“Yes, but…” sounds good to me. That is, in a “perfect system” (a helicopter rises without the pilot needing to compensate for the wind and transition – maybe we could just say a balloon that rises to a certain altitude in a zero-wind situation?) the Earth will not rotate out from under the helicopter (or balloon). The “but” being the need to A) keep the aircraft stable, B) to compensate for drift caused by the anti-torque rotor, and C) zero wind.
Actually, the thrust is to the left, which would push the aircraft to the right. But you knew what I meant. 
You guys are evil.
Things on the earth move at the same speed as the earth’s rotation, obviously. The atmosphere moves with the earth, too, so there won’t be friction between the helicopter (still moving due to the momentum gained on the ground) and the atmosphere.
In other words, yes, the damn thing will stay still relative to the earth AND relative to the air.
with all do respect to participating members to this post, but **Really Not All That Bright ** has the right way that most people would like these questions answered. answer simply. of course you could factor in the wind and how exactly a helicopter works, but i’m sure the poster wasn’t exactly thinking about all of the factors that go into how a helicopter works or how the atmosphere works. you all know what the poster wants, so why not give it to them simply?
i’m really sorry for this, but i have seen too many overcomplicated answers to questions that could easily be anwered in one or two sentences.
i have total respect to everyone here, and if i have done an evil deed by posting this, i am sorry.
I am also sorry for the hijack of this post
Now I’m confused.
Forget the helicopter specifics for a moment (that’s the part I do understand).
To simplify, let’s also assume zero atmosphere.
If an object went from hovering at 10 feet to hovering at say, 50 miles, wouldn’t it begin to drift “backwards” relative to the earth? The earth would continue spinning at the same rate, but the object would begin to drift backwards because its rotational inertia would remain constant, but the size of the circle it inscribes would get bigger. So, like ice skaters opening their arms, it would make fewer revolutions per unit time and would slow down relative to another body spinning at a constant rate… wouldn’t it?
igloorex,
Yes, you’re right. What you’re describing is ORBITAL motion, i.e. what satellites and moons experience.
For them, they start with X velocity tangential to the surface before liftoff. Just like you and I are now hurtling eastward at 500-1000mph depending on our latitude.
As they ascend, the required eastward velocity to remain over the same spot increases, but their own eastward velocity does not, at least assuming the launch is exactly straight up.
So they fall behind, drifting westward relative to the launch point.
Notice though, that the effect is pretty small. Right now you’re 4000 miles fromthe center of the Earth. IOW, you’re in a 4000 mile orbit. If you climb 50 miles, to a 4050 mile orbit, you’ve made a pretty small increase in orbital radius percentagewise. So the westward drift wioll not be huge.
igloorex, what you describe is the Coriolis effect. Yes, there will be a slight drift due to it, but the size of the effect depends on the change in distance from the center compared to the distance from the center. In our helicopter example, the heli might be changing its height by perhaps a mile or two. But this is all happening a few thousand miles from the center of the Earth. A few miles is very small compared to a few thousand miles, so the Coriolis effect will be very small, also.
I always took the “All Stop” command to mean stop all the engines, not the ship. The ship would, I suppose, continue to coast along its present course.
I remember at least a few episodes where the ship encountered som strange object or other in space, the captain would call “Helm, all stop!” and the ship and the object would then be stationary relative to one another. I assume “stop” means stationary with respect to whatever fixed point they measure the ship’s speed against, if that makes any sense.
That’s the leehelm 
Right. The one that’s not on the windward side. Gotcha. Which wind, again?
[hijack=Star Trek] I was under the impression that the warp/impulse engines work by manipulating space (or spacetime, if you must)… so turning off the engines brings you to a stop relative to local space.[/hijack]
The first thing I though of after I got through the “ignoring the wind, how real helicopters work, etc.” stage to the “two perfectly spherical objects in a vacuum, one significantly larger than the other, etc.” stage… I pretty much thought of conservation of angular momentum, like igloorex.
However, I was thinking about it in terms of ballistics (IE the sphere has some inital momentum at Time=0 to get it to some altitude) and I’m not sure if it being maintained at some constant altitute by a force would behave quite the same way…
I recommend that the OP (and everyone else, for that matter) go to the source for the answer to this fascinating question: Galileo’s Dialogue Concerning the Two Chief World Systems. (I like Stillman Drake’s translation.)
The question of relative motion is discussed in the Second Day of the dialogue.
