# Time travel for airplane

I've seen across the internet the explanations of why (ignorind wind) the time an airplane takes to travel from one place to another on Earth is the same regardless of its direction of flight.

The explanations usually rely on using reference frames. But I thought of one that I think is more simple, because it doesn't use reference frames explicitally.
Just consider that the mean time is the mean distance traveled over the Earth divided by the speed relative to the Earth.

If the Airplane has a speed V and travels a distance D the time is D / V, and if it travels in the opposite direction its speed is -V, but the same distance is -D, which shows that the time is the same.

Is this explanation valid?

So of course by talking about "speed" and "distance" I'm implically using reference frames. But I don't need to bring them up to the discussion. So I think this is more easy for people who don't know physics to get with.

Gold Member
Using the phrase "relative to Earth" is necessarily invoking the use of frames or reference.

kent davidge
tnich
Homework Helper
I've seen across the internet the explanations of why (ignorind wind) the time an airplane takes to travel from one place to another on Earth is the same regardless of its direction of flight.

If the Airplane has a speed V and travels a distance D the time is D / V, and if it travels in the opposite direction its speed is -V, but the same distance is -D, which shows that the time is the same.

Is this explanation valid?

So of course by talking about "speed" and "distance" I'm implically using reference frames. But I don't need to bring them up to the discussion. So I think this is more easy for people who don't know physics to get with.
I don't understand the premise. How can the flight time from one place (e.g. San Francisco) to another (e.g. Denver) be the same flying east or flying west? I think you probably mean, given that the surface of the earth is moving east at almost 800 mi/hr at that latitude, why is the flight time from San Francisco to Denver about the same as the flight time from Denver to San Francisco.

If that is what you mean, then your proposed solution, by implicitly using reference frames, ignores the issue completely. It works for vacation planning because we don't usually think about earth's rotation explicitly, though it is a factor in figuring out arrival times. (If I take off from San Francisco at 2PM PST, I arrive in Denver at 5PM MST. On the return trip I take off at 2PM MST and arrive in San Francisco at 3PM PST. This difference is directly related to earth's rotation.)

On the surface of the earth, it is easy to consider ourselves to be in an inertial reference frame. Doing our calculations in that reference frame works for most day-to-day activities. But mentally stepping into space and looking down on a rotating earth puts you in a difference reference frame. To understand the flight-time question in that reference frame, you have to explicitly think about reference frames and converting velocities between reference frames.

kent davidge
I'm not sure I understand the answer. I was not wanting to consider specific cities, I'm not a US citizen, and know nothing about location of US cities. I was only talking about covering a given distance D to the east and then the same distance D to the west.

sophiecentaur
Gold Member
I'm not sure I understand the answer. I was not wanting to consider specific cities, I'm not a US citizen, and know nothing about location of US cities. I was only talking about covering a given distance D to the east and then the same distance D to the west.
Fair enough but why would d that matter?
What is your reasoning to suggest the time in one direction would be different from the time in the other? If you were to sprint along a railway carriage in one direction (smooth track, no air flow, of course) why would your max speed be different one way from the other? You are taking the same number of the same sized steps in the same time. If the windows were blacked out, the only measure of speed would be in the train's frame.

tnich
tnich
Homework Helper
I'm not sure I understand the answer. I was not wanting to consider specific cities, I'm not a US citizen, and know nothing about location of US cities. I was only talking about covering a given distance D to the east and then the same distance D to the west.
Pick any points on the earth you want. The point is that the confusion arises when you start considering earth's rotation. If you decide to do your calculations in a reference frame that is stationary (non-rotating) with respect to earth, and you don't convert the velocity of your airplane to your new reference frame, you are going to get the wrong answer.

kent davidge
Sorry, now I re read post #3, specifically
(If I take off from San Francisco at 2PM PST, I arrive in Denver at 5PM MST. On the return trip I take off at 2PM MST and arrive in San Francisco at 3PM PST. This difference is directly related to earth's rotation.)

So @sophiecentaur 's post
What is your reasoning to suggest the time in one direction would be different from the time in the other? If you were to sprint along a railway carriage in one direction (smooth track, no air flow, of course) why would your max speed be different one way from the other? You are taking the same number of the same sized steps in the same time. If the windows were blacked out, the only measure of speed would be in the train's frame.

Based on @tnich's answer I would say that the time in your situation is the same in either direction I choose to go. But in the case of the airplane and the Earth, because it's a rotating frame, we have the difference in time. Right?

PeroK
Homework Helper
Gold Member
2021 Award
Sorry, now I re read post #3, specifically

So @sophiecentaur 's post

Based on @tnich's answer I would say that the time in your situation is the same in either direction I choose to go. But in the case of the airplane and the Earth, because it's a rotating frame, we have the difference in time. Right?

By your logic a helicopter could hover and watch the Earth spin at 1500 km/h, or whatever, below it?

The flaw in your logic is that the air is essentially moving with the Earth.

kent davidge
Janus
Staff Emeritus
Gold Member
Sorry, now I re read post #3, specifically

So @sophiecentaur 's post

Based on @tnich's answer I would say that the time in your situation is the same in either direction I choose to go. But in the case of the airplane and the Earth, because it's a rotating frame, we have the difference in time. Right?
Note that in tnichs answer has you leaving and arriving San Francisco in Pacific time and leaving and arriving in Denver in Mountain time. You are moving from one time zone to another. MST is one hour ahead of PST, so taking off in at 2PM PST and arriving in Denver at 5pm MST means you had a 2 hr flight.
Taking off at 2 pm MST and arriving at 3 PM PST also equates to a 2 hr flight.
If you were to turn around immediately after arriving in Denver at 5 PM MST, you would get back to San Francisco at 6 PM PST, 4 hrs after you left SF, 2 hrs both ways.
To avoid the complications caused by changing time zones, In aviation you generally use UTC or "ZULU" time. Thus in the trip above, you would leave San Francisco at 2100 Z, arrive at Denver at 2300 Z, and then return to San Francisco at 0100 Z.

kent davidge
sophiecentaur
Gold Member
Based on @tnich's answer I would say that the time in your situation is the same in either direction I choose to go. But in the case of the airplane and the Earth, because it's a rotating frame, we have the difference in time. Right?
How is the Earth any different from the inside of a train? Take a very fast train (1000mph) moving through space. That is precisely the same situation as the surface of the Earth moving at the same sort of speed. You seem to agree with the train scenario so why not with the moving surface of the Earth?
You are moving from one time zone to another.
That would not affect the stopwatch of the guy on the plane, which would objectively measure the time of the flight.
If you want awkwardnesses due to time zones, no one needs to travel anywhere - if I just try to ring my Son, in New York, at his lunchtime when it's lunchtime here in the UK. Lazy devil is still laying in bed at lunchtime. !

kent davidge
Janus
Staff Emeritus
Gold Member
That would not affect the stopwatch of the guy on the plane, which would objectively measure the time of the flight.
If you want awkwardnesses due to time zones, no one needs to travel anywhere - if I just try to ring my Son, in New York, at his lunchtime when it's lunchtime here in the UK. Lazy devil is still laying in bed at lunchtime. !
That was exactly the point I was making, that tnich's example of leaving at 2 pm and arriving at 5 pm going one way and leaving at 2 pm and arriving at 3 pm going the other did not mean that the trip took 3 hrs going one way and 1 hr going the other.

kent davidge and sophiecentaur
sophiecentaur