An Earth of Bacteria
Pretty much out-the-box Gemini 2.5 Flash, summarizing a conversation I had while making a cup of coffee.
Could a Tiny Bacterium Out-Mass the Earth in Two Days? Let’s Run the Numbers!
Ever stumble upon a mind-bending factoid that makes you tilt your head and go “Wait, really?” I recently did. The claim was that if a humble E. coli bacterium, with its speedy reproduction cycle of around 23 minutes (some can even manage 20), were to reproduce with perfect efficiency for just two days, the resulting mass of bacteria would exceed that of earth.
My inner mathematician and actuary immediately perked up. Frank fact or fantastical fiction? Let’s run the numbers.
Start with some basics. A single E. coli is tiny, weighing in at roughly \(1 \times 10^{-12}\) grams. The Earth, on the other hand, is a bit heftier at approximately \(5.972 \times 10^{27}\) grams. Our little bacterial friend can double its population in about 20 minutes under ideal conditions.
Over two days (48 hours), that 20-minute doubling time occurs \(48 \text{ hours} \times 60 \text{ min/hour} / 20 \text{ min/doubling} = 144\) times.
Starting with one bacterium, after \(n\) doublings, we’d have \(2^n\) bacteria. So, after 144 doublings, that’s \(2^{144}\) bacteria, a number with about 44 digits.
To get the total mass, we multiply this number by the mass of a single bacterium: \[ 2^{144} \times 1 \times 10^{-12} \text{ g} \approx 1.6 \times 10^{31} \text{ g}. \]
And guess what? \(1.6 \times 10^{31}\) grams is indeed significantly larger than the Earth’s mass of \(5.972 \times 10^{27}\) grams. So, mathematically, the claim holds up under the assumption of perfect, continuous reproduction.
Hold on a Minute…What About the Fuel and the Power?
Okay, so the math checks out in this idealized scenario. But as anyone who’s ever kept a sourdough starter alive knows, continuous growth requires a constant supply of food (nutrients) and the removal of waste. And when we’re talking about scaling up to the mass of the Earth in just two days, the logistics become…well, astronomical!
Let’s think about the sheer amount of “food” we’d need to feed this ever-growing bacterial behemoth. If we assume perfect conversion of food mass to bacterial mass (which is wildly unrealistic, but let’s roll with it for a moment), we’d need to supply at least the final mass of bacteria, \(1.6 \times 10^{31}\) grams, in just two days. The rate at which we’d have to pump in this “food” would be mind-boggling.
And then there’s the energy required for all that biological synthesis. A rough estimate suggests around \(10 \text{ kJ}\) of energy is needed to produce 1 gram of bacterial biomass. So, to get to Earth mass, we’d need a total energy of about \(1.6 \times 10^{32} \text{ kJ}\). Spread over 48 hours, the power requirement averages out to an already staggering \(9.26 \times 10^{29} \text{ Watts}\), thousands of times the power output of the Sun!
The Grand Finale: What Happens in the Last 20 Minutes?
But here’s where things get truly bonkers. Remember that exponential growth? It means that half of all the growth happens in the very last doubling period. In our case, the mass doubles from half the Earth’s mass to the full Earth’s mass in the final 20 minutes!
Let’s zoom in on that last 20-minute window. In this brief period, we need to produce \(8.0 \times 10^{30}\) grams of new bacteria. Assuming our \(10 \text{ kJ/g}\) energy requirement, that’s \(8.0 \times 10^{31} \text{ kJ}\) of energy needed in just 20 minutes.
The power required for this final burst? A truly eye-watering \(6.67 \times 10^{31} \text{ Watts}\).
To put that into perspective, the Sun’s power output is around \(3.8 \times 10^{26} \text{ Watts}\). Our little bacterial factory in its final 20 minutes would need the equivalent power of roughly 175,000 Suns!
Stepping up the cosmic scale, even the most luminous phenomena in the universe, quasars (the energetic cores of galaxies), have power outputs that can reach up to \(10^{42} \text{ Watts}\). While our factory’s demand is less than that, it’s still in the realm of extremely energetic astrophysical objects.
Frank vs. Fiction: Definitely Fiction
So, while the initial mathematical curiosity of a bacterium out-massing Earth in two days holds true under perfect conditions, the sheer logistical and energetic requirements make this scenario firmly in the realm of science fiction. The need to transport matter and energy at such phenomenal rates, not to mention dealing with the inevitable waste products, presents insurmountable challenges. And we’ve been ignoring gravity…
It’s a fantastic thought experiment, though, highlighting the incredible power of exponential growth while ignoring the fundamental limits imposed by the laws of physics. Next time you see an E. coli under a microscope, you can marvel at its reproductive prowess, just be glad it’s not aiming for planetary domination anytime soon.
