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| Deletions are marked like this. | Additions are marked like this. |
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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 | 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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| MoreLater | $ 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. | 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. 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. |
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.
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.