Estimating the diameter of all earths biomass as a sphere

Question

I am working on an educational resource about the relative scales of resources compared to the earth. Attached is an example of the earth compared to various elements contained within, as well as a comparison of the sum of all water and the atmospheric gasses (at standard temperature and pressure). Such images are great for creating a sense of scale and displaying the fragility if earth's ecosystems.

I want to also include spheres of wet biomass, which I would imagine to be quite small. However, when I do math I am faced with spheres that seem too small. I did the math based on 3 methods and got answers of 6km to 14km diameter, based on Metric Ton estimates of 1.57E+12, 1.38E+11, 1.00E+12

Two folks on Quora estimate 10km diameter. My questions are these:

1. What is the appx wet biomass of all plant and animal life on earth?
2. What is the appx density of it such that it can be converted to volume?
3. Is it really so small? I am used to being humbled, but this seems excessive. :-)

Answer 1

The following calculation is a very rough estimation to get the sensible order of magnitude.

In 2018, Bar-On et al. estimated the total biomass on Earth at about 550 billion tonnes $(5.5\times 10^{14} \, {\rm kg})$ of carbon, $80 \,\%$ of which is stored in plants. $70 \,\%$ of this plant biomass are stems and tree trunks, which are mostly woody.

Let's assume that the whole biomass is in the wood. The authors suggest that the carbon content should be multiplied by a factor of 2 to get the dry biomass. Assuming that the average water content of wood is $80 \,\%$. This gives us the total wet biomass of $$m=2\times(1/80 \,\%) \times 5.5\times 10^{14} \, {\rm kg} = 1.4 \times 10^{15}\,{\rm kg}$$

Approximating the density of wood at $800 \, {\rm kg/m^3}$, we calculate the diameter of equivalent sphere:

$$ V = \frac{4\pi}{3}\left(\frac{D}{2}\right)^3 \implies D = \left(\frac{6}{\pi} V\right)^{1/3} = \left(\frac{6}{\pi} \frac{m}{\rho}\right)^{1/3} = 15 \, {\rm km}$$

Even though we made quite a few assumptions, note that the diameter of sphere scales with mass as $D \sim m^{1/3}$. This means that if we over- or underestimated the total wet biomass by a factor of 2, the diameter of the sphere changes only by a factor of $2^{1/3}\approx 1.26$, e.g. $D \approx (12-19)\, {\rm km}$.

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There is also another interesting visualization published, although it uses the carbon biomass and not wet biomass.