Physics graduates: Aircraft keeps up with earth's rotation

Hi Physics graduates,

Could someone with a degree in physics please answer this for me.  I did some physics 30+ years ago, and I'm having a disagreement with a friend on this issue, and would like some clarification, please.

If an aircraft was to take off from some point (say my house) on the earth, thrust directly upwards to say 30,000 feet, then hover there for say 5 minutes, then come directly down, then in theory, assuming no "winds", it would land (approximately*) back at my house.  Right?

I would have thought the main contributor to the "horizontal" movement (which keeps the aircraft above my house), is the momentum it had before it even took off.

I would have thought a secondary contributor would be the "horizontal" movement of the atmosphere which is basically rotating with the earth. As the craft gains altitude, it would require more “horizontal” velocity for it to keep up with the earth, because its radius from the centre of the earth would have increased. The movement of the atmosphere would act like a wind against it, and help to carry it along (though there could be some delay in the aircraft catching up, hence the "approximately*" above).

My questions are:

    Q1. Does gravity contribute to anything to the "horizontal" movement of the aircraft, or does it only prevent the momentum of the aircraft from making it continue in a straight line out into space, because gravity's force is simply towards the centre of the earth, thus providing a downward component which would tend to make the aircraft rotate around the earth.

    Q2. If gravity does contribute to the "horizontal" movement of the aircraft, how does it do that?

    Q3. Would you agree that the "horizontal" movement is due mainly to the momentum of the aircraft, and secondarily to the movement of the atmosphere?

    Q4. What degree do you have in physics, and when and where did you get it from?

Please number your answers accordingly for clarity.

I've done some web searches on this, and some were helpful, but I'm wanting specific answers to the above questions only from people with university degrees in physics, please.

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Suggestion. Cast the problem in a rotating frame of reference (fixed with earth). Neglecting atmosphere. You have already said " no wind". Atmosphere introduces too many uncertainties.
Q1 you are mixing ideas from a rotating frame with ideas from an inertial frame.    gravity is simply toward center of earth in either frame
Q2. It does not
Q3. In your initial question you " assumed no wind speed". Do not introduce them here. Primarily only.
Q4. PhD.    jHU.     1966
The secondary effect you speak of is the Coriolis effect, which for a 5 minute trip, is a small effect compared to atmospheric influences.

Q1 gravity acts vertically, not horizontally, which is how we tend to define "vertical" and "horizontal"

Q2 gravity allows density variations to contribute to air circulation, which contributes to winds, which can affect aircraft motion.
gravity also holds the atmosphere in place, which makes aircraft motion possible

Q3 I think I would say that the first 1000 feet would be mainly the momentum of the aircraft, but by 30000 feet, it would be mainly the momentum of the atmosphere, though I have not done the calculation to compare their contributions.

Q4 none

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Thibault St john Cholmondeley-ffeatherstonehaugh the 2ndCommented:
I would have enjoyed answering this, but no degree. Sorry.
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tel2Author Commented:
Hi Mike,
Well it didn't stop ozo, and despite my requirements his answer was good to have, so you have my permission.  There doesn't seem to be a lot of "qualified" experts answering yet, anyway.

Thanks aburr and ozo for your great answers.
Sorry, I missed noticing the requirements.
However, (aburr notwithstanding) a physics degree is not necessarily a guarantee against being clueless in applying physics to a practical situation
Thibault St john Cholmondeley-ffeatherstonehaugh the 2ndCommented:
If ozo calculates that something is going to land directly on your house then you'd better move :7)

I'll have a bash at the question.

