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Sticks
2009-May-07, 01:30 PM
This one is inspired by large bridges, such as the Humber Bridges.

It is known that in real long bridges, they need to take into account the curvature of the Earth, so that the distance between the towers at the top is greater than the distance at the base of the towers.

Suppose you had a very long bridge and so the towers are built so that they are both tracing a line that leads to the centre of the Earth. (See attached diagram)

The bridge deck however is perfectly flat and frictionless. In fact it is parallel to a line through the centre of the Earth. It is perpendicular to the line from the centre of the Earth that bisects the angle the line of the bridge towers make at the centre of the Earth.

The Bridge deck is made of a fictional material that keeps it rigid and does not flex. Assume no issues with weather.

Again see diagram.

Now suppose we take two bowling balls, place one at a bridge tower and one at the centre of the bridge deck.

When the balls are released, will the ball at the tower roll, if so which direction. It is the same question with the one in the centre of the bridge deck. The one in the centre of the bridge deck, if it is placed slightly off centre in either direction towards either tower, what should happen.

Explain your reasoning.

I will wait before I give my thoughts.

:think:

a1call
2009-May-07, 01:54 PM
This concept dates back to the days before there was internet. I am curious to see if anyone can find a reference to it on the www. The concept of a tunnel between Paris and London was originally proposed as a straight tunnel through the Earth. In such a tunnel you would not need to spend any energy to get from one side to the other(except for compensating the loss due to friction). The train would ride downhill and the inertia would keep it rolling on the other side reaching to the top.

The theoretical concept is quite valid.

HenrikOlsen
2009-May-07, 02:18 PM
From what I remember, it can be shown that if the earth is a perfect sphere, with a perfectly homogeneous mass distribution, ignoring that the earth is rotating and assuming the ride is perfectly frictionless and that movement is at essentially non-relativistic speeds, then a straight-line point-to-point gravity-driven connection will result in the same traveling time between any two points on the surface.

Ken G
2009-May-07, 02:43 PM
Yes that theorem would hold pretty well here, even though the bridge is outside the mass of the Earth, if the bridge was not too tall. The time would be the same as the time to fall all the way through the center of the Earth and out the other side. I get that this time is root(2)*pi times the radius of the Earth (6378 km) divided by the escape speed (11 km/s), so that's a time of about 43 minutes. The assumption here is that the bridge is not so high that the escape speed from the top of the bridge is significantly below 11 km/s-- and note that it would be a challenge to climb up such a bridge anyway (the tunnel idea works better except for the drilling through magma part).

Torsten
2009-May-07, 03:11 PM
I remember learning this in my first year physics class. The professor referred to it as the "Universal 45 Minute Subway", or something like that. The train, in this case, would accelerate down the slope from one portal, and after passing through the middle point of the journey, slow as it climbed to the opposite portal, and come to a complete stop at the other end. Now of course, gravity would pull it back to the beginning point again.

I believe the first part of the lesson was a description of the idea with a hole "drilled" through the centre of the earth, and then it was extended to examine the concept for a hole connecting any two points on the Earth's surface. But it was >30 years ago and that's all I remember.

I had wondered if anyone would ever bring up this idea again. Thanks Sticks.

Sticks
2009-May-07, 03:51 PM
I also posted this on a non science based forum (http://www.thisisbigbrother.com/forum/viewthread.php?tid=112539), to see what the response would be, and the post right after my OP got it spot on, with the reasons included and even brought up that the ball would behave like a pendulum :dance:

How does this rate as a thought experiment for getting people to think about physics?

Hornblower
2009-May-07, 04:46 PM
A ball released from any off-center point will roll toward the midpoint. If it is frictionless, thought-experiment style, it will roll past the center, stop as far on the other side as the distance from which it was released, and roll back. It will oscillate indefinitely. In a real world case, friction will damp the oscillation and the ball will come to rest at or near the center.

For small angles, the oscillation will be a close approach to simple harmonic motion, as in the manner of a pendulum.

My reasoning: As we move away from the midpoint, the straight deck rises above the curved surface at the same rate that a concave circular-arc deck or a pendulum bob would rise above a hypothetical flat Earth surface. Not perfect, but a vanishingly small discrepancy for small angles. The elevation in either case is proportional to the square of the distance from the midpoint. Gravity does the rest.

