Big big planet

[Giallo]

Assuming it would not collapse and become a black hole, would it be possible to make a planet of the dimensions of a (big) star? (After all if it were composed of, say, gold, there should be no nuclear fusion and thus it should not ignite into a star, right?)

[Meteoric]

Well, roughly a single solar mass, without the outward pressure from fusion, will collapse and form a white dwarf. A solid planet, especially if it's made of a dense material like gold, will have considerably more mass than a ball of gas the same size (my napkin math for gold is saying about a hundred solar masses for a planet the size of the sun). I'm no astrophysicist, but I think it would collapse and also form something resembling some kind of stellar remnant.

A hollow star-sized shell - kind of like a Dyson sphere - would still be ridiculously difficult to make, but maybe not literally impossible.

[yurell]

It'd be interesting to see if you got a quark star i.e. neutron degeneracy isn't enough, but it can't collapse into a black hole because of the scenario.

[ElWanderer]

Well, roughly a single solar mass, without the outward pressure from fusion, will collapse and form a white dwarf. A solid planet, especially if it's made of a dense material like gold, will have considerably more mass than a ball of gas the same size (my napkin math for gold is saying about a hundred solar masses for a planet the size of the sun). I'm no astrophysicist, but I think it would collapse and also form something resembling some kind of stellar remnant.

Comparing the (room temperature, admittedly) density of gold to the average density of our sun, I get 15 solar masses. My first calculation used solar mass and diameter rather than average density, and got 100 solar masses as the answer. I wonder if the average density figure I'm using is wrong. Either way, that's well above the 3-4 solar mass limit for being able to form a neutron star - i.e. collapsing into a black hole seems a good bet, though almighty wikipedia suggests you could potentially get something really weird like a quark star.

Ah, I see yurell suggested that already!

A hollow star-sized shell - kind of like a Dyson sphere - would still be ridiculously difficult to make, but maybe not literally impossible.

You just need to mine enough light, superstrong Unobtanium and command a fleet large enough to assemble it all quickly. A hollow sphere would have the advantage of not having as high surface gravity as a lump of solid gold (the sun has surface gravity of about 27g and if you increased its mass 100-fold without changing its size, you'd get up to 2700g), assuming both are actually stable.

[starslayer]

You would get a black hole. The maximum mass for a neutron star is something around three solar masses, and this gold star would be significantly more massive than that, with no thermal support to hold it up. I doubt there would even be much mass lost here; the star would be in near freefall the whole way down. There would be a couple of pauses along the way (once when it hits electron degeneracy, then again when the "core" hits the Chandrasekahr mass), but complete collapse is almost certainly inevitable.

[Giallo]

Yeah, I didn't want necessarily a golden planet... Any element over Fe₅₂ should work...
So, what would be the limit dimension for a planet with solid surface?

[selfassembled]

Stars can exist with 300 solar masses, but they are correspondingly more energetic, providing pressure to counteract gravity. The thing is, the heat produced by collapse due to gravity provides that pressure, in a sort of negative feedback loop. I would expect that the atomic number won't matter too much as something this massive will likely cross the pressure threshold for degenerating to neutron matter or even quark matter (matter essentially melts at certain pressures and temperatures). The question is whether or not collapsing matter will cross the Schwarzschild radius and become a black hole, I can't answer that.

[Diadem]

Well, roughly a single solar mass, without the outward pressure from fusion, will collapse and form a white dwarf. A solid planet, especially if it's made of a dense material like gold, will have considerably more mass than a ball of gas the same size (my napkin math for gold is saying about a hundred solar masses for a planet the size of the sun). I'm no astrophysicist, but I think it would collapse and also form something resembling some kind of stellar remnant.

Comparing the (room temperature, admittedly) density of gold to the average density of our sun, I get 15 solar masses. My first calculation used solar mass and diameter rather than average density, and got 100 solar masses as the answer. I wonder if the average density figure I'm using is wrong. Either way, that's well above the 3-4 solar mass limit for being able to form a neutron star - i.e. collapsing into a black hole seems a good bet, though almighty wikipedia suggests you could potentially get something really weird like a quark star.

