Sister question to: What's the smallest reasonable natural planet or moon with Earth-like surface gravity? - I want us to have both upper and lower bounds for 1g planets available for authors.

We all know that equation for surface gravity is

$$ g = \frac{4\pi}{3} G \rho r $$

So if we want Earth-like surface gravity of $ g = 9.81 \frac {m}{s²}$, then the equation for radius is

$$ r = \frac{3 g}{4\pi G \rho } $$

where $\rho$ is mean density of a planet.

So what's the biggest radius, or lowest density we can reasonably find in space to give us Earth-like gravity and still have a solid surface? By reasonable, I mean that it does not have to be common or even normal. I mean that:

  • It could, theoretically, occur naturally
  • First reaction of scientists should be "what a coincidence!" and not "it's an alien construct!" or "we have a serious problem with our methodology, this can't be!"

Sadly, on the "above 400km" table here bodies are either smaller than Earth or gaseous in nature, and I don't know how could we get less dense, but bigger.

Note: I'm avare of other questions about big planets but here I don't care for life, tectonics, civilizations etc. I want baseline, canonical answer about biggest size for given gravity.

  • $\begingroup$ Does the planet have to be liveable? I'm wondering if a smaller ice giant might fit the bill. $\endgroup$ – Spencer Nov 5 '17 at 19:49
  • $\begingroup$ @Spencer no. See the very least paragraph, after the horizontal line. $\endgroup$ – Mołot Nov 5 '17 at 19:55
  • $\begingroup$ Just be aware that the "surface gravity of Uranus" is 8.69 m/s^2, less than Earth's. $\endgroup$ – Spencer Nov 5 '17 at 19:55
  • $\begingroup$ I know, but it is gaseous. $\endgroup$ – Mołot Nov 5 '17 at 19:58
  • $\begingroup$ Yeah, I think a preliminary question about composition has to be answered. $\endgroup$ – Spencer Nov 5 '17 at 20:13

10 Answers 10


A water planet.

Water has a much lower density than any rock you could make a planet out of and is nearly incompressible. However, some funny stuff happens when you try to make an entire planet out of it. For the sake of easy calculation, my planet is going to be a balmy 350K, at least for now.

What we're going to do is run through a range of pressures, and water changes forms as that happens. Take a look at the phase diagram of water as I talk you through it. (From this very helpful website)

Phase diagram of water

We're staying on the 350K line for now and moving vertically upward as we travel toward the center of the planet. We start at around 100 kPa at the surface, convert to ice VII at ~2 GPa, and convert to ice X at ~50 GPa, where we stay all the way to the core, which should be around 500 GPa.

Respective densities: liquid water, at 1g/cm$^3$; ice VII, at 1.5g/cm$^3$; and ice X, at 2.5g/cm$^3$.

However, these densities also increase with depth according to the bulk modulus equation.

$$B = \frac{\Delta P}{\Delta V / V}$$

From this website and this article (paywall, sorry), I managed to find the bulk modulus ($B$) of water and ice VII. I couldn't find one for ice X, so I'll assume it's similar to ice VII.

Liquid water has a bulk modulus of 2.2 GPa and it takes 200km of water to reach 2GPa, according to the classic conversion 101kPa/10m. Thus, we can solve for final density with this equation:

$$\rho_f = \frac{(\Delta P + B)*\rho_i}{B}$$

where $\rho_f$ is final density, $\Delta P$ is the change in pressure, and $\rho_i$ is the initial density of water (1g/cm$^3$). This gives us a water density at the bottom of our ocean of 1.9g/cm$^3$. For the rest of the math, I'll use the average value of 1.5g/cm$^3$.

The same equation can be used for the ices, but it's already been done by this graph, made by people (paywall, sorry) far more qualified than I: Density of ice

As you can see, the density of ice VII starts at something like 1.5g/cm$^3$ at 2GPa and is projected to increase to something like 3g/cm$^3$ (7cm$^3$/mol) around 500GPa (which would be our core). I'll use an average density of 2.3g/cm$^3$ for the rest of the math.

