The pseudo force depends only on the acceleration of the observer frame (which is why if the acceleration of the observer is zero (in the inertial reference frame case) there is no pseudo force.)

Therefore, given that the acceleration of the observer is $A_2$ the pseudo force acting on any body (which has mass $m$) observed will be $-m \cdot A_2$.

**No**, we do not take the relative acceleration of the observer and the body observed but the acceleration of the observer as measured by any inertial frame.

Let's look at the definition:

An inertial frame of reference in classical physics and special relativity possesses the property that in this frame of reference a body with *zero net force* acting upon it does not accelerate; that is, such a body is at rest or moving at a constant speed in a straight line. An inertial frame of reference can be defined in analytical terms as a frame of reference that describes time and space homogeneously, isotropically, and in a time-independent manner. *Conceptually,* the physics of a system in an inertial frame have no causes external to the system

Italics mine.

The crucial word is *conceptually*. It carries after it the whole concept of measurement, and physics is about experiments and measurements, and the theories and definitions are tools to describe and then mathematically model the observations, so that one gets a predictive theory.

Measurements come with experimental errors, and thus how complicated the theoretical model one is using depends on these errors. One takes the simplest assumptions, it makes no sense to use the galactic reference frame ( we are also rotating around the galactic center) when measuring a force on bodies on earth, and also the measurement will depend on our measuring instruments.

For example, for usual engineering uses we accept that the earth is flat, the errors of the curvature of the earth to the details of a building are so small that they are within measurement errors.

We accept that the earth is rotating, a non inertial frame, when calculating the coriolis force and the distances planes travel etc, because there the force from the rotation effect is larger than the instrument errors.

So it depends on what you are measuring, whether you can use/assume that the earth is in an inertial frame within measurement errors or not. It depends on the problem at hand.

## Best Answer

The way to properly answer your question is, for lack of a better word, by enumeration. Consider all the ways the Earth moves, compute the fictitious forces corresponding to them and check that they are negligible for the problem at hand. "All" is a strong word, but we can do quite well by looking at what goes into the expressions so that we can tell which ones are the most relevant.

The various motions can be all interpreted as compositions of rotations, so we can use the expression listed at the end of this Wikipedia section: the three kinds of fictitious force acting on a body of mass $m$, moving at velocity $v$, at a distance $r$ from the origin of the rotation (which happens at angular velocity $\Omega (t)$) are

I'm saying "maximum" so we have an upper bound, which is attained when the vectors appearing in the vector products are perpendicular. Now, what is left is to take the system at hand, estimate the forces, and finally see how they compare to the precision of your experiment.

As an example, consider the rotation of the Earth around its axis. An upper bound for $r$ is the radius of the Earth, $r \approx 6400 \mathrm{km}$. The angular velocity is typically $\Omega \approx 2 \pi / (1 \mathrm{d}) \approx 7\times 10^{-5} \mathrm{Hz}$. The typical value of the variation of $\Omega$ is harder to estimate - it fluctuates a lot, from a quick look at the data here it seems like the variation of the length of a day to the next can be of the order of $100\mathrm{ms}$ at most. This relates to a variation of angular velocity of the order of $ 2 \pi (100 \mathrm{ms}) / (1 \mathrm{d})^2 \approx 10^{-10} \mathrm{Hz}$, therefore the quantity $\dot{\Omega}$ will be $\approx 10^{-10} \mathrm{Hz / d}$.

The velocity is dependent on your specific experiment: for concreteness' sake, let us assume we are considering things moving at $10 \mathrm{m/s}$ at most (as much as a fast cyclist, for instance). With these numbers, let us compute accelerations instead of forces (removing the $m$, so that our considerations generalize for any mass):

This procedure can be applied for any other kind of rotation you are worried about --- for example, the rotation of the Earth around the Sun will have $\Omega \approx 2 \pi / (1 \mathrm{yr})$ and $r \approx 1 \mathrm{AU}$; this leads to a Coriolis force of the order of $\approx 4 \times 10^{-6} \mathrm{m/s^2}$ and a centrifugal force of the order of $\approx 5 \times 10^{-3} \mathrm{m/s^2}$.

From this analysis we can already tell that the most significant force to consider is the centrifugal one from the Earth's rotation, followed by the centrifugal one from the Earth's revolution, et cetera. If you had large velocities (for example, if you were doing ballistics) then the Coriolis contribution might be the largest.

So, take the experiment you want to perform, make some Fermi estimates like these, and see whether you can experimentally detect these effects! If it turns out you can, you should look at how exactly they affect your object since they are quite different: the full aforementioned expression shows how to determine their direction as well, and sometimes you might not be able to disentangle their contribution from something else. For example, if you are on the Equator the centrifugal force points radially outward, so it is equivalent to a decrease in the effective gravity at that point.

On the other hand, the Coriolis force deflects the motion of a particle, so it is easier to detect - it is the driving principle behind the Foucault pendulum. This experiment also illustrates the point that, while the accelerations are generally small, if you let them act for a long time their cumulative effect can be rather large.

This procedure (estimating the typical size of the corrections) is quite general: it can also be applied to check whether it makes sense to use Newtonian mechanics in your case, or if you should account for general relativity, quantum mechanics or some other more complicated theory.