# Airplane pulling on the earth?

Ok...Just a thought model

When an aircraft is in flight, the earth pulls on the airplane (weight). The wings generate lift to support the weight.

However, the aircraft is also pulling up on the earth with the same force. So, techincally the earth has momentum imparted to it in the upward direction when the aircraft flies overhead. I know that the "change" in the earth's velocity will be ridiculously small (there would be no way to measure or feel it) due to the earth's massive size.

But don't we impart momentum to the earth evreyday? When I drive my car, I push backwards on the earth, even when we all walk, we push backwards on the earth. I realize that it probably all cancels, at some point in time, as for every person walking eastbound, there is probably someone walking westbound or stopping.

But what about the airplane? Do we indeed impart momentum to the earth, in the upward direction when an aircraft flies overhead, and there is no downwash reaching the earth because it has been dissipated?

Raptor01601 said:
Ok...Just a thought model
When an aircraft is in flight, the earth pulls on the airplane (weight). The wings generate lift to support the weight.
However, the aircraft is also pulling up on the earth with the same force. So, techincally the earth has momentum imparted to it in the upward direction when the aircraft flies overhead. I know that the "change" in the earth's velocity will be ridiculously small (there would be no way to measure or feel it) due to the earth's massive size.
But don't we impart momentum to the earth evreyday? When I drive my car, I push backwards on the earth, even when we all walk, we push backwards on the earth. I realize that it probably all cancels, at some point in time, as for every person walking eastbound, there is probably someone walking westbound or stopping.
But what about the airplane? Do we indeed impart momentum to the earth, in the upward direction when an aircraft flies overhead, and there is no downwash reaching the earth because it has been dissipated?
Don't forget that things that go up tend to come down again and people or vehicles in motion tend to stop eventually. So while you might start driving by pushing off to the west, you'll eventually stop by pushing off to the east. Except for a few special cases like rockets and meteors, the total net change to the Earth will be zero. Even in those few cases where it's not zero, you're not going to move the Earth by some small fraction of a millimeter. Instead, since the Earth isn't completely rigid, the change might show up as a small wholly insignificant increase in heat which will be rapidly distributed through the entire Earth.

Grogs said:
Don't forget that things that go up tend to come down again and people or vehicles in motion tend to stop eventually. So while you might start driving by pushing off to the west, you'll eventually stop by pushing off to the east. Except for a few special cases like rockets and meteors, the total net change to the Earth will be zero. Even in those few cases where it's not zero, you're not going to move the Earth by some small fraction of a millimeter. Instead, since the Earth isn't completely rigid, the change might show up as a small wholly insignificant increase in heat which will be rapidly distributed through the entire Earth.
Sure, as I stated that there is usually someone going east when you are going west. But, that is not guaranteed. But that means that there is no time constraint, as what goes up, does not have to go come down right away. Such as an airplane, which can stay aloft for hours, and land thousands of miles away.

In addition, there are a number of situations where I dont use the earth to stop. For instance, I accelerate in my car, pushing back on the earth. I come on to a frozen lake, almost zero friction, I use aerodynamic drag to come to as stop (braking parachute). My momentum in this case will have been transfered to the air.

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Danger
Gold Member
Well, there's a viscous coupling between the atmosphere and the Earth, but it's irrelevant here. The main consideration is the difference in mass between Earth and anything on it. You could set a couple of thousand jet engines on test stands all facing the same way, and they wouldn't make any noticeable change in planetary rotation over a period of years.

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HallsofIvy
Homework Helper
Jules Verne wrote a remarkably silly story about some men who went around the earth getting nations to sell them all their possible claims to the north pole (may have been titled "The Man who bought the North Pole") for trivial amounts of money. They were going to fire cannons on a mountain in Africa to tilt the earth so all the ice on the north pole was melted and then claim the land!

There appears to be little or no concern for the fact that millions of people would be killed (especially in the African nation). The upshot of it all is that at the last minute the "brilliant professor" who calculated how much powder this would take discovers he has accidently dropped a few decimal places- you would have to more than cover the earth in "cannons" to produce enough force!

After all that, one feels it would be unfair to point out that there is NO land under the north pole ice cap!

By the way, am I the only one old enough to remember wondering what would happen if all the Chinese were to jump in the air at the same time?

russ_watters
Mentor
So I guess the answer to the question is yes, but you'll need a lot more decimal places on your calculator to calculate the effect of these things.

