I know there are similar questions on StackExchange but I think it is different and detailed.

The earth is spinning 465 meters/second so why don't we jump and land on a different location?

I googled about this question and I got some answers:

  1. When we jump, we (and the atmosphere itself) also spin along with the earth so we don't land at a different location. (But why we are also spinning along with earth ?)
  2. The earth is so big and we are very small relative to earth so the tiny jump won't make any difference.

But according to answer 1, We are also spinning with the earth. But the question is, "why are we also spinning along with earth" ??

Regarding the answer 2, it won't make any difference but logically it doesn't make sense or I just don't understand it.

Here is the followup question,

"When we jump why don't we thrown out of earth due to centrifugal force of the spinning earth ??"

Please provide some detailed answer to these questions. With logic if possible.

This phenomenon is not possible I know, but even if it exists, then traveling will be so easy, we just have to hang on space (with help of helicopter etc.. ) and we can travel the whole world in 24 hours :D

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    $\begingroup$ short answer: your feet touch the ground, i.e. they have a horizontal velocity = to the Earth's rotation speed. When you jump, you still have this component and you therefore move in the same direction as the Earth. $\endgroup$ – SuperCiocia Sep 25 '14 at 11:44
  • $\begingroup$ Well technically air resistance should slow you down a bit, so you might end up a very small distance before the place you jumped from. But then again, air is moving with the Earth as well, so it gives you a forward push... I don't know how these two effects balance $\endgroup$ – SuperCiocia Sep 25 '14 at 11:46
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/48287/2451 , physics.stackexchange.com/q/80090/2451 , physics.stackexchange.com/q/1193/2451 and links therein. $\endgroup$ – Qmechanic Sep 25 '14 at 12:19
  • 2
    $\begingroup$ Subquestions effectively don't count, since people often only read the title. Besides, one should only ask one question per post. $\endgroup$ – Qmechanic Sep 25 '14 at 12:21
  • $\begingroup$ I've worked this out in some detail here: physics.stackexchange.com/questions/227391 $\endgroup$ – user854 Jan 4 '16 at 15:55

Actually, you do land a little west of where you jumped. However, that distance is so miniscule for a human jumping that you don't notice it.

The reason something projected straight up doesn't fall back onto the same spot is that its radius from the center of rotation increases as it gets higher. Initially, the projectile and the surface of the earth are moving at the same speed horizontally. As the projectile goes higher, it moves further from the center of rotation and would have to move faster laterally to keep above the same spot on the ground below it. It doesn't, so appears to move west to a observer fixed to the ground. Note that this effect is proportional to sin(90° - latitude). If the projectile was launched straight up at either pole, it would fall back to the same spot with the same orientation.

For a human jumping, the radius change is so small and the flight time so small that the effect is so tiny as to be swamped by many other sources of errors.

  • $\begingroup$ 'As the projectile goes higher, it moves further from the center of rotation and would have to move faster laterally to keep above the same spot on the ground below it. It doesn't, so appears to move west to a observer fixed to the ground.' What happens when the projectile falls back? Isn't it the opposite, the projectil comes back to his start point? $\endgroup$ – HolgerFiedler Sep 25 '14 at 15:03
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    $\begingroup$ As it climbs, the velocity difference between the object and the surface increases. When it descends, the velocity difference returns to zero. But it never gets the displacement back (unless it could descend underground). $\endgroup$ – BowlOfRed Sep 25 '14 at 16:07
  • $\begingroup$ @Holger: As the projectile falls back down, the discrepancy between its lateral speed and the ground's lateral speed decreases again, but the discrepancy still exists until the projectile hits the ground. The west offset distance is the time-integral of this lateral velocity discrepancy. The down portion of the flight contributes the same west offset as the up portion. $\endgroup$ – Olin Lathrop Sep 25 '14 at 16:43
  • $\begingroup$ Could you have attention on this question, please and perhaps is it possible for you to give an answer for this question? $\endgroup$ – HolgerFiedler Oct 4 '14 at 5:27
  • $\begingroup$ Just to add, this isn't merely a theoretical issue - I gather that long-range artillery - eg naval gunnery - trajectory calculations do have to take this into account. $\endgroup$ – peterG Mar 29 '15 at 14:05

As in the comments, because you are initially stationary relative to the Earth's surface, your initial velocity is exactly the same as that of the ground. The reason why is friction and air resistance: if you weren't so (perhaps you'd just dropped in from space, maybe from Betelgeuse Seven to warn Arthur Dent of a disaster in the offing, and you hadn't calculated your Earth spin velocity quite right), you'd be dragging along the ground and you'd also feel a strong headwind against your velocity (as you move relative to the atomosphere). These forces would only cease when you were moving along with the ground, i.e. stationary relative to the ground. So when you jump off the ground, your horizontal velocity component does not change, and so you track alongside the point you pushed off. It's much like two spaceships in orbit initially flying side by side with a spring between them. Imagine that one pushed off the other, and the spring thereafter brought them back together. Their horizontal velocity components don't change over the course of the thought experiment.