P.S. I wrote this before looking at any other responses.

rommel543
2009-May-07, 05:25 PM
The bowling balls would roll to the center of the bridge.
If you look at the distance to the center of the planet the center of the bridge is actually the closest spot regardless that the deck is a straight line. Gravity would cause the bowling balls to be attracted to the lowest spot, which is the center of the bridge.

robross
2009-May-07, 06:48 PM
The bowling balls would roll to the center of the bridge.
If you look at the distance to the center of the planet the center of the bridge is actually the closest spot regardless that the deck is a straight line. Gravity would cause the bowling balls to be attracted to the lowest spot, which is the center of the bridge.

Yes, but see the post right above yours. After the ball reaches the mid-point, it would continue rolling past it. Gravity would slow it until it stopped at the same distance on the other side of the bridge as its starting distance from the center, then roll back, oscillating forever (the surface is frictionless remember.)

Rob

Jeff Root
2009-May-07, 07:11 PM
Oh, look-- a nit!

If the surface of the bridge is frictionless, the ball would slide, not roll. :)

-- Jeff, in Minneapolis

Ken G
2009-May-07, 07:20 PM
How does this rate as a thought experiment for getting people to think about physics?I think it's pretty good on that scale. It has a kind of "sci fi" component where you might imagine actually doing this (I don't think it's feasible), 43 minutes is an interesting number from the point of view of human travel. It also connects to the orbital time around the planet in interesting ways, and it connects to pendulum motion in interesting ways. So there's a lot of unified physics relating to gravity lurking there, and some human interest too.

JohnBStone
2009-May-07, 07:58 PM
Is there a minimum length required for this to work? Does this also work across the deck as well as along it?

For the ball at the tower, does the gravitational attraction of the tower exceed the imparted downward gravitational force on the ball on the deck for small bridges?

Another nit. Bowling balls have an irregular mass distribution and shape due to the finger holes, so if they do roll (rather than slide) wont they quickly roll off the edge of the deck?

If the ball does slide, and it appears it should, then the ball should keep its orientation to the center of the Earth - as if it was attached to a straight line passing through the Earths center. So will rotate through the same angle as the angle between the towers.

If it was a brick rather than a ball sliding down the frictionless surface, where would the rotational energy come from which changed its orientation relative to the center of the earth? Would there be Cherenkov radiation?

hhEb09'1
2009-May-07, 09:11 PM
It seems like this idea crops up every time we get the question about falling all the way through the center of the earth, but I just checked a half-dozen of those threads, and I only found a couple, from 2002 (but there's a lot more):
A levitated train in this tunnel would fall at increasing speed, pass the bottom of the arc and start upward, slowing all the way. At the other end it would come to a stop after a 45 minute trip.


The amazing point is that no matter what city one starts from, (whether one drops directly through the center of earth or cuts a chord thru it), the time of transit to Sidney will be exactly the same irrespective of distance or mass of the vehicle -- everyone reaches Sidney in 42.1 minutes. It should be great for tourism.
So start digging. /phpBB/images/smiles/icon_biggrin.gif
G^2

Hornblower
2009-May-08, 12:00 AM
Is there a minimum length required for this to work?In a frictionless thought experiment, no. With real bridges and balls, we would need to go a considerable distance off center to overcome the static friction. Just how far depends on the amount of surface roughness and the deformation characteristics of the ball and the pavement.

Does this also work across the deck as well as along it?
Yes.


For the ball at the tower, does the gravitational attraction of the tower exceed the imparted downward gravitational force on the ball on the deck for small bridges?In theory maybe, depending on the geometry. In a thought experiment we could calculate the minimum slope away from the tower needed to overcome the tower's gravity.


Another nit. Bowling balls have an irregular mass distribution and shape due to the finger holes, so if they do roll (rather than slide) wont they quickly roll off the edge of the deck?That would depend on how wide the deck is, how the mass is distributed in the ball, and how that distribution affects the direction of roll, if at all. I would expect a good ball to be balanced by a dense counterweight between the holes. I don't know whether or not premium bowling balls are made that way.


If the ball does slide, and it appears it should, then the ball should keep its orientation to the center of the Earth - as if it was attached to a straight line passing through the Earths center. So will rotate through the same angle as the angle between the towers.No, the ball's inertia will keep it in a fixed orientation in an inertial frame of reference. It will be virtually fixed with respect to the stars. There is nothing to force it into a synchronous rotation with respect to Earth's center, at least not in a short time. Given enough time, a real ball in orbit around Earth would become tidally locked, but that probably would take billions of years.