Yeah, I have been plugging some numbers...and I came out with the following

The Mass of the earth is 4.105 X 10^23 slugs

A 100,000,000 pound force acting on the earth for 100,000,000 years would change the velocity of the earth by .6 miles per hour

So, Theoretically, wouldn't a bowling ball fall to the ground faster than an object that is smaller than it but still aerodynamically proportional, because the earts pull on the objects is tha same, but the larger object pulls back harder. the difference would be microscopic, but it would still be there right

Danger
Gold Member
I'm not exactly sure that I understand the question, Ki. In the absence of air, all objects will drop simultaneously. Gravity must overcome more inertia in the heavier one. In air, the one with the higher drag will fall slower.

Well I guess, next year at Worlds jumping day we can test the ideas. I suppose that the conservation laws are prevailing here.

russ_watters
Mentor
Danger said:
I'm not exactly sure that I understand the question, Ki. In the absence of air, all objects will drop simultaneously.
Well, kinda - two objects with mass actually fall toward a common center of gravity. So while the earth will accelerate any object toward that center of gravity at g, how much earth is accelerated toward that center of gravity depends on the other object's mass.

the earth pulls each object the same, but the larger one pulls back harder than the smaller one

Danger
Gold Member
Hey Russ;
Yeah... I was a bit sloppy there and just treated the Earth as if it were stationary. What I meant was that 2 different objects in vacuum will both fall toward the centre of the Earth at the same rate and impact the surface at the same time. There will, of course, be an immeasurable convergence of paths due to the spherical shape of the gravity field along with the 2 objects' mutual attraction.

World Jumping day? You have got to be kidding....

First of all, when you jump, you push down on the earth, and in theory the earth moves opposite (there would be no way to detect such a small motion)

However, also from theory, when you come back down, the earth acelerates up to meet you. Again, because of the massive size of the earth you would be talking about a number so small, there would be no way to measure it.

The Mass of the earth is 4.105 X 10^23 slugs
If it takes a 100,000,000 pound force 100,000,000 years to chanage the velocity of the earth by .6 miles per hour.

What kind of goofy gag is that.....world Jump day

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russ_watters
Mentor
Danger said:
What I meant was that 2 different objects in vacuum will both fall toward the centre of the Earth at the same rate and impact the surface at the same time.
I don't think that's true. Ie:
There will, of course, be an immeasurable convergence of paths due to the spherical shape of the gravity field along with the 2 objects' mutual attraction.
Just because it's small doesn't mean it isn't there. If one object is a paper clip and the other is the moon (better yet, Jupiter), the difference in time to impact will be very measurable.

Danger
Gold Member
russ_watters said:
I don't think that's true.
Haven't you ever seen the demonstration where they drop a ball bearing and a feather in an evacutated cylinder? They accelerate identically.

In a Vacum all objects have the same aceleration..that of gravity..32.2 fps^2 (on earth)

I believe that Newton showed that Inertial mass and gravitational mass are the same thing. an object that has more mass in a gravitational field also has more force (weight) A greater force acting on a greater mass means the aceleration is the same as a lesser mass being acted on by a lesser force (weight) Once you throw in air resistance this all changes.

On one of the moon missions the dropped a hammer and a feather, from the same height, they both reached the surface at the same time

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Tide
Homework Helper
Momentum is imparted on the Earth only while the airplane is ascending or descending. While the plane is in level flight the force that is holding the airplane up is also pushing the Earth downward - balancing the attractive force between the two objects.

There is a second order effect that involves the curvature of the Earth. If the flight is "level" then the center of mass of the "two body" system follows a circular path, i.e. the two bodies orbit each other.

Tide said:
Momentum is imparted on the Earth only while the airplane is ascending or descending. While the plane is in level flight the force that is holding the airplane up is also pushing the Earth downward - balancing the attractive force between the two objects.
I agree. I know that Prandtl showed that the earth will feel a "footprint" of th weight of the aircraft when the aircraft flies overhead.
He also showed that this "footprint" will be an extremely small pressure increase over a huge area (F=PXA). Since this "footprint" is not a "bulk" movement of the fluid, I assume that it's origin is the "swaying" of the statistical element of pressure.

I found a worked out example of this effect. It showed that a 3,000 pound airplane, 100 feet off the surface, would change the pressure (assuming STP of 14.7 psi ) to only 14.7004 psi. This is obviously the reason that humans cain't feel (but hear) the passing aircraft.
Any feedback on these thought would be appreciated !!!

Okay, folks... let's put this one to bed.

