Will a penny ever stand still in the water at a certain depth?

Question

Let’s say I drop a penny in the deepest part of the ocean having a certain depth. Would the penny become buoyant enough to stand still in the water, since the density of water increases with depth? Since the buoyancy of objects becomes greater as the density of water increases, would a penny ever come to a stand still, and if so, what would be an estimated depth of the water? I asked my science teacher this and she couldn’t answer it.

Answer 1

The answer is no. Water, being a liquid, is nearly incompressible, meaning that the density changes very little with increasing pressure. In the very deep ocean, the pressure can approach $10^{8}$ Pa (about a thousand times greater than standard atmospheric pressure of $1.01\times10^{5}$ Pa). However, the bulk modulus $B$ of liquid water (the reciprocal of the compressibility) is even larger, at about $2\times10^{9}$ Pa. ($B$ actually varies with the pressure and temperature with the water, but not enough to make a practical difference.) The ratio of the ambient pressure $P$ to the bulk modulus $B$ gives you roughly how much fractional change there will be in the density of water when it is under the pressure $P$.

This ratio is about $0.05$, so even at the greatest depths of the ocean, the changes to the density of the water will be, at most, about five percent. Actual differences in ocean density often have more to do with differences in the salinity of the water (since the dissolved sodium and chloride ions adds extra mass); however, these changes are also small, at the level of a few percent at most. To determine whether a penny will float or sink, we only need to compare the density of the penny to the benthic density of the water. Modern pennies are made mostly of zinc, while before 1982, they were mostly made of copper. The specific gravities of these metals are $7.0$ and $9.0$, respectively (under atmospheric pressure; they would be slightly greater at $10^{8}$ Pa), making them both several times more dense than seawater (specific gravity of $1.0$, plus or minus a few percent). So a penny would sink to the bottom. Only an object that is very close to the density of water at the surface can find an eventual equilibrium point where its density is equal to that of very deep water.

Answer 2

Yes, it will, but in a different way than one might imagine. In the Earth's oceans it is definitely impossible.

From the phase diagram of water, one can see that assuming a constant temperature of 300 K, water becomes solid ice VI at about 1 GPa. This kind of pressure would be encountered at a depth of about 100 km (actually a bit less because of the increasing density). By comparing this pressure to the bulk modulus of water (about 2.2 GPa) we can conclude that at 1 GPa the density of water will be about 1500 $\mathrm{kg m^{-3}}$. So if we had a (sweetwater) ocean deep enough, the penny eventually comes to rest at this depth on a layer of solid ice, even though its density will be less than that of the penny.

At the same time, we must not forget the penny itself is compressible. However, for zinc, bulk modulus is about 55 GPa, so at 1 GPa the change in volume will be only about 2%, which can be neglected.

However, if we insist on having liquid water, we can also increase the temperature. According to the diagram, at the critical temperature of 650 K we can continue increasing the pressure to about 15 GPa before water solidifies. This looks promising, but the bulk modulus of water is not constant. It increases with pressure, so it is probably still not possible to reach the density of zinc, 7000 $\mathrm{kgm^{-3}}$. So with liquid water it still is not possible.

At even higher temperatures, water goes supercritical. If we find a larger phase diagram, at 1000 K it should be possible to have supercritical water at some 50 GPa, which should theoretically be enough to compress the water while keeping the penny relatively uncompressed. Unfortunately, since bulk moduli are not constant, it is likely that this still will not be enough.

We should also remember that the melting point of zinc is fairly low at standard pressure. Fortunately it gets much higher with increasing pressure, and at 50 GPa it is already above 2500 K.

I am not able to find any experimental data for this extreme region, but it still seems remotely possible that somewhere there actually is a combination of temperature and pressure at which water is supercritical, zinc does not melt yet and the ratio of compressibilities remains low enough that the penny might remain afloat.

Answer 3

No, the density of water will not reach near $7 \;\text{g}\,\text{cm}^{-3}$, the density of zinc.

Drilling petroleum wells results in pressures at least as high (as the Marianas Trench). The steel drill collars still sink — put weight on the drill bit. Without looking it up, the densities of zinc and steel are both near $7 \;\text{g}\,\text{cm}^{-3}$, in round numbers. And, in some wells the drilling mud density may be over $1.92 \;\text{g}\,\text{cm}^{-3}$ ($16 \;\text{lb/gal}$), compared to sea water at roughly $1.02 \;\text{g}\,\text{cm}^{-3}$ ($8.5 \;\text{lb/gal}$). There is a buoyancy effect on the drill string for which drillers compensate (beyond anything I know). Background: the drill string is in tension so weights (collars) are put on at the bottom to provide force for the drill bit against the rock.

Answer 4

TL;DR in theory a weak yes, in practice a resounding no.

In theory

From your comments I infer that you wish to model the question as simply as possible.

Let us assume

• the water to be a stationary compressible inviscid fluid
• in a single phase
• in uniform constant gravity
• and the penny to be an incompressible point mass of different density
• moving with near zero velocity $^1$
• and the laws of physics to be classical and non-relativistic.

Already these assumptions seem tough to swallow. Anyways, in this model what are the possible densities a vertical$^2$ fluid column can take? They are$^3$

$$\rho(h)=\frac{\rho_0}{1-\rho_0gh/B}\tag{1}$$

where

$$ \begin{array}{llr} \rho(h)&\text{liquid density at depth h}\\ \rho_0&\rho(0)\\ B&\text{Bulk modulus of liquid}\\ g&\text{acc. due to gravity} \end{array} $$

As you can see all values of densities are allowed within the depth $[0,B/\rho_0g]$. So eventually the coin will become neutrally buoyant at some depth.

In practice

This is a very bad model of reality.

1. liquids are usually quite incompressible. As a result their density hardly varies with depth. To claim that the liquid density increases sufficiently enough to match that of a solid object is absurd.
2. Even if the density of the liquid was to increase significantly, our model of an inviscid monophase fluid would fail miserably. Its most likely that the liquid solidifies ever before reaching the density required for neutral buoyancy. Even then it may not match that of a metal penny.
3. Even with all our assumptions, the model predicted a singular density and max depth. This is aphysical. Moreover, at pressures high enough to solidify a liquid, temperature and other parameters of material state cannot be ignored.
4. Numerically$^4$, the predicted depths for neutral buoyancy are untenable on earth. Not only that, for such vast liquid columns, currents, concentration and thermal gradients, solubility, drag etc. become more significant in affecting the equilibrium motion of a sinking coin than density variation ever would.

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$^1$So it won't overshoot the equilibrium point.
$^2$ wrt. direction of gravity
$^3$ I used $$ \begin{align} dP/dh&=\rho(h) g\tag{A.3.1}\\ B&=-VdP/dV\tag{A.3.2}\\ d\rho/\rho&=-dV/V\tag{A.3.3} \end{align} $$ to get $$\frac{d\rho}{dh}-\frac g B\rho^2=0\tag{A.3.4}$$

$^4$ With $B=2\times10^9 \text{Pa},\rho_0=10^3\text{kg m}^{-3},\rho_{penny}/\rho_0=7,g=10\text{m s}^{-2}$ $$ \begin{align} \rho(10\text{km})/\rho_0&=1.052\tag{A.4.1}\\ h_{\rho=\rho_{penny}}&\approx 171\text{km}\tag{A.4.2}\\ h_{\rho=\infty}&=200\text{km}\tag{A.4.3}\\ \rho_0 g/B&\approx10^{-6}\tag{A.4.4}\\ \rho(h)&\approx\rho_0(1+\rho_0gh/B)\tag{A.4.5} \end{align} $$