Say I'm flying from Sydney, to Los Angeles (S2LA), back to Sydney (LA2S).

During S2LA, travelling with the rotation of the earth, would the flight time be longer than LA2S on account of Los Angeles turning/moving away from our position?

Or, in the opposite direction, would the flight to Sydney be faster since the Earth turns underneath us and moves Sydney closer?


  • Please ignore jet stream effects and all other variables; this is a control case in an ideal environment.

  • By "dramatically" I suppose I mean a delay of 1 hour or more.

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    $\begingroup$ During the flight, you need to get up to use the restroom. There's one 10 rows in front of you, and another 10 rows behind you. Does it take longer to walk to the one that's moving away from you at 600 mph than the one that's moving towards you at 600 mph? $\endgroup$ – Keith Thompson Oct 31 '11 at 18:08
  • $\begingroup$ @Keith Thompson - it seems you understood best what I was asking and your answer is the closest to the answer arrived at by further research (aerospaceweb.org/question/dynamics/q0027.shtml). Whack it up as a solution and you've earned an accept. $\endgroup$ – Ben Nov 2 '11 at 13:18
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/1193/2451 and links therein. $\endgroup$ – Qmechanic Jun 26 '13 at 8:45

During the flight, you need to get up to use the restroom. There's one 10 rows in front of you, and another 10 rows behind you. Does it take longer to walk to the one that's moving away from you at 600 mph than the one that's moving towards you at 600 mph?

No, because you're moving at 600 mph right along with it -- in the ground-based frame of reference. In the frame of reference of the airplane, everything is stationary.

Similarly, the airplane is already moving along with the surface of the Earth before it takes off. The rotation of the Earth has no direct significant effect on flight times in either direction.

That's to a first order approximation. As others have already said, since the Earth's surface is (very nearly) spherical and is rotating rather than moving linearly, Coriolis effects can be significant. But prevailing winds (which themselves are caused by Coriolis and other effects) are more significant that any direct Coriolis effect on the airplane.

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    $\begingroup$ Then why does it take an hour longer to fly Amsterdam to NY than NY to Amsterdam? $\endgroup$ – Pieter Geerkens Apr 13 '13 at 8:18
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    $\begingroup$ @Pieter: The 100 MPH (160 km/h) jetstream affects North Atlantic tracks. $\endgroup$ – RedGrittyBrick Apr 13 '13 at 12:05
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    $\begingroup$ @PieterGerrkens - or, hangovers make time slow down. $\endgroup$ – Ben Sep 20 '13 at 11:41
  • $\begingroup$ @Locutus exactly! $\endgroup$ – berserk Jul 16 '16 at 4:37

When an airplane starts in any direction, its velocity with respect to any reference frame automatically gets the contribution from the moving Earth's surface.

Equivalently, you may look at the whole situation from the Earth surface's viewpoint and then the Earth's rotation is invisible and can't influence the speed and timing of flights, because of the principle of relativity. In this idealized description, there's no difference. This conclusion would be right when we neglected the atmosphere, mostly for e.g. the rockets that spend most of the time outside the atmosphere.

However, there exists the atmosphere and it has winds - whose average speed ultimately depends on the Earth's spinning as well but the dependence is indirect. In moderate zones, the westerlies dominate – winds from the West


and because the airplane flies in the atmosphere and wants to reach a particular speed relatively to the air mass, it's clear that the speed of westerlies helps you when to speed you up when you fly from the West, and slows you down when you fly to the West. That's why flights from America to Europe (or from Sydney to California) are about 1 hour faster than the opposite flights.

  • $\begingroup$ The frame of reference (earth) is rotating, and that you cannot simply ignore. $\endgroup$ – Kris Van Bael Oct 31 '11 at 7:18
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    $\begingroup$ The prevailing westerlies are surface winds, as are the trade winds. A flight from Sydney to Los Angeles (or the reverse) spends most of its time over areas where the trade winds dominate at surface level. This suggests that LAX to SYD should be faster than SYD to LAX. But it's not. The reason is that winds aloft are typically very different from surface winds. It's the westerly jet streams that make SYD to LAX about 50 minutes faster than LAX to SYD. The same applies to flights between America and Europe. $\endgroup$ – David Hammen Sep 3 '17 at 4:28

The rotation of earth causes two effects: Centrifugal force and coriolis force.