Even a quark star couldn't become that massive. Nor any other hypothetical structure that may exist beyond the density of quark stars (which themselves are already hypothetical).

You can easily derive an upper bound on the mass of any compact object, without using any assumption at all about the nature of the object. The only three assumptions you need to make are that no part of the object is within its own Schwarzschild radius (which is necessary for it not to be a black hole already), that pressure increases with density (but if that weren't the case matter couldn't be stable to begin with), and that the speed of sound in the object is lower than the speed of light. Using those three constraints you can derive that no non-thermal non-rotating compact object can be more massive than 3.6 solar masses. To get bigger you need a source of outward pressure. Rotation is one such source. By allowing rotation you can increase this upper bound to about 4.3 solar masses. Allowing the object to be hot (since heat generates outward pressure) again increases the upper limit by a bit. I don't recall exactly how much, but it wasn't a lot.

So anything more massive than 4.3 solar masses must either:
a) Not be solid. One can imagine a huge superstructure built in space over the volume of several parsecs, so that gravitational interaction remains small while the total mass becomes very big.
b) Have a source of heat. This means they are stars, since fusion is the only significant source of heat in these regimes. Fusion can generate huge amounts of outward pressure, and thus increase the maximum size by a lot. The maximum size of stars is about 300 solar masses, beyond which they become unstable. But I'm actually not sure if they become unstable because gravity becomes too strong or too weak.
c) Be a black hole.

[starslayer]

But I'm actually not sure if they become unstable because gravity becomes too strong or too weak.

The latter. The Eddington luminosity, where radiation pressure overcomes gravity, is given by

L_{Edd} = \frac{4\pi G M c}{\kappa}

M is the mass of the object, and \kappa is the opacity. This also assumes spherical symmetry.

For stars above one solar mass, the empirical luminosity-mass relationship runs as roughly L \propto M^3, so radiation pressure is going to win out at some point. If you assume a star made of fully ionized hydrogen, so that the dominant opacity source is electron scattering, you come up with a mass of ~100 M_sun. 300 solar mass stars come from a time when the metallicity of the universe was essentially zero, so the luminosity-mass relation I gave no longer holds.

[idobox]

I have absolutely no idea how to do the math, but if the planet is cold, a significant portion of it should be solid. You could see that as infinitely many Dyson spheres, and should reduce the pressure.

Also, if this was for some reason engineered, it would make sense to structure the material, removing a lot of mass at the cost of a little strength,like in those 19th irons bridges. The deeper structures would of course need to be much stronger, so denser.

[Meteoric]

Yeah, I didn't want necessarily a golden planet... Any element over Fe₅₂ should work...
So, what would be the limit dimension for a planet with solid surface?

Elements heavier than iron can still undergo fusion, they just produce less energy from fusion than is required to start the reaction. As I understand it, that's why elements past iron are produced in supernovas - the collapse causes a bunch of fusion, but it's consuming more energy than it produces, so there's still no net outward pressure to halt the collapse. So, I'm not sure it would be particularly helpful to make your planet out of a heavy element, and would have the drawback of being heavy.

Now, if you could make a planet out of strong but lower-density material, you could probably make a much larger one. This would also maybe let you keep surface gravity from becoming too absurd.

I have nooooo idea how to calculate an actual upper limit, though.

[PM 2Ring]

I have absolutely no idea how to do the math, but if the planet is cold, a significant portion of it should be solid. You could see that as infinitely many Dyson spheres, and should reduce the pressure.

Also, if this was for some reason engineered, it would make sense to structure the material, removing a lot of mass at the cost of a little strength,like in those 19th irons bridges. The deeper structures would of course need to be much stronger, so denser.

Nice idea, but the density would need to be extremely low to avoid gravitational collapse, and I doubt that such a structured sphere that size would have the strength to support itself, no matter what fancy fractal structure you used. As you note, a structured body will be weaker than a solid one of the same size, but a solid cube of iron around (100 km)³ will collapse into a sphere from self gravity. Maybe a very tenuous structure based on some kind of aerogel-like material could work, though, I guess.