Tide:
Momentum is imparted on the Earth only while the airplane is ascending or descending. While the plane is in level flight the force that is holding the airplane up is also pushing the Earth downward - balancing the attractive force between the two objects.
There is a second order effect that involves the curvature of the Earth. If the flight is "level" then the center of mass of the "two body" system follows a circular path, i.e. the two bodies orbit each other.
Your momentum argument depends on what your reference system is... if it is the Earth, then yes, the Earth loses a small bit of angular momentum as the plane ascends, but gains it back when the plane descends. If you consider the plane to be in the system, then the total angular momentum of the system remains constant.
You are way off on the second point, though. The airplane does not "orbit" the Earth. The airplane flies in the atmosphere above the Earth. Orbiting implies that gravitational forces are the primary forces acting on the system to keep things going. In the case of a plane, aerodynamic forces are what keep the plane aloft. Gravity is trying to smash it back into the ground.

Raptor01601:
In a Vacum all objects have the same aceleration..that of gravity..32.2 fps^2 (on earth)
Not exactly... On Earth, in the Earth reference frame, acceleration of an object in freefall apart from any other (drag) forces is proportional to the square of the distances between the centers of mass of the Earth and the object. So, at mean-Earth sea-level, (r=6378.1363 km), an object would accelerate downward at (a = G*M/r^2) ~9.80 m/s^2. At an altitude of 10 km, the object would accelerate downward at 9.77 m/s^2. A small, but noticeable difference.

russ_waters:
Just because it's small doesn't mean it isn't there. If one object is a paper clip and the other is the moon (better yet, Jupiter), the difference in time to impact will be very measurable.
In classical dynamics, if you consider both to be point masses, no it won't. The acceleration of both objects will be G*M/r^2 toward the center of the Earth in the Earth reference frame. Integrating over the fall, they will both land at the same time. Now, if you have both the bottom of the paperclip and the bottom of the Moon at the same distance above the Earth, the paper clip will actually land sooner, due to the fact that the distance between the centers of mass of the Earth and paperclip actually start closer together.
Now, in reality, with something with as large a mass ratio as the Earth and Moon, there will be general relativistic issues that will deviate from the classical solution.
---
In general, if we take the system of the Earth and everything in a finite local sphere around it, the total energy, mass, and momentum do not change (in classical terms) except when something leaves or enters the system surface (ie satellites on an hyperbolic orbit going out or asteroids/solar particles/UFOs :) coming in). In the grand scheme of things, nature likes balance. Hope that cleared up some confusion.
Cheers...

Tide
Homework Helper
LunchBox,

I only meant orbit in the sense that they are going around each other which I thought was evident from the context.

Raptor01601:
Not exactly... On Earth, in the Earth reference frame, acceleration of an object in freefall apart from any other (drag) forces is proportional to the square of the distances between the centers of mass of the Earth and the object. So, at mean-Earth sea-level, (r=6378.1363 km), an object would accelerate downward at (a = G*M/r^2) ~9.80 m/s^2. At an altitude of 10 km, the object would accelerate downward at 9.77 m/s^2. A small, but noticeable difference.
Yes, I'm familiar with the inverse square law of gravitation, I don't think that was the "crux" of the question.
However, I'm still waiting to see if someone could explain to me how Prandtl fiqured out that an aircraft passing overhead will leave a "footprint" on the earth equal to it's weight. I know that it is an extremely small change in pressure over a huge area, but I would like someone to explain to me the "transport mechanism" down from altitude to the surface of th earth.

I know that it is not a "bulk" movement of fluid. So I have to conclude that the pressure is transmitted by molecular motion. Since pressure = Density X temperature, with a statistical element, there must be some kind of momentary change in the statistical average as the aircraft flies over.
If anyone could explain this effect to me better, please do.

Raptor01601 said:
I'm still waiting to see if someone could explain to me how Prandtl fiqured out that an aircraft passing overhead will leave a "footprint" on the earth equal to it's weight.
en.wikipedia.org/wiki/Hydrostatic_pressure

russ_watters
Mentor
Danger said:
Haven't you ever seen the demonstration where they drop a ball bearing and a feather in an evacutated cylinder? They accelerate identically.
I have. But that isn't what we're talking about. In that case, both objects are so much smaller than the earth that the difference isn't noticeable. It's so small, in fact, that we cut it out of the equations because the difference gets lost in the significant digits. But that doesn't mean the difference isn't there.

Perhaps a new thread in GP...

EnumaElish