Some Flight Calculations and the Motion of the Earth

[JohnMartin2009]

A Problem is posed for a flight from Melbourne to Cairns.

Using the formula for rotational speed at latitude -

1) The latitude of Melbourne is approx 38 degrees south with a rotation velocity of 820 miles/hr
2) The latitude of Cairns is approx 17 degrees south with a rotation velocity of 994 miles/hr

The above two locations have an Earth west to east bound rotation velocity difference of 174 miles/hr. The cruising velocity of a jetstar A321 is 536 miles/hr. How does the aeroplane land when the airport at Cairns is moving west to east at 174 miles/hr under the north traveling flight? The cruise speed is already 536 miles/hr when traveling north, therefore when the plane turns east, the flight must increase velocity by 174 miles/hr, to change the west to east velocity of 820 miles/hr to 994 miles/hr and land under a relatively stationary airport. Yet the additional 174 miles/hr cannot be obtained from a plane already traveling at maximum velocity.

Brian Mullen discusses another version of the problem - Flying on a Spinning Ball.

[flat Earth link deleted]

The above problem may be translated into a more general question. How does the aeroplane take off and land on a spinning Earth with a rotation velocity which varies with latitude? Every north to south, or south to north flight must require an acceleration or deceleration of the aeroplane to equate the west to east rotational velocity of the plane's original take off latitude to the west to east velocity of the plane at the airport's latitude at the end of the flight. I suspect the standard answer will be the plane must navigate the rotating earth, but the standard answer probably will not properly account for the velocity difference between the velocities at latitudes of the two airports.

JM

 

[russ_watters]

The cruising velocity of a jetstar A321 is 536 miles/hr. How does the aeroplane land when the airport at Cairns is moving west to east at 174 miles/hr under the north traveling flight?

...Yet the additional 174 miles/hr cannot be obtained from a plane already traveling at maximum velocity.

Welcome to PF.

536 mph is the cruising speed with respect to the air around the plane, not necessarily with respect to the surface of the earth. The plane just doesn't have to care about the rotation of the earth.

[edit]
But since you want to care about the speed difference: Melborne and Cairns are about 1,600 miles apart and the flight takes about 3 hours. A change in speed of 174 mph in 3 hours is 0.05% of a g. It's nothing to a plane.

 

[JohnMartin2009]

You've assumed a relationship between the atmosphere and the rotating Earth which must be explained. Also, the cruise speed of the plane is relative to the Earth for the plane to take off, fly over and land on the Earth again. If you detach the cruise speed from the Earth you cannot land the plane without reference to the planes landing velocity relative to the moving airport. The problem remains unresolved.

 

[russ_watters]

You've assumed a relationship between the atmosphere and the rotating Earth which must be explained.

Well, the atmosphere's relationship with the Earth is both continuously observed and well explained. I would have hoped this would be something you are well aware of... what part of that relationship are you not aware of? What do you think is happening between the atmosphere and the earth?

Also, the cruise speed of the plane is relative to the Earth for the plane to take off, fly over and land on the Earth again.

The plane does not take off or land at cruise speed, and no, the takeoff and landing speed are most emphatically not measured with respect to the surface of the Earth. Probably the most important instrument any airplane has is the airspeed indicator. The plane's performance including the takeoff and landing speeds are measured against the air. And it's not even (in the most basic planes) the actual speed against the air, but a measured speed that varies with the weather and altitude due to the measurement method (using pressure).

Moreover, even if the speed were measured with respect to the surface of the Earth, it would vary slowly as the plane travels. The problem of a step-change in speed wouldn't exist.

The problem remains unresolved.

The problem is quite obviously not a real problem as it is dealt with thousands of times a day without even putting any effort into it. The actual problem here is that you don't understand the resolution, which is fine, but could you please try to put more effort into this. It isn't a difficult problem.

[edit]
Also, I noticed the link was to a flat earther video. I've deleted it. If that's what this thread is intended to be about, we'll have to lock it. We don't do flat Earth debunking here.

 

[Dale]

Also, the cruise speed of the plane is relative to the Earth for the plane to take off, fly over and land on the Earth again.

No. The cruise speed is relative to the air. This is why on long flights a tailwind or a headwind can change your travel time. All the important speeds are relative to the air: takeoff, cruise, stall, etc., not relative to the ground.

If you detach the cruise speed from the Earth you cannot land the plane without reference to the planes landing velocity relative to the moving airport. The problem remains unresolved.

It is not an unresolved problem but it is actually an effect that is explicitly considered in the design of an airport. Whenever possible airports are built with the runway going into the prevailing wind. By moving into the wind, the airspeed is increased for the same ground speed, making both takeoff and landing occur at a lower ground speed.

