Does Position Change With the Earth's Rotation?

Relative to what? If the helicopter stays in the same spot relative to the universe, when it lands the earth will be long gone on its orbit around the sun.

If you fix your position relative to the azimuth of the sun, then yes, the earth will have rotated beneath you.

If you fix your position by GPS, or terrestrial latitude and longitude, then no, the earth will not have moved.

so you say the location will shift....so if you take off from say india and remain there for a considerable period then you might land in america?..

and does travelling from a plane say from india to USA and vice versa take the same time ..because earth is rotating from west to east and while travelling form india to california you are also travelling from west to east with the earths rotation and while travelling from california to india ie., east to west opposite to earths spinning direction so will that take a shorter time???

Let's assume for a moment your helicopter is sitting on the ground right on Earth's equator. Given that the Earth's circumference is roughly 24,000 miles (anal-retentive purists may here take a hike), and given that the Earth rotates once in 24 hours, your parked helicopter is moving at 1000 MPH [relative to a point on the equator where a line extending between the center of the Earth and the center of the Sun exits the Earth's surface on the first day of Spring - which, for clarity's sake, happens to be the date in our little scenario.)

I know you folks in India are strictly metric, but in keeping with miles the numbers work out very nicely here, so humor me.

Now you climb in your helicopter and go straight up. Strangely, your parking spot remains beneath you and does not immediately zip off to the horizon. Why? Because you're still moving 1000 MPH around the Earth - with the Earth - just as you were when you were parked.

Now here's a challenge for you: geosynchronous TV satellites appear to remain at a fixed location in the sky. You know this to be true if you happen to have one of those satellite dishes bolted to the side of your apartment building, or if you don't but know someone who does. You notice that you don't have to keep re-pointing the dish to keep the TV signal. You don't have to do anything with the dish, in fact, because the satellite appears to be 'nailed to the sky.'

What keeps the satellite from from falling straight to Earth? (This question is related to your helicopter question, by the way.)

-e

Does the atmosphere keep the helicopter moving at 1000 mph? I never really thought of this before, but you would think that the helicopter would slow down relative to the earth's surface.

Same as if you were to jump upward from a moving object and slow down while the object keeps moving.

The atmosphere is also moving at 1000 MPH at the equator (the actual speed at points north and south of the equator varies as the equatorial rate X the cosine of the lattitude). If the atmosphere was not moving at that speed but was "stationary" with respect to the earth's surface, wouldn't there be a tremendous wind from the standpoint of the parked helicopter? But since there isn't such a wind, you'd have to conclude that the atmosphere is also moving with the earth at comparatively the same speed as the surface. There might be a breeze, but its causes are different than being the result of differential rotation.

What forces would make the helicopter slow down relative to the earth's surface? Think about one of Newton's laws: an object at rest remains at rest unless put into motion by a force. Likewise, an object in motion tends to remain in motion unless acted upon by a force. In both cases a force is involved. What force is acting here? None.

The earth's atmosphere is like the air on a bus. If the bus is moving at 60 MPH, isn't the air inside also moving at 60 MPH? If it wasn't, you'd most certainly feel it wouldn't you?

"The earth's atmosphere is like the air on a bus. If the bus is moving at 60 MPH, isn't the air inside also moving at 60 MPH? If it wasn't, you'd most certainly feel it wouldn't you?"

Ok. So now my question would be, what keeps the atmosphere moving? The "trapt" air in the bus has no where to go but in the direction of the bus. Given that the windows are up. The bus is essentially pushing the air. What's pushing the atmosphere?

What keeps it going? The atmosphere's motion is coupled with the Earth's surface through a large-scale kind of "friction" (you might call it) from trees, mountains, terrain, and all that - not to mention its own momentum. It may sound strange to speak of air this way, but the atmosphere has a tremendous amount of mass. And since this mass has been in contact with the Earth's rough surface for at least 5 billion years, it's been moving with the surface for quite some time - not to mention the fact the when the atmosphere itself formed way back then, its raw materials were moving at ground speed as well. There was no time, in fact, when the atmosphere wasn't moving with the Earth's surface as far as anyone knows.

