Can a frame still be an inertial frame if its center varies with time relative to a “true” inertial frame?

Question

Picture a seat on a Ferris wheel. Neglecting any rocking, is the seat of a Ferris wheel an inertial frame?

My guess is that yes it is right? The frame itself isn’t rotating or accelerating relative to the center (where a “true” inertial frame can be placed). The XYZ vector basis of the chair will still be parallel to the XYZ vector basis of the center, and the distance between the chair and center is also constant; thus, the chair's frame is neither accelerating nor rotating.

I am asking this question in order to get an analogy for the Earth J2000 frame (which is inertial relative to the stars) and the sun

Answer 1

Ok I think I answered my question.

No Ferris wheel seat is not inertial. If you add a rotating intermediate frame and perform the transport theorem you find that it will not be inertial.

For the earth J2000 equation, if you perform a similar analysis, the Earth fixed frame is not inertial. However, you can perform vector addition from the inertial point to Earth and then from Earth to the object of interest, to find the vector from inertial point to the object of interest. Apply the transport theorem to these and then you can find the inertial properties of the object relative to Earth.

In other words, earth J2000 is just a coordinate system, but all operations should still be performed inertially.

Answer 2

A Ferris wheel goes in a circle, so (neglecting any linear acceleration of the entire apparatus*) it is an accelerated frame with an acceleration of $\vec a(t) = \dfrac{-\gamma^2 v^2}{r} \hat r \approx \dfrac{-v^2}{r}\hat r$ for $v \ll c$, where $v$ is the relative velocity to the center of rotation. That nonzero value is what an accelerometer on the Ferris wheel would read, adjusted for gravity.

Accelerated frames are not inertial frames. However, for most measurements, the acceleration of a Ferris wheel at real Ferris wheel velocities and sizes is so small that it can be reasonably approximated as zero for most measurements. Certainly the difference in acceleration between the Ferris wheel frame and the stationary-with-respect-to-Earth frame at Earth's surface is negligible.

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*- for instance, because the whole machine is attached to the ground, which applies a force equal to its weight, accelerating the whole machine including cars and passengers at $g$ relative to an inertial (falling) frame.