From: willner@cfa.harvard.edu (Steve Willner)
Newsgroups: sci.space
Subject: Re: SPS fouling astronomy
Date: 13 Aug 92 17:58:39 GMT
Organization: Smithsonian Astrophysical Observatory, Cambridge, MA,  USA


In article <1992Aug12.044959.19501@dartvax.dartmouth.edu>,
Frederick.A.Ringwald@dartmouth.edu (Frederick A. Ringwald) writes: 
> W(out)/W(in) = aA/r^2
> It's easy enough, from this ratio, to estimate the apparent
> magnitude of the satellite, given V(Sun) = -26.7.

> More realistically, take a = 0.07, that of Lunar basalt. A Solar power
> satellite is supposed to absorb energy, after all: it'd be silly to
> make one as reflective as a mirror. This brings it to V = -3.0.

The calculation looks right to me, but we better look again at the
assumptions.  Making the albedo as small as 0.07 is _hard_; even the
best black paints are around 5% or so.  Uncoated silicon has an
albedo of around 40%.  (The index of refraction is about 3.5.)
Practical solar cells have some kind of coating, but it has to be
cheap if you want square kilometers of it.

> Now, a Solar power satellite would probably be blackened on purpose, to
> absorb more energy, so as a lower limit on albedo, take a = 0.01.

This is about the albedo of a laboratory blackbody, i.e. an absorbing
cavity.  No way will a large structure even approach this!

> 13.4 GW. So, ignoring losses - large assumption here - you'd need about
> 1000 of them to run the world, assuming rather profligate energy use,

US electric consumption is about E12 W, so about 100 satellites are
needed to supply that.  World energy use per capita is about 1/5 that
of the US, which has about 1/20 of the world's population, so
replacing electrical generation worldwide would take about 400
satellites at present rates and up to 2000 if the rest of the world
were to consume at the present US per capita rate.  Multiply by 2 if
you want to replace a good fraction of oil usage, by 3 if you want to
replace oil, coal, and gas.  Adjust for population growth and
conservation according to your best guess.  (Figures are from 1985
and come from "A Physicist's Desk Reference.")

Bottom line: the estimate of 1000 is a good round number for
discussion purposes, but we ought to keep the assumptions in mind.

>  So, the total integrated brightness is about V = -9, about 3% that
> of a Full Moon, for a necklace about the celestial equator.

I get -9.5 for albedo = 7% and -11 for a more reasonable albedo of
30%.  The latter is 0.2 of a full Moon.  Not necessarily a problem if
you are a bright time observer, as I am.  :-)

Seriously, if we are building enough space infrastructure for SPS's,
it should be easy for professional astronomers to have access to
space instruments.

> Also, the objection that
> "there'd be nowhere on Earth you could get away from the handiwork of
> humans - no remaining true wilderness" isn't really true: try Alaska

I give up.  The lattitude is only 60 to 70 degrees or so, and there
are plenty of flat places where the horizon is visible.  And the
aurora is seldom bright enough to obliterate the bright planets.
Maybe you were there during a _really_ bright display. :-)
-- 
Steve Willner            Phone 617-495-7123         Bitnet:   willner@cfa
Cambridge, MA 02138 USA                 Internet: willner@cfa.harvard.edu
  member, League for Programming Freedom; contact league@prep.ai.mit.edu
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