So, we now have a planet with 200km deep global surface oceans and a thick core of dense ice. Let's get an actual radius for this thing. Our equation in this case will be something like

$$g = \frac{G*M_{planet}}{r^2} = \frac{G*(V_{core}*\rho_{core}+V_{ocean}*\rho_{ocean})}{r^2}$$

Substituting and solving gives us a radius of

15,000 kilometers

Whew. Of course, I handwaved a lot in there, with my biggest one being the assumption of constant temperature. To account for that, the vertical line we used on the water phase diagram would curve to the right as we increase the pressure. This means we wouldn't pass through the transitions as quickly, which would actually increase our radius, not decrease it, because we'd have more of the lighter stuff (water and ice VII). Additionally, being forced to average out the densities with respect to depth annoyed me, but I didn't want to work with nasty differential equations.

If the "solid surface" requirement truly means solid, we've also got an easy solution- freeze it! Instead of a temperature like 400K, a planet near 200 or 100K would have a frozen surface and similar radius- remember that ice 1h (normal ice) actually has a lower density than water.

As far as creating such a planet goes, I wouldn't be surprised if we found one in the universe somewhere. There's a lot of water around, and one hypothesis for Earth's water is comets. Smash a bunch of comets together and you've got a water planet. As other answers have pointed out, this is implausible, but not impossible. There would likely be a solid core of some other substance and would raise a few scientific eyebrows if it was made of pure water.

Other options

Other answers have pointed out some good ideas, but I still think that water is the ideal material. Substances like liquid hydrogen or organic molecules (hexane, for example) do indeed have lower densities but they have MUCH higher bulk moduli, which was really the deciding factor in this whole equation. See below for a similar graph from here (again, paywall)- and note the difference in axes, where $H$ has a much more dramatic change with pressure. I wasn't able to find a similar one for hexane, but it'd be between the two based on its bulk modulus alone (paywall. sad.).

H2 PV Curve

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    $\begingroup$ The constant temperature assumption is actually probably a good one. The Earth's high temperature core has a lot to do with radioactive decay, with an all-ice planet, there is no radioactive decay, and therefore no heat generation over time. The primoridal heat will radiate off into space over time and you will end up with a cold ice core and a surface at whatever temperature it is maintained by sunlight. $\endgroup$ – kingledion Nov 15 '17 at 15:04
  • $\begingroup$ @kingledion Yes, radioactive decay will be less in a water planet, however, that does not mean that temperature won't rise as you go deeper: When planets form, one of the main heat sources is potential energy. As the planet shrinks, its own mass moves down within the gravity well it creates itself. This energy gets trapped in the core, so the water will get warmer as you go down. For the liquid part, the rate will be that of adiabatic compression (move some water up/down, and the change in density due to the pressure difference matches the changed environmental density -> avoids convection). $\endgroup$ – cmaster Nov 15 '17 at 20:42
  • $\begingroup$ @cmaster The heat formed by compression upon planetary formation is called 'primordial heat.' I argue that it will all escape over the billions of years between planetary formation and potential intelligent life. A billion years is a long time for planetary temperatures to stabilize at a near constant throughout. $\endgroup$ – kingledion Nov 15 '17 at 23:02
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    $\begingroup$ @Dallaylaen I actually still think it'd be better to have the water be as pure as possible- any dissolved organic materials (direct result of organic layer on top) or salts are going to raise the water density a lot. $\endgroup$ – Dubukay Jan 31 '18 at 0:12
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    $\begingroup$ This. This is my favorite answer in the site now. $\endgroup$ – Renan Sep 25 '18 at 1:07

Ammonia is reasonable

Dubukay's answer is pretty great, except he's starting with the wrong substance. Water is indeed not very dense, but ammonia is less dense at pretty much all temperature and pressure combinations (that I could find, at least).