If it was a brick rather than a ball sliding down the frictionless surface, where would the rotational energy come from which changed its orientation relative to the center of the earth?
On a straight surface, no torque is needed to keep the brick parallel to the surface. The brick's inertia keeps it that way, regardless of the direction to Earth's center.

If the brick was sliding down a spherically curved surface, then the energy needed to accelerate its rotation would come from gravity. The effect would be to slow down the forward progress slightly.

Would there be Cherenkov radiation?Totally irrelevant. Cherenkov radiation is the result of charged particles travelling near the speed of light and passing through insulating material faster than light through the same material. It has nothing to do with gravitational action on slow-moving objects such as our thought-experiment bowling balls and bricks.

Ara Pacis
2009-May-08, 12:00 AM
In avoiding any issues with weather, are we also avoiding any issues with air resistance? I don't recall mention of a vacuum.

Hornblower
2009-May-08, 12:03 AM
In avoiding any issues with weather, are we also avoiding any issues with air resistance? I don't recall mention of a vacuum.

See post #7. Air resistance would be one source of the real-world friction I discussed.

Sticks
2009-May-08, 04:53 AM
I wonder, if you could stand, at a tower and look towards the other tower, assuming you are tethered to the tower, what the scene would look like.

Any graphic artists out there?

JohnBStone
2009-May-08, 07:15 AM
No, the ball's inertia will keep it in a fixed orientation in an inertial frame of reference. It will be virtually fixed with respect to the stars.
So the rotation of the ball would depend on rotation of the planet with respect to the stars and presumably the angle of the bridge to the direction of rotation.

So unless the bridge was directly aligned with the planet's rotation and on the equator the ball will follow an inertial path and will roll/slide off the deck. A bit like a Foucault pendulum.


No, the ball's inertia will keep it in a fixed orientation in an inertial frame of reference. It will be virtually fixed with respect to the stars.
On a straight surface, no torque is needed to keep the brick parallel to the surface. The brick's inertia keeps it that way, regardless of the direction to Earth's center.
The brick will be rotated in the inertial frame of reference in order to stay flat on the bridge.


Totally irrelevant. Cherenkov radiation is the result of charged particles travelling near the speed of light and passing through insulating material faster than light through the same material. It has nothing to do with gravitational action on slow-moving objects such as our thought-experiment bowling balls and bricks.
Heh, I was thinking there was some relativity impact here and trying to recall the gravitational waves or somesuch that must happen when gravity acts alone on an object in an inertial frame. The frictionless surface means gravity appears to be the only force acting to rotate the brick which is a bit weird. Something odd must happen.

This is an interesting thought experiment.

Hornblower
2009-May-08, 10:22 AM
So the rotation of the ball would depend on rotation of the planet with respect to the stars and presumably the angle of the bridge to the direction of rotation.In my previous remarks I was assuming no spin of the Earth. Let's now add Earth's rotation to the thought experiment, while keeping the system totally frictionless.

Suppose we held the ball stationary with respect to the bridge and gave it no torque when releasing it. The ball will be rotating at the same rate as Earth as seen in an inertial frame, and their spin axes are parallel. They will remain that way regardless of the ball's translational motion along the bridge deck.


So unless the bridge was directly aligned with the planet's rotation and on the equator the ball will follow an inertial path and will roll/slide off the deck. A bit like a Foucault pendulum.Yes, indeed. The Coriolis effect will kick in and the ball will veer off course and fall over the side.


The brick will be rotated in the inertial frame of reference in order to stay flat on the bridge. Correct. The bridge is rotating at a constant rate, and the brick got that same rotation from the bridge before it was released. No further torque on the brick is needed to keep it parallel with the surface of the bridge.


Heh, I was thinking there was some relativity impact here and trying to recall the gravitational waves or somesuch that must happen when gravity acts alone on an object in an inertial frame. The frictionless surface means gravity appears to be the only force acting to rotate the brick which is a bit weird. Something odd must happen.

This is an interesting thought experiment.
The brick was rotating in parallel with the surface when it was released. Again, no force is acting to maintain that rotation.

The Earth and the brick are undergoing accelerated motion under their mutual gravitation. In theory there should be gravitational radiation carrying energy away from this system, but the effect is vanishingly small. If I am mistaken here, I welcome comments from general relativity experts.

