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Size: 1995
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Size: 2509
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| Deletions are marked like this. | Additions are marked like this. |
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| $ 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 15: | Line 15: |
| The total mass $ m $ is: | The total mass $ M $ is: |
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| $ m = { 4 \over 3 } \pi \rho r^3 $ | $ \large M = { 4 \over 3 } \pi \rho r^3 $ |
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| The surface gravity is $ G m $, and the escape energy per unit mass is $ G m r $. | The surface gravity is $ G ~ M $, and the escape energy per unit mass is $ G ~ M ~ r $. |
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| $ 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 ~ = ~{ \Large { G ~ M } \over r } ~ dM = ~ G ~ \left( { 4 \pi \rho r^3 } \over r \right) ~ \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 $ |
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| Solving for time: | Solving for time and integrating: |
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| $ 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 $ |
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| $ t = \Large \left( { 4 \pi G {\rho}^2 r^5 } \over { 15 A \sigma T^4 } \right) $ | $ \large t = \Large \left( { 4 \pi G {\rho}^2 r^5 } \over { 15 A \sigma T^4 } \right) $ |
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| $ \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) $ | $ \Large t = \LARGE \left( { 9 \rho } \over { 2000 {\pi}^2 } \right)^{\large 1/3} \left( { G ~ M^{5/3} } \over { A \sigma T^4 } \right) $ . . . . ''( isn't THAT the damnedest-looking equation ???)'' |
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| 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! | So - big objects take longer to disassemble than small objects, and the mass removal rate is slower ( rate proportional to $ M^{-2/3} $ ) for big objects than small ones. Since there is more total mass in small !TransNeptunian objects (TNOs) 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! There is a nice escape clause, though. If, like Pluto, they have a large moon in isosynchronous orbit, and are mutually tidelocked, a gigantic-area space elevator net can be constructed between them, of arbitrarily large area. This can be fed by space power satellites, and dissipate a lot more power than a mere planetary surface. |
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 ~ = ~{ \Large { G ~ M } \over r } ~ dM = ~ G ~ \left( { 4 \pi \rho r^3 } \over r \right) ~ \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 and integrating:
\large dt = \Large \left( { 4 \pi G {\rho}^2 r^4 } \over { 3 A \sigma T^4 } \right) \large dr
\large t = \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 { 2000 {\pi}^2 } \right)^{\large 1/3} \left( { G ~ M^{5/3} } \over { A \sigma T^4 } \right) . . . . ( isn't THAT the damnedest-looking equation ???)
So - big objects take longer to disassemble than small objects, and the mass removal rate is slower ( rate proportional to M^{-2/3} ) for big objects than small ones. Since there is more total mass in small TransNeptunian objects (TNOs) 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!
There is a nice escape clause, though. If, like Pluto, they have a large moon in isosynchronous orbit, and are mutually tidelocked, a gigantic-area space elevator net can be constructed between them, of arbitrarily large area. This can be fed by space power satellites, and dissipate a lot more power than a mere planetary surface.