Boost Your Knowledge on Frame of Reference

The reason the galaxy does not spin around me (relative to me) at a trillion times the speed of light when I do a pirouette is the inertial frame of reference to which I am confined along with the rest of humanity.
Where does this frame of reference end?
To the best of my limited understanding, gravity doesn't just stop at some point. Anything that has mass has gravity; gravitational waves (or whatever) extend from the mass spherically. They weaken as they get further away from the source, they get pushed around by other, more powerful, gravity waves, they become progressively weaker as the distance increases, but do they ever vanish completely, i.e. get reduced to absolute zero? Or do they become weaker and weaker forever, without zeroing out? And if so, where does a frame of reference end?

Suppose we are given a mechanical frame consisting of two points. How can we prove that assuming any initial conditions there is an inertial frame of reference in which these points will be in a static plane?

Can the astronaut tell at what speed is he moving? and how do you define speed in that case when you have no outside references. Can you refer to the speed relative to the empty space?

A reference frame on a rotating wheel is not an inertial frame of reference because it is not homogenous (depending on where we are, the body feels different inertial forces acting on them).
Could you provide me a similar example of a reference frame that is not isotropic?

Whether Compton effect depends on the frame of reference or not. The wavelength change is a thing, but does not depend on frame of reference since it depends only on θ. But if I travel at a speed comparable to velocity of light, will there be any changes?

A local inertial frame of reference rocks. A little later it is said that Lorentz frames rock. Are they identical or is one a special case of the other?

This is the common problem of a charged particle moving in a static electric and magnetic field.
Say $\mathbf{\text{E}}=(E_x,0,0)$ and $\mathbf{\text{B}}=(0,0,B_z)$
In the inertial frame of reference, the equation of motion is (1):
$\frac{d\mathbf{\text{v}}}{dt}=-<!--\; -\; -->\frac{q\mathbf{\text{B}}}{m}\times <!--\; \times \; -->\mathbf{\text{v}}+\frac{q}{m}\mathbf{\text{E}}$
We can find equations for $v_x$ an $v_y$ and see that the resulting motion is a circular orbit with a constant drift velocity $v_d=\frac{E_x}{B_z}$.
Surely I should get the same answer if I solve the problem in a rotating frame of reference?
I know that (2):
$\frac{d\mathbf{\text{v}}}{dt}{|}_{Inertial}=\frac{d\mathbf{\text{v}}}{dt}{|}_{Rotational}+\mathit{\omega <!--\; \omega \; -->}\times <!--\; \times \; -->\mathbf{\text{v}};$
If I use Eq. (1) as the LHS of Eq. (2), and choose $\mathit{\omega <!--\; \omega \; -->}=-<!--\; -\; -->\frac{q\mathbf{\text{B}}}{m}$, then I get (3):
$\frac{d\mathbf{\text{v}}}{dt}{|}_{Rotational}=\frac{q}{m}\mathbf{\text{E}};$
How do I obtain a constant drift velocity (as mentioned before) from this? Have I used any formula incorrectly? Does the electric field E=$\mathbf{\text{E}}=(E_x,0,0)$ change form in the rotating frame?

"If the frame of reference is translated or rotated, the vector doesn't change."
Although the length of the vector won't change, the angle that this vector makes with the positive direction of the newly defined x-axis changes, no? Hence, how is it possible to state that the vector doesn't change? Have I misunderstood the basic definition of a vector?

Can you identify the direction of your movement? Let's say ship moving nose forward, or tail forward? Is there a sense of "direction of movement" in such an inertial frame of reference?

I have one vector and its components given a particular frame of reference ($A_x,A_y,A_z$). I have a second vector's components that is computed in a different frame of reference ($B_x,B_y,B_z$) and I need the dot product between these two vectors. I can't simply take the dot product (i.e.,$\mathbf{A}\cdot <!--\; \cdot \; -->\mathbf{B}=A_x{B}_x+A_y{B}_y+A_z{B}_z$) because the two vector's components are computed with different orientations in mind.
Given that there is no way to orient one vector to match the frame of reference of the other vector, is there a way I can still get a dot product?

"Strictly speaking, Newton’s laws of motion are valid only in a coordinate system at rest with respect to the 'fixed' stars. Such a system is known as a Newtonian, or inertial reference frame. The laws are also valid in any set of rigid axes moving with constant velocity and without rotation relative to the inertial frame; this concept is known as the principle of Newtonian or Galilean relativity."
Why should the inertial frame of reference not spin?

I was studying about potential energy, and I suddenly thought that is there any relation between potential energy and frame of reference?
For example, we say for an object raised to a height $h$, potential energy is $mgh$. But this is so when we are talking about its distance from Earth. If a person is holding a suitcase then for the person shouldn't be the P.E. of suitcase be zero?

A ball is thrown at a relative velocity of $3m/s$ in $+ve$ $x-<!--\; -\; -->$ direction in a rocket in gravity free space. The rocket has a constant acceleration of $2m/s^2$ also in $+ve$ $x-<!--\; -\; -->$ direction.
Whether uniform acceleration equations of motion would apply in above case when using the rocket as a frame of reference. As an example, would the equation $s=ut+.5a{t}^2$ apply if rocket is taken as frame of reference?