Ara Pacis
2009-May-08, 12:18 PM
Wait, if (assuming no air resistance) the ball or brick is set on the deck with the stars as the reference frame, wouldn't it fall to the bridge that rotates away underneath it, land (assuming no bouncing along with no friction), then slide down the deck to the mid-point by gravity but then be accelerated up the other side and flung off that end at around Earth's rotational velocity due to centrepetal force of the bridge that rotates underneath it? I suppose the results would vary depending on where on the bridge the ball was released, but the result would be the same but with different launch velocities.

NEOWatcher
2009-May-08, 01:06 PM
I wonder, if you could stand, at a tower and look towards the other tower, assuming you are tethered to the tower, what the scene would look like.
I doubt anything would be noticable. You'd see a sagging deck, but it takes almost 2 miles for the deck to sag 1 foot.
You'd need an air mask if you want to see some curvature, because it would be the same as seeing the curvature of the Earth.

Why would you need to be tethered?

Sticks
2009-May-08, 03:32 PM
Why would you need to be tethered?

Since it is a frictionless surface I did not want to slip and fall off :whistle:

Anyway I forgot about the rotation of the Earth

Maybe you could run this with the bridge over the pole, then at the Equator in the direction of rotation of the Earth and then Parallel.

galacsi
2009-May-08, 04:37 PM
A ball released from any off-center point will roll toward the midpoint. If it is frictionless, thought-experiment style, it will roll past the center, stop as far on the other side as the distance from which it was released, and roll back. It will oscillate indefinitely. In a real world case, friction will damp the oscillation and the ball will come to rest at or near the center.

For small angles, the oscillation will be a close approach to simple harmonic motion, as in the manner of a pendulum.

My reasoning: As we move away from the midpoint, the straight deck rises above the curved surface at the same rate that a concave circular-arc deck or a pendulum bob would rise above a hypothetical flat Earth surface. Not perfect, but a vanishingly small discrepancy for small angles. The elevation in either case is proportional to the square of the distance from the midpoint. Gravity does the rest.

P.S. I wrote this before looking at any other responses.

I agree with your answer but with two nitpicking :

The gravity at the tower is less than at the center of the bridge , but it does not change anything IMO.

Beware ! If the two balls are released in the same time we risk a collision ! Then even with elastic balls things become messy !

Sticks
2009-May-08, 04:50 PM
BTW Over on this other forum (http://www.thisisbigbrother.com/forum/viewthread.php?tid=112539) where I raised this thought experiment one poster, ange7 seemed to be on the ball, if you pardon the expression. What would be a nice idea would be if some here went over there.

We could have a forum cross over / exchange. :)

tdvance
2009-May-08, 08:04 PM
I wonder, if you could stand, at a tower and look towards the other tower, assuming you are tethered to the tower, what the scene would look like.

Any graphic artists out there?

I'd imagine, you'd see a bridge narrow to a point, one that appears a few miles away and on the celestial sphere, since to the eyes, that's essentially "infinity".

mugaliens
2009-May-10, 12:18 PM
Why would anyone build an exceptionally long bridge deck where the middle is at sea level and the ends are higher than Everest?

I contend the bridge deck would follow the curvature of the Earth. I further contend that for increasingly long distances, successive towers would be used. I finally contend that beyond a certain distance, ferries are more economnically feasible. Case in point: Not too long ago I was looking into making a trip to England and discovered that the rate for going through the Chunnel was prohibitively expensive, whereas I could take an overnight ferry, complete with sleeping quarters, and breakfast, no less, cutting both my costs and travel time signficantly over taking the Chunnel and overnighting in a hotel somewhere.

Alas, I was unable to take the trip as my vehicle had to be shipped before a certain date!

Argh...

hhEb09'1
2009-May-10, 04:16 PM
I contend the bridge deck would follow the curvature of the Earth.From the OP:

It is known that in real long bridges, they need to take into account the curvature of the Earth,

Not too long ago I was looking into making a trip to England and discovered that the rate for going through the Chunnel was prohibitively expensive, whereas I could take an overnight ferry, complete with sleeping quarters, and breakfast, no less, cutting both my costs and travel time signficantly over taking the Chunnel and overnighting in a hotel somewhere.