 

[JohnMartin2009]

Well, the atmosphere's relationship with the Earth is both continuously observed and well explained. I would have hoped this would be something you are well aware of... what part of that relationship are you not aware of? What do you think is happening between the atmosphere and the earth?

The rotating Earth model requires the atmosphere to rotate with the Earth, whilst also including local motions within the atmosphere. The atmosphere's relationship with the Earth is both continuously observed and attempts are made to explain the observations within a model.

Starting from Melbourne, if the atmosphere rotates with the Earth, and the plane flies north to Cairns, the local west to east atmosphere velocity around the plane will increase by 174 miles/hr. The problem of the rotating Earth may then be translated to the problem of the rotating atmosphere above the rotating Earth. The atmosphere is constantly accelerating with a change in latitude and the plane must be constantly accelerating west to east, whilst having a constant cruise velocity.

The plane does not take off or land at cruise speed, and no, the takeoff and landing speed are most emphatically not measured with respect to the surface of the Earth. Probably the most important instrument any airplane has is the airspeed indicator. The plane's performance including the takeoff and landing speeds are measured against the air. And it's not even (in the most basic planes) the actual speed against the air, but a measured speed that varies with the weather and altitude due to the measurement method (using pressure).

The problem is quite obviously not a real problem as it is dealt with thousands of times a day without even putting any effort into it. The actual problem here is that you don't understand the resolution, which is fine, but could you please try to put more effort into this. It isn't a difficult problem.

I remain unconvinced. The Earth's rotation and the standard assumption of a rotating atmosphere above the rotating Earth must include accelerations within the atmosphere relative to a north bound flight from Melbourne. The airspeed indicator is apparently an indicator of speed of the plane moving through the atmosphere that sits above a rotating Earth. However, if the atmosphere is rotating at a variable velocity that changes with latitude, the air speed indicator must indicate an air speed of an atmosphere that is constantly accelerating west to east whilst the plane flies north. To deny the air acceleration from west to east relative to the north bound plane is to deny the rotation of the atmosphere relative to the rotating Earth which have variable rotation velocities with latitude.

To further highlight the problem I may ask what is the rotational velocity of the atmosphere at the Earth's surface at Melbourne? Answer - 820 miles/hr. What is the rotational velocity of the atmosphere at the Earth's surface at Cairns? Answer - 994 miles/hr. Therefore, for the plane to land, the airspeed indicator must account for the change in atmosphere's rotating velocity from west to east. To explain away the problem is to artificially compartmentalise the problem and ignore the accelerations within the atmosphere which are required of a rotating Earth with a rotating atmosphere.

The problem is further highlighted again by noting another example of a flight from the south pole to the equator. The west to east velocity difference of the atmosphere between the south pole and the equator is about 1,000 miles/hr. The velocity difference is indicative that no plane could fly through an atmosphere that has a velocity perpendicular to the flight path of 1,000 miles/hr.

If the plane flies with an airspeed indicator, what indication is there that the north bound flight encounters an ever accelerating atmosphere with change in latitude?

JM

 

[JohnMartin2009]

Also, the cruise speed of the plane is relative to the Earth for the plane to take off, fly over and land on the Earth again.

No. The cruise speed is relative to the air. This is why on long flights a tailwind or a headwind can change your travel time. All the important speeds are relative to the air: takeoff, cruise, stall, etc., not relative to the ground.

The cruise speed is relative to the air connected to the rotating Earth within the rotating Earth model. The cruise speed is a measure of the plane's velocity through an atmosphere that rotates with the Earth. The cruise speed is therefore indirectly related to the Earth. If the Earth rotates, the atmosphere also rotates with the Earth. If the Earth has a variable rotation velocity with latitude, then the atmosphere must likewise also have variable rotation velocity with latitude. To detach the cruise speed from the Earth rotation velocity is to make an artificial division in one's mind which is inconsistent with the rotating Earth model.

JohnMartin2009 said:

If you detach the cruise speed from the Earth you cannot land the plane without reference to the planes landing velocity relative to the moving airport. The problem remains unresolved.

It is not an unresolved problem but it is actually an effect that is explicitly considered in the design of an airport. Whenever possible airports are built with the runway going into the prevailing wind. By moving into the wind, the airspeed is increased for the same ground speed, making both takeoff and landing occur at a lower ground speed.

Your answer is irrelevant to the problem posed simply because the Earth and the atmosphere must have a variable west to east rotation velocity with change in latitude.