As for the tiny jump relative to the big Earth bit, I think you mean something like the following. On the ground, our freefalling acceleration towards the Earth centre owing to gravity alone is $g$, and the ground resists this by thrusting upwards on us. This is an accurate picture for a stationary Earth. Now for a spinning Earth: to accelerate us on a circular path so that we stay stationary relative to the ground calls for an acceleration equal to $\omega^2 \, r$, where $\omega$ is the angular speed of the Earth and $r = R_\oplus cos\theta$ our orthogonal distance to the axis of Earth rotation, $R_\oplus$ the Earth's radius and $\theta$ our latitude. Therefore this acceleration at the equator is:

$$\omega^2\,R_\oplus = \left(\frac{2\pi\text{ rad}}{24\times 3600s}\right)^2 \times 6.4\times 10^6m$$

or about $3.5\times 10^{-3}\,g$. Thus, this reduces the force that the ground must push us up with by about 0.35% - it is nothing like what is needed to reduce this force to nought. Another possible meaning is that our escape velocity i.e. the upwards velocity that would make our kinetic energy equal to the shift in gravitational potential energy $G\,M_\oplus\,m/R_\oplus$ (where $m$ is our mass) needed to reach a point infinitely far from Earth is about 11 kilometres a second. Pretty impressive compared with any plausible "jump" velocity.

  • $\begingroup$ so when I jump off why is my horizontal velocity the same? earth is not dragging me with its spinning any more right? my feet are off the ground $\endgroup$ – m4ngl3r Mar 17 '17 at 8:34

IT's not particularly noticeable when you jump, because you didn't get very high. But if you, say, shoot a model rocket a few hundred meters up, barring interactions w/ the air, it'll land to the west of its launch point. For an analogy, suppose you're in a car going at constant angular speed in a circle of some radius. If you then run right behind the car at the same angular speed (and linear speed of course), you'll keep up. But if you move to a larger radius without changing linear speed, you'll fall behind because the circumference is greater.

  • $\begingroup$ Noticeable point ... $\endgroup$ – Bhavesh G Sep 25 '14 at 12:10
  • $\begingroup$ It is not noticable, because air is moving 200 mph at higher altitudes, so that should be the case $\endgroup$ – m4ngl3r Mar 17 '17 at 8:24

When planes fly in the west direction they are basically doing what you suggest: the position of the Sun stays almost fixed and the Earth rotates below them. Still to do that they burn a lot of fuel: first they need to come at a stop with respect to the Sun, which means getting some speed with respect to the Earth, then they have to keep this speed winning the air resistance and preserving the altitude.

As when you jump you do nothing of the above, that's why you land in the very same position.


If you jump insite a train (which is not accelerating nor decelerating at this moment) where do you land? At the same position. Why? Because to be decelerated or to be accelerated every body needs a force which acts on him. Without force no change of the bodies velocity. Why you will be dcelerated when you jump out from a (slow :-)) moving train? Becaus the air is the force that will stop you.

The big idea of Newton was it to understand - against the everyday intuition - that without friction body will moves with his given velocity without ending.

I have to expand the answer a bit.

Imagine there is a tower hundred meters high. Will the tower feal a acceleration from the earths rotation? The answer is yes and we can proof this by "disconnecting" the head of the tower from the "rest" of the tower. Depending of the high of the tower the head will rise higher (for very high towers, there the centrifugal force is higher the gravitational force) or fall down (for not so high towers and of course if the placed the head nearby the rest). The rising head goes slower during his rise, a falling head goes faster. But if you influent of the rising head and push him back with a force that is directed to the centre of the earth, the head returns to the tower in the same position from which it starts!

But there is one point more. A rotating mass influent on the gravitational potential. This potential is twisted through the rotating body. And because this body is not alone in the space other masses influent too. Than higher you jump than more you "feel" the other masses and you will get less rotated. Jump on a mass without atmosphere and you will land in some distance from your start point.

  • $\begingroup$ Except that you are wrong. $\endgroup$ – Carl Witthoft Sep 25 '14 at 11:58
  • $\begingroup$ @CarlWitthoft's comment was before my edit. $\endgroup$ – HolgerFiedler Sep 25 '14 at 14:02
  • $\begingroup$ But it's still wrong. I disagree with each paragraph. $\endgroup$ – Carl Witthoft Sep 25 '14 at 14:42
  • $\begingroup$ "The big idea of Newton was it to understand" Galileo had it first. Newton worked out a proper mathematical formulation for dynamics, but relativity and inertia were both already known. $\endgroup$ – dmckee --- ex-moderator kitten Sep 27 '14 at 18:58
  • $\begingroup$ when you jump in the train, you should land on another place... like in the train, so on the earth so therefore earth is not spinning $\endgroup$ – m4ngl3r Mar 17 '17 at 8:35

we move along with the earth because when we jump, our velocity is same as that of earth. But due to air resistance we will land away from our original position but the difference is negligible.


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