If I were to jump one meter in the air and hang for one second, would I fall back down in the same spot or would the earth rotate ever so slightly under me, causing me to land a short distance away from my original point of departure?
I am conflicted on this. If I look at the equation for angular velocity, I see that $w = v/r$ where $w$ is the angular velocity, $v$ is the linear velocity, and $r$ is the radius of the object I am on (in this case the Earth). There is another version of this stating that $w$ = $2\pi/t_{rev}$ where $t_{rev}$ is the time it takes to complete one revolution.
Being on the Earth, I have a certain linear velocity, $v_{G}$. When I jump in the air 1 meter, I am not applying any force except for vertically so I do not believe my linear velocity would change. However, I am increasing the distance I am from the center of the earth so now my angular velocity would be $w_{A} = v_{G}/(r+1)$. Therefore, it seems that my angular velocity in the air ($w_{A})$ would be slightly less than my angular velocity on the ground ($w_{G}$). If I were to recalculate the linear velocity given this discrepancy in $w$, I would get $v_{G}=w_{G}r$ and $v_{A}=w_{A}r$. This shows I get a small difference unless the radius in $v_{A}$ must account for the jump ($r+1$), in which case I end up with $v_{A} = v_{G}$.
I've gone around in circles trying to decide if one would actually move. This top answer seems to think you would: Earth moves how much under my feet when I jump?
Any insight would be appreciated. Thank you!