by the shell on itself will be $$g_mathrm s=Gfrac{M_text{shell}}{R^2}=Gfrac{4pi R^2T rho}{R^2}=4pi GT rho$$
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 $W_mathrm s=frac12 g_mathrm s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be distributed over is only $A=2pi Rcdot T$. 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 $6.6cdot 10^{-11} mathrm{m^3 kg^{-1} s^{-2}}$.
Putting the numbers in we get that, for the densest and more resistant EPS we get $$TR < frac{40cdot10^6 mathrm{Pa}}{pi cdot (640 mathrm{kg/m^3})^2 cdot 6.6cdot 10^{-11} mathrm{m^3 kg^{-1} s^{-2}}}=4.71cdot 10^{11} mathrm{m^2}$$
Coincidentally, $10^{11} mathrm m$ is the order of magnitude of the Earth-Sun distance.
answered Feb 2 at 18:56
L.Dutch♦L.Dutch
91.4k29211439
$endgroup$
$begingroup$
Technically, you've calculated the local value of g (gs as you've noted), which is acceleration, rather than force.
$endgroup$
– WhatRoughBeast
Feb 2 at 19:38
1
$begingroup$
@WhatRoughBeast, correct, but the "own weight" is a consequence of the "own gravity"
$endgroup$
– L.Dutch♦
Feb 2 at 20:03
2
$begingroup$
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.
$endgroup$
– Loren Pechtel
Feb 2 at 22:58
4
$begingroup$
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 $frac12 g_s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be dirstributed over is only $2pi Rcdot T$. So even if you keep $T, rho$ constant, the lateral pressure the shell needs to withstand to keep itself up grows in proportion to $R$.
$endgroup$
– Henning Makholm
Feb 3 at 2:54
$begingroup$
@HenningMakholm, good point, I will expand my answer using that. Thanks
$endgroup$
– L.Dutch♦
Feb 3 at 5:17
|
show 3 more comments
$begingroup$
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 $V approx 4pi R^2T$ and a mass $M= 4pi R^2T rho $.
The gravitation pull exerted by the shell on itself will be $$g_mathrm s=Gfrac{M_text{shell}}{R^2}=Gfrac{4pi R^2T rho}{R^2}=4pi GT rho$$
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 $W_mathrm s=frac12 g_mathrm s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be distributed over is only $A=2pi Rcdot T$. 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 $6.6cdot 10^{-11} mathrm{m^3 kg^{-1} s^{-2}}$.
Putting the numbers in we get that, for the densest and more resistant EPS we get $$TR < frac{40cdot10^6 mathrm{Pa}}{pi cdot (640 mathrm{kg/m^3})^2 cdot 6.6cdot 10^{-11} mathrm{m^3 kg^{-1} s^{-2}}}=4.71cdot 10^{11} mathrm{m^2}$$
Coincidentally, $10^{11} mathrm m$ is the order of magnitude of the Earth-Sun distance.
answered Feb 2 at 18:56
L.Dutch♦L.Dutch
91.4k29211439
$endgroup$
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 $V approx 4pi R^2T$ and a mass $M= 4pi R^2T rho $.
The gravitation pull exerted by the shell on itself will be $$g_mathrm s=Gfrac{M_text{shell}}{R^2}=Gfrac{4pi R^2T rho}{R^2}=4pi GT rho$$
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 $W_mathrm s=frac12 g_mathrm s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be distributed over is only $A=2pi Rcdot T$. 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 $6.6cdot 10^{-11} mathrm{m^3 kg^{-1} s^{-2}}$.
Putting the numbers in we get that, for the densest and more resistant EPS we get $$TR < frac{40cdot10^6 mathrm{Pa}}{pi cdot (640 mathrm{kg/m^3})^2 cdot 6.6cdot 10^{-11} mathrm{m^3 kg^{-1} s^{-2}}}=4.71cdot 10^{11} mathrm{m^2}$$
Coincidentally, $10^{11} mathrm m$ is the order of magnitude of the Earth-Sun distance.
answered Feb 2 at 18:56
L.Dutch♦L.Dutch
91.4k29211439
edited Feb 3 at 10:10
answered Feb 2 at 18:56
L.Dutch♦L.Dutch
91.4k29211439
answered Feb 2 at 18:56
L.Dutch♦L.Dutch
91.4k29211439
answered Feb 2 at 18:56
L.Dutch♦L.Dutch
91.4k29211439
91.4k29211439
$begingroup$
Technically, you've calculated the local value of g (gs as you've noted), which is acceleration, rather than force.
