On Jun 4, 2009, at 11:37 PM, Mark Iverson wrote:
"You are making it up as you go by the look of it. Now consider a
column through which the photons
of S_T are passing through the atmosphere they have a momentum what
happens to that momentum due to
interaction with the particles in the amosphere on the way out? Try
to stay focused on the
question."
Photons require about 2.94x10^9 watts per kg of thrust, or about 3
million square meters per kg of thrust at maximum insolation. See:
http://www.mtaonline.net/~hheffner/PhotonThrust.pdf
Using earth's radius r as 6400 km, the area is Pi*r^2 = 1.3x10^14
m^2. Total solar insolation thrust on the earth is thus (1x10^3 W/
m^2)(1.3x10^14 m^2)/(2.94x10^9 W/kgf) = 4.4x10^7 kgf. In other
words, about 4.4 metric tons of force.
Using Newton's F=m*a, or a=F/m, we have, for a 6x10^24 kg earth, that
a = (4.4x10^7 kgf)/(6x10^24 kg) = 7x10^-17 m/s^2.
Light shining on the earth for a year thus would accelerate the earth
by a whopping 2.2x10^-9 m/s, or about 22 angstroms per second, a
truly nanospeed.
You may want to check my work. I didn't.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
