In the mobile game "The Path To Luma", the main hero travels to very tiny planets. Let's not think about where is the sun and how we leap through such small planets. What I want to ask is, if there is a spherical planet with a radius of 50 meters, will there be enough gravity for humans to live there?

Specifically, that planet...no...asteroid consists of the same materials as Earth. Also, the atmosphere is enough and air, ground (er... soils and rocks?) density is the same as Earth.

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    $\begingroup$ The simple answer is no. A fifty metre diameter planet wouldn't enough gravity to keep any air let alone support human life. There needs to be a secret gravity generator at the centre of these micro-planets for it to work. :) $\endgroup$
    – a4android
    Feb 17 '17 at 13:00
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    $\begingroup$ @Gallifreyan Questions seeking to understand how an existing fictional world works can be on topic, if appropriately scoped and assuming it contains sufficient information to be answerable without requiring referencing additional material (there should be no need, for example, to buy the game referenced and analyze it in order to answer the question). This question pretty clearly meets at least the second of these criteria. Compare Are questions based on movies okay? on Worldbuilding Meta. $\endgroup$
    – user
    Feb 17 '17 at 13:08
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    $\begingroup$ This sounds like it's based on The Little Prince by Antione de Saint-Exupery as that had micro-planets of about this size in it. Yes, with air and plants too. It's a charming fantasy. Beautifully written too. $\endgroup$
    – a4android
    Feb 17 '17 at 13:11
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    $\begingroup$ @a4android and OP : XKCD analyzed it in what-if.xkcd.com/68 : "It would feel like you were stretched out on a curved rubber ball, or were lying on a merry-go-round with your head near the center" $\endgroup$
    – Goufalite
    Feb 17 '17 at 14:03
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    $\begingroup$ You can change universe... en.wikipedia.org/wiki/Raft_(novel) (If you read one book in the Xeelee sequence, read this one). $\endgroup$
    – Rmano
    Feb 17 '17 at 17:09

Our known quantities are:

  • Radius of the body: 50 metres
  • Density of the body: same as Earth's, 5515 kilograms per cubic metre

This is enough to calculate the acceleration due to gravity on the surface of the body. We multiply the density $\rho$ and the volume $V$ to get the mass, multiply it by Newton's universal gravitational constant $G$, and divide by the square of the radius $r$ of the body:

$$ g' = \frac{\rho \cdot V \cdot G}{r^2} $$

To spare the reader lengthy calculations and descriptions of constants, I did my calculations in Wolfram Alpha - it is a handy tool, as it recognises natural language and automatically plugs in constants.

My calculations show that the acceleration due to gravity will be $ 7.7 \cdot 10^{-5} \;m/s^2$. For reference, right now we are experiencing an acceleration about $ 9.8 \;m/s^2$, so the magnitude differs by more than hundred thousand.

Thus, your rock will not able to hold air, and even itself, together, let alone let you walk on it.

To solve this, you could increase the density of this object by a factor of (up to) ten thousand - how you do this is up to you, but if my other calculations are correct, this is the limit after which the radius of the planet becomes closer to its Schwarzschild radius, which you definitely don't want in your life.

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    $\begingroup$ I believe you meant 9.8 m/s^2, not 9.6 m/s^2. The commonly used approximation is 9.82 m/s^2 for Earth. $\endgroup$
    – user
    Feb 17 '17 at 13:12
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    $\begingroup$ With gravity like that a single sneeze would launch the average person faster than escape velocity. Pack plenty of tissues. $\endgroup$
    – a4android
    Feb 17 '17 at 13:14
  • $\begingroup$ @MichaelKjörling thanks, that's my bad. I'm still polishing, since I also have to learn MathJax in process :P $\endgroup$ Feb 17 '17 at 13:17
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    $\begingroup$ Try MathJax basic tutorial and quick reference on Mathematics Meta. It's densely written, but very informative. $\endgroup$
    – user
    Feb 17 '17 at 13:21
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    $\begingroup$ @Gallifreyan sorry for that, I personally do not distinguish between them, that's spellchecker plugin, for some reason it decided I prefer American English over British English. But I noticed that a week ago and I'm working in the direction to distinguish them. Fixing fraction was not my idea it's from edit queue, proposed edit was also not commonly used(but valid, but style also a preference of taste). Thanks for raising my awareness about BEng and AEng. $\endgroup$
    – MolbOrg
    Feb 19 '17 at 9:33

First matter first: to have a body in a spherical shape, you need to exceed a certain radius, dictated by the material. Most likely with 50 meters you will have a potato shaped object.