$endgroup$
– WhatRoughBeast
Feb 2 at 19:38
1
$begingroup$
@WhatRoughBeast, correct, but the "own weight" is a consequence of the "own gravity"
$endgroup$
– L.Dutch♦
Feb 2 at 20:03
2
$begingroup$
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.
$endgroup$
– Loren Pechtel
Feb 2 at 22:58
4
$begingroup$
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 $frac12 g_s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be dirstributed over is only $2pi Rcdot T$. So even if you keep $T, rho$ constant, the lateral pressure the shell needs to withstand to keep itself up grows in proportion to $R$.
$endgroup$
– Henning Makholm
Feb 3 at 2:54
$begingroup$
@HenningMakholm, good point, I will expand my answer using that. Thanks
$endgroup$
– L.Dutch♦
Feb 3 at 5:17
|
show 3 more comments
$begingroup$
Technically, you've calculated the local value of g (gs as you've noted), which is acceleration, rather than force.
$endgroup$
– WhatRoughBeast
Feb 2 at 19:38
1
$begingroup$
@WhatRoughBeast, correct, but the "own weight" is a consequence of the "own gravity"
$endgroup$
– L.Dutch♦
Feb 2 at 20:03
2
$begingroup$
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.
$endgroup$
– Loren Pechtel
Feb 2 at 22:58
4
$begingroup$
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 $frac12 g_s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be dirstributed over is only $2pi Rcdot T$. So even if you keep $T, rho$ constant, the lateral pressure the shell needs to withstand to keep itself up grows in proportion to $R$.
$endgroup$
– Henning Makholm
Feb 3 at 2:54
$begingroup$
@HenningMakholm, good point, I will expand my answer using that. Thanks
$endgroup$
– L.Dutch♦
Feb 3 at 5:17
$begingroup$
Technically, you've calculated the local value of g (gs as you've noted), which is acceleration, rather than force.
$endgroup$
– WhatRoughBeast
Feb 2 at 19:38
$begingroup$
Technically, you've calculated the local value of g (gs as you've noted), which is acceleration, rather than force.
$endgroup$
– WhatRoughBeast
Feb 2 at 19:38
1
1
$begingroup$
@WhatRoughBeast, correct, but the "own weight" is a consequence of the "own gravity"
$endgroup$
– L.Dutch♦
Feb 2 at 20:03
$begingroup$
@WhatRoughBeast, correct, but the "own weight" is a consequence of the "own gravity"
$endgroup$
– L.Dutch♦
Feb 2 at 20:03
2
2
$begingroup$
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.
$endgroup$
– Loren Pechtel
Feb 2 at 22:58
$begingroup$
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.
$endgroup$
– Loren Pechtel
Feb 2 at 22:58
4
4
$begingroup$
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 $frac12 g_s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be dirstributed over is only $2pi Rcdot T$. So even if you keep $T, rho$ constant, the lateral pressure the shell needs to withstand to keep itself up grows in proportion to $R$.
$endgroup$
– Henning Makholm
Feb 3 at 2:54
$begingroup$
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 $frac12 g_s Trho cdot pi R^2$ whereas the area of the equatorial cross section this force needs to be dirstributed over is only $2pi Rcdot T$. So even if you keep $T, rho$ constant, the lateral pressure the shell needs to withstand to keep itself up grows in proportion to $R$.
$endgroup$
– Henning Makholm
Feb 3 at 2:54
$begingroup$
@HenningMakholm, good point, I will expand my answer using that. Thanks
$endgroup$
– L.Dutch♦
Feb 3 at 5:17
$begingroup$
@HenningMakholm, good point, I will expand my answer using that. Thanks
$endgroup$
– L.Dutch♦
Feb 3 at 5:17
|
show 3 more comments
$begingroup$
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.
Gravity = G * Mass(M1) / radius2
F = mA or, Weight = Mass(M2) * Gravity
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 hard-science 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.
answered Feb 2 at 19:09
JBHJBH
48.3k699228
$endgroup$
$begingroup$
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
$endgroup$
– chasly from UK
Feb 2 at 19:29
1
$begingroup$
Ah-hah. I apologize that I jumped to a conclusion, but I hope and believe my answer addresses your original intent.
$endgroup$
– JBH
Feb 2 at 19:30
1
$begingroup$
1.57x10⁸ km, you say? That's within 5% of the distance between earth and sun. So you have a polystyrene Dyson sphere, nice.
$endgroup$
– Fabian Röling
Feb 2 at 20:59
1
$begingroup$
@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.
