# If I jump will I land in the same spot? [duplicate]

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!

• you are moving at the same speed with the earth, if you want to move, then jump forward. Its the same as jumping on a moving train, you still land on the same spot Oct 8 '13 at 23:03
• That makes sense, but is it because the change of radius is affecting both the angular velocity and linear velocity? ie. $v_{A} = w_{A}(r+1) = (v_{G}/(r+1))(r+1) = v_{G}$ So effectively your angular velocity adjusts to compensate for the height differential? If I just look at $w_{G} = v_{G}/r$ and $w_{A} = v_{G}/(r+1)$, it seems $w_{A}$ would have to adjust for the increased radius. Oct 8 '13 at 23:31
• I understand it might be a duplicate - I linked that question in my original answer. However, the answers to that question did not seem to give a definitive response. If the answer given by Johannes is correct, then I am set. The original poster did contest it though. Edit: In response to your answer on the linked question, the effect is very small for the one second jump. However, if I hovered a helicopter 1 meter above the ground for around 5 hours, would I be able to see a noticeable effect (a little shy of a meter movement, for example)? Oct 8 '13 at 23:38
• Possible duplicate: physics.stackexchange.com/q/58154/2451 Oct 8 '13 at 23:58
• @PrimeWaffle If you hovered in a helicopter (or any other hovering device), you would be exerting an additional force to counteract gravity, the amount you would move relative to the Earth would be highly dependent on the direction and magnitude of this force. Oct 9 '13 at 1:08