Picture a solid lead ball. Apart from the imaginary case where it's emerged in an infinite volume of water (in which case the water will not form a black hole), in which it surely will float (the gravity of the water pulls on the ball spherical symmetrically), maybe it can in another situation.

Assume the ball finds itself in a thick layer of water. The ground the water is resting on is flat and has an equal temperature everywhere, above zero degrees Celcius. The water is in thermal equilibrium with this ground, and, assuming the water surface is flat too, it's also in thermal equilibrium with the air above the water, which has a constant temperature which is lower than the bottom temperature. It is clear that a (non-static) equilibrium develops. Heat is moving from the ground to the water surface.

But my main question is: Can a lead ball in these circumstances float in this massive and deep water or be accelerated upward at a certain depth?

Here one can find info about the compressibility of lead.
Here one can find the compressibility of water.

This is NOT a homework question.

  • $\begingroup$ Probably related physics.stackexchange.com/questions/238060/… $\endgroup$
    – Kyle Kanos
    Commented Jun 5, 2017 at 10:12
  • $\begingroup$ Why do you think the ball might float? $\endgroup$ Commented Jan 26, 2020 at 9:57
  • 1
    $\begingroup$ Because at a certain depth the lead might be compressed that much that the upward force of the water exerted by the water on the ball becomes bigger than the force on the ball by gravity. $\endgroup$ Commented Jan 26, 2020 at 15:24

2 Answers 2


There's currently an answer which observes that the density of water increases with pressure, and speculates that at some depth the density of liquid water becomes higher than the density of lead. That speculation is unwarranted. The linear approximation that liquid water's density increases by a few parts per million per atmosphere doesn't account for phase transitions. At temperatures below the liquid-vapor critical point, water becomes a solid at pressures above $10^4$ or $10^5$ atmospheres:

Phase diagram of water.svg
By Cmglee - Own work, CC BY-SA 3.0, Link

The "hot ice" phases, which occur at high pressure, have approximate densities of

  • ice VII: 1.50 g/cm$^3$
  • ice X: 2.46 g/cm$^3$
  • ice XI: 0.92 g/cm$^3$

All of these densities are much closer to the STP density of water (1 g/cm$^3$) than to the STP density of lead (11 g/cm$^3$) so it seems highly unlikely that the density of liquid water just on the low-pressure side of that phase transition is high enough to support lead. You could probably confirm that using the NIST Fluids Webbook.

So a sufficiently deep ocean on an ice-giant planet would develop a floor made of ice-VI or ice-VII, depending on the temperature there, and your lead ball would sink to this floor. If geological/hydrological/cryological dynamics made this hot-ice floor unstable, your lead ball would tend to sink into it as its surface evolved.

Also note that the highest-pressure phase, ice-XI, is less dense than ice-X. That inversion probably drives convection processes in the mantles of ice-giant planets, which would transport heavy metals like lead or iron down into the cores of those planets.


Though it appears you are seeking an answer that makes use of the increase of density with depth, a small enough mass of lead could "float" on water because of surface tension (check out this link to surface tension).

According to the link the two major assumptions are:

(i) The object's surface must not be wettable;

(ii) Its weight must be low enough for the surface tension to support it.

When an object is placed on a liquid, its weight depresses the surface. It then becomes possible for surface tension forces (which are parallel to the surface of the water) to act in a direction opposite to the weight. For a small enough mass, these can balance each other.

According to Wikipedia, for a rod of density $\rho$, length $L$ and cross-sectional area $A$ the condition for equilibrium is $$\rho A L g = 2 \gamma L \sin \theta,$$ where $g$ is the acceleration due to gravity and $\gamma$ is the surface tension. For a water-air interface $\gamma \approx 75$ m$N$/m, while for a water-mercury interface $\gamma \approx 400$ m$N$/m.

The maximum value of $\sin \theta$ is unity so this gives an estimate for $A$ as $$A \approx \frac{2 \gamma}{\rho g} \approx 10^{-8} m^2$$ so that rods of radius of order 0.1 $mm$ can float on water.

In fact, since the length $L$ cancels out it appears, in principle, that large masses can float, as long as the cross sectional area is small enough.

The included pictures are taken from Wikipedia and show some small objects floating on water (paper clip and small coin):

enter image description here

enter image description here

  • 2
    $\begingroup$ An order of magnitude for "how small" would be interesting here. The radius of curvature for a typical water meniscus isn't very much smaller than the lead "sinkers" used by fishermen (which are mostly no longer lead these days, but a less poisonous metal). $\endgroup$
    – rob
    Commented Jan 25, 2020 at 22:42

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