I've read that if you dip an orange into soapy water, the film of water that clings to the orange is as thick, compared to the orange, as the Earth's atmosphere is to the Earth. Spin an orange and the film of water moves with it. (the water has to be soapy here because the oils in the orange's skin tend to be hydrophobic.)

I had thought about the terrain to be the reason for the atmosphere to keep moving, but I figured the effect on the rotation of a gas would be small.

Another good ratio between the Earth & it's atmosphere is an apple. I was told the skin of the apple represents the atmosphere.

Thanks for the info

No doubt, It will work like a geo stationary satellite and will land at exactly same point.

Throw a stone vertically up, doesn't it land exactly same point?

You know that, and I know that. But it wasn't our question.

-e

If you ignore winds, yes it will return at the same location if you do not move the craft horizontally, just vertically.

The reason is because the aircraft is in relative motion with the Earth's rotation.

In reality there will always be a crosswind component the pilot must deal with and you will get some drift. Another issue is that it really is darn near impossible to move a helicopter only in the vertical direction without some horizontal movement introduced by the pilot.

Hey this is just for my understanding its a hypothetical situation...

then how do u explain the above post about flying against the earths spin n along it...

You are flying relative to the land. Imagine riding in a bus. You walk up and down the aisle. It takes no more effort to walk up the aisle than it does to walk down it. You move the same distance relative to the bus, even though the bus is moving. Same thing - you move the same distance relative to the land, even though the land is moving.

I agree that for small distances off the ground, a helicopter taking off and then landing vertically would land in the same place, regardless of how long it was in the air. But for large heights, I think the helicopter would land in a different spot.

Again, it's a "question of relative to what?" What is vertical?

By definition, if you travel up a vertical axis and then back down, you end up at the same point, regardless of how high you travel. If you set your vertical axis as a line from the center of the earth to the point at which you take off, you will arrive back at that same point. If you set your vertical axis as a line that extends from the center of the earth to the center of the sun, then the earth will rotate below you and you'll land in a different spot.

You wrote:

"By definition, if you travel up a vertical axis and then back down, you end up at the same point, regardless of how high you travel."

I disagree. Moment of Inertia changes as height increases as can be seen by the formula:

where I is the moment of inertia, m is the mass, and r is the distance from the axis of rotation (in our case, the middle of the Earth).

Kinetic Energy of Rotation is given by the formula;

where I again is moment of inertia and ω is rotation frequency (rotations per second, see diagram at bottom).

So basically you're saying that regardless of r, ω won't change. If that's true, then kinetic Energy would increase the higher you travel, essentially violating conservation of Energy.

So because of the reasoning provided above, I don't believe that it would land in the same spot regardless of height. For low heights (h<<r) I think we'd be ok because I is basically constant.

Nice explanation. I think that is technically correct. However, the maximum altitude of a helicopter isn't very significant compared to the radius of the Earth. So you would expect a very small westerly drift, but I would think that you would need to be in the air longer than you have fuel to make it measureable.

I agree, the change would be very small and hard to notice.

You're confusing math with physics. What I said was that if you define a vertical axis, then ascend and descend that axis, you will always end up at the same spot.

In other words, if I erect a flag pole 1000 miles high and climb it, then slide back down, I will always end up at the bottom of the pole.

In even other words, to maintain your location directly over a spot on the surface of the earth, ω must stay constant, and your tangential velocity and inertia must increase, so yeah the energy goes up as you increase in altitude. If this wasn't the case we wouldn't need rockets to get into orbit.

A 30kg geo-synchronous satellite has higher kinetic energy than my 30kg dog sleeping on the porch, because it's further from the center of the earth, even though it stays directly over the porch.

As I said, it's all a matter of how you define "vertical". If you shoot a bullet exactly straight up (i.e., perfectly perpendicular to the tangential plane at the equator of a perfectly round rotating sphere), and neglect the effects of air, it will, as you say, lose ω as it rises and will fall back in a different spot. But that's defining "vertical" independent of the rotating surface of the earth.