$endgroup$
– JBH
Feb 2 at 21:29
$begingroup$
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.
$endgroup$
– Henning Makholm
Feb 3 at 2:40
|
show 4 more comments
$begingroup$
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.
Gravity = G * Mass(M1) / radius2
F = mA or, Weight = Mass(M2) * Gravity
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 hard-science 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.
answered Feb 2 at 19:09
JBHJBH
48.3k699228
$endgroup$
$begingroup$
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
$endgroup$
– chasly from UK
Feb 2 at 19:29
1
$begingroup$
Ah-hah. I apologize that I jumped to a conclusion, but I hope and believe my answer addresses your original intent.
$endgroup$
– JBH
Feb 2 at 19:30
1
$begingroup$
1.57x10⁸ km, you say? That's within 5% of the distance between earth and sun. So you have a polystyrene Dyson sphere, nice.
$endgroup$
– Fabian Röling
Feb 2 at 20:59
1
$begingroup$
@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.
$endgroup$
– JBH
Feb 2 at 21:29
$begingroup$
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.
$endgroup$
– Henning Makholm
Feb 3 at 2:40
|
show 4 more comments
$begingroup$
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.
Gravity = G * Mass(M1) / radius2
F = mA or, Weight = Mass(M2) * Gravity
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 hard-science 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.
answered Feb 2 at 19:09
JBHJBH
48.3k699228
$endgroup$
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.
Gravity = G * Mass(M1) / radius2
F = mA or, Weight = Mass(M2) * Gravity
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 hard-science 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.
answered Feb 2 at 19:09
JBHJBH
48.3k699228
answered Feb 2 at 19:09
JBHJBH
48.3k699228
answered Feb 2 at 19:09
JBHJBH
48.3k699228
answered Feb 2 at 19:09
JBHJBH
48.3k699228
48.3k699228
$begingroup$
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
$endgroup$
– chasly from UK
Feb 2 at 19:29
1
$begingroup$
Ah-hah. I apologize that I jumped to a conclusion, but I hope and believe my answer addresses your original intent.
$endgroup$
– JBH
Feb 2 at 19:30
1
$begingroup$
1.57x10⁸ km, you say? That's within 5% of the distance between earth and sun. So you have a polystyrene Dyson sphere, nice.
$endgroup$
– Fabian Röling
Feb 2 at 20:59
1
$begingroup$
@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.
$endgroup$
– JBH
Feb 2 at 21:29
$begingroup$
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.
$endgroup$
– Henning Makholm
Feb 3 at 2:40
|
show 4 more comments
$begingroup$
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
$endgroup$
– chasly from UK
Feb 2 at 19:29
1
$begingroup$
Ah-hah. I apologize that I jumped to a conclusion, but I hope and believe my answer addresses your original intent.
$endgroup$
– JBH
Feb 2 at 19:30
1
$begingroup$
1.57x10⁸ km, you say? That's within 5% of the distance between earth and sun. So you have a polystyrene Dyson sphere, nice.
$endgroup$
– Fabian Röling
Feb 2 at 20:59
1
$begingroup$
@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.
$endgroup$
– JBH
Feb 2 at 21:29
$begingroup$
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.
$endgroup$
– Henning Makholm
Feb 3 at 2:40
$begingroup$
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
$endgroup$
– chasly from UK
Feb 2 at 19:29
$begingroup$
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
$endgroup$
– chasly from UK
Feb 2 at 19:29
1
1
$begingroup$
Ah-hah. I apologize that I jumped to a conclusion, but I hope and believe my answer addresses your original intent.
$endgroup$
– JBH
Feb 2 at 19:30
$begingroup$
Ah-hah. I apologize that I jumped to a conclusion, but I hope and believe my answer addresses your original intent.
$endgroup$
– JBH
Feb 2 at 19:30
1
1
$begingroup$
1.57x10⁸ km, you say? That's within 5% of the distance between earth and sun. So you have a polystyrene Dyson sphere, nice.
$endgroup$
– Fabian Röling
Feb 2 at 20:59
$begingroup$
1.57x10⁸ km, you say? That's within 5% of the distance between earth and sun. So you have a polystyrene Dyson sphere, nice.
$endgroup$
– Fabian Röling
Feb 2 at 20:59
1
1
$begingroup$
@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.
$endgroup$
– JBH
Feb 2 at 21:29
$begingroup$
@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.
$endgroup$
– JBH
Feb 2 at 21:29
$begingroup$
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.
