The polar plane problem

[Ribera]

Hello everyone.

 

Even simple situations can be not so easy to deal with. I ask myself a simple question and I haven't found an easy explanation.

 

Imagine a plane taking off from the north pole, and flying at 1000 km/h to the equator. When it arrives there it turns "right", in direction of the west, that is in the opposite direction of earth rotation (it is still flying at same proper speed of 1000 km/h).

The plane doesn't land during that time.

The very simple question is : what will be its speed relatively to the ground ?

 

At north pole it will have no angular momentum due to earth rotation, because it is located at the axis of earth sphere. Its speed relatively to the ground will then be 1000 km/h. During the travel to the equator, the ground beneath the plane is going faster and faster (due to earth rotation). So at the equator after turning to the west it will get a total speed of 1000 km/h (due to his propulsors) + a relative rotational earth speed of 1670 km/h = 2670 km/h !

 

That would make the plane to crash instantly, which is something that doesn't happen.

The main problem I see is that it seems that the earth has no mean to "communicate" its rotational speed to the plane which keeps flying at its original polar speed, that is 1000 km/h.

 

 

 

 

 

 

[J.C.MacSwell]

Hello everyone.

 

Even simple situations can be not so easy to deal with. I ask myself a simple question and I haven't found an easy explanation.

 

Imagine a plane taking off from the north pole, and flying at 1000 km/h to the equator. When it arrives there it turns "right", in direction of the west, that is in the opposite direction of earth rotation (it is still flying at same proper speed of 1000 km/h).

The plane doesn't land during that time.

The very simple question is : what will be its speed relatively to the ground ?

 

At north pole it will have no angular momentum due to earth rotation, because it is located at the axis of earth sphere. Its speed relatively to the ground will then be 1000 km/h. During the travel to the equator, the ground beneath the plane is going faster and faster (due to earth rotation). So at the equator after turning to the west it will get a total speed of 1000 km/h (due to his propulsors) + a relative rotational earth speed of 1670 km/h = 2670 km/h !

 

That would make the plane to crash instantly, which is something that doesn't happen.

The main problem I see is that it seems that the earth has no mean to "communicate" its rotational speed to the plane which keeps flying at its original polar speed, that is 1000 km/h.

 

 

 

 

 

 

Starting off, every horizontal direction is due South. If you pick one and maintain it at 1000 km/hr you would leave the Earth at a tangent. If you followed a uniform curve to the equator at the same speed wrt a non rotating Earth frame, shortest path just above the surface, you would not be maintaining due South wrt the rotating Earth frame we generally consider, but would experience an ever increasing "apparent" vector in the Westward direction. This combined with the 1000km/hr Southern vector would put your speed well above 1000 km/hr. This would be "communicated" to the plane by the Earth's atmosphere, and require an exceptional amount of extra power to maintain against the increased drag

 

Without the atmosphere it would be different. The only power required that would be that to produce a vertical (radial) thrust to balance the weight of the aircraft against gravity as without the atmosphere it would have no drag but no lift.

 

You can use vectors to resolve the speed wrt the ground. 1000km/hr South plus 1670 km/hr West when it reaches the equator.

[studiot]

 

 

The main problem I see is that it seems that the earth has no mean to "communicate" its rotational speed to the plane

 

You are right to ask this as a classical physics question.

 

Classically the plane is still subject to the Earth's gravity.

Escape velocity is some 40 times as fast as your aircraft speed.

 

It is also a good idea to revise the difference between air speed and ground speed, you seem to have blurred this distinction.

[Strange]

opposite direction of earth rotation (it is still flying at same proper speed of 1000 km/h).

 

 

Q 1. What are you measuring its speed relative to?

 

 

 

The very simple question is : what will be its speed relatively to the ground ?

 

See the answer to Q 1.

[John Cuthber]

The simple answer is that, in order to fly "straight" from the North pole, over London and down to the equator following the line of longitude, you need to use the rudder.

"straight" lines are poorly defined on a rotating sphere.