Also, I noticed the link was to a flat earther video. I've deleted it. If that's what this thread is intended to be about, we'll have to lock it. We don't do flat Earth debunking here.

I'm not a flat Earther and have no interest in the topic. The thread is about the problem associated with a rotating Earth model. The problem remains unresolved.

JM

 

[russ_watters]

The rotating Earth model requires the atmosphere to rotate with the Earth, whilst also including local motions within the atmosphere. The atmosphere's relationship with the Earth is both continuously observed and attempts are made to explain the observations within a model.

Starting from Melbourne, if the atmosphere rotates with the Earth, and the plane flies north to Cairns, the local west to east atmosphere velocity around the plane will increase by 174 miles/hr. The problem of the rotating Earth may then be translated to the problem of the rotating atmosphere above the rotating Earth. The atmosphere is constantly accelerating with a change in latitude and the plane must be constantly accelerating west to east, whilst having a constant cruise velocity.

This is all correct. Why you think this is a problem, though, you don't say.

I remain unconvinced.

Unconvinced of what? Your original formulation included a plane having an airspeed above its cruising speed due to a sudden need to change reference frames upon reaching its destination. This is obviously unnecessary as you now seem to realize, since constant acceleration (relative to a hypothetical stationary Earth) can keep the plane's airspeed and speed relative to the real rotating Earth constant. It sounds to me like you've found your error in understanding and already corrected it.

To further highlight the problem I may ask what is the rotational velocity of the atmosphere at the Earth's surface at Melbourne? Answer - 820 miles/hr. What is the rotational velocity of the atmosphere at the Earth's surface at Cairns? Answer - 994 miles/hr. Therefore, for the plane to land, the airspeed indicator must account for the change in atmosphere's rotating velocity from west to east.

No, it's not the airspeed indicator that accounts for the issue, it's the atmosphere itself. Have you looked at an actual airspeed indicator as a plane takes off or lands? Have you seen what the numbers are that it reads? Can you google what the current wind speeds are at Melborne and Cairns?

If the plane flies with an airspeed indicator, what indication is there that the north bound flight encounters an ever accelerating atmosphere with change in latitude?

You mean what indication is given in the plane? None is given or needed. It's simply a built-in feature of the Earth and atmosphere that has no noticeable impact on the plane (save for normal weather variations, which Dale mentioned, and are smaller than the "issue" you think you see). I calculated for you how small the impact is in my first reply.

For airplanes and cars, going out for a walk or sitting on the couch, we choose to model the Earth as stationary (that we are rotating with it) because it's convenient and the impact of the rotating Earth is too small to be relevant. This isn't the case, though, when launching/navigating a spacecraft . For that, the stationary earth-centered frame (in which the spacecraft is detached from and not rotating with the Earth) is needed/more convenient.

The thread is about the problem associated with a rotating Earth model. The problem remains unresolved.

Are you saying you believe the Earth is round, but not rotating? That's a take I've never encountered before, but it would also be against our forum rules. You are encouraged to learn accepted models here, but active promotion of non-standard models is not acceptable. And arguing against accepted models instead of trying to actually learn them is the same as promotion of a non-standard model.

 

[JohnMartin2009]

[edit]
But since you want to care about the speed difference: Melbourne and Cairns are about 1,600 miles apart and the flight takes about 3 hours. A change in speed of 174 mph in 3 hours is 0.05% of a g. It's nothing to a plane.

The planes cruising velocity is constant so any increase in west to east velocity cannot be accounted for at all. There has to be an upper limit to what accelerations are possible for a north bound plane. If the 174 mile/hr west to east acceleration is real, the flight must account for the velocity at Cairns. One such solution to the problem is to have the plane fly at a velocity of cruise minus a calculated velocity and slowly accelerate the plane east with the atmospheres change in velocity of 174 miles/hr. I doubt that this is ever done, so the problem remains.

I have a speculative prediction. There will not be any resolution to the problem of the natural variability in west to east velocity of a north bound flight within the rotating Earth model. The current answers have been unable to grapple with the full force of the problem, and will probably not resolve the problem. Long distance, North to south flights in the northern hemisphere and south to north flights in the southern hemisphere are problematic in the rotating Earth model.

JM

 

[A.T.]

The current answers have been unable to grapple with the full force of the problem

Nope, you have been unable to grapple with the answers.

 

[russ_watters]

The planes cruising velocity is constant so any increase in west to east velocity cannot be accounted for at all. There has to be an upper limit to what accelerations are possible for a north bound plane. If the 174 mile/hr west to east acceleration is real, the flight must account for the velocity at Cairns. One such solution to the problem is to have the plane fly at a velocity of cruise minus a calculated velocity and slowly accelerate the plane east with the atmospheres change in velocity of 174 miles/hr. I doubt that this is ever done, so the problem remains.

