# How large could a planet be yet still approach 1 Earth gravity and support life as we know it?

Life on Earth has developed with one Earth-gravity, more or less, as a constant for epochs. Our bones and organs both are adapted to this amount of pull, (or push, depending on how sciency you want to put it, right?) but could a larger planet have a similar pull (at the surface) based on various factors such as less planetary density, slower planetary rotation, or a counter-pull from another structure, like being surrounded by a dense shell?

Based on this conception that it's possible to explain a larger planet with one Earth gravity at the surface supporting life as on our planet, how large could that other planet be, and how could that happen?

• Gravity increases linearly with masses multiplied, and decreases with distance squared. Newton's law of universal gravitation: $F = G \frac{m_1m_2}{r^2}$, $G$ is constant. For any given element, mass scales linearly with amount (obviously). So if you want a larger planet that has Earth-like gravity, it must be made up of lighter elements. At some point, you have a gas giant instead of a rocky planet (like Earth). You have to decide where along that continuum it's no longer "Earth-like enough" for your taste. – user Jun 17 '16 at 19:44
• Note that Newton's law only applies to point gravitational sources, but where the differences in mass between the two objects is large, it's usually good enough and you don't need to reach for general relativity to explain how the universe works. (For comparison, I recall reading somewhere that in Apollo, NASA checked the accuracy of Newton's equations on a trip to the Moon. The spacecraft, near the Moon, was out of position by something like 10 cm if you used Newton's equations instead of Einstein's. Good enough for most purposes.) – user Jun 17 '16 at 19:49
• Do you want it to be a naturally-formed body, or could it be an artificial structure? As I said in this answer, if you use a Dyson sphere made of exotic "strange matter" and surrounding a more massive gravitating body, you can have 1 g acceleration at the surface with a very large sphere. – Hypnosifl Jun 17 '16 at 22:15
• As for the final note in your question, you really should ask about that on Worldbuilding Meta now that you have the reputation to do so. – user Jun 18 '16 at 16:59

## 6 Answers

For a spherically symmetric planet, surface gravitation is determined by just two quantities: The average density, $\rho$, and the radius, $R$. In particular, due to spherical symmetry you can consider the whole planet's mass to be concentrated in the center, and then you get for the gravitational acceleration at surface:

$g = G\frac{M}{R^2}$ with $M = \rho V = \frac{4\pi}{3}\rho R^3$

and therefore

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

This means if you want the same gravitation on the surface for a larger planet, you need to reduce its average density by the same factor.

The problem is that you cannot arbitrary reduce the density, as sufficiently away from the surface, the pressure will make sure your material is compressed to have no significant holes, and therefore it will only the substance itself that determines the density.

One possibility is to not have an iron core. However that would in turn mean there's no magnetic field to protect you from cosmic radiation and star wind. According to this page, the core (inner and outer combined) has a radius of 3400 km (2100 + 1300), and makes up 32% (or roughly 1/3) of the earth mass. With an earth radius of about 6400 km, it is slightly more than half the earth radius, which means it has a bit more than 1/8 of the earth's volume. Thus replacing it with mantle material would reduce earth's density by roughly 25% (actually less because the core is under larger pressure and therefore denser). Assuming this density doesn't change significantly when making the planet larger, this would mean you would end up with a planet whose radius is about 4/3 the earth's radius (and the surface is therefore 16/9 the earth's surface, or 77% more than earth's surface).

Since the mantle is also denser than the crust, I think another density reduction (and thus radius increase) should be possible (but maybe at the cost of continental drift/geologic activity; those might have played a major role in the emergence of life).

Note that a spherical shell around the planet would not affect the gravitation on the surface at all, since such a shell doesn't cause a gravitational force in the inside.

A fast rotation would reduce the apparent gravitation near the equator, but any rotation fast enough to make a noticeable difference would also make the Coriolis force quite strong. Also, it would do nothing for the polar regions, so you'd get a steep gravity gradient over latitude. I'm not sure that would be very life friendly.

A "counter-pull" from another structure is known as tidal force. I'm pretty sure that before it would make a noticeable difference on the gravitational pull (note that you don't feel the tidal force of the moon on Earth, you only see it from the tides) the tidal forces would tear the planet apart.

• Added some relevant Wikipedia links and fixed a few typographical mistakes that I noticed. – user Jun 17 '16 at 22:21
• I think that by the time you are looking at enclosing a planet in an outer spherical shell (as suggested) would, besides the little detail of building something like that, be ridiculously hard to keep stable. The slightest perturbation and eventually it hits the planet. It's kind of got all the downsides of a Dyson sphere, with none of the upsides (that I can see, at least). – user Jun 17 '16 at 22:23
• @Michael Kjörling - If the Dyson sphere was enclosing a non-rotating gas planet and was built at around the radius of its upper atmosphere, then in the case of drifting, the pressure of the gas would increase on whichever side started to approach the planet--perhaps this would be enough to check the drift? Alternatively you could have some kind of artificial system of rockets on the inner walls, perhaps using particles they collect that have streamed out from the central mass (solar wind in the case of a star) to replenish their fuel. – Hypnosifl Jun 18 '16 at 16:27
• @Hypnosifl I don't know, and haven't thought very deeply about this, but intuitively, that would suggest that the density of the Dyson sphere surrounding the planet is lower than the density of the atmospheric gases of the planet (because otherwise the Dyson sphere would be pushing the gases out of its way, not the other way around). Not likely. – user Jun 18 '16 at 16:57
• @Michael Kjörling - But keep in mind there is no net gravitational force on the sphere even if it becomes off-center--so any unbalanced force, however small, should be continuously accelerating it in the direction it needs to go to become more centered (i.e. accelerating the closer side away from the center and the farther side towards it). – Hypnosifl Jun 18 '16 at 17:26

If your planet is made of water, you could have a radius around four times that of Earth with about the same gravity, using no additional forces or artificial means of lessening the force of gravity.