[Ribera]

Starting off, every horizontal direction is due South. If you pick one and maintain it at 1000 km/hr you would leave the Earth at a tangent. If you followed a uniform curve to the equator at the same speed wrt a non rotating Earth frame, shortest path just above the surface, you would not be maintaining due South wrt the rotating Earth frame we generally consider, but would experience an ever increasing "apparent" vector in the Westward direction. This combined with the 1000km/hr Southern vector would put your speed well above 1000 km/hr. This would be "communicated" to the plane by the Earth's atmosphere, and require an exceptional amount of extra power to maintain against the increased drag

 

Without the atmosphere it would be different. The only power required that would be that to produce a vertical (radial) thrust to balance the weight of the aircraft against gravity as without the atmosphere it would have no drag but no lift.

 

You can use vectors to resolve the speed wrt the ground. 1000km/hr South plus 1670 km/hr West when it reaches the equator.

 

I don't see from where this "ever increasing "apparent" vector in the Westward direction" could come from.

 

If you are speaking about the atmosphere drag, a very simple vectorial computing gives us (in 2 dimensions)

angle (with the southward direction) = inv tangent (1670/1000) = 59°

speed = SQRT(1000² + 1670²) = 1886 km/h which would make the plane crash.

 

Whatever it is, the atmosphere drag is not a solution, because if the plane flies straight ahead to the equator, the atmosphere displacement would mean a lateral wind which very fastly would make it to crash.

[swansont]

Whatever it is, the atmosphere drag is not a solution, because if the plane flies straight ahead to the equator, the atmosphere displacement would mean a lateral wind which very fastly would make it to crash.

Atmospheric motion and wind are not the same thing. Wind is motion relative to the earth.

[J.C.MacSwell]

 

I don't see from where this "ever increasing "apparent" vector in the Westward direction" could come from.

 

If you are speaking about the atmosphere drag, a very simple vectorial computing gives us (in 2 dimensions)

angle (with the southward direction) = inv tangent (1670/1000) = 59°

speed = SQRT(1000² + 1670²) = 1886 km/h which would make the plane crash.

 

Whatever it is, the atmosphere drag is not a solution, because if the plane flies straight ahead to the equator, the atmosphere displacement would mean a lateral wind which very fastly would make it to crash.

As John Cuthbert implied with his comment with regard to the need for use of the rudder, the plane would need to turn (59 degrees if your calculation is correct...seems right) while flying "straight" (geodesic and shortest route wrt a non rotating Earth frame) to the equator

 

It would then be flying straight into the apparent wind, and not experience any lateral component. It would have to be capable of 1886km/hr, but forward, not sideways.

[sethoflagos]

 

I don't see from where this "ever increasing "apparent" vector in the Westward direction" could come from.

 

If you are speaking about the atmosphere drag, a very simple vectorial computing gives us (in 2 dimensions)

angle (with the southward direction) = inv tangent (1670/1000) = 59°

speed = SQRT(1000² + 1670²) = 1886 km/h which would make the plane crash.

 

Whatever it is, the atmosphere drag is not a solution, because if the plane flies straight ahead to the equator, the atmosphere displacement would mean a lateral wind which very fastly would make it to crash.

 

You are not being specific about what you mean by 'south'

 

There are essentially two extremes to choose from:

 

1) Earth based, following a line of longitude.

2) Astronomically based eg following the stars as waypoints.

 

Either way you will experience significant lateral forces (in the real world at least).

 

In the earth based case, it is the coriolis 'force' (or rather 'effect') which manifests itself as an apparent acceleration westward as you move further away from the axis of rotation - hence the need for left rudder as mentioned by John Cuthber.

 

In the astronomically based southern heading you will encounter an ever increasing sidewind tending to drive you eastwards from the air which is broadly coupled to the earth's rotation. You would need serious supersonic capability to beat that one.

