Is there a theoretical maximum size for rocky planets?

Question

As title suggests, what is the size limit by volume for a rocky planet, if any? Some ideas are discussed here, but I'm wondering if there is more concrete, expert evidence to point either way. If possible, it should be at least large enough to host one or more entities that eat Earth-sized planets (a future question in itself). Surface temperature/surrounding atmosphere/other livable conditions and its mass are not restricting factors at the moment, as long as the planet can exist for an arbitrary amount of time.

Answer 1

Estimates vary, but I'll be cautious and say that a radius of roughly two Earth radii is most likely the upper limit for rocky planets.

There are many studies, both theoretical and empirical, that have tried to attack the problem. I'll attempt to summarize the results of a few of them:

• Lammer et al. 2014: This group focused on planets losing their "hydrogen envelopes" - gaseous layers of hydrogen that they may accrete during the early parts of their lives. Their calculations indicate that planets of less than one Earth mass ($M_{\oplus}$) would accumulate envelopes of masses between $2.5 \times 10^{16}$ and $1.5 \times 10^{23}$ kilograms. The latter is about one-tenth of Earth's mass. Planets with masses between $2M_{\oplus}$ and $5M_{\oplus}$ could accumulate envelopes with masses between $7.5 \times 10^{20}$ and $1.5 \times 10^{28}$ kilograms - substantially more massive than Earth! This is the peak envelope mass, though; the group calculated that planets with masses of less than $1M_{\oplus}$ would lose their envelopes within about 100 million years. They found that planets with masses greater than $2M_{\oplus}$ will keep their envelopes, and so become "gas dwarfs" or "mini-Neptunes."
• Lopez & Fortney 2013: Lopez and Fortney worked off of data from Kepler and modeled the radii of planets. They determined that planets with radii of less than $1.5R_{\oplus}$ will become super-Earths, and planets with radii of greater than $2_{\oplus}$ will become mini-Neptunes. That suggests a radius limit of $2R_{\oplus}$, though most terrestrial planets will probably be under $1.5R_{\oplus}$.
• Seager et al. 2008: This group tied mass and radius together based on theoretical calculations. They eventually came to the equation $$M_s \approx \frac{4}{3} \pi R_s^3 \left[1+ \left(1-\frac{3}{5}n \right)\left(\frac{2}{3} \pi R_s^2 \right)^n \right]$$ where $n$ is a certain given parameter and $M_s$ and $R_s$ are the mass and radius scaled by composition-dependent values. It is therefore possible to compare the papers by Lammer et. al. and Lopez and Fortney if $n$ is known. The resulting values are dependent on the material the planet is made of (see Table 3 for examples), but it seems that a pure silicate planet would have an upper limit of $3R_{\oplus}$, while an ocean world could reach $4\text{-}5R_{\oplus}$.

I'd go with about $2R_{\oplus}$ as the upper limit for terrestrial planets, though there may be exceptions in certain extenuating conditions.

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That's for planets that form as terrestrial planets from the start. Curiously enough, gas planets can become terrestrial planets by having their outer layers blown away by their parent star, leaving behind an object called a chthonian planet. These "planets" aren't much more than the core of the gas planet. No chthonian planets have been confirmed to exist, but they're possible.

I should add that Samuel also proposed the $2M_{\oplus}$-limit in his answer below.

Answer 2

Since we're talking about a planet, and not a star, we can compute the upper bound based on the maximum possible mass an object can have and still be made of atoms. The transition away from atoms being atom will take place when the force holding the atoms apart is overcome by the force of gravity. Once gravity is too great, our atoms will collapse into degenerate matter, forming a white dwarf.

The last opposing force, after the intermollecular forces forming the usual solid structure of an atom is electron degeneracy pressure. The amount of electron degeneracy pressure that exists is based on the average molecular weight per electron, which is $\mu_e$ in this equation for the Chandrasekhar limit:

$$M_{\text{limit}}=\frac{w^0_3\sqrt{3\pi}}{2}\left(\frac{\hbar c}{G}\right)\frac{1}{(\mu_em_H)^2}$$

Ignoring everything else, all of which is constant with respect to the material the object is made of, we can see that the mass is inversely proportional to ($\mu_e$).

Since the Chandrasekhar limit is about 1.39 for stars which have an iron core, which is to say that the core of the star will begin to degenerate when the star exceeds this mass, we can use the relative electron density of iron vs. our terrestrial element of choice to figure out how big our object can be. Silicon is about the best we can do, with 14 electrons and an atomic weight of 28. We may be able to do better with some lighter isotope, but then we'd have to worry about electron collapse stripping away too many of our electrons and collapsing our planet into a neutron star. Comparing this to iron, the core of most stars that we see going supernova (iron doesn't fuse and the stuff that does fuse is held apart by fusion pressures), which has an atomic number of 26 and an average atomic mass of 55.8, we can compute the effective mass per electron as 86.8% the electron, giving us a maximum mass for a silicon planet of 1.60 sols.

