Air flight and Earth's rotation I read a response asking why flights of equal distance east and west take roughly the same time (disregarding wind actions). I have trouble visualizing part of the answer; "the speed of the rotation of the Earth is already imparted to the aircraft, and the Earth matches that speed during the entire flight. (Of course, in the case of spacecraft, these speeds become very important.)" 
This left me wondering why if the speed of the rotation is imparted then why wouldn't the direction (speed and direction of spin) also be imparted? Also at what distance from the earth does this imparted speed become irrelevant?
I've been searching around this and other sites for something that could help me visualize how this works. On the Aerospaceweb.org site I found a question titled Launch speeds and the Earth's Rotation, which discussed the speed penalty or speed bonus depending on whether launching to the east or west. "the Shuttle does not need to accelerate from 0 to 17,000 mph, but only from 915 to 17,000 mph. In other words, the Shuttle only has to accelerate by 16,085 mph (25,880 km/h) to reach its orbital speed because that extra 915 mph is provided by the Earth itself." Then the answer went on to say; "a launch towards the west. In this case, the Shuttle would experience a speed penalty of 915 mph (1,470 km/h). It would now have to accelerate to 17,915 mph (28,825 km/h) to reach orbit because it has to overcome the initial velocity imparted on it by the rotation of the Earth." 
I can see this, it makes sense to me. What I am struggling to grasp is how come the rule doesn't work for east west flight times. It seems as if somehow gravity tags an object at rest on the planet and when it lifts up and moves in any direction, that gravity tag keeps it from being affected by the direction of the planets rotation and by the speed of rotation. Why does it matter for the shuttle but not for an aircraft? 
 A: There is a slight effect on the lift requirement for an aircraft.  A "stationary" object on the equator is actually traveling in a circle, at one earth-circumference per sidereal day velocity.  That lessens its apparent weight because it is accelerating toward the center of the earth (centripetal acceleration).
The effective gravitational constant (g minus centripetal acceleration)  for that object at rest is
g = 9.8 m/s^2
but traveling eastward at 250 m/s in a 747
g_east_at_equator = 9.753m/s^2
and traveling west
g_west_at_equator = 9.8266 m/s^2
It's about 4 parts per thousand, so you might argue whether a 40 lb checked-bag
overweight charge should really be set at 39.8 lb or 40.2 lb.
There's no 'distance from the earth' reason to care, but spacecraft need
to get about an extra 465 m/second velocity if they go into polar orbit,
as opposed to using the Earth's surface rotational velocity to help them into an eastbound equatorial orbit.   Orbital 'effective' gravity acceleration
is zero, of course (gravitational constant equals centripetal acceleration).
A: What may be confusing you is that the passage you quote about the Space Shuttle is talking about speed relative to a fixed frame of reference: one fixed relative to the distant stars. On the other hand, when you think about aircraft flying through the air (or people walking along the ground), you think about a co-rotating frame of reference: fixed relative to the Earth itself.
Each of these frames of reference makes sense on its own, but mixing the two together makes a mess.
In the fixed frame of reference, the Shuttle needs to orbit at 17,000mph. This speed is the same whether it is orbiting west, east, north, or south. In this frame of reference, the Earth is rotating at 915mph eastwards - which also means that the Shuttle, just before take-off, is moving at 915mph eastwards. Consequently "915mph eastwards to 17,000mph eastwards" requires less effort than "-915mph westwards to 17,000mph westwards".
In the co-rotating frame of reference, the Shuttle needs to orbit at 16,085mph if it is orbiting eastwards or at 17,915mph if it is orbiting westwards. In this frame of reference, the Earth is stationary, and so is the Shuttle just before take-off. "Stationary to 16,085mph" requires less effort than "stationary to 17,915mph".
The fixed frame of reference makes more sense for spacecraft because it makes all orbital speeds the same.
For aircraft, both frames of reference are again possible, but in this case the co-rotating frame of reference makes more sense because aircraft travel through the air, and the air rotates along with the Earth. For simplicity let's fly along the Equator.
In the co-rotating frame of reference, the aircraft flies at 560mph eastwards or westwards, above a stationary Earth. To get to a destination 560 miles away, it flies for an hour.
