# 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?

• The air is also travelling with the earth (disregarding wind). As the plane itself starts with the rotational speed of the earth, there is nothing to change that. Or do you expect the plane to suddenly lose the rotational speed as its wheels leave the runway? – hdhondt Jun 18 '16 at 23:15
• If you are traveling North-South, then it becomes (somewhat) important. A plane at either pole doesn't have this velocity component, while a plane at the equator travels with approx. an additional 500m/s. During the flight the plane does have to change its East-West velocity by this amount, i.e. there is an additional force acting on it. – CuriousOne Jun 19 '16 at 0:44
• hdhondt, I don't have any expectations here, I am trying to understand something. That is why I asked. – Luke.H Jun 19 '16 at 9:57
• CuriousOne, I mentioned posting this question to my wife and she asked me about going north or south and I told her I would try and find out. So your answer is much appreciated. – Luke.H Jun 19 '16 at 10:16
• Suppose you walk from point A to point B, and then at the same speed you walk back. Same time, right? Now take a boat from A to B and back. If the water's not moving with respect to the land, it's the same, right? Now if you take a blimp (an air-boat) and the air is not moving with respect to the land - same? Now take an air-plane (doesn't float, it surfs). Same? It's all relative to the land, which happens to be zooming through space at fantastic speed, but you never notice. – Mike Dunlavey Jun 25 '16 at 19:05

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.

• Martin Kochanski, I checked this as answered, I realize that there were other answers that probably did answer my question with mathematics. Unfortunately, I don't have the background and, frankly, the ability to grasp that kind of an answer. Your response, in particular, the fixed frame of reference portion helped me to visualize the solution. Now I am going to have to start a new question about why atmospheric drag doesn't slow the earth's rotation. Thank you. – Luke.H Jun 22 '16 at 8:48

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).

• Whit3rd, I'm brand new on this site and not experienced with the math. I'm more visual and verbal but I am able to follow your response within my rookie limits. Your points about polar and equatorial satellites showed me that a satellite has to be eastbound to hold position over the same place on earth. This really helped me visualize part of my question. I'm still trying to see east and west respective flight times and how gravity changes in relation to direction. – Luke.H Jun 19 '16 at 11:03

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.

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 $infinity$