How dense can a human get until they "sink" through the ground?

Question

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?

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!