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Full Version: (17B) (19B) (27S) Percentage of Sphere's Surface Visible for Height Above Surface
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For example:

Radius of Earth = 6378 km
Distance of Apollo 17 from Earth's surface at 1972-12-07T10:39Z = 29000 km
Visible percentage of Earth's surface in 1972 "Blue Marble" photograph = 40.99% (41%)
I was just wondering whether this formula holds up if the height were negative (for example, if the GPS shows you in a valley).

I'm thinking in such a case the physics would suggest 100% visible (for a hollow sphere or Earth), or 0% (for an opaque sphere or Earth), rather than a negative value. Geometrically, I think I'd choose 100% - except its the interior of the surface you can see, not the outside. Perhaps - 100%? With the negative flagging an interior view? I'm not sure what result would be preferred.
For a negative height, the simple formula will give a negative percentage visibility, but I have no idea how that should be interpreted geometrically!

I only posted the formula because it turned out to be so simple and elegant compared to how I was expecting it to turn out.
An interesting formulation from 1913

as well as an un-referenced web source …

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