Differences between revisions 5 and 7 (spanning 2 versions)

NEO Settlement

Start with this claim in Near-Earth Objects by Yeomans: 1000 NEOs larger than 1000 meters, 1e6 NEOs larger than 30 meters. Assume a power law of quantity versus size (assume spherical, a lousy assumption).

diameter	radius	cumulative	area A	volume
> m	m	count	hectare	V m³
31.6	15.8	1000000	0.314	1.66e4	from Yeomans
100	50	100000	3.14	5.24e5
316	158	10000	31.4	1.66e7
1000	500	1000	314	5.24e8	from Yeomans
3160	1580	100	3140	1.66e10
10000	5000	10	31400	5.24e11
31600	15800	1	314000	1.66e13

The quantity is the cumulative integer count of all objects larger than that radius (or area or volume). Assume a power law for the cumulative count versus area A in hectares: N(A) = 314000 hc /A, and dN/dA = n(A) = -314000 hc /A² = -K₀/A², where K₀ = 314000 hc. Assume a minimum usable area of A₀ (perhaps 1/4 of 0.314 hc or 0.0785 hc) , and a maximum area A₁ for 314000 hectare for 1 large NEO asteroid.

Total area:

A_{tot} ~=~ \int_{A₀}^{A₁} ~A ~n(A) ~dA ~=~ \int_{A₀}^{A₁} ~-K₀ / A ~dA ~=~ K₀ \ln( A₁ / A₀ ) ~=~ 314000 ~ \ln{ 400000 } ~=~ 4e6 ~ hc

Total volume:

A ~=~ 4 \pi r^2 ~~~~ r ~=~ \sqrt{ A / 4 \pi } ~~~~~~ V ~=~ 4 \pi r^3 / 3 ~=~ A^{3/2} / 6 ~ \sqrt{ \pi }

V_{tot} ~=~ \int_{A₀}^{A₁} ~V ~n(A) ~dA ~=~ \int_{A₀}^{A₁} ( ~-K₀ / A² ) ~A^{3/2} / 6 ~ \sqrt{ \pi } dA

V_{tot} ~=~ ( K₀ / 3 \sqrt{ \pi } ) ( \sqrt{A₁} - \sqrt{A₀} ) ~\approx~ ( K₀ / 3 ) ~ \sqrt{ A₁ / \pi } ~=~ 104720 ~ \sqrt{ 100000 } ~ hc^{3/2} ~=~ 3.31e7 ~hc^{3/2} ~=~ 3.31e13 ~m³

At an average density of 2000 kg/m³, that is 6.6e16 kg of near earth asteroid to play with.

Assume a cylindrical small habitat, 25 meters radius, 25 meters across, 6 RPM and 1 gee. Surface area is 2π*(25m)² = 3930m² = 0.97 acres; assume day/night cycle rotation of solar illumination and green photon recycling as red to produce the photosynthetic equivalent of 5 acres. Assume 4 meters of shielding, about 16000 m³ of asteroid material made into one habitat.

There's enough material in the NEOs alone to build 2e9 acres of agricultural/living habitats, perhaps the agricultural production equivalent of 1e10 acres, and we haven't even got started on the rest of the asteroid belt, with vastly more material. Total world agricultural acreage is about 4e9 acres feeding 7 billion people; assuming a combination of better practices and better nutrition, we might be able to feed 20 billion people out there, in the NEOs alone.

For comparison, a hypothetical tent-terraformed Mars has a Sun-facing area of

We will need to harvest water and other semi-volatiles from comets and the outer/heavier asteroids (like Ceres). The limiting material may be nitrogen; the nitrates in the asteroids are probably not enough, and the planets big enough to collect ammonia are deep gravity wells a long long ways away.

-  ⇤ ← Revision 5 as of 2017-12-08 01:20:34 → 
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+   ← Revision 7 as of 2017-12-08 05:15:43 → ⇥
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-||<)>   > m ||<)>     m ||<)>  count ||<)>hectare||<)>     m³ ||
+||<)>   > m ||<)>     m ||<)>  count ||<)>hectare||<)>   V m³ ||
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-The quantity is the cumulative integer count of all objects larger than that radius (or area or volume).  Assume a power law for the cumulative count versus area A in hectares: N(A) = 314000/A, and dN/dA = n(A) = -314000/A².  Assume a minimum usable area of A₀ (perhaps , and a maximum area A₁ for 314000 hectare for 1 large NEO asteroid.
+The quantity is the cumulative integer count of all objects larger than that radius (or area or volume).  Assume a power law for the cumulative count versus area A in hectares: N(A) = 314000 hc /A, and dN/dA = n(A) = -314000 hc /A² = -K₀/A², where K₀ = 314000 hc.
Assume a minimum usable area of A₀ (perhaps 1/4 of 0.314 hc or 0.0785 hc) , and a maximum area A₁ for 314000 hectare for 1 large NEO asteroid.
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+Total area:
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+$ A_{tot} ~=~ \int_{A₀}^{A₁} ~A ~n(A) ~dA ~=~ \int_{A₀}^{A₁} ~-K₀ / A ~dA ~=~ K₀ \ln( A₁ / A₀ ) ~=~ 314000  ~ \ln{ 400000 } ~=~ 4e6 ~ hc $
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+Total volume:
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+$ A ~=~ 4 \pi r^2 ~~~~ r ~=~ \sqrt{ A / 4 \pi } ~~~~~~ V ~=~ 4 \pi r^3 / 3 ~=~ A^{3/2} / 6 ~ \sqrt{ \pi } $

$ V_{tot} ~=~ \int_{A₀}^{A₁} ~V ~n(A) ~dA ~=~ \int_{A₀}^{A₁} ( ~-K₀ / A² ) ~A^{3/2} / 6 ~ \sqrt{ \pi } dA $

$ V_{tot} ~=~ ( K₀ / 3 \sqrt{ \pi } ) ( \sqrt{A₁} - \sqrt{A₀} ) ~\approx~ ( K₀ / 3 ) ~ \sqrt{  A₁ / \pi } ~=~ 104720 ~ \sqrt{ 100000 } ~ hc^{3/2} ~=~ 3.31e7 ~hc^{3/2} ~=~ 3.31e13 ~m³ $ 

At an average density of 2000 kg/m³, that is 6.6e16 kg of near earth asteroid to play with.
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 Assume a cylindrical small habitat, 25 meters radius, 25 meters across, 6 RPM and 1 gee.   Surface area is 2π*(25m)² = 3930m² = 0.97 acres;  assume day/night cycle rotation of solar illumination and green photon recycling as red to produce the photosynthetic equivalent of 5 acres.  Assume 4 meters of shielding, about 16000 m³ of asteroid material made into one habitat.
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-Fit
+There's enough material in the NEOs alone to build 2e9 acres of agricultural/living habitats, perhaps the agricultural production equivalent of 1e10 acres, and we haven't even got started on the rest of the asteroid belt, with vastly more material.  Total world agricultural acreage is about 4e9 acres feeding 7 billion people;  assuming a combination of better practices and better nutrition, we might be able to feed 20 billion people out there, in the NEOs alone.  

For comparison, a hypothetical tent-terraformed Mars has a Sun-facing area of 

We will need to harvest water and other semi-volatiles from comets and the outer/heavier asteroids (like Ceres).  The limiting material may be nitrogen; the nitrates in the asteroids are probably not enough, and the planets big enough to collect ammonia are deep gravity wells a long long ways away.

Diff for "NEOsettlement"

NEO Settlement