tides v3
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Phil's working note (version 3) on using his earlier "frames doc" to treat Earth tides. He treats the gimbals situation as a special case with frame rotation ω = 0, uses the effective-force equation ma' = F - mS, and takes S = -Ω²b. Balancing gravity from the attracting body at the Earth's center gives mΩ²b = GM1m/r12², and he says he will then consider other points on the Earth as Butikov does. A gimbals frame picture is mentioned but not captured in the text.
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How to apply frames doc to the Earth tides problem.
In the gimbals situation, the vector b is not soldered to Frame S'. My entire frames doc never considers this possibility! Well, I guess really this would be a case of ω = 0 where ω is that of frames doc. Therefore I will use Ω below.
This situation can be regarded as Special Case #2 with ω = 0 and axes aligned. I have this completely general result valid in the general case.
ma' = F'eff = F – mS – mω x (ω x r') – 2m ω x v' – m x r' (8.4a)
For Special Case 2 with ω = 0 and = 0 this says
ma' = F'eff = F – mS (8.4a)'
For the tide on earth case, we know that
S = -Ω2b // the change is always to the center
Therefore
ma' = F'eff = F + m Ω2b
where for a particle of mass m at the center of the earth,
F = - GM1m/r122
Meanwhile, we have, for any particle at rest on the earth, a' = 0 in Frame S'. Then
ma' = F'eff = - GM1m/r122 + m Ω2b = 0
and we get our basic result that
m Ω2b = GM1m/r122
We can then go on to consider particles at other locations on the Earth as Butikov does. Here is my gimbals Fame S' picture
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