Tether brightness

Rob Matson (Rob_Matson@cpqm.saic.com)
26 Feb 1996 19:27:55 U

                      Subject:                              Time:  18:47
  OFFICE MEMO         Tether brightness                     Date:  96/02/26

I may not have all the numbers right, but here's a first cut which others can
modify/correct as needed.  I'll use Bjorn Gimle's figure of 3 arcminutes for
human visual resolution (though it seems to me a 9mm diameter, dark-adapted
pupil should do much better than this), assume a range of 400 km, a tether
surface reflectivity of 0.8, and inband solar illumination of 177 W/m^2.  3
arcminutes is 873 microradians, which at a range of 400 km translates to 350
meters.  Thus, the 0.1" wide tether (2.54 mm) produces an equivalent visual
response to an area of 0.89 square-meters (350m x 2.54 mm).

.89 m^2 x 0.8 x 177 W/m^2 = 126 W.  Assuming the tether is approximately
Lambertian, this 126 W will be reflected into a hemisphere:

Intensity = 126 W / (2 pi steradian) = 20 W/sr.

At a range of 400 km, the resulting irradiance is 20 W/sr divided by the range
squared, which gives 1.25 x 10^-10 W/m^2.

Assuming the star Vega is magnitude 0, and Vega's power in the visual band is
3.34 x 10^-9 W/m^2, the tether is 26.7 times dimmer than Vega. This would give
the tether a visual magnitude of about 3.6:

2.5 x LOG10(26.7) = 3.57.

Of course, if you go back and change the visual resolution to 1 arcminute, the
tether would be three times dimmer, or magnitude 4.8.
Of course, this calculation doesn't take atmospheric absorption into
consideration; nevertheless, the tether should be quite visible.

My calculations show southern California will get a chance to see it the
morning of March 6th, assuming it stays up that long.  It's a low pass in the
south, passing under Sagitarius...