If we start with a very simple system we can add complications like wind later.
We'll use a large billiard ball the size of the earth so it can be round and the centre of gravity at the centre of the ball. Use the white spot ball. The spot is where your house is and the launch site. 'Up' will be defined as movement away from the centre of gravity. 'Straight up' will be defined as following a straight line from the centre of the ball, through your house and away into space, perhaps through a geostationary satellite.
The ball is turning relative to a datum point in space, spinning on its own axis.
The plane at rest is revolving with the surface of the ball, held there by gravity and prevented from sliding around by a mixture of friction and lack of any lateral forces.
Now start the engines, point it Straight up and go. As the plane leaves the ground it stops begins to follow a straight line in a direction that is tangental to the point where it left the earth and at 90 degrees to Straight up. As you alluded to in the question, if the plane follows this line it will fall behind your house. A line drawn between the plane and the centre of the ball will no longer pass through the launch point.
But you want the plane to go Straight up, and you have given it thrusters. So continue in a line that is Straight up, the direction of thrust must continually alter, forwards if you like, to keep the plane on that line between your house and the geostationary satellite.
If you follow the natural tangental path the landing will be some distance from where it launched, but if you follow the 'Straight up' path, Straight down will bring it back to your house.
In the good old days it was easy to demonstrate that an object leaving a spinning surface followed a tangent to the edge of the disk by resting small objects on a record player turntable and watching them roll off. Unfortunately turntables are not so common now so you can either believe what you read or will have to improvise some sort of turntable.
Thibault St john Cholmondeley-ffeatherstonehaugh the 2ndCommented:
Sorry, missed part of the question. That's what not having a degree does.
Having gone Straight up (please note the use of capitals is deliberate, the phrase is referring to my earlier definition), you want to hover 'there' for five minutes. Definition of there is required. It could be at that exact location relative to the datum point in space that I mentioned briefly, but I think you mean hover directly above your house. You will need to use your thrusters to maintain this position. They will need to be firing anyway to keep it from falling, but you will add lateral thrust to keep the plane above your house.

If you reintroduce the atmosphere that I removed before then that just adds extra forces which the thrusters will overcome. If the atmosphere is wind free and acting as a stationary fluid rotating with the earth you won't need so much forward thrust to stay above your house, if there are random winds blowing, the direction of thrust will need to counteract them to remain stationary or maintain its course.

The question really hinges on your definition of 'travelling straight up'. I suspect you and your friend are swapping between 'above your house' and 'where I would have been with no lateral thrust'. Definition of lateral thrust here would be any direction except directly toward the centre of the ball.
I like your discussion. Here is a related question:
Let's shrink the radius of the Earth from 3959 to 3820 miles.  This makes the circumference of the Earth 24000 miles.

Next move your house to the equator and get rid of the atmosphere.

Helicopters don't work now, so we will go with an antigravity drive which can only work vertically.

Your house is moving East to West at 1000 mph, but the point 6 miles (31680 ft) above your house is moving a little faster 1001.6 mph.

Now if you go up six miles very fast, maintain that altitude for 5 minutes, and come down fast, you will find yourself approximately     1.6/12  = 0.133 miles = 440 ft east of your starting point.
tel2Author Commented:
Thanks for that explanation, d-glitch.

I hear your requests, but please give me some time, because the wife is a bit (or even a byte) reluctant to move house to the equator just for this discussion, and we have both become quite fond of the atmosphere.  But given time, we'll get used to breathing water, I guess.

> "Your house is moving East to West at 1000 mph, but the point 6 miles (31680 ft) above your house is moving a little faster 1001.6 mph."
How did you calculate 1001.6 mph?

> "You will find yourself   1.6/12  = 1.33 miles = 440 ft East of your starting point."
I'm no mathematician, but I thought 1.6/12 = 0.133ish, so which of those figures is wrong, and where did the 1.6 and 12 come from?

If you are able to give summary answers to Q1 - Q4, that would also be great.

1.6/12 = 0.133 = 440 ft     You sa my typo before I corrected it.

1001.6 = (3820 + 6)*2*Pi/24    
The point above your house moves through a bigger circle than your house.

The 1.6 is the difference in the speed of your house and the point above it in mph.
5 minutes is 1/12th of an hour.

