So, we've gone off to another star system and decided to build a Dyson bubble--a continuous statite, supported by radiation pressure, surrounding a star--with a surface gravity of 1g to live on.

Now, clearly, the bubble has a total surface area of hundreds of thousands of Earths--plenty of space! Except... we don't actually get to use all of that. Since the ground is held up by radiation pressure, we can't exceed a certain maximum areal density. Colonies have to be built in widely-separated dimples, with unobtainium cables spreading their weight over an enormous surrounding area. So, to figure out what kind of living area we'll actually have available, it's not sufficient to just know how much total surface area we have to work with--we need to know how many multiples of Earth's biosphere's mass we can support.

And to do that, we need to know how much the ecosphere--or at least the portions of it necessary to support human life, if that's less than all of it--actually weighs.

So, if we were to leave all of the mantle rock and metal behind and just strip off the life-supporting surface of the planet, how much would it weigh?

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    “The biosphere” is usually used to refer to all living things, but here it sounds like you really just mean the crust. Can you clarify what’s included in “the biosphere”? – Dubukay Nov 10 at 23:59
  • @Dubukay Well, that's kinda the question, isn't it? It should be obvious that we can't just take all of the animals without the land they live on or water they live in, or all the plants without the soil they grow in, and get a viable biosphere. Some amount of substrate material is necessary to support life. So, how much? Do we need the whole crust? Could we get away with, say, stripping off just the top 10 meters? – Logan R. Kearsley Nov 11 at 0:56
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    Words have meanings. This question uses the word "biosphere" with a meaning it does not have. – AlexP Nov 11 at 1:14
  • @AlexP Can you suggest a better one? Your criticism is reasonable, but I wasn't able to think of a better option that would fit in the space available for a question title. – Logan R. Kearsley Nov 11 at 1:20
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    Ecosphere. – AlexP Nov 11 at 1:33

The weight of the earth's biomass is estimated around 1.8 E15 kg by several sources summarized here https://hypertextbook.com/facts/2001/AmandaMeyer.shtml

Bedrock is not very biologically active. Therefore we will set an upper bound to the amount of crust needed by considering soils and sediments. A survey of depth to bedrock is given below

https://agupubs.onlinelibrary.wiley.com/doi/full/10.1002/2016MS000686

The world average depth to bedrock is about 13m. Assuming a soil density of 2 g/cm^3 (high estimate) we can estimate a world soil mass of 4E18 kg.

Water is also important to life. Most of the world's water is in the oceans, and most freshwater is frozen. The hydrosphere weighs 1.38E21 kg, but only 2.5% of this is fresh and 31% of that is liquid. This component weighs 1E19 kg. This is almost all groudwater.

https://water.usgs.gov/edu/earthwherewater.html

This gives a total "ecosphere mass" upper bounded at 1.4E19 kg for the Earth, or about 93 tonnes/m^2 in intensive units. If we remove groundwater (which mostly doesn't interact with the biosphere) then we have much less. This gives a mass of 4.1E18 kg or 27 tonnes/m^2

I would guess this is still quite high and we could make do with 2m of soil or less. In any case, actual biomass is inconsequential.

For fun, this is 4E13 W/m^2 of radiation pressure to balance. Taking the sun as a point source, we achieve this at 6E-6 AU = 900km which invalidates the assumption. The sun's core is about 100,000 km radius, and thermal pressure dominates here.

Mirror films naively seem like the best support option because they have 1. double the momentum transfer per unit energy and 2. twenty times less radiative absorption than a blackbody. On the outside face, they conduct heat to a blackbody which emits into space. However, reflectance drops rapidly at emitting temperatures, so we're limited to something like 1000K. This allows a flux of 1E6 W/m^2, so you need 10^7:1 support to loaded area. This assumes the film weighs nothing.

However, it turns out the flux is so sensitive to max temperature that a blackbody with higher temperature tolerance is better. If you use something like a high performance ceramic at 2300K, you can tolerate 1.5E6 W/m^2.This however makes cooling the biomass much more difficult.

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    Great answer, but I don’t think you can skip the oceans as you do. Many, many creatures live in and require the salt water and actively interact with it. – Dubukay Nov 12 at 15:52
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    Fair enough. I guess my answer only includes land. Including oceans adds 3 orders of magnitude, so I think we get farther from the answer OP is looking for (minimum needed to sustain humans). – Bill Nov 12 at 17:16

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