Yes, it is the sidereal day that should be used but to the accuracy required this is equivalent to the solar say.

]]>I confess that the sign of the oblateness effect isn’t immediately obvious to me, but I think you must be right that it makes the fraction smaller. In the extreme case of maximal oblateness, the Earth becomes a disk, and the satellite orbits in the same plane as the disk, so the fraction tends to zero.

]]>It was inevitable that the answer would be 42 (%).

]]>And this comment was supposed to be attached to my previous comment. I’m really not doing well today.

]]>And adding to the quibbling traditional here, the fraction of the _surface_ that is actually _visible_ is much smaller, because (a) cloud makes much of the surface not observable (at visible and NIR wavelengths, at least) and (b) it is night for nearly half the surface. It therefore all depends on what is meant by “visible” (does illumination by the Moon count,…?) ]]>

You can also ask what fraction is never visible throughout the satellite’s orbit. I guess that’s two spherical caps with angle (pi/2 – theta), so the answer is 1 – sqrt(1-(r/a)^2), or about 1%.

Santa Claus can’t watch satellite TV.

]]>(As I always tell my students, it’s probably worthwhile to check by considering extreme cases. As a tends to r, the answer tends to zero, and as a tends to infinity, it tends to 1/2. Both sound right.)

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