#format jsmath
 
= 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 β (higher for high efficiency and high black body emissivity) times the black body radiation from a sphere of radius $ r $:

$ \large P ~ = ~ dE / dt ~ = ~ β 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 = { \Large G \LARGE \left( { 4 \pi \rho r^3 } \over { 3 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^4 dr ~ = ~ 4 \pi r^2 β \sigma T^4 dt $

Solving for time and integrating:

$\large dt = \Large \left( { 4 \pi G {\rho}^2 r^2 } \over { 3 β \sigma T^4 } \right) \large dr $

$ \Large t \LARGE ~ = ~ { { 4 \pi G {\rho}^2 r^3 } \over { 9 β \sigma T^4 } } ~ = ~ { { G \rho M  } \over { 3 β \sigma T^4 } } $
² ³
 .G = 6.674e-11 m³ / kg s²
 .σ = 5.67e-8 kg / s³ K⁴

 . power per kg at 2500 m²/kg .  Sun is 3.86e26 W.

|| Object    ||   Mass    ||  Radius || kg/m³ || T (K) ||       Time || Distance || W/m² ||  W/kg || power W ||
|| Earth     || 6.0e24 kg || 6371 km ||  5500 ||   400 ||  16M years ||     1 AU || 1366 || 3.4 M || 2.0e31  || use less mass ||
|| Moon      || 7.3e22 kg || 1737 km ||  3344 ||   400 || 120K years ||     1 AU || 1366 || 3.4 M || 2.5e29  || use less mass ||
|| Mars      || 6.4e23 kg || 3390 km ||  3900 ||   400 || 1.2M years ||   1.5 AU ||  607 || 1.5 M || 1.0e30  || use less mass ||
|| Asteroids || 3.0e21 kg ||  470 km ||  2100 ||   400 ||   3K years ||   2.0 AU ||  342 || 850 K || 2.5e27  || use 15% of mass ||
|| Pluto     || 1.3e22 kg || 1187 km ||  1860 ||   250 ||  80K years ||    50 AU || 0.55 || 1.4 K || 4.8e24  ||
|| "TNO"     || 6.0e19 kg ||  200 km ||  1860 ||   250 ||  360 years ||    50 AU || 0.55 || 1.4 K || 8.0e22  ||
|| "TNO"     || 6.0e16 kg ||   20 km ||  1860 ||   250 ||   4 months ||    50 AU || 0.55 || 1.4 K || 8.0e18  ||

So - big objects take longer to disassemble than small objects - duh.  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, with much larger area, and cooling on both sides.  This can be fed by space power satellites, and dissipate a lot more power than a mere planetary surface.

Most dwarf planets do not have such moons - but the first mass lifted can be used to make an artificial moon, with the addition of material from other smaller bodies maneuvered into position.