In the fixed frame of reference, the eastbound aircraft flies at 1,475mph eastward, above the Earth, which is rotating eastward at 915mph. After an hour, the surface of the Earth has moved 915 miles, so the aircraft is $1475-915=560$ miles ahead of it. The westbound aircraft flies at $915-560=355$mph eastward, above the Earth, which is rotating eastward at 915mph. After an hour, the surface of the Earth has moved 915 miles east, the aircraft has moved 355 miles east, so the aircraft is above the point on the Earth's surface which is $355-915=-560$ miles east (in other words, 560 miles west) of the starting point.
What makes the air move at the same angular speed as the rotation of the Earth? Simply this: that if one layer were rotating faster or slower than the other, drag would speed up the slower layer and slow down the faster one. 
A: It is all fine with all your math calculations answer me one thing the plane capacity to fly a maximum speed is constant now I will use a sinario where we use a escalator up one side and down the other side. The escalator movement represent the earth's movement. The distance is fixed point A to point B. Now if I walk at the same speed on both sides which will be shorter in duration. 
A: You seem to be asking more than one question, but I'll limit my answer to the explicit question at the end, namely, "Why does [the rotation of the Earth] matter for the shuttle but not for an aircraft?" The short answer is that it matters in one situation and not the other because the two situations are utterly different. You're comparing apples and oranges that involve two very different frames of reference, namely, outer space and the mass of air through which aircraft fly. Spacecraft are launched to leave the atmosphere as soon as possible and establish a terminal velocity after which no thrust is applied to the spacecraft. Once out of the atmosphere and moving at the correct velocity, the spacecraft merely "falls" around the Earth in an elliptical orbit (a circle being a special case of an ellipse) utterly unaffected by the Earth's atmosphere (generally speaking!). The spacecraft's path relative to the Earth's surface thereafter depends only on its altitude and its inclination relative to the equator. For example, the physics of geosynchronous and geostationary orbits are the same (geostationary being a subset of geosynchronous), except for the position of the satellite relative to the ground, which oscillates north and south at the same longitude in the case of a non-geostationary geosynchronous orbit. The reason the U.S. launches from southern Florida and Europe launches from French Guiana is to take advantage of the west to east velocity of the surface of the Earth relative to the final velocity (speed and direction) of a successfully orbiting spacecraft IN SPACE, OUTSIDE THE ATMOSPHERE. At the equator, the spacecraft is already traveling 1000 mph in a due easterly direction relative to outer space, so less fuel is required to reach a terminal velocity of, say, 17,000 mph in a due easterly direction OUTSIDE THE ATMOSPHERE, because the launch vehicle only needs to accelerate the spacecraft from 1000 mph to 17,000 mph in order to reach orbital velocity OUTSIDE THE ATMOSPHERE. Oh yes, there is a lot more we could discuss, but let's move on to aircraft. Ignoring gliders, and unlike spacecraft in orbit, aircraft in flight require constant thrust to overcome gravity via lift generated by the flow of air over the wing and to overcome the drag of the air in order to move forward at a constant speed. The most important thing to note is that aircraft move relative to the mass of air in which they fly. The speed of the Earth at the equator (1000 mph) RELATIVE TO OUTER SPACE is irrelevant to the physics of flight IN AIR. To get from point "A" to point "B" on the surface of the Earth, the movement of the air relative to the surface of the Earth must be taken into account, but, again, the speed of the Earth's surface relative to outer space is irrelevant. By the way, the amount of air you have to fly through is much more important (because of fuel consumption and schedule planning) than the distance over ground. If I'm flying at the equator, and the air is stable (i.e., there is no wind), the amount of air I will fly through is the same whether I fly east or west (or in any other direction, for that matter). The speed of the surface of the Earth relative to outer space is, again, irrelevant. If there is a westerly wind (i.e., wind from the west), then that means the mass of air I'm flying through is moving from west to east. If I want to fly 100 miles to the west, I'll have to fly through a lot more air than if I fly 100 miles to the east, and this will have a huge effect on flight time and fuel consumption, but it has nothing to do with the speed of the Earth's surface relative to outer space. Does this explain why the rotation of the Earth matters for the launch and flight of the shuttle but not for an aircraft? As for speed being "imparted" from one thing to another, that's really misleading. Both aircraft and spacecraft on the ground at Cape Canaveral are traveling at the same speed as the surface of the Earth relative to outer space. This matters when your frame of reference is outer space, as in the case of spacecraft, but it means nothing when your frame of reference is the atmosphere. It's really just about inertia. Imagine standing on an 50' platform on the equator at the western end a large, firm (not sandy), perfectly flat desert. Now imagine the Earth stops rotating instantaneously. You (and the air around you) would instantaneously accelerate to 1000 mph in a due easterly direction relative to the ground. Assuming the air moves right along with you in a due easterly direction (for a few seconds at least), you would hit the surface of the desert after 1.77 seconds, about 2600 feet to the east, with a vertical speed of about 38.6 mph and a horizontal speed of 1,000 miles per hour. Recall, the air is moving with you, so you suffer no wind damage during your flight. Here's my question: Would you skip?
