On May 3, 2007, at 2:35 AM, Michel Jullian wrote:
Hi Horace,
I don't deny that gravimagnetism exists (it's an obvious
consequence of gravity propagating at a finite speed, if the term
means what I think it means i.e. the gravitational Lorentz force)
but when you say "the ambient gravimagnetic field in the vicinity
of Earth required to account for the precession of the Earth", are
you suggesting the observed precession rate is not, or not
entirely, accounted for by the official explanation that this
precession is due to the gravitational torque exerted by the Sun on
the Earth's equatorial bulge?
http://en.wikipedia.org/wiki/Precession_of_the_equinoxes#Explanation
The official theory works nicely though, I remember I had to derive
the precession rate as a physics exercise when I was a student many
years ago, assuming the Earth was an homogeneous ellipsoid of the
right dimensions, and it came out strikingly close to observations.
Regards,
Michel
You make an excellent point Michel. I seem to have have a major flaw
in reasoning here! At very least, I need to make a correction, and
at most dump the whole idea.
First, let's (hopefully) dispense with the moon. (I haven't convinced
myself this is can be done yet.) It appears the earth's polar motion
due to the moon nets to zero, so we can eliminate it as a cause for
precession of the equinox. For example, see:
http://www.pietro.org/Astro_Util_StaticDemo/
MethodsNutationVisualized.htm
Now, we still in any case have to deal with differential gravitation
on the earth's bulge from the sun as a source of precession. I was
under the impression that, ignoring nutation, the pole precessed at a
fairly constant rate throughout the year. If the differential
gravity effect on the bulge ring were a valid explanation for all the
precession, then it would have to come to a complete stop twice a
year at the equinoxes, when the net torque from the sun is zero,
especially if the sun and moon are aligned. I was under the
(apparently wrong) impression this does not happen. However, I have
not located a chart yet today that nets out the polar motion due to
the sun, or even shows intuitively this is not so.
The net precession rate is (360 deg/ 25765 yr)*(3600 secsarc/deg) =
50.3 secs arc per year average. During solstices, the precession
rate has to be twice this if the precession rate is zero at the
equinoxes.
I computed Q_earth = 4.26E21 N m to effect the average precession
rate. This would mean the torque would have to be twice that at the
solstices, or 8.52E21 N m. To simplify the model lets assume the
excess mass 2m is in the form of a barbell with m mass at each end,
and with length = the diameter of the earth, and on a 23 deg. angle
with the sun. Let m1 be the mass closest, m2 be the mass furthest.
Let distance to the sun r be 1 AU = 1.5E11 m, and earth radius be
6378 km = 6.4E6 m. Earth diameter is 12.8E6 m. Due to the 23 deg.
tilt, the difference in distances to the sun d for m1 and m2 is d =
cos(23)*(12.8E6 m) = 11.8E6 m. Msun = 2E30 kg.
Now for some seat of the pants calcs. The forces on the masses are:
F1 = G (m Msun)/r^2
F2 = G (m Msun)/(r+d)^2
F1 - F2 = [G (m Msun)/r^2] - [G (m Msun)/(r+d)^2]
F1 - F2 = G m Msun [1/r^2 - 1/(r+d)^2]
F1 - F2 = G m Msun [7E-27 m^-2]
But, the torque on the axis is sin(23)*(6.4E9 m)*(F1-F2), so torque
Q = sin(23 deg.) (F1-F2)
Q = sin(23 deg.) (6.4E9 m) G m Msun [7E-27 m^-2]
m = Q / (0.39 (6.4E9 m) G Msun [7E-27 m^-2])
m = (8.52E21 N m) / (2.1E3 m^2/s^2)
m = 4.1E18 kg
The mass of the earth is 6E24 kg, so the ring to account for the
torque is only a millionth the mass of the earth. So a rough
approximation to the 43 km thick bulge ring mass is 8E18 kg. This
8E18 kg seems way too low for the bulge mass.
The density of earth is about 5.5 g/cm^3 (too high for crust, but OK
for this.) The volume of the ring must be 8E18 kg/(5.5 g/cm^3) =
1.45E15 m^3. Using 2 pi (6.4E9 m) as length of ring, and 43 km as
thickness, we get
ring width = (1.45E15 m^3)/[2 pi (6.4E9 m) (4.3E4 m)] = 0.8 m
I have something major wrong above. Likely, I'll have to go back to
the original torque calc. It's almost 6AM now, so I'm too tired to
resolve this right now.
Regards,
Horace Heffner