Ammonia is reasonable because it is common in the solar system. Nitrogen is the 5th most common element in the solar system, at about 1/5 the abundance of oxygen, and ammonia is also the most common nitrogen bearing compound. Ammonia is also relatively more common the farther out from the sun you go. Wyckoff, S., et al. 1991 show that ammonia to water ratio in comets increases with distance from the sun. Other sources show comets that are up to 50% ammonia. We haven't really explored the Kuiper belt yet and haven't even confirmed the existence of the Oort cloud, so there is a possibility that majority ammonia objects could exist out there, in contrast to the majority water moons that we have found among the outer planets.

Ammonia's density - liquid

First off, there is a liquid ocean phase. At 300 K, Ammonia will phase transition to solid around 1 GPa. I was able to find an Isothermal density vs. Pressure plot here at nist.gov. The curve indicates that density will increase from 35 mol/l to 50 mol/l from standard pressure to 1 GPa. Using the molar density of ammonia (0.017031 kg/mol), that range becomes 600 kg/m$^3$ to 850 kg/m$^3$. Based on the convex shape of the curve, a good estimate for average density would be 750 kg/m$^3$.

The depth of an ammonia ocean will depend on the pressure at which ammonia phase transitions to solid, which is 1 GPa. From the hydrostatic pressure equation, we can estimate this depth from $\Delta p = \rho gh$ as $$h = \frac{\Delta p}{\rho g} = \frac{ 1 \text{GPa}}{750 \text{ kg/m}^3 \cdot 10 \text{ m/s}^2} = 130 \text{km}.$$ This is using the assumption of 'Earth-like' surface gravity, and since the depth isn't too great that assumption will be fine.

Ammonia's density - solid

Ammonia ice at standard pressure and -80 C is 817 kg/m$^3$; compare that to ater ice at 917 kg/m$^3$ at 0 C and standard pressure. I'll use the numbers from Dubukay's answer to make a comparison calculation. I found ammonia ice density characteristics in Fortes, A., et al, 2003. This work only covers solid Ammonia phases I and IV, so we'll concentrate there. I was unable to find an accurate phase diagram on the internet (i.e. that matches what is in the paper) so you'll have to use your imagination. Phases I, II, and III all transition to IV by about 2 GPa, so I'll argue that the Phase IV data is most important to calculating the ultimate density of the planet.

If you are looking at the paper, you will notice that the density is stated in molar volume $\text{mol_vol}$. This is converted to density by the following transformation $$\frac{1}{\text{mol_vol} \text{ cm}^3\text{mol}^{-1}\cdot \frac{1 \text{ mol}}{0.017031 \text{kg}}\cdot\frac{1\text{ m}^3}{1000000 \text{ cm}^3}} = \frac{17031}{\text{mol_vl}}\, \frac{\text{kg}}{\text{m}^3}$$

Dubukay's charts show that Ice VII at 2 GPa has a density of 1500 kg/m$^3$; From Fig 4 of the paper, Ammonia will be at 1000 kg/m$^3$. Ice X at 50 GPA will be 2500 kg/m$^3$; from Fig 5, Ammonia will be at 1900 kg/m$^3$.

Copying the estimate core ratio from Dubukay's answer, we can use 1700 kg/m$^3$ (instead of water's 2300 kg/m$^3$) as an estimate of core density.

Calculating radius

Surface gravity is $$ g = \frac{G\left(\rho_{core}\cdot V_{core} + \rho_{ocean}\cdot V_{ocean}\right)}{r^2}. $$ The volume of the core is a sphere of radius $r - 130000$, our calculated oceanic depth, while the volume of the ocean is $\frac{4}{3}\pi\left(r^3-[r-130000]^3-\right)$. Plugging those in we get $$10 = 6.674\times10^{-11}\frac{4\pi}{3r^2}\left(1700\cdot[r-130000]^3 + 750\cdot r^3 - 750\cdot[r-130000]^3\right).$$ Wolfram Alpha solves $r$ at 21,000 km.


