In sci.physics, Brad Guth

wrote
on Tue, 27 Feb 2007 18:30:52 +0000 (UTC)
lgate.org:
"The Ghost In The Machine" wrote in
message

Misplaced a decimal point in the Venusian radius. I now see total
occultation.

Thanks much for that honest info.

I think we're both off by some +/- factor, as I'm not at all that
certain that VL2 = 100% occultation unless you're giving something
better than 150 km worth of added radius as due to that Venus cloud
layer, and otherwise only taking into account for the actual solar
surface.

VL2 is also a bit of a halo station-keeping orbit, and thereby the outer
portions of the flaming solar atmosphere should be giving VL2 at least
some degree of direct solar influx.

I guess that I'll have to redo my best swag and report back.
-
Brad Guth

Must I? Oh well...reprising my previous calculation with
more precision and a corrected decimal point:

If one again assumes

r =~ R * cuberoot(M2/(3*M1))

where M1 is presumably 1.998435 * 10^30 kg, M2 4.8685 * 10^24 kg,
and R 1.08208926000 * 10^11 m. This gives r = 1.0954 * 10^9 m.
At that distance the angular displacement of Venus, which has
diameter about 1.2102 * 10^7 m, will be about 1.1988 * 10^-2 radian.
The angular displacement of Sol, which has diameter 1.392 * 10^9 m,
will be 1.2864 * 10^-2 radian.

Therefore, 86.84% -- the square of the ratio of angular displacements.

If one includes the 150km cloud cover the percentage only ups to 91.20%.

(These are approximate because I'm confusing pie wedges with triangles
for simplicity in the calculation.)

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