We can calculate this based on celtschk's equation for gravity: $g=\frac{4πG}{3}ρR$

To use this to calculate the size of our planet, we need to know roughly the density of water, which will depend on the pressure it is under. Water, at pressures it encounters in Earth's ocean, is a roughly incompressible liquid, but at sufficiently high pressures it can change phase into a variety of different ices. For our water planet, let's assume that the pressure is mostly similar to the pressure found at the base of Earth's mantle, or within an order of magnitude of that pressure. This gives a pressure of around 140 GPA, which corresponds to a form of ice called ice X.

Unlike run of the mill ices we encounter on Earth's surface, which are less dense than liquid water, ice X has a density of around 2.5 $\frac{g}{cm^3}$. This is roughly a quarter of the density of the rock that makes up Earth, so based on our equation for gravity, we should be able to have a planet with about four times the radius of Earth with the same gravity, if it were made mostly of water. This planet could support life as we know it if "life as we know it" can be taken to mean "fish".

Building on the other two answers, the rough answer is about the size of earth, plus maybe 10 to 12%. This assumes no magical building materials - just rock.

As celtschk has pointed out, the surface gravity (for a uniform sphere) is equal to the product of radius and density. The average density of the earth is about 5.5. While the core is denser than the crust, which contradicts the leading assumption, the core isn't all that large, so let's go with the average.

The specific gravity of magma runs about 2.2 to 2.8, or a bit less than half of 5.5. While it's true that the density of rock increases with pressure (so the core will be denser even without any iron) this figure https://en.wikipedia.org/wiki/Structure_of_the_Earth#/media/File:RadialDensityPREM.jpg

shows that the core (while twice as dense as the outer layers, only makes up about 1/4 of the total mass of the earth. The core's radius is about 1/2 that of earth, so its volume is 1/8. Swapping iron for rock will only reduce the total mass (and average density) by 1/8, so a best-case number would allow an increase of about 12% in the radius of the earth.

Of course, with no heavy core, there is no magnetic field and no plate tectonics. The source of the earth's internal heat is radioactive decay of heavy elements. With no tectonics there will be no mountain building (orogeny) with the result that all land above the ocean surface will be eroded away, and an otherwise earth-like planet will be a water world.

Gas giant size., although it would take a lot of future tech to realize humans living on top of huge hot-gas balloons floating in the upper atmosphere of Saturn. I'm not sure that it is physically possible: the failure mode is just too catastrophic and the weather is ... difficult. You'd certainly need multiply redundant fusion power ststems and super strong materials.

But if you could solve the technical issues, gravity on the surface of such a platform would be Earthlike.

Gas giants are much the same size across a vast range of masses. The more mass they have the more it is compressed at the core. They range from planets like Uranus up to brown dwarf not-quite-stars. You get Earthlike gravity at the low end, like Saturn.

Several previous answers provide you with values for Rocky (about Earth sized), Water (about 4x Earth sized), and Gaseous (about Saturn sized) planets when just gravity is considered.

But you could make a significantly more massive planet that maintains a surface gravity of 1 g if you put it under extreme rotation. As celtschk mentions, such a planet would possess a surface gravity that increased tremendously with latitude.

A planet such as this is limited to certain maximum spin properties or it would tend to fly apart. An article I read indicated that a 1 rotation every 2.5 hours was about the maximum. Using this as my limiting factor I provided an answer to a similar question a while back.

When you add in spin, you could get your rocky planet's size up to

• mass $\approx 1.25 \times M_{Jupiter}$
• radius $\approx 10 \times r_{Earth}$
• equatorial gravity - $g_{equator} = 1 g$
• polar gravity - $g_{polar} = 16.6 g$
• gravity variation by latitude - $g \approx 1 + \sin\left(latitude\right) \times 15.6$

Only a relatively thin strip along the equator would be inhabitable to surface dwelling humans. However, humans adapted for aquatic life could probably use the whole planet. Buoyancy would also vary by latitude and this could make an interesting story plot.

There is no limit

If you are looking for only natural gravitational force than you need to use the Law of Universal Gravitation and the planet can have less and less mass until it can no longer maintain Earth-like properties (primarily geothermal properties). However, you can substitute gravity with other forces if you will allow your planet to partially artificial, most notably centripetal force.

• I think you'll run into a limit due to the holographic principle. Also note that the density of a black hole decreases with size. – JDługosz Jun 18 '16 at 6:46