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https://en.wikipedia.org/wiki/2014_MU69
43 AU from the Sun
Radius ~20km (-ish) or 3.35×10exp13 m3 (-ish)
@(say) 1.5 metric tonnes (MT) per m3 = 50 trillion MTs.
Paul Birch estimated about 40 MT per m2 of habitat living area for all the "hardware" in "spheres" as follows: toposphere - the shape structure, lithosphere (sculpted fractal hollow relief), hydrosphere ("seas" and "oceans" of several meters deep - visualize https://www.google.bg/search?q=maldives … 24&bih=648 ), atmosphere - ( earthlike of 1 Bar N2-O2 mainly), luminosphere - magnifying soleta of nano-thin metalized sheets (aluminium?) . Ok, lets give more generous 50 MT/m2.
= ONE TRILLION square meters of habitable heavenly 1G (centrifugal) LAND.
= one million square kilometres prime realty. ( a Bishop ring size or a really long spaghetti habitat , or most probably something in-between)
LUMINOSPHERE -- at this distance it is under a Watt per sq.m. natural insolation.
It would be necessary every square meter of LAND to be provided directly or indirectly (for "natural" and industrial energy supply plus inefficiencies, wastes ... ) with about 1000 times wider photonic capture optics.
i.e. the soleta area must be in the order of 1 billion sq.km. or 1 000 000 000 000 000 m2.
A circle ( or square, or swarm / cloud of vacuum bubbles ... or whatever ) with diameter of 40(ish) thousand kilometres must tap and focus all the solar light on the habitat.
Soleta of solar sail material would be about 0.1-1 g/m2
A tonne covers a sq.km.
A billion tonnes the whole thing only.
@50 MT per 1 sq.m. of habitat floor ( LAND ) area this corresponds to only 2% of the initial mass to be invested into luminospheric hardware.
( Freeman Dyson have calculated that - but I do not know his base for habitat areal density , looks similar - that such way, via simple optics a habitat can be run with Earth-level of natural stellar light insolation on distances from several fusor / stellar radii (deflector soleta) to several light years ...)
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