Alas, I was unable to take the trip as my vehicle had to be shipped before a certain date! I'll have to remember that. :)

Hornblower
2009-May-11, 12:14 AM
Why would anyone build an exceptionally long bridge deck where the middle is at sea level and the ends are higher than Everest?

I contend the bridge deck would follow the curvature of the Earth. I further contend that for increasingly long distances, successive towers would be used. I finally contend that beyond a certain distance, ferries are more economnically feasible. Case in point: Not too long ago I was looking into making a trip to England and discovered that the rate for going through the Chunnel was prohibitively expensive, whereas I could take an overnight ferry, complete with sleeping quarters, and breakfast, no less, cutting both my costs and travel time signficantly over taking the Chunnel and overnighting in a hotel somewhere.

Alas, I was unable to take the trip as my vehicle had to be shipped before a certain date!

Argh...
This is a thought experiment for pondering the laws of inertial and gravitational motion, not an argument for actually building a bridge of such a shape.

mugaliens
2009-May-11, 01:16 AM
Well, ok folks... Interesting thought experiment, Sticks!

publiusr
2009-May-18, 06:54 PM
I'd imagine, you'd see a bridge narrow to a point, one that appears a few miles away and on the celestial sphere, since to the eyes, that's essentially "infinity".

Haze would degrade the view anyway

Kaydeb
2009-May-22, 01:21 PM
I also posted this on a non science based forum (http://www.thisisbigbrother.com/forum/viewthread.php?tid=112539), to see what the response would be, and the post right after my OP got it spot on, with the reasons included and even brought up that the ball would behave like a pendulum :dance:

How does this rate as a thought experiment for getting people to think about physics?
:clap:
I have absolutely no background in physics, but I found it an intriguing question and even got it right (if I understand all of the subsequent posts, of which there is no guarantee!).

Gandalf223
2009-May-22, 04:59 PM
You'd see a sagging deck, but it takes almost 2 miles for the deck to sag 1 foot.

I disagree. The bridge has been postulated as being perfectly flat. Photons travelling from distant parts of the bridge to the viewer's eyes would not be deflected far enough by the earth's gravity to change the bridge's appearance. It would appear perfectly flat; the lines along the edges of the bridge deck would appear perfectly straight and would recede toward infinity.

Since the bridge (as postulated) would be extremely long, it's also reasonable to suppose that the lines would converge so far that the human eye could not separate them, meaning the bridge would actually recede to infinity (visually.)


Haze would degrade the view anyway

Since the bridge deck, at least at the towers, would be above most of the Earth's atmosphere, that might not be a problem. If the height were such that the midpoint of the deck were tangent to the planet's surface, the view would appear to recede to the surface, meeting the Earth at the horizon. As I mentioned above, if it were long enough it could be reduced to a point (visually) before it got there. Of course, that would depend on the width of the deck. If the bridge were higher than that (tangential to the surface) it might be entirely above the hazy atmosphere.

NEOWatcher
2009-May-22, 05:22 PM
I disagree. The bridge has been postulated as being perfectly flat...
Yes and no. I was speaking from the casual observers notion of the surface of the Earth being interpreted as flat. So; Yes, sag is an unfortunate word, but I was trying to get some concept of the middle of the bridge appearing closer to the Earth's surface than the ends.

So; let me clarify that my comment is only perception and could probably be that way due to optical illusion effects, and is totally dependent on the ability to percieve the curvature of the Earth.

jj_0001
2009-May-22, 05:36 PM
So, allow me to add another initial condition, since I think the original problem is quite nicely handled.

Now, take the bowling ball, still on the frictionless surface, and give it a very strong spin, say in a clockwise direction, at 75% of the angular rotation that it can withstand due to centripital accileration, with the spin axis (originally) pointed straight through the center of the earth.

What path will the ball take? :)

Hornblower
2009-May-22, 10:56 PM
So, allow me to add another initial condition, since I think the original problem is quite nicely handled.

Now, take the bowling ball, still on the frictionless surface, and give it a very strong spin, say in a clockwise direction, at 75% of the angular rotation that it can withstand due to centripital accileration, with the spin axis (originally) pointed straight through the center of the earth.

What path will the ball take? :)If there is no friction between the ball and the bridge, the spin will have no effect on the ball's trajectory. The spin axis will remain pointing in the same direction in space, so it will move away from Earth's center as the ball moves over the surface, and even more so as the bridge turns with the Earth's rotation.