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Let's assume that there was some mechanism by which we could remain suspended in air. By this I mean that our feet is not in contact with the ground. One possible way of doing this would be by means of a jetpack. If we could remain suspended in air for a while in this manner, will we be in a different place when we come down, due to the Earth's rotation? When I thought of this problem in the micro scale using a ball (which rotates), it appears like I'll be in a different place when I come down from my suspended state. Am I overlooking something here?

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maybe you should read up on the Foucault pendumum : en.wikipedia.org/wiki/Foucault_pendulum –  anna v Feb 3 '12 at 6:21

4 Answers 4

If you're standing on the equator then you're moving, along with the earth's surface, at about 1,000 miles per hour. So if you fly up e.g. 100 meters with your jet pack you're still moving sideways at the same speed so the spot you took off from remains below you.

Well, not quite. Because you're 100m above the surface you're moving in a circle that has a 100m greater radius than the Earth, so it has a bigger circumference than the earth's surface. That means that the earth will rotate slightly faster than you will, so you'll see your takeoff spot gradually move away to the East. If you're a distance $d$ meters above the surface your orbit is $2\pi d$ meters larger than the Earth's surface, so in the example above your take off spot will move away at about 630m a day or about 26m per hour. The chances are that the winds or random movements of your jetpack would be more than this so you probably wouldn't notice.

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So let's say we're on a planet surface where there is no atmosphere. This is a bit of a tricky proposition, because you need thrust to keep from falling. If you began from the surface and simply produced thrust upward, then your trajectory would not be "straight up" - true. But for a planet surface with the atmosphere, the gas already has that correction made. Aside from weather, the Jovian gas moves at a rotational speed that the radius would predict. I mean, the velocity vectors are the same as if it were solid. –  Alan Rominger Feb 2 '12 at 19:06
@Zassounotsukushi: You can have rocket thrust without an atmosphere, that is how most spacecraft manoeuvre in fact. In any case, it is not the atmosphere that Rennie is concerned with when he states that you are still moving sideways at the same speed so the spot you took off from remains below you. He is concerned with your initial sideways velocity, imparted upon you by the ground your were standing on before launch. Likewise, if you throw a ball out of a car window perpendicular to the car, the ball will move at an angle to the ground, its forward component equal to the speed of the car. –  dotancohen Feb 3 '12 at 2:38
@Zassounotsukushi: in general planetary atmospheres do not move at the same speed as the surface of the planet. Have a look at the Wikipedia article on Jupiter's atmosphere. On Neptune the max wind speed is 2,100km/h! Even Earth has the jet stream that circles the Earth at 200km/h. –  John Rennie Feb 3 '12 at 7:06

The air in the lower layer of the atmosphere is dragged by the Earth with a velocity that is pretty much the same as that of the surface of the Earth, otherwise there would have been ~400 m/s winds at the Earth surface. However, on the average, there is some difference between the velocity of the air in the lower layer of the atmosphere and the surface of the Earth, as exemplified by the so called trade winds. So yes, if suspended in air, you'll move to a somewhat different place (on the average).

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When the helicopter starts out, it's sitting on the ground and the ground (being part of the Earth) is rotating at one revolution per day, as we know. Since the helicopter is also sitting on the ground, it's also inside this moving reference frame, and has the momentum that goes with it so the helicopter is also moving at one revolution per day. In fact, so is the air! Now, when the helicopter takes off, it flies straight up to some height above the Earth's surface. But though the helicopter has exerted a force (through the use of its rotors) to lift it straight up, it hasn't exerted a force in the horizontal direction to counter the motion (momentum) it already had that one revolution per minute! So though the helicopter is no longer touching the ground, unless the pilot purposely exerts a force against the helicopter's initial momentum, the helicopter will continue to move at one revolution per day, and thus remain above the same spot on the Earth's surface from where it took off. The momentum that the helicopter started with is the same as what it ends with that's conservation of momentum! The same is true on a smaller scale when you jump in the air if you jump straight up, you'll land exactly where you started, because in every other direction (except up and down), your momentum is the same (try it out the next time you're on a plane: the plane acts like a miniature version of the Earth, and when you jump, you land right where you were, even though the plane's going 500 miles an hour!). On a larger scale, rocket scientists have to account for the motion of the Earth before they launch a satellite. In order to put the satellite into a specific orbit, they can't just shoot it straight up from the Earth's surface. They have to apply horizontal forces as well, in order to counter the Earth's rotation and get the satellite into the correct orbit.

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There are different ways of "suspenending yourself in air". The main question is, what are you suspended relative to? If it is the air you are suspended relative to, then you will obviously move with the wind/rotation of earth. On the other hand, if you are using a jetpack, you will keep the tangential component of your velocity as you go up. Thus you will rotate slightly slower than the earth, and on landing you will find yourself in a different place.

Note that I neglected centripetal forces (which one can't do). Actually, you may rotate at different rates, depending on the force you apply through the jetpack. Assuming that you stay at the same height above the earth, changing your jetpack force will slow down your rotational speed as well, as $\frac{m\omega^2}{r}=mg-F_{jetpack}$. Then it really depeds upon your mechanism.

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