I'm not confusing anything. A flagpole is attached to the ground, so it always moving the same rotational speed as the Earth. The same cannot be said of a helicopter, which has the same initial rotational speed as the surface of the Earth, but starts to lose some of that rotational speed as soon as it leaves the ground, simply because it's r is getting larger. It will take longer to complete one circular rotation than the surface of the Earth will, period. The higher it goes, the longer it will take, which means that a different spot will be underneath it when it lands. It won't be a very large distance, probably on the order of meters, but it won't be the same spot.

Unless, of course, it is accelerating, gaining energy - which is what it must do in order to stay in position over a moving earth. Which is to say, maintaining the same rotational speed as the earth.

So, is "vertical" maintaining the same rotational speed as the earth, or is "vertical" maintaining the directional vector that it had at the moment it left earth?

Tell a helicopter pilot "go straight up and come straight down". He'll land where he started. Tell the pilot "generate force in only the "up" direction, and he'll come down in a different place. It's all a matter of semantics.

I'm taking vertical to mean radially outward from the center. I think that is a reasonable definition.

Radially outward from the center and rotating with the sphere, or just radially outward from the center?

I know, I'm just being an ass, but that's wherein lies the difference. When I say "vertical", I mean a direction that's exactly the same an hour or a day or a year from now -"up". Your vertical, 12 hours from now, will be my "down".

Let me try to be more clear, in the coordinate system I'm using, vertical means radial motion with constant dθ/dt. If dθ/dt for the helicopter changes (either an increase or a decrease) then the motion of the helicopter cannot be said to be vertical.

Keep in mind, there is no such thing as negative radius. So your statement that if you wait twelve hours you'd get a -r is incorrect. I'm using a polar coordinate system with origin at the center of the Earth.

r minus 2r = -r.

But you are correct. Lets call it r at θ = Π radians.

"If dθ/dt for the helicopter changes (either an increase or a decrease) then the motion of the helicopter cannot be said to be vertical."

YAY! We agree.

This sounds like a difference of angular velocity (i.e., radian per unit time) verses linear or even orbital velocity in meters per second.

As you ascend your angular velocity goes down. From the perspective of the helicopter pilot the vertical axis is between his craft and the geocenter of the Earth.

For round numbers, the Earth's radius is about 6378 km at the equator. If you take a helicopter at the equator up to 21,000 feet you are about 6.8 km above the geoid of the Earth. That is 0.1% of the total radius.

Assume that the craft maintains the same velocity as the Earth's crust as it rotates. If the crust moves at 1000 mph the helicopter needs an easterly velocity of 1001 mph to maintain position of the same spot on the Earth he took off from. That would give the Earth and the helicopter the same angular velocity at that altitude.

Somebody please stop the world. I wanna get off.

-e

I think one needs to look at the issue of everything moving at the speed of the spin of the earth, including the atmosphere. So a helicopter rising within the atmosphere would encounter a "wind" that blows at the same rate as the earths rotation.

So unless the helicopter gains enough altitude to leave the atmosphere "not possible", it would be held in position by the atmosphere and without other factors like lateral movement induced by wind induced by weather or pilots action, the helicopter would land where it took off.

It's true that the atmosphere rotates with the Earth, but not perfectly. The Atmosphere rotates (very) slightly slower than the Earth. Even if the atmosphere did move exactly at the same speed as Earth, it's a gas, which means that it will only slow the effect due to height on the helicopter (increasing r). Furthermore, the atmosphere becomes less dense as you go higher, meaning it would have less of an effect, even as increasing r is having a large effect on the helicopter.

From a practical standpoint the helicopter moves with the Earth. But if we want to be precise we should mention that it slowly becomes out of step with the ground as time or height or both increases.

And of course, there's the wind.

Since the sun "rises" in the east and "sets" in the west, the surface of the earth is moving from west to east. For winds aloft, the prevailing direction, at least in this part of the world, is "westerly," or from west to east. Doesn't this mean that the atmosphere is moving slightly faster than the surface of the earth in this part of the world?