Edited August 14, 2016 by sethoflagos

[Sriman Dutta]

Your plane starts from North pole and moves towards the equator. But, since the earth is moving, your plane will experience a sideward force called Coriolis force whose magnitude is given by 2mvw(m is the plane's mass, v is its velocity and w is earth's angular velocity). This force makes the plane to move in a direction which is against the direction of motion of earth, I.e., towards the west. So the relative velocity of the plane starts to increase right from the pole( although it is negligible at the pole). The relative velocity of plane=linear velocity of plane+w/r. r is the radius

One thing should be clear- if we consider only the plane's reference frame, its velocity, wheresoever it may be, is 1000kmph. But it's RELATIVE speed wrt the earth's reference frame is 2670kmph.

[studiot]

Since you didn't reply to my answer to your question which was basically where does the extra angular momentum/energy come from that is gained by the plane on its southwards journey, I hesitate to bother further.

 

However think on this.

 

The energy comes from the aircraft engine having to work harder.

As others have said it will need to adopt a suitable heading to counter/keep up with the westward movement of the Earth's surface and fly at an air speed (did you look this up as suggested, there is plenty on BigG) which is greater than that of the southward progress.

This greater speed will result in the engines working harder.

[Ribera]

Your plane starts from North pole and moves towards the equator. But, since the earth is moving, your plane will experience a sideward force called Coriolis force whose magnitude is given by 2mvw(m is the plane's mass, v is its velocity and w is earth's angular velocity). This force makes the plane to move in a direction which is against the direction of motion of earth, I.e., towards the west. So the relative velocity of the plane starts to increase right from the pole( although it is negligible at the pole). The relative velocity of plane=linear velocity of plane+w/r. r is the radius

One thing should be clear- if we consider only the plane's reference frame, its velocity, wheresoever it may be, is 1000kmph. But it's RELATIVE speed wrt the earth's reference frame is 2670kmph.

 

This Coriolis force is the problem. As you know it isn't a real force, but a geometrical effect due to rotation of earth beneath a flying object which has its own inertia. This object is "independent" of the earth and that's why its path relative to the ground is like a circle, due to earth rotation. So this "force" cannot modify the speed or the direction of the flying plane, which keeps is own polar speed in a context of increasing ground speed during his flight to the equator.

[Strange]

So this "force" cannot modify the speed or the direction of the flying plane, which keeps is own polar speed in a context of increasing ground speed during his flight to the equator.

 

You seem to be ignoring the fact that the plane is flying through the atmosphere. And the atmosphere is moving with the planet.

[Reaper79]

it seems to me, that if the velocity of the plane without outside influences is 1000 km/h, then it stands to reason, with all things considered the velocity of the plane is equal to 1000 km/h +/- Planet Rotation, +/- windspeed.

 

From a N/S perspective the earth is rotating west, which would lower your total land speed, because the shortest distance between two points is a straight line. however air speed will not effected since the north pole and the equator are an equal distance apart from any two points, even though the turn of the earth will cause you to travel more distance than following a line of longitude.

 

What angle is the wind striking your plane? are you being pushed forward or pulled back?

 

when you reach the equator and turn westward, the earths rotation will increase your land speed distance, but you will still be traveling at 1000 km/h in relation to your starting position since the turn of the earth has no correlation with bodies not attached to it, but obviously gravity will have an impact on the calculations.

 

If your question is how much does the turn of the earth effect the distance traveled from NPa to EQb, and respectively from EQx to EQy while traveling in an aircraft at 1000 km/h,

 

1000 / ((NPa + EQb) - (Earth Rotation -/+ wind speed))

 

1000 * ((EQx + EQy) + (Earth Rotation -/+ Wind Speed))

 

I'm not 100% on the math, but I think it should look something like that.... others will clarify

[Jonas Taelman]

 

You are right to ask this as a classical physics question.

 

Classically the plane is still subject to the Earth's gravity.

Escape velocity is some 40 times as fast as your aircraft speed.

 

It is also a good idea to revise the difference between air speed and ground speed, you seem to have blurred this distinction.

You still need to take the Coriolis effect into account.

[studiot]

You still need to take the Coriolis effect into account.

 

Was there something in my post you did not understand?

 

It was a direct response to the OP assertion that the Earth has no (classical mechanics) connection to the plane.

 

It was not an exhaustive analysis of all the forces acting, real or imaginary.