How big can a planet be ( like Jupiter or Saturn)and have an "earth-like" gravity. i. e can à planet be as big as Jupiter be rocky and have oceans and an oxygen rich atmosphere..

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    $\begingroup$ Clarify "big". Do you mean mass or radius? And define "earth-like", because the things that make a world "earth-like" in the sense that exo-planet searchers are talking about is lot broader than what would support something like Earth, or do you mean Rocky planet? $\endgroup$
    – Durakken
    Commented Oct 31, 2016 at 3:00
  • $\begingroup$ As in; can a planet as big as Jupiter or Saturn be a rocky world with some lakes/oceans and an oxygen atmosphere. $\endgroup$ Commented Oct 31, 2016 at 9:20
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    $\begingroup$ There was a lot of questions about size of planets. For example about sustaining life: worldbuilding.stackexchange.com/q/158/17556, about size: worldbuilding.stackexchange.com/q/15062/17556 Questions on this site are really getting repetitive. You can enter [planets] [gravity] into search bar to see all questions with this topic. $\endgroup$ Commented Oct 31, 2016 at 13:39
  • $\begingroup$ sigh @JohnSmith When you say "big" no one understands what you mean because it can refer to how much a planet weighs or the distance from its center to its surface. Super-Earths are said to be able to get up to 10x bigger. They mean Massive (how much it weighs), and most of these Planets are only 2x radii larger than Earth. But you can change density and make the a much larger radius or a much smaller radius. $\endgroup$
    – Durakken
    Commented Oct 31, 2016 at 14:58
  • $\begingroup$ If you want it to be rocky, then this may be a duplicate of worldbuilding.stackexchange.com/q/9948/627. $\endgroup$
    – HDE 226868
    Commented Oct 31, 2016 at 16:42

6 Answers 6


There are all sorts of fun variants.

The hollow earth, an unstable life section from a melting and outgassing large mass can be huge with gravity in some regions made by spin. Transportation, going 'up the wall and across', makes sailing interesting. The world could be between thousands and millions of years old and threatened when idiots start drilling.

Lots of unstable shapes. For example, Mars could have sustained life for a long time until the water and atmosphere slowly boiled off. Non-spherical shapes or even accreting shapes in the intermediate between asteroids and accretions body would be interesting to live in. Or imagine a solar system with dozens of habitable planets.

Unstable worlds probably don't have billions of years to evolve life, but any injection of multi-cellular life would be plenty. Having a crashed ship could move science from an experimental basis to a form of archeology.

What bizarre cultures!

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    $\begingroup$ How does this answer the question? $\endgroup$
    – kingledion
    Commented Oct 31, 2016 at 19:30
  • $\begingroup$ It's interesting to people looking to build worlds with big areas. Have you tried the narrow-literal-answers StackExchange? $\endgroup$ Commented Nov 1, 2016 at 14:07

I assume that you want to planet to have a solid surface; so let us say that this planet is completely composed of ice with a density of 1000 kg/m$^3$. That is really light; less dense than Uranus or Neptune, and less than any of the large moons in the solar system.

If the planet has a density of 1000 kg/m$^3$, then it can have a radius of 35000 km and still have a surface gravity of 1g. That is larger than Uranus and Neptune (both about 25000 km radius).

If we go with a potentially more realistic density of 1800 kg/m$^3$ (about the same as Ganymede, Callisto, and Titan), then the radius to give 1g surface gravity is 19500 km.

If the planet is Earth-like in density, then its radius will have to be Earth-like to get Earth-like surface gravity.

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    $\begingroup$ Probably can't be quite that big, as ice will transform into denser crystal structures at high pressures. $\endgroup$
    – jamesqf
    Commented Oct 31, 2016 at 5:26
  • $\begingroup$ @jamesqf is right, ice will get denser down there. $\endgroup$
    – Mołot
    Commented Oct 31, 2016 at 6:46
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    $\begingroup$ @jamesqf But even some exotic crystal structure of ice isn't going to get much more dense than regular ice. The elements oxygen and hydrogen are low mass. The densest known ice is "ice XV" at 1300 kg/m3 $\endgroup$
    – Bohemian
    Commented Oct 31, 2016 at 10:39
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    $\begingroup$ @Yakk But isn't the nuclear reaction in the sun causing an outward pressure to counteract the force of gravity? If fusion stopped, the core would collapse to something much denser. $\endgroup$
    – Trenin
    Commented Oct 31, 2016 at 15:23
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    $\begingroup$ @Trenin Sure. My point is that a big pile of H2O shouldn't be assumed to stay in that molecular form. The density of compressed O or H could be higher than H2O. The H is going to be sub-fusion, and so would the O (as, iirc, it takes more pressure to fuze heavier elements (?), and pure-H stars don't start that light). $\endgroup$
    – Yakk
    Commented Oct 31, 2016 at 15:37

So what I think you're asking about is the surface gravity, which for Earth is about 9.8 m/s² (source).

Let's look at this explanation. As we can see, we can simply fill in 9.8 for the "gravitational acceleration of planet" variable, fill in any arbitrarily large r (but not infinite) and find the right mass for that planet to have.

So hypothetically, the planet can be arbitrarily big; if you're asking about what kind of planet would realistically form, well, that's an entirely different ballgame. Artificial planets are definitely a possibility though.

