[ExI] upon pondering your next million years

Lee Corbin lcorbin at rawbw.com
Tue Jul 15 07:07:34 UTC 2008


Spike writes

> Assume the sun radiates energy at about 5000 degK

It's better to use 6000 according to http://www.solarviews.com/eng/sun.htm.

> and we know that the sun subtends and angle of about half a degree, so its
> radius is about a quarter of a degree and a radian is about 57 degrees, so
> the earth's orbit radius is about 57 over a quarter or about 200 solar radii
> and change.

I take the Earth to be at 215 solar radii:   865,000/2   into 93,000,000.
Where I remember that the sun is about 865,000 miles in diameter.

> Since blackbody radiation is constants times T^4 and the
> radiating surface increases as the square of the radius, then the blackbody
> temperature of the shell decreases as the inverse square root of the shell's
> radius.

That's quite right.  I'd put it like this: If T is the temperature of
the sun, then T^4 is the power emitted by one square meter
of the sun's surface. If now the radius of the sun is increased
by a factor of r  (or we erect an inside of a Dyson shell at
distance r), then the same power passes through r^2  (square)
meters. (I.e., the surface has gone up by r^2.)  So a new
temperature

T^4
----    =   (T')^4
r^2

or   T' = T/(sqrt(r))

A neat application of this is to take r = 215 solar diameters, or the
distance from the sun to the moon.  (The moon works very well on
this because one side faces the sun so long it gets into thermal
equilibrium.)  Using   6000/(sqrt(215)), we get  T = 409 degrees.
But that's Kelvin, so subtracting 273, we get T = 136 Celcius.
And from http://www.enchantedlearning.com/subjects/astronomy/moon/
"TEMPERATURE The temperature on the Moon ranges from
daytime highs of about 130°C = 265°F to nighttime lows of
about -110°C = -170°F."

So this thermodynamic VVRBOTEC (Very Very Rough Back Of The
Envelope Calculation) comes within 6 degrees of the actual value!!!

> If an M-Brain had a Neptunian orbital radius, then (if I recall correctly)
> Neptune's orbit is about 30 AU, then that's about 6000 solar radii for
> single digit precision BOTECs, square root is about 80, so a Neptune radius
> M-Brain that big would still radiate photons at about 5000/80 or about 65
> degrees Kelvin,
>
> ***absolutely regardless of what the individual nodes are doing***
>
> assuming the first and third laws apply in an M-brain universe.
>
> Amara and other astronomy fans, do you follow my reasoning on why a
> Neptune-orbit sized M-Brain would radiate photons at about 65 degK?

I get using   T = 6000/sqrt(r), with r for Neptune being 30*215,

T = 6000/sqrt( 6450 )   =   6000/80   =   600/8  =  300/4  =  75 degrees K.

I assume I'm just using a hotter sun than you were.

Lee




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