Moreover, to have a decent gravity you need more mass. Just as a reference, Ceres has a radius of 473 km, a mass of 0.00015 Earth masses and a surface gravity of 0.029 G.

This means that the same effort you would exert here on Earth to jump 1 meter high would make you jump 34 meter high (if you don't reach the escape velocity of 0.51 km/s you will then return on the ground).

Then, with such a flimsy gravity, forget about keeping any atmosphere or liquid water.


Of course if the planet was made of a very strange matter or had a singularity at the center then it might be possible, but I think the tidal forces would create issues. Your head would probably experience significantly less gravity than your feet for example.

  • $\begingroup$ No. While you could do something like degenerate matter contained in handwavium to get a reasonable surface gravity the escape velocity would still be very low--it couldn't hold atmosphere. (On the flip side you can consider a megastructure that has low enough gravity to be considered free-fall and yet retain an atmosphere.) $\endgroup$ Feb 17 '17 at 21:08
  • $\begingroup$ Not sure if that is correct. If you are talking about holding an atmosphere for thousands of year sure but at least short term or with replenishment it should be possible, although the atmosphere would be far thinner than a larger planet's atmosphere. $\endgroup$
    – Drenzul
    Feb 20 '17 at 15:58
  • $\begingroup$ Check this calculator: stardestroyer.net/Empire/Tech/Beam/Calculator.html 50m radius, 1g at the surface. Escape velocity is .03km/sec (Note: The page rounds this to zero. If you speak HTML it's easy enough to edit the rounding formula.) That's vastly less than the thermal velocity of our atmosphere: en.wikipedia.org/wiki/Thermal_velocity $\endgroup$ Feb 20 '17 at 23:37
  • $\begingroup$ Yep, which means the atmosphere is slowly going to bubble off. How quickly is the question. .03 km/s is still 30m/s which is a fair velocity so its not just going to go POP and disappear off into space. I reckon if someone dumped an earth like atmosphere on a magic rock like that, you'd probably be able to survive for at least a few years before the atmosphere boiled off.... assuming you could survive the low atmospheric pressure. $\endgroup$
    – Drenzul
    Feb 22 '17 at 9:26
  • $\begingroup$ That 30m/s is far less than the average velocity of an air molecule on Earth. The atmosphere would leave almost as fast as if you simply dumped it in space. $\endgroup$ Feb 23 '17 at 4:09

Let's assume that by "enough gravity for humans to live there" you mean "enough gravity for humans to stick to the surface (due to gravity)". This implies you're asking a Volumetric Mass Density problem since gravity is a function of how Mass Density influences surrounding space.

Specifically, you're asking a question about how massive this small planet would have to be given its radius (so how dense it must be).

Lets call small planet 'Small'; and earthly planet "Earth".

For there to be "enough gravity for humans to stick to the surface" lets assume the mass of planet "Small" has exactly the same mass as the planet "Earth", regardless of volume, since human's stick to the surface of Earth due to gravity pretty well.

Volumetric Mass Density is defined as its mass per unit volume: $$ \rho = \frac{m}{V} $$

So let's represent Earth's Volumetric Mass as: $$ m_e = \rho_e \cdot V_e $$

Likewise let's represent Small's Volumetric Mass as: $$ m_s = \rho_s \cdot V_s $$

Since (we're saying) humans stick equally well to both planets why not make Small's mass (regardless of radius) the same as Earth's Mass since we stick pretty well to Earth. That means: $$ m_s = m_e $$

$$ \rho_s \cdot V_s = \rho_e \cdot V_e $$

But we want to solve for Small's Volumetric Mass Density $\rho_s$ so we can see what it's made of (Earth is mostly molten nickle and iron).

