Superbird A thoughts

Rob Matson (
2 Oct 1996 14:55:23 -0800

How fast would the sweeps drift south in this scenario?  Assuming a satellite
elevation angle of 30 degrees from the ground (to simplify the math), and a
range of 39,000 km, 2.3 deg/hour is the same as 0.038 deg/minute, and .038 deg
at that range corresponds to a ground footprint of 52 km.  So Robert Sheaffer
in northern CA and Ron Lee in CO should see their peak flashes several minutes
before I do in southern CA.  This is testable.

But even more importantly, if this description is accurate, then flashes will
again be visible on the ascending node!!  Superbird A is back at 1.7 degrees
south latitude at around 5:44:30am PDT; fortunately, it is still dark in both
CA locations at this time, though Colorado may be out of luck.  Certainly
worth a look-see.  Furthermore, the flashes should be 3 times brighter due to
the vastly improved geometry.  Beta in the evening is 67 degrees for SoCal;
beta in the morning is 151.  Brightness is proportional to the square of
SIN(Beta/2).  (Beta being the angle from the sun to the earth to the

Of course, all this assumes the fast sweeps are roughly east-west oriented. 
If instead they are north-south, then it is the sun's relative motion and not
the satellite's that dictates the duration of the flashes. But I think I can
prove that isn't the case.  The sun is moving roughly 15 degrees per hour
relative to the earth-satellite line.  That means the specular reflection
moves a half-degree per minute.  So, unless the specular image is considerably
larger than a half-degree wide, flashes should could not possibly last 10

This is good news, because it means there's a good chance that flashes are
actually visible on the ascending node...  --Rob