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?
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.
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).
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.
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?
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