To rearrange this equation and solve for Small's mass-density $\rho_s$ lets divide both sides by $V_s$ leaving:
$$ \rho_s = \frac{\rho_e \cdot V_e}{V_s} $$

But recall that the top term in this fraction $\rho_e \cdot V_e$ is really just $m_e$? So lets simplify by replacing the top term $\rho_e \cdot V_e $with $m_e$ making our equation:
$$ \rho_s = \frac{m_e}{V_s} $$

This says the mass-density of planet Small must be equal to the the Mass of the Earth divided by the Volume of planet Small. So lets figure it out!

  • If the Earth's Mass $m_e$ is: $5.9721986×10^{21}$ metric tons; and
  • The radius $r_s$ of planet Small is 50m;
  • And the Volume $V_s$ of planet Small is calculated from its radius using: $$V_s=\frac{4}{3} \cdot \pi \cdot {r_s}^3$$
    (which works out to be 523599 $m^3$)

Through substitution $\rho_s$ must be: $$ \rho_s = \frac{5.9721986×10^{21}t}{523599m^3} $$ $$ \rho_s = 1.14061×10^{16}t/m^3$$

Answer: Asking Wolfram Alpha what has this density we indeed get the answer that this planet would be more dense than a neutron star ( $8x10^{13}$ - $2x10^{15}$ ) putting it into the range of exotics such as gravastars, objects that exist inside the Schwarzschild radius of an Earth-mass object.


If you need to explain it for some reason, there could be a little ball of superdense matter in the middle, like a little black hole. I'm assuming since you're asking that you want to get technical, so there are problems a with that or any way of amping up the gravity on a small object. One is that earth gravity draws a certain amount of space dust, between 5 and 300 metric tons of it. We don't notice it on Earth because even at the maximum, that is about 5e-20 of the earth mass. If your 50-meter planet has earth gravity it's not going to stay 50 meters for long.

Another thing is the atmosphere. Even if you have some dense matter gravity source at the center the air can only be as dense as air under a given pressure and temperature. So if it is similar in pressure, temperature, composition, and humidity to that of earth it will be dense up to about 16 km and will extend out to about 500 km. So your planet is really a gas planet with a tiny ball of matter in the middle. What would this do to weather? I don't know. Weather is usually affected by solar heat rising from the surface, air flow blocked by land mass, and water evaporating from the seas, just to name a few. With mostly a big ball of air over 10 times the size of the land inside I don't think it would behave the same way at all. What would happen to the land mass when all that air absorbed or released water? For that matter would you still have clouds and rain with such a small surface to cause temperature disturbances to cause rain and evaporation? If there were the kind of weather patterns that would develop are unpredictable.

I think it wouldn't be long before you had a lot of debris orbiting, but around a much smaller center than earth. There's no reason that it would draw in any less debris that earth does but it could orbit as low as 300 miles from your core. If it did enter your atmosphere it would easily add significant acceleration to any matter that wasn't part of the superdense core, since there isn't much mass on those parts. I think with all this space dust, meteors, and atmospheric anomalies you would have a very unstable surface.


Wil McCarthy's The Collapsium and sequels have miniature planets, called "pianettes", held together by the gravity of artificial neutronium at their cores... and even then, he makes the smallest habitable one (inhabited by a single human) about 600m in diameter. That is already small enough for several inconvenient effects — a building with 90° angles can't be very large (or else gravity and the floor would point in noticeably different directions near the edges), the air pressure is already noticeably lower a few meters above the surface, and anything any distance away takes on a disorienting tilt. At 100 meters diameter all of these problems would be magnified.

Using conventional materials, it's impossible; even osmium isn't dense enough.


If you have a 100-meter sphere, the only practical thing to do is to make it hollow, spin it up to generate centrifugal pseudo-gravity, and live on the inside surface. Even at that, it's a bit small for the pseudo-gravity to be entirely comfortable.

  • $\begingroup$ Hi, welcome to Worldbuilding! This is a valid answer, but personally, I would like to see some maths (or a reference, or something similar) to back up your claim that 'it's a bit small for the pseudo-gravity to be entirely comfortable'. Thanks $\endgroup$ Feb 18 '17 at 13:53

for gravity, YES, it is possible. Imagine a 50 meter building to be a planet. Can you walk on top of a 50meter building? The answer is yes! Gravity generated by that building is negligible, but the one from earth is still there. also the one from the Sun.

so just by looking at spec of one planet is not enough to answer the question. there could be one side of that planet that you could stand on!


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