But my instincts tell me that any object that large which isn't actively counteracting gravity will collapse, and a body that massive will collapse like the iron core of an old star, i.e. it will go supernova and the remnant will form a neutron star or black hole.

Note that before the metal mega-planet goes degenerate it will not be cold - even in the initial stages of collapse large amounts of heat will be produced.

[idobox]

I have absolutely no idea how to do the math, but if the planet is cold, a significant portion of it should be solid. You could see that as infinitely many Dyson spheres, and should reduce the pressure.

Also, if this was for some reason engineered, it would make sense to structure the material, removing a lot of mass at the cost of a little strength,like in those 19th irons bridges. The deeper structures would of course need to be much stronger, so denser.

Nice idea, but the density would need to be extremely low to avoid gravitational collapse, and I doubt that such a structured sphere that size would have the strength to support itself, no matter what fancy fractal structure you used. As you note, a structured body will be weaker than a solid one of the same size, but a solid cube of iron around (100 km)³ will collapse into a sphere from self gravity. Maybe a very tenuous structure based on some kind of aerogel-like material could work, though, I guess.

But my instincts tell me that any object that large which isn't actively counteracting gravity will collapse, and a body that massive will collapse like the iron core of an old star, i.e. it will go supernova and the remnant will form a neutron star or black hole.

Note that before the metal mega-planet goes degenerate it will not be cold - even in the initial stages of collapse large amounts of heat will be produced.

The big idea here is that if your 'planet' is made of something solid, rather than fluid, a significant part of the weight of a part will not be imparted on the lower parts, reducing the pressure in the center.
A Dyson sphere doesn't impart any pressure on what's inside. The whole structure could be built as a series of concentric Dyson spheres with dampeners between layers to ensure stability.
Or you could use pilars between layers to support part of one sphere's weights, to lower the stress one layer imparts on itself.

[ElWanderer]

The big idea here is that if your 'planet' is made of something solid, rather than fluid, a significant part of the weight of a part will not be imparted on the lower parts, reducing the pressure in the center.

I don't think gravity cares whether the material making up a solid (geometric) planet is solid (state), liquid or gas. The centre is still supporting the entire mass of the planet.

A Dyson sphere doesn't impart any pressure on what's inside.

True there's no gravitational interaction inside - instead there will be a massive amount of pressure on the shell (hence the need for light, superstrong construction materials). I suspect the moment you build any supporting structure inside the sphere, that will take up some of the load. If you connect supporting structure all the way through the centre of the sphere, you'll transfer all the load to the centrepoint and you're back to square one, albeit with a lower mass planet than if it were fully solid (geometric).

The multiple hollow spheres with support pillars sounds like the shellworld in Iain M. Bank's Matter

[idobox]

Gravity pulls stuff whether it's solid or not, but if it's solid, a part (up to 100% in the case of a Dyson sphere) of the weight can imparted to neighbors rather than on what is beneath.
As you pointed out, a true Dyson sphere is very difficult to build, especially close to something as heavy as a "big big planet", that's why I think the best solution would be an hybrid, where part of the weight of a layer is supported by the lower layer, and part by itself.
I doubt this kind of structure could reach the size of a star, but it could still reach pretty amazing proportions.

[mfb]

You can easily derive an upper bound on the mass of any compact object, without using any assumption at all about the nature of the object. The only three assumptions you need to make are that no part of the object is within its own Schwarzschild radius (which is necessary for it not to be a black hole already), that pressure increases with density (but if that weren't the case matter couldn't be stable to begin with), and that the speed of sound in the object is lower than the speed of light. Using those three constraints you can derive that no non-thermal non-rotating compact object can be more massive than 3.6 solar masses. To get bigger you need a source of outward pressure. Rotation is one such source. By allowing rotation you can increase this upper bound to about 4.3 solar masses. Allowing the object to be hot (since heat generates outward pressure) again increases the upper limit by a bit. I don't recall exactly how much, but it wasn't a lot.