Anything of planetary size is very unlikely to be a pure solid, unless artificial. While formation in the far Kuiper belt would reasonably exclude heavy rocks and metals from the formation of a planet, I am not aware of any mechanism that would exclude water and other volatile compounds from being swept up into our nascent ammonia planet.

A truly reasonable estimate would be a planet that was partly water ice and partly ammonia. This would differentiate into a mostly icy core and a mostly liquid ammonia ocean. There would likely be carbon dioxide, carbon monoxide, and methane scattered about too. So perhaps an estimate halfway between 15000 km for water and 21000 km for ammonia would be the most reasonable of all.

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    $\begingroup$ This is a great answer, and deserves more upvotes. Ammonia does indeed seem to be superior to water in every way except abundance, and that's up to the OP or storywriter to decide the prior probabilities and willingness to suspend disbelief. Cheers! $\endgroup$ – Dubukay Nov 15 '17 at 21:58

Hyperion is the least dense moon in our system. It does look spongy.

spongy hyperion


Slightly more than half as dense as water, Hyperion's composition is still a mystery. Porous water ice may account for the difference, as could the inclusion of lighter materials such as frozen methane or carbon dioxide. The existence of such materials would be consistent if a number of smaller ice and rock bodies had drawn together, or accreted, to form the moon, making Hyperion similar to a rubble pile. An example: A 2012 Icarus study of the surface suggests Hyperion is mostly made up of water ice with some "additional materials", such as carbon dioxide. The carbon dioxide does not appear to be pure ice, but a more complex structure such as a clathrate (where molecules of one substance are trapped inside the ice of another)

The density of water is 1 gm/cm3 The density of methane clathrate is 0.9 gm/cm3

How can Hyperion have a density 54% of water? It is porous.


We have also determined Hyperion's size and mass, and calculated the mean density as 544+/-50kgm-3, which indicates a porosity of >40 per cent.

Maybe a large component of Hyperion was volatile substances like water or ammonia, and over time these have been lost to space. I can imagine a larger body with more gravity also losing its volatiles to space if it were this porous - Mars lost its atmosphere over the millennia, presumably stripped away by the solar wind. Certainly nothing as light as this would have any metals with which to generate a protective magnetic field.

Once you allow porosity, you can make your body arbitrarily big - a huge filmy lattice. But let us use Hyperions density as an observed possible density for a celestial body. What size would an object of this composition and earths gravity be?

Thank you Eric James Stone for your fine gravity calculator!! http://www.ericjamesstone.com/blog/home/gravity-calculator-for-astronomical-bodies-based-on-radius-and-density/

With Hyperion's density of 0.54 g/cc I found that a body of 64,000 km radius would have 99% of Earth's gravity. Earth is 12,742 km.

Still not even as big as Saturn. But if you made it hollow...

Here is the question regarding Hyperions pores and their crushability from the planetary exploration stack. https://space.stackexchange.com/questions/23626/how-large-could-hyperion-be-and-stay-porous 10 votes and a bucketload of comments but no answers as of 11/15/17.

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    $\begingroup$ I see one problem. At such big size, is porosity still possible? Planets tends to be at hydrostatic equilibrium. Porous bodies don't. How could we avoid collapse of pores at bigger-than-Earth scale? $\endgroup$ – Mołot Nov 6 '17 at 7:11
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    $\begingroup$ I read about hydrostatic equilibrium. Then I read it again. Third time the charm? I am still not sure. Maybe I will put that question on the planetary science stack. $\endgroup$ – Willk Nov 6 '17 at 12:55
  • $\begingroup$ Sounds like a good idea. I can't figure out if your idea would work on such scales. I want it to work. I couldn't fint a solid reason for it not to. But somehow it feels off - if forces are so big they pull big bulges of rock down to form a sphere, why wouldn't they crush pores as well? And I do hope there is a way they wouldn't! $\endgroup$ – Mołot Nov 6 '17 at 12:57
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    $\begingroup$ I have been pondering this: the crush ability of deep pores. Many comments but no answers on planetary science stack. I think pores (at least central pores) must be full of some uncrushable low volatility substance. Naphtha? $\endgroup$ – Willk Nov 12 '17 at 19:43
  • $\begingroup$ Could you please link to relevant question on sister site, or article, or sth? So we know that for now we don't have scientific answers available. $\endgroup$ – Mołot Nov 15 '17 at 16:06

