If a human standing on bedrock could indefinitely increase their mass without increasing in volume, how much mass could they add to themselves before they start causing damage to the ground? How massive can they get before they sink into the ground?

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    $\begingroup$ I suggest you google the term "compressive strenght" - it gives you a rough estimate as to when damage to the ground occurs. Calculating the mass of the person pretty much trivial then. $\endgroup$ Jun 27, 2020 at 10:41
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    $\begingroup$ I think it would be kind of undignified. Your legs would sink in but then you would hang up at the crotch because of the greater crossectional area. You would not sink further until you were considerably denser and until then you would be in the rock up to the crotch, waving your arms and grunting. \ $\endgroup$
    – Willk
    Jun 27, 2020 at 14:29
  • $\begingroup$ Standing on rock, not soil? $\endgroup$
    – smci
    Jun 27, 2020 at 18:29
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    $\begingroup$ Can you clarify the difference being a human being makes as opposed, for instance to a statue, which wouldn't drag questions of life in with it? $\endgroup$ Jun 28, 2020 at 13:44

1 Answer 1


Ooh, this is an interesting question.

Assuming granite bedrock, you have a compressive strength $\sigma$ of 130 MPa. That's the pressure needed to start pushing through the rock. We also need to assume a certain area of the human's feet or shoes - one way to do this is to use some sort of average total foot area, such as by summing the surface areas of the sole regions on page 1241 here to get 178 $cm^2$ per foot, but if our human is wearing shoes, it might go up to 190 $cm^2$, for a total area of 380 $cm^2$. Converting to square meters, we get 0.038 $m^2$.

edit: I'm also going to specify that the shoes have to be strong enough to hold up our incredibly dense human. Otherwise, proceed with the following calculations but for a barefoot human (0.0354 $m^3$) instead.

Assuming your human is standing on Earth, and not jumping around or leaning on something, or getting blown by the wind, the pressure they exert is equal to $$ mass_\text{human} \times \text{gravitational acceleration}_\text{Earth} \div \text{Area}_\text{shoes} $$ (that is to say, the gravitational force divided by the area that force is exerted over). It's when the gravitational force, spread over their shoe area, exceeds the compressive strength of the rock that they start sinking: $$\sigma = m_\text{human} \times g_\text{Earth} \div A_\text{shoes}$$

Solving for the mass and substituting the values, we get: $$m_\text{human} = \frac{A_\text{shoes} \times \sigma}{g_\text{Earth}}$$ $$m_\text{human} = \frac{0.038 m^2 \times 130 MPa}{9.807 m/s^2} \times \frac{10^6 kg\,m^{-1}s^{-2}}{MPa}$$ (conversion of MPa is just $10^6$ Pa)

Dividing it out, you get: $$m_{human} = 503,000 kg$$

A normal human has a density of about 1,010 kg/$m^3$, according to this somewhat sketchy website, so if you assume a normalish weight of 80 kg you get a "human volume" in cubic meters of 80/1,010, or about 0.08 $m^3$.

To get the density of this superdense human, you divide their mass by their volume to get $$503,000 kg/(0.08 m^3) = 6,400,000 kg/m^3$$ Just to be on the safe side (in case it's an especially strong granite), you'd probably want to bump that up to $$6.5 Gg/m^3$$ That's much denser than most things we know on Earth, about 40 times denser than the core of the sun and denser than all of the known elements. It doesn't nearly measure up to a neutron star or an atomic nucleus (without electron cloud), though!

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    $\begingroup$ This is a great answer! $\endgroup$ Jun 27, 2020 at 14:22
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    $\begingroup$ Yes! Please visit more questions and spread your ethic of "show your math". Because then I learn some math. $\endgroup$
    – Willk
    Jun 27, 2020 at 14:32
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    $\begingroup$ The shoes would need to have a greater compressive strength than granite, otherwise you'd just sink through them first, but +1 anyway as it's not much different. $\endgroup$ Jun 27, 2020 at 22:45
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    $\begingroup$ It doesn't even get into the ballpark of a white dwarf, let alone a neutron star. $\endgroup$ Jun 28, 2020 at 0:48
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    $\begingroup$ This answer works until they get up past the legs, since after that the pressure will be spread out a little more due to the pelvic area and torso. Great answer, though, I like the equations and explanations. $\endgroup$
    – c1moore
    Jun 28, 2020 at 14:30

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