For the uninitiated, a Fermi Problem is a seemingly unanswerable problem, which you can actually estimate reasonable by breaking the problem down into smaller parts. They're really fun, and I intend to post more in the future.
The question at hand is: How far would the oceans sink if we took all the fish out?
I'll answer in two very different ways.
Preliminaries
In both cases, I'll answer how much the ocean depth would decrease by first calculating the total volume of fish in the ocean. Why is this helpful? Because for a spherical shell, you can estimate its volume by just taking its surface area and multiplying by the thickness, i.e.
\[ V \sim 4 \pi R^2 \Delta R \]
So, if I can estimate the total volume of fish in the ocean, if I take all of those fish out, change in the height of the ocean will just be this volume divided by the surface area of the ocean, which I will take to be the surface area of the earth times 70% or so.
Now I just need to estimate the volume of all of the fish in the oceans...
Energy Budget
First I'll estimate the volume of fish in a rather general way. I'll try to do it on energy grounds. I know how much solar energy hits the earth per meter squared on average (340 W/m^2).
I'm going to assume that fish get their energy from plankton, and plankton get their energy from the sun, both with about 10% efficiency. I'll also assume that life occupies as much space as possible, probably about half of the ocean surface counts as liveable. From this I get the total energy available to make fish. How many fish does that allow? I'll assume that fish use about as much energy per kilogram as humans do. I know that humans have to take in about 2000 food Calories or 2,000,000 calories a day to survive. From this I get the total weight of all of the fish in the ocean, and for their volume I assume they're the density of water (fish are bouyant).
My calculation:
\[ \underset{\text{ \tiny mean solar flux} }{\left( 340 \frac{\text{W}}{\text{m}^2} \right)} \cdot \underset{\text{ \tiny earth surface}}{4 \pi \left( 6 \times 10^6 \text{ m} \right)^2} \cdot \underset{\text{ \tiny frac ocean}}{(0.70)} \cdot \underset{\text{ \tiny frac liveable} }{\left( \frac 12 \right)} = 5.4 \times 10^{20} \text{ W} \text{ (ocean life energy budget)} \]
\[ \cdot \underset{ \text{\tiny plankton eff}}{(0.10)} \cdot \underset{\text{\tiny fish eff}}{(0.10)} = 5.4 \times 10^{18} \text{ W} \text{ (fish energy budget)} \]
\[ \cdot \underset{\text{\tiny energy budget of man}}{\left( \frac{ 75 \text{ kg} }{ 2 \times 10^6 \text{ cal/day} } \right)} \cdot \underset{\text{\tiny cal $\leftrightarrow$ Ws}}{\left( \frac{ 1 \text{ cal} }{ 4 \text{ Ws} } \right)} \cdot \underset{\text{\tiny s $\leftrightarrow$ day}}{\left( \frac{ 60 \cdot 60 \cdot 24 \text{ s} }{ 1 \text{ day} } \right)} = 4.4 \times 10^{14} \text{ kg} \text{ (mass of fish)} \]
\[ \cdot \underset{ \rho_{\text{fish}} = \rho_{\text{water}} }{\left( \frac{ 1 \text{ m}^3 }{ 1000 \text{ kg} } \right)} \cdot \underset{ \text{\tiny ocean surface}}{\left( \frac{ 1}{ 4 \pi (6 \times 10^6 \text{ m})^2 (0.7) } \right)} = \text{ change in depth (m)} = 1.38 \times 10^{-3} \text{ m} \approx 1 \text{ mm} \]
So in the end I get about a millimeter change in the ocean height. Honestly, this is a lot larger than I expected. The ocean is a big place. I suppose there are also a lot of fish in the world. If you look, if we assume the average fish is about 10 kg, I've calculated that the population of fish in the ocean is 4.4E13 fish, or 4 million billion fish. That's a lot of fish.
I'm not so sure about my estimate, so I'll do it again from a different direction...
Fishing
Next, I'll try and estimate again, this time guessing based on how much we fish fish. I know that overfishing is a problem, which means that for some fish species we fish more than fish make little fishies, so if I can estimate how much fish we eat, I can estimate how much we fish, so I can estimate how many fish there are, and then I can estimate how much the ocean depth changes. This calculation is very rough, I took some very basic order of magnitude guesses at some of the parameters. In particular, I had to guess how many fish species we fish, which I took to be 1/100 for no very good reason. My calculation is below:
\[ \underset{\text{ \tiny fish eaten per person}}{ \left( \frac{ 1 \text{ fish}}{ 1 \text{ week}} \right)} \cdot \underset{\text{ \tiny fish weight}}{\left( \frac{1 \text{ kg}}{1 \text{ fish}}\right)} \cdot \underset{ \text{ \tiny week $\leftrightarrow$ year}}{ \left( \frac{ 52 \text{ weeks}}{ 1 \text{ year} }\right)} \cdot \underset{\text{\tiny people who eat fish}}{\left( 10^9 \right)} =5.2 \times 10^{9} \text{ fish/year fished} \]
\[ \cdot \underset{\text{\tiny fish we eat}}{ \left( \frac{100 \text{ tot fish}}{1 \text{ fish eaten}} \right) } \cdot \underset{\text{\tiny fish lifetime}}{\left( 10 \text{ years} \right)} = 5.2 \times 10^{13} \text{ kg (total fish)}\]
\[ \cdot \underset{ \rho_{\text{fish}} = \rho_{\text{water}} }{\left( \frac{ 1 \text{ m}^3 }{ 1000 \text{ kg} } \right)} \cdot \underset{ \text{\tiny ocean surface}}{\left( \frac{ 1}{ 4 \pi (6 \times 10^6 \text{ m})^2 (0.7) } \right)} = \text{ change in depth (m)} = 1.6 \times 10^{-4} \text{ m} \approx 0.2 \text{ mm} \]
This time I got 0.2 mm or so, which is in relatively good agreement with my other number. At least its not several orders of magnitude off.
Geometric Average
Honestly I trust my first number more than the second, but I'm going to average my two results in order to come up with the number that I'll submit to the contest. But, I'm not going to average my two numbers arithmatically:\[ \mu = \frac{1}{2} ( x_1 + x_2 ) \]
as you normally do to average numbers, instead I am going to average them geometrically, i.e. I'm going to take the square root of their product:
\[ \mu = \sqrt{ x_1 x_2 } \]
My Answer
This kind of averaging is logarithmic in nature, and my experience has been that it is much more successful average to use when you are doing fermi problems. Doing this on my two numbers I obtain my entry:\[ \Delta x = \sqrt{ 0.164 \text{ mm} * 1.38 \text{ mm} } \approx \frac{1}{2} \text{ mm} \]
About half a millimeter.
I'll be sure to let everyone know how I do in the contest. Drop us an email or leave a comment to let me know how good you think my guess is. Happy Fishing.
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