What about charged objects? The required charge would be so large that this object could not be stable (here: keeping its charge) within our universe, but maybe within a perfect vacuum?


A non-rotating Dyson sphere around the sun (1 solar mass) with the diameter of 1 AU would require extremely strong materials. A smaller sphere (with r_s as 1 solar radius) would have a mass which is lower by r_s/1AU, but gravitational force which is larger by (1AU/r_s)^2, so it would require materials which are even stronger by ~2.5 orders of magnitude.
This does not look like a good idea. In addition, it would reduce the problem to the white dwarf / neutron star / quark star / whatever system. Extremely strong material with pressure in every direction. But without the possibility to get beyond ~3 solar masses.



A perfect sphere with radius r without a central object and area density \rho has a mass of M=4\pi r^2 \rho, giving an outer surface gravity of g=\frac{MG}{r^2} = 4\pi \rho G and an average gravity (inside the material) of 1/2 of that. Therefore, if one would split the spheres into two half-spheres, the attracting force between them is F = c\,M\,g = c 4^2 \pi^2 r^2 \rho^2 G with some constant c<1/2 given by integration over the half-spheres. This is distributed over an area of 2 \pi r \delta_r with the thickness \delta_r, which gives a pressure of

p = \frac{F}{2\pi r \delta_r} = c 8 \pi r \rho^2 G \delta_r^{-1}

To calculate some numbers, assume c=1/2, \rho=1000kg/m^2, \delta_r=50cm and p=100MPa (random number). In that case, I get ~60 million kilometers as radius, using a mass of 5*10^25 kg (10 times the mass of the earth), which gives a surface gravity of 8*10^(-7) m/s^2 and a surface area of 5*10^16km^2, which is about 100 million times the surface of the earth.
I think this is large .

[Sockmonkey]

So how big could you build a hollow iron sphere? I know it's going to vary depending on shell thickness but what would be the rough numbers on this?

[Giallo]

Wait... shouldn't be the gravity inside an empty sphere equal to zero?

[gmalivuk]

Yes, but I think mfb meant the attractive force between the two hemispheres

[Giallo]

Ok. By the way, would a Dyson sphere around a star be stable? By intuition I would say that any perturbation from perfect equilibrium would lead to destructive consequences... am I wrong?

[Diadem]

Well the mass of a spherical shell is proportional to its surface, which goes up as r^2. Gravity goes down with distance as r^2. So they cancel out and the surface gravity on a dyson sphere is constant with distance.

So that's actually not a problem. The strain on the dyson sphere does not increase with distance, so you can make it as big as you want. Well, almost. If you make it too big it'll be contained within its own Schwarzschild radius and thus become a black hole. But other than that, no problem.

Ok. By the way, would a Dyson sphere around a star be stable? By intuition I would say that any perturbation from perfect equilibrium would lead to destructive consequences... am I wrong?

Define 'stable'? No system with more than 3 bodies is, strictly speaking, stable. Eventually you'll see chaotic behavior which most likely ends with one of the bodies being ejected or destroyed. Even systems with 2 bodies will decay over time. A planet orbiting a star, even with nothing else around, will over time lose orbital energy and ultimately fall into the star.

So no, a dyson sphere will not be completely stable. But the corrections you need to make to keep it in place are trivial for any race advanced enough to build one.

[gmalivuk]

am I wrong?

Yes, because by the same argument that proves the force of gravity inside a hollow spherical shell is zero, it can be shown that the net gravitational force a mass somewhere inside that shell has on the shell itself is also zero.

[Giallo]

Oh, right... dumb me

[Charlie!]

Hm, interesting. So each hemisphere has mass 2*pi*r²*thickness * density, and a center of mass at 3/8*r. So the gravitational force between them is G*256/9*pi²*r²*thickness²*density². What determines the (highest possible) breakdown size is when the compressive strength of your object (in force / area) is equal to to the force between hemispheres divided by the contact area 2*pi*r*thickness.