A: There has to be a certain amount of slippage that occurs between the air (which is not a solid mass but has a density that fluctuates in respect to humidity as gravity pulls it) and the surface of the earth causing friction. The further you get from the surface the less affect the spinning mass has on the air surrounding it. Even water (which is much more dense and highly effected by gravity) will have a certain amount of slippage that is caused by the air forcing against it.
So the real question is how much air is being dragged by the spinning effect and how much air is being pulled by gravity, knowing gravity weakens as you move away from the center of the earth and also knowing air is less dense (thinner) as you travel away from the earth. Add into the entire equation, the fact that gravity affects solid objects more than gases and what remains is the humidity in the air that acts to pull it by gravity.
Once you get to a point above the clouds (as airplanes will travel) the amount of humidity has been greatly reduced and the pulling effect of gravity has weakened considerably, this should allow a greater ability to resist gravity and friction caused by the spinning of the Earth. We know the earth is not pulling the moon by Earth's surface rotation and there is no amount of air movement once outside the atmosphere.
There has to be a point where the spinning friction loses it grip and the only factor that's holding air in place is a very small gravitational pull. Once you move away from the equator that spinning has much less bite on the air since the surface is moving at a much slower speed due to the distance it is travelling relative to the equator which also reduces the centrifugal force. So this again dilutes the entire thought of friction being a major force in the upper atmosphere. Plus without mountains and the like while over an ocean the amount of friction pulling air along losses even more power allowing slippage. Meaning while the earth spins at one speed, things that aren't attached are not necessarily moving at the same speed! Now move the entire flight path to fly from Canada to the UK with a plane flying above the clouds.
What kind of times could be expected with the plane flying at the same speed. I do realize that there is a limit to how high a plane came fly since it requires air density to stay in air and move. But I also know planes can fly above clouds when the flight is long enough to make it worth the cost of fuel it takes to get there.
This question has come up between me and  my friends about 40 years ago and we agreed that there has to be some gain in time (but a possible loss in fuel) while flying opposite the direction based on the altitude you're flying.
Of course if the Earth was flat, then everything that was said above would certainly be true (minus the rotation of earth which is the factor that we're all trying to incorporate into an understandable model) and times are not affected by anything but winds caused between hot and cold air movement.
As you see these are the types of explanations that fuel the flat earthers, no easily explained answer can be found rationally but if you flat earth it, it makes sense and is very easily explained. I myself think that there's a better explanation than what's being posted, almost seems as though the answers are coming from a prewritten text that eludes the question entirely. You can't gain 915 miles in speed in a model that has a small amount of speed loss relative to the ground, the only way that's seen is when that object is hurled out of the atmosphere where the ground speed is not affecting the space around that object, even then, facing the other direction shouldn't matter much if it's traveling perpendicular to gravity the only obsticle to overcome us a slight off axis push on one side that diminishes as it travels outward. The earth should move to one side slightly as it ascends but the loss in speed can't be exactly proportional to the speed of the earth, that makes no logical sense.
That's like saying if you travel the opposite direction of the earths spin you'll end up falling to earth which I can't imagine happening unless you're in the atmosphere with no centrifugal force meaning you are no longer powered with a force strong enough to over come gravity.
That brings up this question. Using a planets gravity to slingshot can only be done if you are traveling with that planets rotation? Gravity works in reverse if you travel against it? It will slow you down? Not likely.
A: Actually it's not like that when Earth is rotating all the objects in it's gravitational influence will also move with the same angular velocity $w$ so even if you are at certain height you are still stuck with your initial point but you can contradict it simply by using another external force which can cancel out your gravitational pull or you can go up to an height equals to $infinity$
I've simply used Newton's gravitational law. I don't know what will happen if we use Einstein's field equation  