This planet, of course, would never form on its own. An object of this size would normally accumulate a thick enough atmosphere to undergo fusion, and would be a small star. Normal stars also don't produce nearly this much silicon unless they're really big, in which case they'll produce it and then rapidly fuse it into iron before going supernova. It is, however, assuming you can gather all that silicon up and put it in one spot without it gathering an atmosphere thick enough to push it over the edge mass-wise and turn it into a neutron star, the biggest ball of terrestrial elements one can possibly make. In other words, it is the theoretical maximum size for a rocky planet.

Answer 3

The problem is the transition from "rocky" to "gas giant" is not well defined. From this article:

“The largest “terrestrial” planet is generally considered the one before you get too thick of an atmosphere, which happens at about 5-10 Earth masses (something like 2 Earth radii). Those planets are more Earth-like than Neptune-like.”

So, after perhaps 10 Earth masses it's going to start looking more like a gas giant than a terrestrial planet. The article you already linked commented on the maximum size for gas giants.

...by crashing about 80 Jupiters together, you’d get the same amount of mass as the smallest possible red dwarf star. And all that mass would compress and heat up the core and it would ignite as a star.

You're unlikely to develop beings on a planet surface that will eat other planets. A better approach might be to grow them in a solar gas torus or simply in a hard vacuum around their home star.

Answer 4

While HDE 226868 provides a good argument for a maximum size of a rocky planet based on it retaining too much gas to be considered terrestrial if it's any bigger this only applies to the formation of a planet.

Lets consider a system with a super asteroid belt orbiting outside a large terrestrial planet (think of Mars/our asteroid belt.) A rogue mass gives the planet a big whallop in it's orbital direction, kicking the high point of it's orbit into the asteroid belt. Over time many of those asteroids give it a beating that makes Earth's heavy bombardment period look like a picnic.

Since they formed as smaller bodies they didn't retain the hydrogen and helium of a gas giant.

Of course this is a low probability scenario but it's not completely impossible, thus you can have a bigger planet.

Answer 5

As the article you linked mentioned, very massive stars can potentially continue fusing elements up to iron. Eventually, the star runs out of material to fuse, and it sheds its "envelope" while the core collapses. According to wikipedia, if a star has an iron core of more than 1.4 solar masses, that core will collapse into neutron-degenerate matter, aka a neutron star. With more than 4 solar masses, it will form a black hole.

So 1.4 solar masses is definitely an upper bound. Iron cores smaller than this will collapse to form white dwarfs, though, and I haven't found a size limit for this yet. But 1.4 solar masses is almost certainly not the least upper bound.

Answer 6

If you take very light materials, like carbon, oxygen, not too much hydrogen, lithium, bor, and so on, you can probably get a very large planet, with a very active core. Maybe even larger than Jupiter.

But it would become unstable if it took in a planet with a large iron core.

If you want something really large and able to use planetary materials, I suggest an empty shell. A Dyson sphere without a star in the middle.

If it has inhabitants mining whole planets, they may long have used up hydrogen for their fusion reactors, so you don't need to bother about that - not to mention that by expanding the empty space in the shell, you can have the gravity at below where it would attract all the hydrogen floating by.

There is no physical limit to how large such a structure could be - except that the shell can only have a certain strength. If it gets too robust, it would heat up like a planet of a similar diameter as the thickness of the shell or parts of it. This means, nearby stars (as when close to Earth) would easily crack such a shell if it was too large.

I won't be able to do the mathematics, but I guess that in the outer parts of the galaxy, a shell with a size a few times our sun could easily get close enough to Earth to swallow it. In the inner parts of the galaxy, where moving around exerts lots more force, it would probably have to be somewhat smaller.

Answer 7

There's the possibility of the planet being large enough to cause the genesis of a black hole. If we go with increasing the size of the Earth to the Sun's radius, it would have a mass of 3.9 solar masses, and the pressure at its core would be 8.75*10^11 atm. The Schwarzchild radius would be 11.5 km, if the entire planet collapsed into a black hole, but we need to make a black hole, no matter how small. However, a beam of light fired from the dead center of the planet will still make it out of the radius. I've chosen Earth as it has an core of predominantly iron, and the nuclear fusion of iron is endothermic, so it'd work to absorb extra heat that would slow down the planet's collapse.

If we double the planet's radius, we'd octuple its mass. Now, its mass is 6.20*10^31 kg, and its Schwarzchild radius is 920 km. This planet, with the mass of about 312 Suns, we'll do the lightbeam test again to find out if a black hole will form there. Still no dice. We need a mass of 1000 solar masses to produce the pressure needed to create a black hole, as with an quasi-star. And any object that big, forget about life being able to exist there.