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!

  • 1
    $\begingroup$ 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 $\endgroup$
    – mcodesmart
    Oct 8, 2013 at 23:03
  • $\begingroup$ 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. $\endgroup$ Oct 8, 2013 at 23:31
  • $\begingroup$ 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)? $\endgroup$ Oct 8, 2013 at 23:38
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/58154/2451 $\endgroup$
    – Qmechanic
    Oct 8, 2013 at 23:58
  • $\begingroup$ @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. $\endgroup$
    – N. Virgo
    Oct 9, 2013 at 1:08

1 Answer 1


Assume that you jump straight up, standing on the equator.

As soon as your feet leave the ground, you are in a highly elliptical orbit around the center of the earth. At that point you have the same angular velocity as the point you jump from. As you rise toward your one and only apogee, conservation of angular momentum requires that your angular velocity decrease very slightly, then increase again as you drop. Of course, the orbital motion will stop when you hit the ground again. Overall, the ground will have rotated slightly further. I doubt that the effect could ever be measured.

At the poles, you land where you jumped from. Mid-latitude jumps are left as an exercise for the reader...


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