Seriously, Galileo is a great read. This book, like several others of his greatest hits, is in the form of a discussion between three characters, one of whom presents Galileo’s views, another of whom (Simplicio) presents the arguments of his opponents and the common “wisdom” of the time. It’s great fun to see Simplicio’s arguments reduced to shreds by the eminently sensible Salviato.
Do not be intimidated. Galileo wrote for the average man, not for specialists or scientists. It’s not hard to understand, and you will enjoy reading it. Watch modern science beting invented right in front of your eyes!
No, it would begin to slow down, until you turned the engines back on.
[I learned me my interstellar physics from Battlestar Gallactica. Does it show, at all?]
Well, it would slow down. Gradually.
Clearly we’re confounding several quite different scenarios. This is fine, since we’re just exploring the physics, but I think we should just go ahead and reduce each to its simplest case to make the factors easier to enumerate.
1. No atmosphere, no gravity effects, no launch
For the sake of argument let’s just assume a massless planet and a massless craft. That may seem far-fetched, but it frees us of having to worry about the possible physics of some imaginary antigravity technology.
Result: if the planet rotates, it rotates “out from under” the craft. There is no appreciable force to convey the motion of the planet to the craft.
2. No atmosphere, no gravity, craft launched from surface
At this point, we have to assume some mass in order for momentum to have meaning, and for trajectories to be predictable. Let’s say the planet has a mass of 1 attogram, so its influence on the craft’s mass/momentum is negligible
The craft would have the instantaneous tangential momentum it had at the instant of launch (let’s say the engine is programmed to precisely cancel whatever force it applies at launch with exact mirror forces at its destination altitude). but the momentum vector of the launch point is constantly changed by the mechanical strength of the planet itself – no corresponding force is acting on the craft.
Result: the path of the craft will be a straight line. It may have a velocity vector of zero, relative to the center of the planet, but that’s still just a special case of a “straight line”. The path of the point under craft will be a circle. The two paths will inevitably diverge.
3. No atmosphere, some gravity/mass
Now the craft is no longer travelling a straight line. Gravity will bend that path into some trajectory-- circular, eliptical, parabolic or hyperbolic, depending on the rotation rate of the planet, and the initial and final radial distance of the craft from the center of gravity of the planet.
However, a new distinction arises: the craft “orbits” due to forces acting [effectively] from center of the planet, but the point on the surface will rotate around the axis of the planet – or if you prefer, it will rotate around the point on the axis closest to it. Unless the craft is launched from the equator, the two motions won’t even be in the same plane. The launch point will orbit in a plane that is perpendicular to the axis, while the craft will orbit in a plane that contains the line connecting it to the center of the planet
Result: except in the one special case of a geosynchronous orbit for a craft launched from the equator, the craft will not hover over its launch point.
4. “ideal” (simple) atmosphere
The craft will remain approximately over the point of launch, but there will be small ancillary effects. The above scenarios prove that this is not due to any force or momentum imparted by launching from the rotating planet, but due to the force of the ‘ideal’ rotating atmosphere that keeps perfect pace with the planet.
This image may help: Imagine a balloon tethered to the ground high over a windless desert (for the sake of argument all air currents are directly upwards). For there to be a displacement, we need a force that we can express as a lateral tension on the tether.
As a practical matter: if a helicopter tried to deliberately hald the same position (relative to the stars), as the earth turned under it, it would experience an airspeed of up to 1000 mph at Earth’s equator. More precisely:
[b*]effective airspeed = (planet circumference/rotation period) * cos(latitude)***
Its airframe and rotors wouldn’t survive, unless it was very near the poles.
For the sake of argument, let’s assume an isolated planetoid in deep space, so we can disregard the velocity of planetary revolution around its sun. Immobility relative to the stars (via active stationkeeping) is the closest we can realisitically come to the geometry of Scenario #1 (no atmosphere, no gravity)
5. Real World
At this point, atmospheric effects (weather, wind currents, etc.) outweigh all other lateral effects like tthe Coriolis Force. However, Coriolis forces are quite critical to maintaining the atmophere in its general pattern of motion
While thinking about this, I stumbled across a formulation of the Coriolis force that I’d never considered before: The Coriolis Force can be expressed as the resolution of the difference between atmospheric rotation (imposed by friction with the planet) and the (conflicting) orbital paths (on different planes) that the mass and velocity of the air at each point would otherwise impose.
The math on that quickly gets too hairy for me to casually do on he back of an envelope, but I’m curious if that formulation is ever mentioned in computational meteorology, where tehy don’t shy away from heavy math – or if, perhaps, it’s known to be completely wrong, maybe there’s some other term or force that encapsulates the resolution of those competing influences.