And another mistake.  Dawn seems to move East to West, but really the Earth rotates West to East under a fixed (for this question) Sun.  So you wind up 440 ft west of your starting point.

Q1  Gravity only works vertically.
Q2  Gravity only works vertically.
Q3  The tangential velocity of the flyer/lifter is entirely due to the motion of the surface of the Earth.
Q4  BSEE (a variety of applied physics)  MIT  1970's
Gravity exerts a force on an object which, in the absence of other forces, will accelerate that object toward the center of the Earth.

For an object on the surface of the Earth, gravity applies the force perpendicular to the surface that gives rise to friction.  Friction between the object and the surface of the Earth is what allows them to match rotational velocities.
tel2Author Commented:
Thanks for that, d-glitch.

> "1001.6 = (3820 + 6)*2*Pi/24"
Where does the 24 come from?  24 hours in a day?  Wouldn't that give you the distance traveled in 1 hour (i.e. 1/24th of a rotation)?  Shouldn't it be:
   (3820 + 6)*2*Pi/24/12
to give the distance traveled in a 5 minute period?

> "...go up six miles very fast..."
The ideal is to get up there (and back down after the 5 mins) instantaneously, to make your calculations as accurate as possible, right?
 (3820 + 6)*2*Pi/24/12
to give the distance traveled in a 5 minute period?
6*2*Pi/24/12 = 1.6/12 = 0.133  would be the difference in distances traveled in a 5 minute period

Or, lets approximate 30000 feet as 9000 meters, and reduce the acceleration of gravity to 0.8m/s^2
so we can toss a vaccumcraft upward at 120m/s, letting it rise to 9000m over 150 seconds and fall back over 150 seconds
I'll make the circumference of the Earth 40,000,000 meters, and put your house at the equator.

The  circumferential velocity of the craft at launch would be 40000000/(24*60*60)m/s
(which is almost 4 times the radial launch velocity, and could make it stay up for a little longer than 300 seconds, but lets ignore that for now)
The  circumferential velocity of the top of the 9000 meter next to your house would be 9000*2*pi/24/60/60 m/s more than that, or about 0.6545m/s
The craft would be in a partially geosynchronous orbit with an apogee about 9000 meters up, and a perigee that it will never reach because it is under ground.
But instead of getting into orbital mechanics, lets make the force of gravity constant with distance, instead of 1/r^2, and look at it from a rotating frame of reference.
Viewed from above the North pole, an observer rotating will the earth will see the path of the craft rotate clockwise under the influence of a fictitious Coriolis force at 360 degrees per 24 hours, or 1.25 degrees over 5 minutes.
While going up, the path would appear to deviate to the west, and while coming down, it would appear to deviate to the east.
A rough numerical simulation steping through a second at a time estimates the net effect to be that it lands about 130 meters to the west,
which is not too far from where it lands under d-glitch's scenario.

If you add an atmosphere, then the lateral drift would probably be limited to less than the diameter of the craft, depending on its density.
(note that drag is non-linear, so lateral drag is amplified by vertical velocity, and the 30000 feet of atmosphere it crosses probably outweighs the craft)

(note too that with constant instead of 1/r^2 gravity, the circumference of the Earth does not affect the answer (until it gets close to 130m) and you can think of the drift as being mainly due to the rotation of the gravity vector over the 5 minutes relative to the initial velocity vector)
tel2Author Commented:
Thanks for that, ozo.  Couldn't have said it better myself (trust me).  Please accept this post as a virtual honourary physics PhD, which means you now meet my original requirements.

However, would this be a good time to break it (gently) to you all that I've since discovered that the earth is not rotating at all?  It came at a bit of a shock initially, but the evidence backs it up:
Furthermore, this explains why I don't feel dizzy all the time, and why I can't get to the shops at 1000mph by just jumping directly up and down.

