Floating Gold Solution

The following is a solution to Puzzle #17 – Floating Gold. Take a look if you haven’t already.


goldspherefeatThe key to the solution of this puzzle is Archimedes’ principle. It says that the buoyant force exerted on a body submerged in a fluid is equal to the weight of the fluid the body displaces.

To recap, we are asked to find the smallest radius of a hollow, gold sphere of thickness \delta = 1 cm such that it will float in water.

The first step is to ask: how much of the sphere is submerged? Remember we have been asked to find the smallest sphere that will float. The buoyant force on the sphere is greater if it displaces more fluid, by Archimedes’ principle. Therefore the greatest buoyant force that can act on the sphere exists when it is entirely submerged. If the sphere does not float when it is entirely submerged, it will not float at all! This means the smallest sphere we can get away with is that which floats when its whole volume is located beneath the surface of the water.

Next we want to form an equation giving us the radius of the sphere. To do so, we stipulate that the weight of the sphere is countered by the buoyant force which acts on it.

The buoyancy force is, from Archimedes’ principle, equal to

\displaystyle f_b=\frac{4}{3}\pi r^3\rho_w

where r is the outer radius of the sphere and \rho_w is the density of water. I hope there is nothing too mysterious about this statement.

What about the sphere’s weight? For simplicity, we’ll assume that the hollow part of the sphere is a vacuum, or at least that we can neglect the weight of whatever’s inside it. In this case, the weight of the sphere is

\displaystyle f_w=\frac{4}{3}\pi(r^3-(r-\delta)^3)\rho_g

where, to reiterate, \delta is the thickness of the sphere and \rho_g is the density of gold. This equation comes from working out the volume of the hollow sphere, which is the difference between the volume of two solid spheres of radii r and r-\delta, and multiplying by the hollow sphere’s density.

If the shell is floating, the two forces are equal in magnitude:


\displaystyle \frac{4}{3}\pi(r^3-(r-\delta)^3)\rho_g=\frac{4}{3}\pi r^3\rho_w

We’ll first dispense with the common pre-factors and take the density of gold to the other side:

\displaystyle r^3-(r-\delta)^3=r^3\left(\frac{\rho_w}{\rho_g}\right)

Next we’ll expand the left-hand side:

\displaystyle r^3-(r^3-3r^2\delta+3r\delta^2-\delta^3)=r^3\left(\frac{\rho_w}{\rho_g}\right)

Now at this point, I should acknowledge that it would be perfectly reasonable to throw away the terms of order \delta2, giving us a simple linear equation in r to solve. This amounts to approximating the hollow sphere as having volume 4\pi r2\delta, the product of its outer surface area and its thickness. As an exercise, however, I’ll choose to retain these small terms, and in doing so will have the opportunity to show you something nice.

We’ll simplify the equation above and divide through by \delta3. Doing so gives

\displaystyle 3\left(\frac{r}{\delta}\right)^2-3\left(\frac{r}{\delta}\right)+1=\left(\frac{r}{\delta}\right)^2\left(\frac{\rho_w}{\rho_g}\right)

Let’s neaten things by defining the quantities

r=\delta x


\displaystyle \frac{\rho_w}{\rho_g}=\epsilon

Rearranging the equation above and using these new quantities, we are presented with the cubic equation

\epsilon x^3-3x^2+3x^2-1=0

The cubic equation above, like all cubics, can be solved exactly. I will not give you an exact solution, but a damn good approximation using perturbation theory.

Forget for a moment the physical meaning of \epsilon; just imagine it’s an arbitrary ‘small’ parameter whose value we can vary at will. Consider first the limit in which \epsilon → 0. Then the cubic equation above reduces to


Where do the (real) roots of this equation occur? A little thought (or some graph plotting software) will tell you that this equation has no real roots; its vertex is situated below the x axis at (0.5,-0.25) and, since the coefficient of the x2 term is negative, it is an ‘upside-down’ parabola, which tends to -\infty as |x|\infty.

Now suppose that \epsilon is a minuscule, but positive number. What happens now? Imagine starting at the vertex (0.5,-0.25), and moving in the direction of increasing x. At first, very little seems to have changed; you are sliding down an inverted parabola, apparently falling without end. But slowly, as x gets bigger and bigger, the initially tiny \epsilon x3 term starts overtaking the -3x\2 term, and the curve begins to level out! Eventually, you will start moving back up towards the x axis and, at some huge displacement from the origin, you eventually meet it; you have reached the root of the equation. The smaller the parameter \epsilon, the longer it takes the curve to re-join the x axis.

