Mirror-ball satellite

ROB MATSON (ROBERT.D.MATSON@cpmx.saic.com)
26 Jan 1998 18:59:13 -0800

Hello all!

Kerry Kirkland asked about the visibilitiy of a 19-inch sphere studded
with 1-inch circular mirrors as seen at an altitude of 300 km.  It may
come as a surprise that such an object would be quite visible, and
very interesting to observe!

Even more surprising is that only one one-inch mirror would be
contributing to the observed brightness of the satellite at any given
time (more on this below).  Also, if the satellite isn't spinning, then
it will flash at a variable rate that increases as it approaches
culmination, and decreases again as it descends.  If the satellite
is spinning, then it will "scintillate" -- flashing approximately 60
times for every revolution.

The reason that only one mirror contributes to the observed brightness
at any given time is that adjacent mirrors are angled 6-degrees from
one another.  The solar image only subtends about a half-degree.  What
this also means is that more than 90% of the time the satellite would
not be directly reflecting the sun.

So how bright are the flashes?  That depends on several factors,
including the satellite range, phase angle and mirror reflectivity, but
I'll give you a ballpark estimate based on the following assumptions:

1.  range of 400 km
2.  phase of 90 degrees
3.  mirror reflectivity of 95%
4.  sun visual magnitude of -26.7

Since the phase is 90 degrees, the individual mirror causing the specular
reflection would be tilted 45 degrees from normal to the observer.  This
reduces the projected area of the mirror from pi*(0.5*0.5 inch^2) to
COS(45) * pi * (0.5*0.5 inch^2) = 0.555 inch^2 = 3.58 cm^2.

The solid angle subtended by the mirror (as seen by the observer) is
the projected area divided by the range squared:

3.58 cm^2 / (400 km)^2 = 2.24 x 10^-15 steradians

By comparison, the sun subtends roughly 6.0 x 10^-5 steradians.
This means that just based on projected area alone, the sun would
be brighter than the mirror reflection by a factor of:

(6.0 x 10^-5) / (2.24 x 10^-15) = 2.68 x 10^10

Including the reflectivity factor of 0.95, the sun is brighter by a
factor of 2.82 x 10^10.  In visual magnitudes, the difference is:

2.5 * LOG10(2.82 x 10^10) = 26.1

Since the sun is magnitude -26.7, this means the flashes would
be magnitude (-26.7 + 26.1) = -0.6.  Most definitely naked-eye
visible!

Now if the satellite isn't spinning, or is spinning extremely slowly,
the flashes may be slightly brighter than this.  The reason is that
the sun is not uniformly bright across its disk -- the center is more
than 3 times brighter than the edge.  If the entire sun had the same
intensity as the center, it would perhaps be as much as one
magnitude brighter.  So the flashes in the above example could
actually rival the star Sirius.

On the other hand, if this "disco ball" is spinning too fast, then the
shortness of the individual flashes will begin to affect their perceived
brightness.  But I can compute a lower limit on the brightness by
assuming an 8.33% duty cycle (0.5 degree sun / 6-degrees between
mirrors) -- i.e. a factor of 12.  That's 2.7 visual magnitudes, so the
-0.6 becomes +2.1.  Still an easy object.

Others out there, please check my math!  It's possible that I've goofed
somewhere along the way, but the result seems reasonable when you
compare with the -8 magnitude flashes of Iridium at twice the range,
but with mirrors 3200 times larger.  --Rob