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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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.
kingledion
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21Probably can't be quite that big, as ice will transform into denser crystal structures at high pressures. – jamesqf Oct 31 '16 at 05:26
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6@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 – Bohemian Oct 31 '16 at 10:39
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@Bohemian According to this graph, at this pressure water is metallic, whatever that means. I would strongly doubt it would manage to maintain it's low density against such pressure. – Tomáš Zato Oct 31 '16 at 13:43
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The sun is slightly denser than water (1.4). Calling what happens to H2O under gas-giant-size pressure "water" shouldn't be taken as assumed. – Yakk Oct 31 '16 at 15:09
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2@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. – Trenin Oct 31 '16 at 15:23
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@Trenin to cause nuclear fusion the temperature and pressure have to be high enough. To achieve that you need a mass of roughly 0.075 solar masses (or roughly 75-80 Jupiter masses). Before that, you have got something called a brown dwarf.
Gravitational collapse is halted by several forces. The first one encountered is electron degeneration pressure. It is what keeps white dwarfs and anything lighter from collapsing further if there is no fusion. Gravity is simply not enough here. But it's still enough to cause a tremendous pressure inside a planets core and even a higher density of matter.
– Adwaenyth Oct 31 '16 at 15:35 -
2@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). – Yakk Oct 31 '16 at 15:37
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@Yakk At 35000 km radius, the planet will have a mass about twice that of Neptune or Uranus, and about a third of Saturn. So the pressures are large but not that phenomenal. I agree that 1000 kg/m$^3$ density will not hold for ice at that pressure. However, the 1800 kg/m$^3$ density that I proposed for the entire object can certainly accommodate high pressure ices or even a rocky core. – kingledion Oct 31 '16 at 15:55
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@jamesqf, a ballpark estimate says that at the core of a 35000-km iceball, the pressure will be around 300 MPa, resulting in either ice IX, with a density of 1.16 g/cm2, or ice II, with a density of 1.17 g/cm2. Either way, not much difference. – Mark Oct 31 '16 at 21:16
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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.
Vaesper
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3You 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. – nijineko Oct 31 '16 at 03:19
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5@nijineko There would be zero net gravity on the inside. http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell2.html – gmatht Oct 31 '16 at 05:05
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@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. – JDługosz Oct 31 '16 at 06:15
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@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... – Zommuter Oct 31 '16 at 07:02
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@Zommuter he was pointing out the mistake in the statement that a hollow planet/dyson sphere could have "a surface gravity of 1g (on both sides!)" of the crust. This isn't true; assuming the shell is uniform in density and thickness the gravitational force on the inside is 0. – Vaesper Oct 31 '16 at 08:41
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@Zommuter you wouldn't need to dig at all: the hollow planet would appear much too large for the materials found at the surface, thus indicating that it's overall density (Mass/Volume) is far below this value - the inner part of a Dyson Sphere has no relevant ammount of mass, but a large ammount of volume. – Trish Oct 31 '16 at 08:46
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@Trish True, though I wonder if digging couldn't reveal that even before someone figures out their planet is not flat and how Newtonian gravity works, for example via ancient mining shafts. – Zommuter Oct 31 '16 at 08:48
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@Zommuter yes, you could find it out that way very easily too, as only the mass in the area below you is taken into account for gravity. If you dig down any ammount, you become lighter fast, untill you reach close to 0G when shoving the shovel through the most inner layer. This is actually part of a question often asked to physics students in their Theoretical classes... – Trish Oct 31 '16 at 08:59
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@Trish Hm, I remember being taught about the linear decay inside a solid sphere and the the vanishing gravity inside a hollow one, but we never talked about the shell. Though for continuity's sake the decay in the shell must of course be fast to reach zero on the inside – Zommuter Oct 31 '16 at 09:01
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1@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. – Tomáš Zato Oct 31 '16 at 13:45
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It would be 1g on both sides. Think about it - if the crust of a dyson sphere (for debate purposes, let's say the sphere is the size of earth's orbirt in diameter) is thick enough to produce 1g of force on the outside, it is producing 1g of force on the inside surface of the sphere. Gravity pulls towards the center of mass, not the center of the sphere. The mass would be the crust of the sphere, and if it was as thick as, say, the earth itself (roughly) then it would be producing 1 g in both directions. – nijineko Oct 31 '16 at 20:17
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No, because the center of mass of the entire crust is still the midpoint of the sphere. Look at it this way; take a hollow ball and point out the center of mass to me. It's a point. If you say it's in the shell, then please point the specific location out to me. – Vaesper Oct 31 '16 at 20:25
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@nijineko: The gravitational force inside a Dyson sphere (or the magnetic force inside a Faraday cage) is 0, and I'm sure there's math to back that up. But intuitively, consider that once you're inside the sphere, there's a certain amount of mass pulling you "up" (back toward the crust you just broke through) and a certain larger — but more distant — mass pulling you "down" (toward the center). These coincidentally balance each other out. Whereas in a solid planet, the larger mass is not more distant — it's right there next to you — so you're pulled strongly "down" and only weakly "up". – Quuxplusone Nov 01 '16 at 01:12
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As the sun is not strong enough to pull us off the earth, i do believe that invalidates your premise. besides, the gravity of a hollow dyson sphere the diameter of the earth's orbit will override anything not equally strong and local. Also, the dyson sphere doesn't have to have a sun at the center. therefore if one was standing on the inside surface of the sphere or on the outside surface of the sphere, hypothetical individuals on either side will experience 1g of gravity pulling them towards the sphere. – nijineko Nov 01 '16 at 01:24
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The gravitational pull from the sun is a lot smaller relative to the distance. Maybe flatland helps you imagine it; a ring-planet in flatland would cause a curve in the geometry outside of it, but inside of it - barring another mass - it would be flat. – Vaesper Nov 01 '16 at 07:29
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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:
$\rho \times r=0.35091\times 10^{11} [\text{kg/m³} \times \text{m}]$
Trish
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2You 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. – kingledion Oct 31 '16 at 12:56
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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:
${v}^{}_{e}=\sqrt{\frac{2GM}{r}}$
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".
stop-cran
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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? – gerrit Oct 31 '16 at 15:16
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@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. – stop-cran Oct 31 '16 at 15:32
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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
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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. – Memming Oct 31 '16 at 14:21
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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!
Charles Merriam
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It's interesting to people looking to build worlds with big areas. Have you tried the narrow-literal-answers StackExchange? – Charles Merriam Nov 01 '16 at 14:07
[planets] [gravity]into search bar to see all questions with this topic. – Tomáš Zato Oct 31 '16 at 13:39