[amsat-bb] ANS-199 AMSAT News Service Special Bulletin - AMSAT Fox-1C Launch Opportunity Announced

[Phil Karn]

On 07/19/2014 09:23 PM, Robert Bruninga wrote:

> I cannot believe that.  The equilibrium of a nominally black (solar panels
> on all sides) spacecraft is something like about 0 to 30 C (32F to 90F) a
> very benign operational range.  The only time you DO have thermal issues is
> when you DO have attitude control and have things that are not equally over
> time seeing the sun and dark sky.

See Dick's paper for the details; I'm just quoting his results. I know
the basic physics of heat transfer in space but I would never call
myself an expert. He is.

But I can do a back-of-the-envelope calculation that tells me he's right.

The solar cells they're using have an absorptivity and emissivity that
is both 0.98, as I recall, so a cubesat covered with them is essentially
a perfect blackbody.

A blackbody cube with one face normal to the sun at 1 AU will reach an
equilibrium temperature of -21.35 C. The problem is that the ratio of
radiating area to absorbing area for a cube is 6:1 (with the sun normal
to one surface). A sphere would be warmer because its ratio of radiating
to absorbing area is only 4:1. A thin flat plate normal to the sun (like
a solar wing) would be even warmer -- 2:1.

And that -21.35 C figure is for continuous sunlight. Throw in eclipses
and things get much worse. Yes, it would be a little better when the sun
shines on a corner rather than normal to a face, and Earth albedo and IR
radiation will warm things a little, but not enough to matter.

--Phil

PS: Temperature of 10 cm blackbody cube at 1 AU:

Area facing sun: .01 m^2
Solar constant: 1367.5 W/m^2
Absorbed power = 13.675 W

Total radiating area: .06 m^2
Emissivity = 1.0 (perfect blackbody)
Stefan-Boltzmann constant = 5.6703e-8 W/(m^2K^4)


T = (13.675 W / (5.6703e-8 * 1.0 * .06)) ** (1/4)
  = 251.8K == -21.35 C