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Re: Asteroid Problem (2)



Umm, not really. As others here have stated, I was always taught and
tell my students that the tides are due to the gradient of the
gravitational field of the Moon and Sun (no, I don't use the word
"gradient" in an introductory astronomy class). The tidal force from
an object varies, therefore, as M/r^3. While the Moon dominates the
tides felt on Earth, the Sun has a noticeable effect, hence the
difference between Spring and Neap tides. Are those what you were
thinking of?

Ahh, but we are told that tides are generated by the difference
between lunar and solar forces, so that the force equivalent to that of
the moon is generated by a satellite having 10^{-3} moon masses at a
distance (1/r_{moon})^{2} =10^{-3)/r^{2}, or
r = 10^{-3/2}r_{moon} = .03x60 earth radii or about 1.8 earth
radii.


================================
Stephen D. Murray
Physicist, A Division
Lawrence Livermore National Laboratory
Email: sdmurray@llnl.gov
Phone: (925) 423-9382
FAX: (925) 423-0925
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