$endgroup$
– Henning Makholm
Feb 3 at 2:40
$begingroup$
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.
$endgroup$
– Henning Makholm
Feb 3 at 2:40
|
show 4 more comments
$begingroup$
L.Dutch calculates the acceleration of gravity just outside a thin hollow polystyrene shell as
$$ g_s = 4pi GTrho $$
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
$$ 2pi R cdot T $$
so the average lateral pressure must be
$$ frac{tfrac12g_s T rho pi R^2}{2pi R T } = pi GTRrho^2 $$
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
$$ 2pi GT^2rho^2 = 1;rm mPa$$
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
$$ pi GTRrho^2 = 12;rm p{.}s{.}i{.} $$
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).
answered Feb 3 at 4:04
Henning MakholmHenning Makholm
84869
$endgroup$
$begingroup$
Thanks. I was wondering about solar wind.
$endgroup$
– chasly from UK
Feb 3 at 10:15
add a comment |
$begingroup$
L.Dutch calculates the acceleration of gravity just outside a thin hollow polystyrene shell as
$$ g_s = 4pi GTrho $$
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
$$ 2pi R cdot T $$
so the average lateral pressure must be
$$ frac{tfrac12g_s T rho pi R^2}{2pi R T } = pi GTRrho^2 $$
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
$$ 2pi GT^2rho^2 = 1;rm mPa$$
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
$$ pi GTRrho^2 = 12;rm p{.}s{.}i{.} $$
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).
answered Feb 3 at 4:04
Henning MakholmHenning Makholm
84869
$endgroup$
$begingroup$
Thanks. I was wondering about solar wind.
$endgroup$
– chasly from UK
Feb 3 at 10:15
add a comment |
$begingroup$
L.Dutch calculates the acceleration of gravity just outside a thin hollow polystyrene shell as
$$ g_s = 4pi GTrho $$
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
$$ 2pi R cdot T $$
so the average lateral pressure must be
$$ frac{tfrac12g_s T rho pi R^2}{2pi R T } = pi GTRrho^2 $$
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
$$ 2pi GT^2rho^2 = 1;rm mPa$$
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
$$ pi GTRrho^2 = 12;rm p{.}s{.}i{.} $$
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).
answered Feb 3 at 4:04
Henning MakholmHenning Makholm
84869
$endgroup$
L.Dutch calculates the acceleration of gravity just outside a thin hollow polystyrene shell as
$$ g_s = 4pi GTrho $$
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
$$ 2pi R cdot T $$
so the average lateral pressure must be
$$ frac{tfrac12g_s T rho pi R^2}{2pi R T } = pi GTRrho^2 $$
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
$$ 2pi GT^2rho^2 = 1;rm mPa$$
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
$$ pi GTRrho^2 = 12;rm p{.}s{.}i{.} $$
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).
answered Feb 3 at 4:04
Henning MakholmHenning Makholm
84869
edited Feb 3 at 6:10
answered Feb 3 at 4:04
Henning MakholmHenning Makholm
84869
answered Feb 3 at 4:04
Henning MakholmHenning Makholm
84869
answered Feb 3 at 4:04
Henning MakholmHenning Makholm
84869
84869
$begingroup$
Thanks. I was wondering about solar wind.
$endgroup$
– chasly from UK
Feb 3 at 10:15
add a comment |
$begingroup$
Thanks. I was wondering about solar wind.
$endgroup$
– chasly from UK
Feb 3 at 10:15
$begingroup$
Thanks. I was wondering about solar wind.
$endgroup$
– chasly from UK
Feb 3 at 10:15
$begingroup$
Thanks. I was wondering about solar wind.
$endgroup$
– chasly from UK
Feb 3 at 10:15
add a comment |
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$begingroup$
Into which environment is this prop placed?
$endgroup$
– L.Dutch♦
Feb 2 at 16:25
$begingroup$
@ L.Dutch - I mentioned that in the last sentence of my notes.
$endgroup$
– chasly from UK
Feb 2 at 16:33
$begingroup$
That's unclear. Are there other bodies? How big and how far are they?
$endgroup$
– L.Dutch♦
Feb 2 at 16:34
$begingroup$
There is no Moon. For the sake of argument let us say that everything else is the same as in our solar system.
$endgroup$
– chasly from UK
Feb 2 at 16:35
1
$begingroup$
Possible duplicate of Building a full-sized Lego Earth - what would it look like at various levels?. While there is quite a lot of difference between ABS and EPS, at this scale it should even out.
$endgroup$
– o.m.
Feb 2 at 18:28