Follow-Up
On my next coffee break I wondered how big a Petri dish we’d need. Here are Gemini’s thoughts.
Assuming the cells are growing in a single, monolayer sheet on a Petri dish, we can calculate the total area and the speed of the expanding circle.
A single E. coli cell is roughly cylindrical with a diameter of about \(1 \ \mu \text{m}\) and a length of \(2 \ \mu \text{m}\). To simplify, let’s approximate the area of a single cell on the surface as that of a circle with a diameter of \(1 \ \mu \text{m}\). Area of a single cell = \(\pi r^2 = \pi (0.5 \times 10^{-6} \text{ m})^2 \approx 7.85 \times 10^{-13} \text{ m}^2\).
From our previous calculation, the number of bacteria after two days is \(2^{144}\), or approximately \(1.6 \times 10^{43}\) cells.
Total area = (number of cells) \(\times\) (area of a single cell) = \((1.6 \times 10^{43}) \times (7.85 \times 10^{-13} \text{ m}^2) \approx 1.26 \times 10^{31} \text{ m}^2\)
For comparison, the surface area of the Earth is approximately \(5.1 \times 10^{14} \text{ m}^2\). The total area of the bacterial colony would be roughly \(2.5 \times 10^{16}\) times the Earth’s surface area.
Speed of Expansion
The area of the circle of cells is given by \(A = \pi R^2\), where \(R\) is the radius of the circle. We know that the area is also equal to the number of cells times the area of a single cell: \(A = N \times A_{cell}\). Thus, \[ \begin{align} R &= \sqrt{\frac{A}{\pi}} \\ &= \sqrt{\frac{N \times A_{cell}}{\pi}} \\ &= \sqrt{\frac{(2^n) \times (7.85 \times 10^{-13} \text{ m}^2)}{\pi}} \\ &\approx \sqrt{2^n \times 2.5 \times 10^{-13} \text{ m}^2} \end{align} \]
The speed of expansion is the rate of change of the radius with respect to time, \(v = \dfrac{dR}{dt}\). However, the number of cells \(N\) is a function of the number of doublings \(n\), and \(n\) is a function of time \[ n = \frac{t}{t_{double}}, \] where \(t_{double} = 20 \text{ min} = 1200 \text{ s}\).
Let’s find the speed at the end of the 48-hour period (\(t = 48 \text{ h} = 172800 \text{ s}\)). At this point, \(n=144\).
First, let’s find the radius at \(n=144\): \[ R_{final} = \sqrt{\frac{1.26 \times 10^{31} \text{ m}^2}{\pi}} \approx 6.34 \times 10^{14} \text{ m}. \]
This is a gigantic radius. For comparison, the distance from the Sun to Pluto is about \(5.9 \times 10^{12} \text{ m}\). The radius of the bacterial colony is over 100 times that distance.
The speed of expansion is not constant. Let’s look at the change in radius during the last doubling (the last 20 minutes). At \(t = 47 \text{ h } 40 \text{ min}\), the number of cells is \(2^{143}\) and the radius is: \[ \begin{align} R_{initial} &= \sqrt{\frac{(2^{143}) \times (7.85 \times 10^{-13} \text{ m}^2)}{\pi}} \\ &\approx \sqrt{2^{143} \times 2.5 \times 10^{-13} \text{ m}^2} \\ &= \frac{R_{final}}{\sqrt{2}} \approx 4.48 \times 10^{14} \text{ m}. \end{align} \]
The distance the circumference expands in the last 20 minutes is: \[ \Delta R = R_{final} - R_{initial} = (6.34 - 4.48) \times 10^{14} \text{ m} = 1.86 \times 10^{14} \text{ m}. \]
The speed of expansion at the edge is: \[ v = \frac{\Delta R}{\Delta t} = \frac{1.86 \times 10^{14} \text{ m}}{1200 \text{ s}} \approx 1.55 \times 10^{11} \text{ m/s}. \]
This speed is over 500 times the speed of light (\(c = 3.0 \times 10^8 \text{ m/s}\)), further demonstrating the absurdity of the scenario when applied to a physical system.
The function for the speed of expansion is \[ v(t) = \frac{dR}{dt} = \frac{d}{dt} \sqrt{\frac{A_{cell}}{\pi} 2^{t/t_{double}}}. \] Using the chain rule, this gives an exponential function for speed, \(v(t) \propto 2^{t/(2t_{double})}\).