I suppose it depends on your lattitude. This link does a nice job explaining it.

http://www.pilotfriend.com/av_weather/meteo/prv_wnd.htm

They all have it wrong. If you assend to any height and stay hovering at the same poossition without any drift you would eventualy notice the earth move beneath you.

This would be 15 deg per hour if you could hold it steady for that long.

The effect would be dependent on how close you are to the pole or equator.

Once airborn you are an independent body with no influence from the ground.

The thing is that no pilot can just go up and hover for this long they automaticaly want to fly some where.

Don't complicate what is a straight forward question, just read it carefuly then think what is the real part. Why is it written as it is? Most people put too much complexity into their answer. KISS.

It will land exactly same point. However for a third person sitting outside of the earth it will be moving in projectile + trapezoidal motion. And the time of projectile + trapezoidal motion will be equal to the time difference between landing and taking off.

Its like you are sitting in train and you throw a stone upside but with in train. For you stone will fall vertically at the same point but for outsider who is watching from window stone will follow projectile path.

It will land in the same place. Why? Just riddle me this: how would it otherwise know that it was staying "aloft in the same place for a while"? See?

Very hashishy...If a point moves, is it the same point? Or has it gotten older?

Nice to see that relativity is still being debated decades after Einstein came up with it.

Everyone has a point, depending on his point of view. If the helicopter was connected to a long pole and made to go straight up and then down (the pole is just there to eliminate drift, this is a hypothetical question anyway), it would land at the same place, relative to a person standing near the flag pole. From this guy's view, the 'copter did fly straight up, came down and landed exactly where it started from.

To a person out in space and staying in a fixed position relative to...now I'm lost...the center of the earth and the sun, the 'copter would seem to describe a curve but STILL LANDED IN THE SAME SPOT ON EARTH.

Now if the 'copter flew up in a straight line (obviously without the pole this time) as viewed by the guy out in space, the guy on the ground would see the 'copter fly off in a westerly direction. When it comes back down, the 'copter would land due west of his original position ON EARTH.

As to the question, "if you're going to fly half-way around the world, is it faster to fly in an easterly direction or a westerly direction?", my guess is it's going to be the same. I'm not counting the effect of wind here. Flying towards the east is faster if you ride the jet stream. Obviously, going against the jet stream will slow you down a lot.

Good grief! Can't people tell anymore when they're being put on? Sheesh.

Can intellect exist in a vacuum?

Its sad in a way that such explanations are needed for people that probably consider themselves otherwise intelligent.....but thanks to humour from CR4 Folks, it is far more acceptable, so thanks to the fun people out there!

You make a boring 'somthing' well worth reading AND the right facts emerge as well.....pretty damn good!

You wrote: "You make a boring 'somthing' well worth reading AND the right facts emerge as well.....pretty damn good!"

I think that, in part, this result is due to Brownian Motion. The remainder is simply the byproduct of a billion monkies hammering away on a billion typewriters whilst trying to serendipitously pen Hamlet.

-e

what's a typewriter?

Something exploding gremlins use to make sonic booms, what else?

Want more information? Just google the word "gullible" and you'll be an instant expert.

-e

"a billion monkies hammering away on a billion typewriters whilst trying to serendipitously pen Hamlet."

Paraphrasing someone: the Internet has proved that is not possible.

Give it time.

-e

Are you sure? Have you not read people's emails? It looks like the Mongolian hordes have descended upon the English language.

Some of em arre to bad too reed!

Some funny stuff here, and some intriguing considerations, but basically, it's all relative. To your point of reference.

To stick with the question, as posed, in a non-mathematical way:

"If one flies a helicopter vertically"
- Reasonable man's interpretation; vertically relative to a ray starting at the Earth's "center" and extending through the point on the crust where you take off.

"and stays aloft in the same place"
- Reasonable man's interpretation; stationary relative to the point on the ground where you took off.

"for a while,"
- in this case, time is irrelevant, since any time spent stationary to a reference point (or ray) is equal to any other point in time.