That makes no sense.

First, do you understand that the definition of "acceleration" is a change in speed or a change in direction? So the plane's nose can be pointed north and the plane can be flying a constant 538 mph even while accelerating east? (which is in fact exactly what happens)

Second, do you recognize that the speed limit on the plane is due to aerodynamics? If it exceeds the maximum speed, the wings rip off due to the aerodynamic forces? So that 820 mph that it is "moving" while sitting still on the ground in Melborne doesn't rip the wings off because that's not the wind speed?

Third, do you realize if the wind is zero at both cities, then plane can be going a constant 538 mph between the two cities with respect to the air, even though its speed over a hypothetical rotating stationary reference frame is changing? If it starts at 538 mph with respect to the air and ends at 538 mph with respect to the air, then the allowable airspeed hasn't been exceeded and nothing has changed that needs to be addressed.

I have a speculative prediction. There will not be any resolution to the problem of the natural variability in west to east velocity of a north bound flight within the rotating Earth model. The current answers have been unable to grapple with the full force of the problem, and will probably not resolve the problem. Long distance, North to south flights in the northern hemisphere and south to north flights in the southern hemisphere are problematic in the rotating Earth model.

Your inability to understand how the physics works does not constitute a flaw in the model. Since you have changed your argument to try to get around an error you made (regarding the step-change in speed when it gets to its destination), I am convinced that you are not here to learn, but rather to argue, so I'm locking the thread.

edit, btw:

How does the aeroplane land when the airport at Cairns is moving west to east at 174 miles/hr under the north traveling flight?

That kind of "problem" actually happens, but not for the reason you believe. The jet stream winds can exceed 174mph west to east, and a plane flying in the jet stream can indeed have a ground speed that substantially exceeds its airspeed. I say it's a "problem" only in that it presents navigation/fuel difficulties, and turbulence, not that it can prevent a plane from landing successfully. A plane flying from west to east with a speed over ground of 710 mph because of the wind happens, and does not present any problems. Here's one from last year that exceeded 800mph:
https://www.forbes.com/sites/jamesa...ght-between-new-york-and-london/#5fd642b52fd9

The only "problem" would be that it has to start slowing down and losing altitude a few minutes earlier than normal.

 

[Dale]

The cruise speed is relative to the air connected to the rotating Earth within the rotating Earth model. The cruise speed is a measure of the plane's velocity through an atmosphere that rotates with the Earth. The cruise speed is therefore indirectly related to the Earth.

No. The cruise speed has nothing whatsoever to do with the motion of the air relative to the earth. It is purely 100% a measure of the plane’s speed relative to the air. Given a specific design of the aircraft the cruise speed is a fixed speed relative to the air. This is true regardless of whether you are in a wind tunnel or in natural wind or in air at rest with respect to the ground.

The cruise velocity is the velocity of the plane relative to the air, and the wind velocity is the velocity of the air relative to the ground. To get the ground velocity of the aircraft you simply add the cruise velocity and the wind velocity as vectors.

If the Earth has a variable rotation velocity with latitude, then the atmosphere must likewise also have variable rotation velocity with latitude.

You seem to be unaware of the standard terminology, so part of your confusion may be that you are getting responses to questions that answer what you are saying rather than what you want to say.

The rotational velocity, also called angular velocity, is constant at all latitudes. It is one rotation per sidereal day pointed toward celestial north. The units for angular velocity are angle/time.

What you are trying to say is the tangential velocity. That is the linear velocity of a given point on the surface of the Earth measured with respect to an earth-centered inertial frame. That quantity points horizontally due East at all points on the Earth and varies by latitude. The units for tangential velocity are distance/time.

To get the velocity of the plane in the Earth centered inertial frame simply add the ground velocity to the tangential velocity. Since the Earth has a variable tangential velocity with latitude, the velocity of the aircraft in the Earth centered inertial frame also varies with latitude.

The problem remains unresolved

There is no unresolved problem here other than an unresolved educational problem. You appear to be confused by the different reference frames and velocities involved. Again, there is the cruise/takeoff/landing velocity which is measured with respect to the air, the wind velocity which is measured with respect to the ground, and the tangential velocity which is measured with respect to the Earth centered inertial frame. Normal vector addition allows one to transform between frames. The acceleration required to increase the velocity in the Earth centered inertial frame results in a completely negligible acceleration