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Sep14-11, 04:39 PM
P: 788
Quote Quote by A.T. View Post
A uniform inertial force field causes uniform acceleration, so it cannot cause or contribute to any deformation (tides).
The moon's gravitational force is nonzero and pointed toward the moon (even if there is a gradient), yet the earth center stays still in the rotating frame, and the "tide-producing force" on the antipodal side of the earth even has opposite sign. Hence the centrifugal force is "necessary" to describe what is observed in the rotating frame.

I have not idea what you mean by "residual vectors" and "taking out the earth rotation"...So again:
In which rotating frame would the centrifugal force vector be the same for A & B (as shown in the picture)?
Let r be the radius of the earth, x the distance from the barycenter to the center of the earth, and [itex]\omega[/itex] the orbital frequency. In the rotating frame, the "centrifugal acceleration" at the center of the earth is [itex]\omega^2 x[/itex] directed to the right, at A is [itex]-\omega^2 (r - x)[/itex], directed toward the left, and at B is [itex]\omega^2 (r + x)[/itex] toward the right again. Since people like to remove the component of centrifugal force which is due to rotation around the earth's axis and put that into the geoid, take out outward accelerations [itex]-\omega^2 r[/itex], [itex]+\omega^2 r[/itex] from the accelerations at A and B, respectively. The remaining centrifugal bits are what I called the "residual" vectors. What are they?