I have a random question that I was thinking about. Consider there are two people $P_1$ and $P_2$ facing each other in Outer space. Person $P_1$ is spinning around a point $Q$ at a constant rate relative to $P_2$ who is completely stationary. Now, imagine you are person $P_1$. You take a penny out of your pocket and simply release it in front of your face and watch it continue to be at eye level because of your reference frame. Being person $P_1$ means you can simply release the penny like this because you are in Outer space. However, person $P_2$ is watching this from a different reference frame. Now, imagine you are person $P_2$. Person $P_2$ must see the penny continue to rotate around $P_1$ right at eye level. Thus, the penny must act as though it is in a circular orbit. How could the penny act in a circular orbit though when there are no forces acting upon it? It should appear as though it is moving in a straight line from $P_2's$ perspective. What actually happens here?
3 Answers
Since person $P_1$ is rotating around point $Q$ there has to be a centripetal force acting on $P_1$, constantly pushing $P_1$ towards $Q$.
Therefore, if the force is somehow only affecting $P_1$ the person, when the penny is released, it'll just coast with the instantaneous velocity that $P_1$ had at the same moment the penny was released.
But thinking about the scenario more realistically, the agency responsible for the force on $P_1$ should also in general be affecting everything in $P_1$'s vicinity, physical forces don't magically pick a specific person or random object to act on. They arise due to coupling with some physical property like charge, current, mass (if you're doing Newtonian gravity at least), etc. Of course, we could just say that $P_1$ is tied by a physical string that's fixed at one end to point $Q$, then the penny will indeed coast away as stated.
But either way, ultimately it will require specifying what precisely is the type and nature of the force that's holding $P_1$ in orbit, to determine what will happen to the released penny.
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Give accelerometers to both $P_1$ and $P_2$. At least one of them must have an accelerometer that reads some non-zero value.
Any observer whose accelerometer reads a non-zero value is a non-inertial observer. Their rest frame is non-inertial and includes fictitious forces. Should they release a penny it will accelerate relative to their frame, subject to the fictitious forces.
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OOps - I have described everything assuming P2 is rotating, not P1 as you stated. So now reversed all my references between P1 and P2.
P1 is in a non-inertial frame. That should be immediately obvious to them for many reasons. If P2 is not positioned on P1's axis of rotation, then P2 will appear to be rotating around P1 (according to P1's point of view). Any other object will also appear, from P1's perspective, to be rotating around them. More distant objects, for example nearby planets or stars, will likely appear to be exceeding the speed of light in their rotation. P1 will also notice that if they release a coin it will appear to fly away from them. If they hold their arms out then, depending on the rotational speed, they may feel an outrwards pull on their arms. P1 needs to understand that the normal rules of $inertial$ frames won't be holding for them.
If P1 holds a penny out and attempts to drop it in a stationary position, it will fly away from them. If they throw it sideways at exactly the right speed, it will seem to them to be orbiting around them (and wil also then appear stationary according to P2).
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