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    $\begingroup$ You could even go with a hollow planet (or dyson sphere), as long as the crust was thick enough to have sufficient mass to produce a surface gravity of 1g (on both sides!), though you might have to plan carefully for mountains and oceans. $\endgroup$
    – nijineko
    Commented Oct 31, 2016 at 3:19
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    $\begingroup$ @nijineko There would be zero net gravity on the inside. hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell2.html $\endgroup$
    – gmatht
    Commented Oct 31, 2016 at 5:05
  • $\begingroup$ @nijineko gmatht is right: the overall density of a dyson sphere could give you 1g on the outer surface, but the inside of the shell will be zero gravity. $\endgroup$
    – JDługosz
    Commented Oct 31, 2016 at 6:15
  • $\begingroup$ @gmatht So what? Inside a homogeneous solid sphere gravity also linearly decays and reaches zero at the centre. The more interesting question would be how much gravity is inside the shell, i.e. how deep would one have to dig in order to figure out the planet is actually hollow... $\endgroup$
    – Zommuter
    Commented Oct 31, 2016 at 7:02
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    $\begingroup$ @gmatht Zero gravity in the inside is a bonus as far as SciFi is concerned, IMHO. I'd worry much more about the lack of magnetic field. $\endgroup$ Commented Oct 31, 2016 at 13:45

Even if "Big" is undefined, we can find the gravitation of a planet on its surface by starting at this term:

$\vec F_G=m\times \vec g=-\frac{G \times m \times M}{d^2}\times|\vec d|$ where G is the gravitational constant, m&M are the masses of two objects and d is their distance and the last argument is the direction from M to m.

Dropping the neglectable mass of the test object, we get

$|g|=\frac{G\times M}{r^2}$ where G is still the gravitational constant, M the planetary mass and r the planetary radius. now, we want to get M from r.

$M=\rho \times V=\rho \times \frac 4 3 \pi r^3$ where $\rho$ is the average density of the planet. Hint: 1000 kg/m³ for water, a Dyson sphere's hollow core does reduce the density quite a lot.

So all in all you want to look for any solution of the following term to get "earthlike" surface gravitation:

$9.81 \text {m/s²} = \frac {4 G \times \pi}3 \times \frac {\rho \times r^3}{r^2}= -\frac {4 G \times \pi}3 \times {\rho \times r}$

You can eaily seen that this is a function that will demand the solution $G \rho r=2.34196 \text{m/s²}$ which is, as we know G is a constant of $G=6.67390\times 10^{-11} \text{m³/kg s²}$, equal to:

$\rho \times r=0.35091\times 10^{11} [\text{kg/m³} \times \text{m}]$

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    $\begingroup$ You aren't dropping the mass of the test object ($m$) because it is neglectable. It is on both sides of the first equation you give, so you divide both sides by $m$ then take the magnitude of both vectors to get the second equation. $\endgroup$
    – kingledion
    Commented Oct 31, 2016 at 12:56

Besides surface gravity, there is another, more strict, upper limit for mass of "earth-like" planet - low enough atmospheric escape to sustain dense hydrogen atmosphere. Such big planets almost inevitably become gas giants.

Very roughly speaking, that's 10${M}^{}_{⊕}$ (or about 2 Earth radii).

Here are some thoughts:

The main mechanism of atmosphere leak to space is thermal escape - any object, including atmospheric particles, that moving faster than the escape speed will leave the planet. The higher the planet mass, the higher the escape velocity is:


Here $M$ is mass of a planet, $r$ is its radius, and $G$ is gravitational constant.

On the other hand, mean speed of atmospheric particles increases with temperature and decreases with the particle mass:

$\overline{v}=\sqrt{\frac{8 R T}{\pi \mu}}$

Where $T$ is temperature, $\mu$ is the molar mass of particle and $R$ is the gas constant.

So light particles (especially hydrogen atoms) are more likely to escape.

If the mean speed in upper atmosphere doesn't exceed $0.2 {v}^{}_{e}$, such atmosphere is treated as stable. In other cases substantial part of molecules will constantly leave the upper atmosphere and the atmosphere (or its particular component) will fastly get depleted.

For atomic hydrogen at 1000°C (exosphere conditions) the mean speed is 5 km/s. So Earth (with escape speed 11.2 km/s) easily loses hydrogen, whereas Saturn (with escape speed 35.5 km/s) almostly doesn't. Hypothetical planet with $10{M}^{}_{⊕}$ and $2{R}^{}_{⊕}$ should have escape speed 25 km/s, which is near the limit.

However, even a light moon like Titan can sustain dense atmosphere because it's cold enough. On the other hand, there could possibly exist so called chthonian planet that are heavier than $10{M}^{}_{⊕}$ and have orbits very close to star. Such planets should have rocky surface, since they have lost their atmosphere because of extremely hot conditions. Too hot though, to treat such planets as "earth-like".

  • $\begingroup$ What if we make the star very active with a strong solar wind, thus stripping a planetary atmosphere of its excess hydrogen even if it otherwise would retain it? $\endgroup$
    – gerrit
    Commented Oct 31, 2016 at 15:16
  • $\begingroup$ @gerrit yes, but such a star should emit severe radiation. Since the planet should have very low magnetic field for solar wind to blow off the atmosphere, all this radiation should reach the surface. However, the planet can be tidally locked and thus face the star with always the same side. This is likely if thr planet is on low orbit near the red dwarf. The "night" side in such case can be cool enough to establish a base. $\endgroup$
    – stop-cran
    Commented Oct 31, 2016 at 15:32

It can be infinitely big, by which I mean it can be a flat infinite world. Same reasoning as https://worldbuilding.stackexchange.com/a/12443/7400

  • $\begingroup$ Hah, that's awesome. Would that still be a planet though? $\endgroup$
    – Vaesper
    Commented Oct 31, 2016 at 14:01
  • $\begingroup$ That depends on the definition of a planet. It has a surface where people can live...I guess you can't really have day/night cycle without some strange tricks. $\endgroup$
    – Memming
    Commented Oct 31, 2016 at 14:21

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