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Re: [Phys-L] field of an ellipsoidal distribution



On Thursday, August 13, 2015 3:28 PM, Bob Sciamanda wrote

Donald,
For your students you might look at:
https://www.youtube.com/watch?v=lNJ2icc1fPk

Thanks Bob. I did take a look and agree with the comments of Carl Mungan who
wrote on Thursday, August 13 3:59 pm:

The YouTube video neglects a small correction by assuming that N points in
the radial direction. This is the old question of "which way does a plumb
bob
point" on a rotating earth point? The correct answer is to do a vector sum
of
g_0 (called g_s in the video) and g_0*A (called w_e^2*R_e in the video).
Then they agree:

dropping the A^2 term in Polvani we have g(theta) = g_0*sqrt(1-
2A*cos^2(theta)) = g_0*sqrt(1-2A*sin^2(lambda)) switching to co-latitude
lambda rather than latitude theta

which is approximately g*0*(1-A*sin^2(lambda)), the same result as in the
video.

Yes, I did the vector sum indicated above and then took the square root of
the resulting vector dotted with itself to get the magnitude of surface g as
a function of latitude. This simple approach (rigid, solid sphere with
uniform density, and constant spin rate) produced agreement with a few
pieces of actual gravity data for the earth (Young and Freedman, "University
Physics", 13th edition, Table 13.1, p 422, 2012) of about -0.1% at the
poles, +0.05% at the equator, and +/- 0.03% at mid-latitudes with the sign
of the errors agreeing with what one would expect from using a sphere to
approximate an oblate spheroid.

When I have time, I'll make use of the measured gravity data John Denker has
referenced on his web site to do a fuller comparison.

Don

Dr. Donald G. Polvani
Adjunct Faculty, Physics, Retired
Anne Arundel Community College
Arnold, MD 21012