"when the copter returns to earth, will it land at precisely the same point where it took off?"
- Yes, since the pilot was expending energy to not only remain aloft, but stationary in relation to his reference.

"Or will it land at some other point since earth is revolving?"
- No, but there will have been some energy used to remain stationary over a reference point on the ground. As the distance from earth grows, the amount of energy required to remain stationary over a point vertically "below" you increases.

As noted in other posts, if you do not expend energy (accelerating against drifting of the air mass) to remain directly vertical "above" your takeoff point, you will instead remain at the mercy of the air mass in which you are hovering, which is never stationary...as you know, since you can feel the wind outside, and see any clouds move, and know there are fairly strong currents at various altitudes which affect weather patterns.

Once you leave the atmosphere, in something other than a helicopter, the few particles that hit you will not be part of the atmospheric gases, so you will stay moving through space at whatever velocity you had when you stopped accelerating, and are now at the mercy of gravitational forces instead (except for debris that may slam you into another orbit).

If you reach any altitude high enough with a thrust vector that is simply "away from the earth's center" then by the time you stop pushing yourself away from earth and allow gravity to drag you back down to the surface, you will have allowed the movement of the air mass you travelled through, which is slower than the earth's rotation but affected by high altitude currents, to affect your position relative to your takeoff point. There is a decreasing density of that air mass, which gradually releases you from gas particles keeping you at a close angular rotational speed to that of the earth's crust, and you will view the earth rotating beneath you at a rate that is of a higher difference than your own the further you are from the surface, and in this case, you will land at a point generally west, depending on the weather patterns that affect you on the way both "up" and back "down" relative to the earth.

Without any atmospheric affects, only the physics apply, and conservation of angular momentum is the only part you have to worry about. In that theoretical case, you will always come "down" directly west of where you went "up," with the distance determined by how high you went and how long you stayed at that greater distance from the earth's center.

There was a movie called Gismo made in the 70's. It was an accumulation of video taken in the 20's through the 50's, of people, who without regard for engineering principals, built contraptions that they thought would do things. Many were flying contraptions. For reasons not fully explained, the inventors would seek out publicity and then do their trial run before the camera, with predictable results. One unusual flying machine, that had a large parasol (umbrella) above it, was supposed to fly by an up and downward motion of the parasol. The inventor carefully explained that he was going to climb to an altitude, where he would hoover and let the earth rotate below him, landing in Paris later in the day. He was bringing along sandwiches to munch on during his trip. He pretty much got everything wrong, including the direction that he theoretically would have been traveling. His contraption hammered itself apart, never rising more than a few feet into the air. He was shaken, but unhurt and undaunted, convinced that his principal was correct, it only needed some fine tuning.

The movie "Gismo" and its airing on Public Television created a renewed interest in Pacobel's Canon in D, used in the final segment where the movie lauded the spirit if not the results of the inventors and showing a skier sailing into the air with a hang glider.

Something to consider at the next wedding you attend.

You wrote: "Something to consider at the next wedding you attend."

Especially if it's my wedding I'm at. I don't ski well and I've no experience with a hang glider, but I'm willing to settle for a strap-on JATO bottle to get me the hell outta there.

-e

I was thinking about Pacobel's Canon in D -- which is now a staple at nearly every wedding ceremony as a result of that show.

Ok, here's a follow up question to see it maybe I can get my point across about moment of inertia being important.

Let's say that you have an planet that is similiar to Earth rotating at 1 rpd (rotation per day). This planet is spherical and has no atmosphere. You've managed to build a 'hovercraft' that can hover like a helicopter except you don't need the air to support it and there is no height limitation.

So if the hovercraft takes off vertically, hovers for a while, and then comes straight down, will it;

a) Land in the same spot as it took off from

b) Land in a different spot as it took off from

c) Hovercrafts aren't real

That all depends on the time interval. If the time is long enough but not a full rotation, then it will land in a different place, regardless of any other effect.