So you don't actually want a good strength to weight ratio - you want a good strength to weight-squared ratio. So let's look at diamond instead of iron, since that will be much much better. Compressive strength of around 3*10^11 pascals, pretty much the highest we know of. If we set the thickness at 100m, just as a reasonable-ish value to resist meteors, then the maximum radius is r = 3*10¹¹/(G*128/9*pi*100*3520²) = 8*10^10 meters, which is about 0.5 AU. So... not bad! A sphere the size of the sun should be no problem, until something slams into it.

[Diadem]

This is confusing. Both my calculation and Charlie!'s are correct, near as I can tell. Yet they give contradicting answers.

[Charlie!]

This is confusing. Both my calculation and Charlie!'s are correct, near as I can tell. Yet they give contradicting answers.

The trick is that it's not the mass that goes into the calculation, but the product of the masses of the two things attracting each other, so there's a factor of m^2, or r^4. And the reason why you have to use something like a hemisphere, rather than a small patch of constant size, is because the force required to hold something up in a sphere also depends on the angle, so if you have a small patch and increase the size of the sphere, you decrease the angle holding it up.

[Diadem]

Ah yes, angles. Of course.

[mfb]

So each hemisphere has mass 2*pi*r²*thickness * density, and a center of mass at 3/8*r. So the gravitational force between them is G*256/9*pi²*r²*thickness²*density².

You cannot contract the half-spheres to their center of mass like this is possible with spheres. You have to integrate the forces over the half-spheres. This is the reason why I introduced the "c" to avoid the actual calculation.


100m thickness needs a lot of material, and carbon is harder to get than large amounts of iron (we talk about disassembling planets anyway, and earth has 32% oxygen, 29% iron, 17% silicon, 16% magnesium, ... and 0.2% carbon).
A big meteor would destroy a large area of this 100m-shell, while a smaller shell might get a hole similar to the size of the meteor itself. Smaller meteors could be evaporated or whatever.

[Charlie!]

So each hemisphere has mass 2*pi*r²*thickness * density, and a center of mass at 3/8*r. So the gravitational force between them is G*256/9*pi²*r²*thickness²*density².

You cannot contract the half-spheres to their center of mass like this is possible with spheres. You have to integrate the forces over the half-spheres. This is the reason why I introduced the "c" to avoid the actual calculation.

Ah, right. So there's a fudge power of 10 in there somewhere.

100m thickness needs a lot of material, and carbon is harder to get than large amounts of iron (we talk about disassembling planets anyway, and earth has 32% oxygen, 29% iron, 17% silicon, 16% magnesium, ... and 0.2% carbon).

Carbon's .5% of the universe, I'm sure we'll find enough eventually.

A big meteor would destroy a large area of this 100m-shell, while a smaller shell might get a hole similar to the size of the meteor itself. Smaller meteors could be evaporated or whatever.

Interesting - so you'd advocate the "eggshell" approach? Heck, with good enough prediction, you could just cut holes for the asteroids to go through ahead of time.

[a2jaume]

Yes, because by the same argument that proves the force of gravity inside a hollow spherical shell is zero, it can be shown that the net gravitational force a mass somewhere inside that shell has on the shell itself is also zero.

It's been a long time since I did such problems, but that argument would be Gauss theorem, and I think you're wrong, else we would not have weight. As I understand it, we only feel the pull of masses that are internal to our position, so the shell has to feel any mass that's inside it. What it cannot perceive is the effect of masses that have a spherical distribution and are on the outside.

[gmalivuk]

That's why I said *net* force. Yes, of course it pulls inward everywhere, and things on the surface of the shell would feel different weights depending on their position relative to the internal distribution of mass.

But I still contend that, treating the shell as a single rigid body, it will not feel any net force in any direction due to matter inside it. After all, the magnitude of gravitational force is symmetrical. If the shell as a whole exerts no force on something inside it, then something inside it exerts no force on the shell as a whole. (Which is to say, all the forces in different directions balance out upon integration over the whole surface.)

