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Reading an old book of classical mechanics (Spiegel's theoretical mechanics) I came across a passage that I would like to clarify. In order to find the equations of motion of free falling point particle that regulate motion according to a laboratory on to the earth's surface, the relative acceleration theorem is exploited. In a step (red arrow) apparent gravitational field is introduced to get (6), and then (7) ($\mathbf{r}$ the "small" vector from laboratory origin to the point, the triple vector product has magnitude something like $h\cdot 10^{-9}$). But later (blue arrow) the apparent field is written as $-g\mathbf{k}$ where $\mathbf{k}$ is the laboratory $z$ versor, orthogonal to the ground, toward the sky. Isn't this a bug? The direction of apparent field is not $\mathbf{k}$. Of course the effect is small, but Coriolis force too is small. If this is not a bug, why? If this is a bug, how to fix It?

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The problem is that $\mathbf{g}$ does not represent the real gravitational field but the direction of the plumb line (this is the meaning of the choice in equation (5)). Therefore it has not the direction of the versor $-\mathbf{k}$. The effect is small but the discussion assumes without justifying it that it is negligible compared to that of Coriolis. How can I justify it?

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  • $\begingroup$ Have you tried comparing the magnitudes of the two forces for h on the order of a few meters? $h\cdot 10^{-9}$ is a fairly negligible number at that height. They do mention immediately under your red arrow that this is done to great approximation near the Earth's surface. $\endgroup$
    – Triatticus
    Commented Apr 19, 2022 at 17:07
  • $\begingroup$ @Triatticus Of course I understand this negligible triple vector product. I thank you for the interest but you misunderstood my question: it is easy to see what happens from (6) to (7) but I can't understand why with blue arrow the apparent gravitational field is supposed to have $\textbf{k}$ direction: this approximation should be justified as I did with the approximation (6) to (7) after the red arrow. $\endgroup$ Commented Apr 19, 2022 at 19:36
  • $\begingroup$ Oh wait I'm looking at it now, no they are correct, because the gravitational field of a perfect sphere is directed perpendicularly into the surface, the deviation comes from the part after $-g\hat{k}$ (since it has components in both the $\hat{i}$ and $\hat{j}$ directions). In the limit $\omega\to 0$ you should recover a perfectly perpendicular direction of g at all angles. $\endgroup$
    – Triatticus
    Commented Apr 19, 2022 at 20:20
  • $\begingroup$ @Triatticus but in $\omega\to 0$ limit, Coriolis force too is zero. The book is ignoring the rotation effect of the not ground orthogonal direction of apparent field but not ignoring the rotation effect of Coriolis force. I don't say this is wrong, but it is not justified. I would like to put a patch on it to fix the proof but I can't do it. $\endgroup$ Commented Apr 19, 2022 at 20:31
  • $\begingroup$ I guess I don't understand your issue, it can't be with the $-g\hat{k}$ term as that part is only dependant on the shape and mass distribution of the object (here a sphere) and has nothing to do with rotation. So it's the rest of the terms (the coriolis term)? I'm a little unsure what your last comment meant to say maybe some letters got jumbled around. Maybe you can edit the question with how you think the derivation should look. $\endgroup$
    – Triatticus
    Commented Apr 19, 2022 at 21:16

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