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Size: 2001
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Size: 2048
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| Deletions are marked like this. | Additions are marked like this. |
| Line 9: | Line 9: |
| $ P ~ = ~ dE / dt ~ A 4 \pi r^2 \sigma T^4 $ | $ \large P ~ = ~ dE / dt ~ A 4 \pi r^2 \sigma T^4 $ |
| Line 13: | Line 13: |
| $ dm = 4 \pi \rho r^2 dr $ | $ \large dm = 4 \pi \rho r^2 dr $ |
| Line 17: | Line 17: |
| $ m = { 4 \over 3 } \pi \rho r^3 $ | $ \large m = { 4 \over 3 } \pi \rho r^3 $ |
| Line 23: | Line 23: |
| $ dE ~ = ~ G ~ m ~ r ~ dm = ~ G ~ \left( { 4 \over 3 } \pi \rho r^3 \right) ~ r ~ \left( 4 \pi \rho r^2 dr \right) $ | $ \large dE ~ = ~ G ~ m ~ r ~ dm = ~ G ~ \left( { 4 \over 3 } \pi \rho r^3 \right) ~ r ~ \left( 4 \pi \rho r^2 dr \right) $ |
| Line 27: | Line 27: |
| $ dE ~ = ~ { \Large 16 {\pi}^2 \over 3 } G {\rho}^2 r^6 dr ~ = ~ A 4 \pi r^2 \sigma T^4 dt $ | $ \large dE ~ = ~ { \Large 16 {\pi}^2 \over 3 } G {\rho}^2 r^6 dr ~ = ~ A 4 \pi r^2 \sigma T^4 dt $ |
| Line 31: | Line 31: |
| $ dt = \Large \left( { 4 \pi G {\rho}^2 r^4 } \over { 3 A \sigma T^4 } \right) \large dr $ | $\large dt = \Large \left( { 4 \pi G {\rho}^2 r^4 } \over { 3 A \sigma T^4 } \right) \large dr $ |
| Line 33: | Line 33: |
| $ t = \Large \left( { 4 \pi G {\rho}^2 r^5 } \over { 15 A \sigma T^4 } \right) $ | $ \larget = \Large \left( { 4 \pi G {\rho}^2 r^5 } \over { 15 A \sigma T^4 } \right) $ |
Disassembling a Planet
How long does it take to disassemble a (cold) planet? That is limited by the heat that must be dissipated to do so, and the efficiency of the process (inefficient processes generate heat). Large objects are far more costly to take apart (per unit mass extracted) than small objects.
The power available P for disassembly is proportional to an efficiency constant A (higher for high efficiency and high black body emissivity) times the black body radiation from a sphere of radius r :
\large P ~ = ~ dE / dt ~ A 4 \pi r^2 \sigma T^4
The increment of mass removed is:
\large dm = 4 \pi \rho r^2 dr
The total mass m is:
\large m = { 4 \over 3 } \pi \rho r^3
The surface gravity is G ~ m , and the escape energy per unit mass is G ~ m ~ r .
The incremental energy dE to remove dr of material is
\large dE ~ = ~ G ~ m ~ r ~ dm = ~ G ~ \left( { 4 \over 3 } \pi \rho r^3 \right) ~ r ~ \left( 4 \pi \rho r^2 dr \right)
Simplifying and making proportional to power:
\large dE ~ = ~ { \Large 16 {\pi}^2 \over 3 } G {\rho}^2 r^6 dr ~ = ~ A 4 \pi r^2 \sigma T^4 dt
Solving for time:
\large dt = \Large \left( { 4 \pi G {\rho}^2 r^4 } \over { 3 A \sigma T^4 } \right) \large dr
\larget = \Large \left( { 4 \pi G {\rho}^2 r^5 } \over { 15 A \sigma T^4 } \right)
Let's make that in terms of mass M:
\Large t = \LARGE \left( { 9 \rho } \over { 2 {\pi}^2 } \right)^{\large 1/3} \left( { M^{5/3} } \over { 10 A \sigma T^4 } \right)
So - big planets take longer to disassemble than small planets, and the mass removal rate is slower ( rate proportional to M^{-2/3} ) for big planets than small ones. Since there is more total mass in small bodies than big ones, we will build most of the stadyshell from small objects before the big objects are disassembled - and we must leave holes in the stadyshell for them (and their retinue of solar power satellites and concentrating mirrors) to orbit through!