The effect of centrifugal force is exactly balanced out by the fact that the earth is non-spherical (it bulges at the equator). The whole surface of earth is an isopotential surface with respect to "gravity plus centrifugal force". The downward force that pulls on everything, that people lazily call "gravity", is really "gravity plus centrifugal force" on earth. There is no other special or surprising effects of centrifugal force on earth. Things get pulled down towards the ground like we intuitively expect. [Update--As pointed out in the comments, gravity is weaker at the equator, but only by a fraction of a percent; I doubt that measurably affects airplane speed.]

Coriolis force has a very important and very indirect effect on air travel because it alters winds, weather, and in particular the direction of the jet stream. (See Luboš's answer.) As for direct effects on the airplane, it is negligibly small. The plane experiences a force pushing it rightward (in the northern hemisphere), about 300X weaker than gravity or more (if I calculated correctly). The pilot steers very slightly leftward to compensate, maybe not even consciously.

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    $\begingroup$ However, equal potential does not mean equal force. Gravity is still less on the equator. $\endgroup$ – Kris Van Bael Nov 1 '11 at 6:57
  • $\begingroup$ I think you are confusing "centrifugal" with "centripedal" force. Last paragraph of this article might help on the note of weighting objects. $\endgroup$ – mathgenius Dec 11 '15 at 12:50
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    $\begingroup$ @mathgenius centripetal force (note spelling) is the force that causes an object to follow a curved path. It is directed toward the center of rotation. Centrifugal force, as noted correctly in this answer, is the apparent force directed away from the center that is experienced by objects in a rotating frame of reference. Adding centrifugal force to gravity makes gravity weaker, because the direction is opposite, which is why the earth bulges outward at the equator. $\endgroup$ – phoog Jan 6 '18 at 14:42

In reality the biggest effect is due to winds (as luboš points out). However the question explicitly asked to ignore wind. So I assume an atmosphere that rotates along with the earth. In that case there are two effects. But I wouldn't be surprised if the effects are very small.

Corriolis: Most pronounced when flying from a pole to the equator. During that flight, the plane will need to build up the extra 1000miles/hour that the earth is moving at the equator. Another way of putting it: the plane propulsion will have to compensate for the corriolis pseudo-forces.

Centrifugal: the fact that the earth is curved also means that the plane has to change its speed in a vertical way. However, the gravity is actually helping the plane. Another way to put it: due to the earth rotation the gravity is slightly less than it would be if earth didn't rotate. So the plane needs to produce less lift.

But there is no difference in flying west or east.

  • $\begingroup$ have you considered the fact that the atmosphere also revolves around the earth at almost the same speed... and generally the airplane speeds are wrt air not ground. $\endgroup$ – Vineet Menon Oct 31 '11 at 11:44
  • $\begingroup$ Of course, these are the remaining effects due to the non-linear movement of the reference frame. $\endgroup$ – Kris Van Bael Nov 1 '11 at 7:00
  • $\begingroup$ Why are you calling "Coriolis" a pseudo-force? When you calculate the derivatives, you clearly see the influence of Coriolis, hence it is a real force, not a pseudo force. $\endgroup$ – ChrisR Nov 1 '11 at 11:42
  • $\begingroup$ Coriolis is definitely a pseudo-force. It's a compensation for the fact that you have chosen a reference frame that is rotating. Consider a spinning disk and a ball moving from the center of the disk to the edge. Suppose the ball moves in a straight line. But for an observer that lives on the spinning disk, it's as if the ball is curbing. As is if is taken off path by a force... coriolis. $\endgroup$ – Kris Van Bael Nov 1 '11 at 16:54
  • $\begingroup$ Dear @Kris, the centrifugal force won't be separately manifested because in all these contexts, only the total of the centrifugal plus the gravitational acceleration will be seen. The Coriolis force won't add energy extra work because it's perpendicular to the direction of the motion - much like in the case of the magnetic force acting on an electric charge. So both of these forces are really inconsequential for the timing in an airplane - they don't speed up or slow down the aircraft - and one could argue that none of them is related to the "big" effect proposed by the OP. $\endgroup$ – Luboš Motl Nov 3 '11 at 10:04

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