A recent study [1] has found that, despite showing ostensible differences in mass and size, a considerable number of the extrasolar planets discovered so far have a surface gravity very similar to that of Earth.

Firstly, the surface gravity of the small bodies in the Solar System and rocky planets smaller than Venus grows with the square root of the mass. Secondly, in the case of gaseous giant exoplanets, the surface gravity linearly grows with the mass. And surprisingly, in the transition zone (between 1 and 100 land masses), we find some sort of plateau that shows a constant surface gravity roughly similar to that of Earth.

So, it seems to exist a correlation between mass and radius of the planets in order to sustain this plateau. although Uranus, Neptune and Saturn are, respectively, 14, 17 and 95 times more massive than Earth, their surface gravities barely vary between 0.9g and 1.1g. So, the answer to your question is: Saturn is the biggest reasonable natural planet with Earth-like surface gravity, or Jupiter if you want do give some concession.

Mass versus surface gravity.

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    $\begingroup$ Welcome to Worldbuilding! This answer explains what the largest body in the solar system with a surface gravity similar to Earth, but in doing so, falls down in two main ways: 1. These numbers are experimentally known, but is there something theoretically larger? and 2. The question is looking for planets which "have a solid surface", which Saturn doesn't, so would you be able to edit your answer to take these into account? Thanks $\endgroup$ – Mithrandir24601 Nov 12 '17 at 18:23
  • $\begingroup$ I asked: "and still have a solid surface?" $\endgroup$ – Mołot Nov 12 '17 at 18:23
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    $\begingroup$ Hi atMalkyor, and welcome to Worldbuilding. I see you have posted what appears to be the exact same answer on another question. While we certainly encourage newcomers to participate, it's also usually better if you tailor the answer to the specific question, even if you are referencing the same source material. $\endgroup$ – a CVn Nov 12 '17 at 18:32

Sadly, on the "above 400km" table here bodies are either smaller than Earth or gaseous in nature, and I don't know how could we get less dense, but bigger.

I'm thinking perhaps if we have a much smaller iron-nickel and heavy metals core, this would decrease the average density. In theory you might have no significant iron core at all - if the original dust cloud from which the planet accreted had a very low iron content, which would happen with Population III dust.

Also, in that case the compression on the core would be less, which would in turn further decrease the core density.

Earth's density is an average 5.5 between a 3.0 for the upper crust and a nucleus thought to be from 9.0 and 13.6, made of an allotrope of iron (STP density = 7.784).

Mars, for example, has a density of 3.9, even if it has an iron core; but the total mass is significantly less, and the core is therefore less compressed (density nearer 9.0 than 13.6). This in turn argues for a surface density nearer to 2.4.

So, if we were to remove the iron core altogether from a Mars-like planet, the remaining material would have an average density (STP) of 2.4, possibly increasing to 4.8 (probably less) at the core. Applying a structure similar to Earth's, we may expect an average density of 3.0, giving a radius of around 11,700 km, or 85% more than Earth's.


No one went with the obvious:

Liquid Hydrogen

has a density of 0.16 mg/cm³

It isn't improbable either as rogue planets could be small chucks of hydrogen. without the mass needed to produce its own internal heat or heat from the sun it would just freeze into a liquid

At this point it technically isn't a gas giant either having a clear physical surface. speaking of as far as I can tell the only qualification for a "surface" is matter existing in a liquid, plasmatic, of solid state since gas means atmosphere.

