The centripetal force on Earth is constantly exposing Earth to the acceleration. Why can't we feel this change of direction?
2 Answers
At the equator, when the effect is the largest, we should feel a centrifugal correction of: $$a=\Omega_\oplus^2R_\oplus\approx\left(\frac{2\pi}{86164\,\text{s}}\right)^2\cdot6378\,\text{km}=0.0339\,\frac{\text{m}}{\text{s}^2}\approx0.00346g$$ This $0.3\%$ correction to $g$ is measurable, but imperceptible.
The centripetal force acts perpendicular to the surface that's why it just changes the direction.
Now you must be talking about centrifugal force, which is responsible for the acceleration you feel while on marry go round,
The centrifugal acceleration is $$a_{cfg} = \omega^2 r$$
Now taking $r=6400000$ m $\omega= 2\pi/ T$ we get $$a_{cfg} =\left(\frac{2\pi}{60×60×24}\right)^2×6400000 = 0.0338\ m/s^2$$
While the acceleration due to gravity is $ g= 9.8\ m/s^2$ completely overshadowing this effect.
If there were no gravity you would have surely felt it, even now you feel it, but you can't distinguish it from gravity as both are fictitious force and have same acceleration for every mass,
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1$\begingroup$ I'm pretty sure gravity isn't a fictitious force $\endgroup$– DanDan面Commented Apr 5, 2023 at 3:26
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$\begingroup$ Does it have all the properties of a fictitious force?? Does it vanish in the inertial frame of reference? $\endgroup$– user324939Commented Apr 5, 2023 at 3:28
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$\begingroup$ Well, for one, it has an action-reaction pair, which fictitious forces lack $\endgroup$– DanDan面Commented Apr 5, 2023 at 3:29
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$\begingroup$ Every fictitious force is same as gravity, and you can't distinguish it by any experiment from a fictitious force. That's equivalence principle . $\endgroup$– user324939Commented Apr 5, 2023 at 3:32
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$\begingroup$ That is true if we're working in GR, but I think OP is working in the Newtonian framework, where gravity is modelled by a real, rather than fictitious force $\endgroup$– DanDan面Commented Apr 5, 2023 at 3:34