[Diadem]

That's correct. And made me come up with an interesting problem.

Imagine a spherical shell of point masses, that do not interact with each other. Such a configuration is stable, since there's no forces anywhere. No place a large mass M inside the shell, but off-centre. The attraction from the mass M causes the spherical shell to fall inwards. But, initially at least, there is no net force on the shell, so its centre of mass does not change. Eventually, the infalling point-masses will impact on the mass M. What I am wondering however is, does the M move before the first point-masses hit it?

In other words: The net force on the point-masses is initially zero because they form a spherical shell. But each point-mass will experience a different acceleration, so as they fall inward they won't remain spherical. Does this give a net-force, or not?

[Jplus]

This is admittedly slightly off-topic, but nature would never produce a star-sized planet out of itself. Or actually it would, but then it would be a star. The larger a celestial object is, the larger the proportion of hydrogen in its mass will be (as a rule of thumb). That's because of the way these objects come into existence. Strictly speaking, between 10 and 90 Jupiter masses there's no clear distinction between stars and planets anyway.

[Giallo]

That's correct. And made me come up with an interesting problem.

Imagine a spherical shell of point masses, that do not interact with each other. Such a configuration is stable, since there's no forces anywhere. No place a large mass M inside the shell, but off-centre. The attraction from the mass M causes the spherical shell to fall inwards. But, initially at least, there is no net force on the shell, so its centre of mass does not change. Eventually, the infalling point-masses will impact on the mass M. What I am wondering however is, does the M move before the first point-masses hit it?

In other words: The net force on the point-masses is initially zero because they form a spherical shell. But each point-mass will experience a different acceleration, so as they fall inward they won't remain spherical. Does this give a net-force, or not?

I think not, since the total momentum will be 0 at any time, by conservation, so the center of mass will remain in its initial point.

[Charlie!]

That's correct. And made me come up with an interesting problem.

Imagine a spherical shell of point masses, that do not interact with each other. Such a configuration is stable, since there's no forces anywhere. No place a large mass M inside the shell, but off-centre. The attraction from the mass M causes the spherical shell to fall inwards. But, initially at least, there is no net force on the shell, so its centre of mass does not change. Eventually, the infalling point-masses will impact on the mass M. What I am wondering however is, does the M move before the first point-masses hit it?

In other words: The net force on the point-masses is initially zero because they form a spherical shell. But each point-mass will experience a different acceleration, so as they fall inward they won't remain spherical. Does this give a net-force, or not?

I think not, since the total momentum will be 0 at any time, by conservation, so the center of mass will remain in its initial point.

You have a wrong premise.

[Diadem]

Well the total momentum of the total system is of course zero. That's trivially true.

But the momentum of individual pieces is not constant. Indeed, all the point masses start moving after t=0.

[mfb]

The point masses closer to the large mass are accelerated faster. Therefore, I would expect that the shell will become asymmetric, and the forces of these close masses should grow quicker than the other forces. I would expect that the central mass experiences a net force to this direction. It moves, until the point masses hit it, afterwards it moves towards the other side, collecting the other masses. Assuming all particles hit the central mass, it is at rest afterwards (in the initial (not inertial ) system of the mass), at the center of mass of the whole initial system.

[curtis95112]

I don't think that the center of mass will move.
There is no net force on the system.

[mfb]

That was never questioned.
However, parts of a system can move, without a movement of the center of mass.

[legend]

Imagine a spherical shell of point masses, that do not interact with each other. Such a configuration is stable, since there's no forces anywhere. No place a large mass M inside the shell, but off-centre. The attraction from the mass M causes the spherical shell to fall inwards. But, initially at least, there is no net force on the shell, so its centre of mass does not change. Eventually, the infalling point-masses will impact on the mass M. What I am wondering however is, does the M move before the first point-masses hit it?

I'm pretty sure it does. The net force on M would remain zero if the sphere would colapse towards it's center. Because it doesn't you get an asymmetry in your configuration which produces a net force on the mass.