I also want to point out the flaws of this question

  • The geological processes/construction of stellar object can greatly influence its ultimate density over its material composition. A spherical space station can be made of diamonds and ultimately be as dense as a dust cloud.

  • Also, pressure has a significant effect on our perceived understanding of chemistry. So when picking planetary compositions with respect to planetary density you have to understand chemistry at planetary pressure scales (which we are just beginning to explore).

  • For instance- at high enough pressures hydrogen is expected to behave more like a metal than a gas. This is believed to happen in Jupiter.

  • $\begingroup$ In the question I wrote "solid surface". This answer probably does not meet this requirement. And things you list as flaws in the question are actually just things a good answer should address. $\endgroup$ – Mołot Nov 11 '17 at 12:12

If the core were replaced by mantle material the earth would be considerably lighter. Exactly how much exactly is hard to say. But very roughly:

Density of the mantle 3300 – 5500 kg/cubic metre
Thickness of mantle 2886km
so density increase with depth is roughly 600kg/cubic metre/1000km
Max density at centre 7800 kg/cubic metre
Density at the top of the mantle 3300 kg/cubic metre
Take the average = 4.5 kg/cubic metre
Put that into your equation gives the new earth’s radius as 7789km

If you assume an even lighter rock, say Granite at 2600 kg/cubic metre, the average density might drop to perhaps 4 kg/cubic metre and the radius then becomes 8762km

I would have thought that anything bigger would be straining credulity, although could not be ruled out. All answers should make allowance for the increase in density with depth, whatever the material is used, as I have attempted to do (not very accurately).


Take the iron out of the core. This will give you at least a 1000 km extra radius. It will also turn off the magnetic field. Which you probably need to do to get the atmosphere down to some reasonable level. (A larger planet has a slower taper on g as you gain altitude, so it's harder for a molecule of air to escape. You have a larger escape velocity too)

Play with the Chemical Rubber Handbook (or as physicists refer to it, "The Book of Random Numbers" and look up density of materials. Then for good candidates, check the following:

  • is it made of abundent elements? (Ytterium fluoride is an unlikely material...)
  • Is it stable at high temperature?
  • Does it compress drastically under high pressure. For many compounds this won't be known.

If you can half the density, you can double the radius. This would correspond to a radius of about 12,000 km.

Postulate that most of the atomic enrichment in this particular star cluster happened due to a cluster of supernova on one edge. As the ionized material streamed forth, the galactic magnetic field acted as a mass spectrometer, sorting the ions by mass/charge ratio, so there is a swath of stars that had very high content of light elements compared to heavy elements. What happens to the density if multiply the atomic abundance of earth by 1/atomic number. Now you have a core made of lithium and berylium and probably a very poisonous crust; fluorine being as common as chlorine, and potassium being much rarer than on earth. The latter would reduce planetary heating. Coupled with the scarsity of radioactives, it may mean that the planet never liquified, so heavy elements didn't segregate to the core. This would allow reasonable abundances on the surface despite the planetary scarcity.

A very useful calculator for mass radius gravity escape velocity issue:



To get the same surface gravity of Earth a planet with Moon composition ($\rho =3346 kg/m^3$)would be $r \approxeq 10500 Km$, so I suppose that's possible, although not common due to specifics of Moon formation.

  • $\begingroup$ ρ of such planet will be higher due to compression under own gravity. This wasn't a problem in smallest question as the effect would only add some margin and make planets proposed even better. It is a problem here, under biggest question, as it can ruin everything. $\endgroup$ – Mołot Nov 6 '17 at 12:39

Bulk allotropic Carbon is basically incompressible and has a specific density of roughly 3.5, giving a solid world with a radius around 10,000km at 9.81m/s^2. Such a planet couldn't form in the modern universe but something similar, with only very small amounts of heavier elements, might have formed when the universe was very young and heavy elements like Iron were much rarer.


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