The diagram below shows the cubic above in gold and the cubic devoid of its highest order term in brown. The slow turnaround is shown for an \epsilon of about 0.5. If this parameter were reduced, the root at x would shoot off to the right.


So the root x we’re interested in gets larger and larger as \epsilon → 0. On this basis, we propose a solution of the form


where \alpha < 0 and x_0 is a prefactor of order 1. Let’s substitute the proposed solution back into the cubic and see what we get:


Somehow we’ve got to make this equation behave itself. To do so, we’ll attempt to ‘balance’ the first two terms in the equation by demanding that their exponents be equal:






What did we do? This doesn’t seem to have helped at all: we’ve just ended up with another cubic! The difference is that, instead of depending on some horrible inverse power of \epsilon, the roots of this cubic can actually just be expanded as a power series in \epsilon:


How do we find the coefficients \{c_n\}? It’s quite easy; we substitute this form of x_0 into the new cubic above, which will give something of the form

\text{apple}\times \epsilon^0+\text{banana}\times\epsilon^1+\text{coconut}\times\epsilon^2+...=0

It turns out that, because the set of functions \{\epsilon^n\}, n\in\mathbb{N}_0 are linearly independent, the only way the equation above can be generally satisfied is if all of our apples, bananas and coconuts are identically equal to 0. In fact, you can get arbitrarily close to the exact solution to x_0 (and hence to x) as long as you’re willing to compute the coefficients \{c_n\}. We’ll only worry about the first three.

First, let’s try balancing terms of order \epsilon0 = 1. That is, we are just saying that


Substituting this into the cubic, and discarding terms of order \epsilon and above, we get


Therefore it must be true that


The solution we’re after is the non-trivial one, c_0 = 3. That’s our first coefficient:


Now we’ll calculate the next coefficient, making our answer a little better. We say that


Substituting this into the cubic gives


We now expand the brackets. Remember that every time we find a term of order \epsilon2 or higher, we throw it out the window and move onto the next term.

27+27c_1\epsilon-27-18 c_1\epsilon+9\epsilon+O(\epsilon^2)=0

The zeroth order terms 27 and -27 cancel out (as they should do if we calculated c_0 correctly), and the first order terms group together to give




So that’s the next piece of the puzzle:


Now we can go one order higher:


The next coefficient c_2 is found using the same procedure above. I’ll spare you the mess and just skip to the answer:

\displaystyle c_2=-\frac{2}{9}

These three terms will be more than enough to get a good answer. Recall our expression for the root x:


Using our values for \alpha and the first three coefficients c_0, c_1 and c_2 gives

\displaystyle x\approx\left(3-\epsilon-\frac{2}{9}\epsilon^2\right)\epsilon^{-1}

\displaystyle x(\epsilon)\approx\frac{3}{\epsilon}-1-\frac{2}{9}\epsilon

Re-expression of this solution in terms of our original parameters yields

\displaystyle r\approx\left(\frac{3\rho_g}{\rho_w}-1-\frac{2\rho_w}{9\rho_g}\right)\delta


Given the values

  • \delta=1\,\text{cm}
  • \rho_g=19\,300\,\text{kg\,m}^{-3}
  • \rho_w=1\,000\,\text{kg\,m}^{-3}

we find

r\approx 56.9\,\text{cm to 3 sf}

The numerical solution is, to six unnecessary decimal places, 56.888178 cm, therefore our answer is pretty good! If we retain only the most important term in our approximation, ie

\displaystyle r\approx\frac{3\rho_g}{\rho_w}\delta

we get

r\approx 57.9\,\text{to 3 sf}

which is, itself, not too bad an answer. This same answer would have been found if we had taken the sensible route and dropped all the \delta2 terms as was stubbornly avoided earlier. Even so, I think this method shows a good example of the power of perturbation theory. It relied on the fact that \epsilon was a small number, about 0.051 for this particular instance of a gold sphere in water. This solution holds for any evacuated sphere made of a material much denser than the fluid in which it floats, and will start to break down once we stray from this regime.


The second part of the puzzle asks: is it possible for a hollow, metal sphere of any thickness to float in air? Well, if the sphere were full of air in equilibrium with the air outside it, the answer is a resounding no. It is akin to asking whether a piece of metal will spontaneously float in air. Archimedes’ principle says that it will not, because metals are denser than air.

If instead the hollow sphere were filled with something less dense than air, the answer is yes. In fact, this incredible feat has been achieved, in the form of foil party balloons! The buoyancy provided by the relatively low density of helium contained therein is enough to counter the weight of the aluminium foil shell. Generally, though, we are restricted to making balloons very thin, such that their weight does not become too great.


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