Why don't you use a cannon? Any powered craft is capable of adding to it's total energy, and moving in different directions, although I ass/u/me that you intend to limit the force vector of your hovercraft to just the vertical.

A cannonball makes for a simpler and more elegant argument.

The answer regarding location simply depends on your definitions of the terms. If "vertical" and "hover" are movement and states that are relative to the rotating coordinates of planet, you will come down in the same place relative to the planet.

I personally don't feel that any of the other interpretations of vertical movement are viable for this - but that is just a matter of opinion. I'll describe my thinking for a couple of them:

First, the extreme oposite interpretation - that they are relative to non-rotating coordinates centred on the C. of G. of the planet - should be not be seriously considered (just think of the horizontal impulse when you take off and land). However, this is the only interpretation under which might you return to the same place in absolute terms (other than by staying aloft for precisely the correct time - before you shoot me down on that)

If the constraint is that the only accelerating force is vertical, obviously the rotation of the planet will leave you behind. However, this means that hoverers ascending vertically with different accelerations will follow different paths - both of which are "vertical". I think that defining "vertical movement" in this way is stretching things - but it would be different if you required "vertical acceleration". Incidentally, I can't see why you need to include a hovering time - it seems clear that the hoverer would be left behind without any hovering period being needed

Finally, your comment "hovercraft aren't real". Certainly, what you describe is not a hovercraft - that is a surface-moving vehicle that rests on a cushion of air. But such craft - using rocketry or the like - do exist.

We are talking semantics here.

"...the hovercraft takes off vertically..." means to me that the craft moves upward at a right angle to the surface, or as someone put it earlier, along a ray extending from the center of the planet through the point of liftoff.

"...hovers..." means, according to Merriam-Webster, to remain stationary directly over a point on the surface.

"...comes straight down..." means to descend at a right angle to the surface, the exact opposite of taking off vertically.

Therfore, the craft would come down at the same point on the surface from which it left. In order for it to land at another point would require redefining either "vertical," "hover," or "straight down."

If you take conservation of angular momentum into account and apply thrust in a purely vertical direction, then as the craft rises from the surface, its angular velocity will decrease in order to preserve angular momentum. This will cause the craft to move horizontally relative to the point on the surface from which it arose; the higher it rises, the faster the relative horizontal motion. However, now the craft is not truly hovering, but moving continuously with reference to the surface.

If we assume that the process is reversible, then as the craft descends, its angular velocity will increase until the angular velocity of the craft exatcly matches the angular velocity of the surface by the time it touches down, and the craft will land without skidding across the surface, albeit at a location different than the location from where it launched.

This is what I assume is being described. The craft climbs without imparting any force tangential to the circumference of the Earth. The craft has a horizontal velocity toward the east determined by the rotational speed of the Earth. As it climbs it has a longer distance to travel than the spot on the surface of the Earth to stay on the same radial as the spot on the ground, hence it falls behind. Arc A-B and A-C, which are the same length, illustrate this. Of course, the distance an aircraft can climb relative to the Earth's diameter is miniscule and wind would have a much greater effect.

i'm glad you said miniscule, because judging from the scale of that drawing i'd say that was one helluva helicopter!

Hey guys thank you all for ur gr8 explanations

Your title says - "Earth's Rotation".

Your question talks about "Earth's Revolution".

I don't get, what is it actually? They're 2 diff things, so don't get them mixed up. And also because the answers to them would be different.

http://dictionary.reference.com/

Your title says - "Earth's Rotation".

Your question talks about "Earth's Revolution".

I don't get, what is it actually?

---------------

For Pete's sake!

Your post comes after no less than fifty-eight other posts whose authors, after having read the context in which the word revolve is misused (a common mistake), have clearly and correctly inferred the word rotate was the word actually intended.

And yet, after reading all these posts you now decide to jump in and announce you don't get it, then proceed to tell this thread not to confuse the two words?

Quite honestly, have you nothing better to contribute at this late stage than to make petty distinctions between words? Just offhand, I'd say it's pretty clear these distinctions have been implicitly understood by others all along.

So what is it, really, that you don't get?

SORRY!!