Working out how to grade your answers may take a university degree, so please hold while I finish off the one I almost finished 30 years I.T.
Assuming this is not an example of Poe's law, stellar parallax is seen, Foucault pendulums do precess, the atmosphere rotates with the earth, and the video seems to confuse statements made in different reference frames.
tel2Author Commented:
The '8)' the end of my comment was intentional, ozo.  But that looked like good scientific analysis in your last post (not that my undergraduate mind can really judge such things - that's why I called you guys in), and I hope you found those last 2 links entertaining.
tel2Author Commented:
Before I grade this...

Hi Mike,
It's about time I thanked you for your post - thank you - sounds reasonable.  Yes, my definition of 'travelling straight up' would be to keep directly above my house.

Hi Viki,
Thanks for that helpful link, which agrees with the 2 things which I had thought were the main contributors, i.e. momentum and atmosphere.  Of course, as ozo has pointed out, we could also say that gravity is a significant contributor because it holds the atmosphere in place.

Hi ozo,
Your answer to Q3 sounds reasonable to me, and weeks ago I had been thinking/guessing along the same lines myself, but I didn't really wasn't sure about why the contribution would move from aircraft momentum to atmospheric movement.  Would you (or anyone else) say that this is because, as more time has been spent in the (moving) atmosphere, there has been more time for the movement of the atmosphere to act on the aircraft, and so we could say that its original momentum would make less difference to its velocity over time?
As we have seen, we land nearby even without an atmosphere, which says that we can account for most of the movement by only considering the momentum of the craft.  (although quantifying "most" depends on whether the "total" movement  is considered with respect to a non-rotating frame in which the house moves 83 miles in 5 minutes, as 60000 feet up and down, as the distance from the house when it lands, or something else)
If the craft were a hot-air balloon, it would be clear that its motion would be strongly influenced by the atmosphere.
On the other hand, if the craft were a 42 lb 6.7 in cannon ball, then the 6.7" diameter, 30,000' column of air (which would be about 7/10 of the atmosphere since the pressure at that altitude is about 3/10 atmospheres) that it passes through would weigh about 360 lbs, over 8 times more than the cannon ball, so its momentum contribution would have to be proportionate.
Craft with wings and propellers would probably interact with a lot more air compared to its mass than a cannon ball, but they can also choose to move independently of the atmosphere, which would make most of its movement dependent on its means of navigation.  If it's inertial navigation, it could be purely the momentum of the craft/instruments.   If it's visual, it may be the motion of the house, and perhaps of clouds. If it's GPS, satellite positions are strongly influenced by gravity and momentum, and even relativistic effects of gravity would have to be taken into account.
tel2Author Commented:
Well explained, ozo.

Thanks very much to all of you for all your helpful posts.

The requirement of having a physics degree was for the sake of the friend of mine who was not accepting the posts I had previously pointed him to, which backed up my analysis of this issue, and he suggested I talk to a physics student (but I didn't know any).  He was saying that gravity was the main contributor, while I thought the aircraft's momentum and the movement of the atmosphere with the earth were the main things.  Anyway, it was good to hear from people without physics degrees, too.

You all deserve more points that I've awarded above, but there are only 500 to go around, and lacking a maths degree, it's hard to decide how to fairly distribute them.  Often university graduates get paid more, so I guess I have to bear that in mind.  8)
tel2Author Commented:
But ozo, I have one more question.  It's about your first comment, which was:
"The secondary effect you speak of is the Coriolis effect, which for a 5 minute trip, is a small effect compared to atmospheric influences."

You seem to be saying that the Coriolis effect would be small compared to atmospheric influences.  Aren't they the same thing?  The secondary effect I mentioned was atmospherics.

Please explain.

The secondary effect I mentioned was atmospherics
Then I misunderstood.
Coriolis effect does influence winds, but since I was assuming a still atmosphere, relative to the 30000 ft anemometer next to your house, the effect of the atmosphere would be to damp the drift caused by Coriolis deflection of the path of the aircraft.
tel2Author Commented:
Thanks ozo.

And thanks again for the links, Viki!
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