Pinemarten

2003-May-14, 08:37 AM

An evacuated hole, pole to pole, a little bigger than the bowling ball.

How would it act?

How would you word the inevitable poll?

How would it act?

How would you word the inevitable poll?

View Full Version : Bowling Ball, Empty Hole, Pole to Pole, and Poll.

Pinemarten

2003-May-14, 08:37 AM

An evacuated hole, pole to pole, a little bigger than the bowling ball.

How would it act?

How would you word the inevitable poll?

How would it act?

How would you word the inevitable poll?

Nanoda

2003-May-14, 09:30 AM

-?

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Pinemarten

2003-May-14, 10:03 AM

I guess I should have explained it. C2C with the BA reminded me of it.

A hole is drilled through the earth, pole to pole; a little larger than a bowling ball.

The hole is Vacuum.

A bowling ball is dropped (placed ?) down the hole.

How would it act ?

How would we word the poll?

Edited to better define a vacuum.

A hole is drilled through the earth, pole to pole; a little larger than a bowling ball.

The hole is Vacuum.

A bowling ball is dropped (placed ?) down the hole.

How would it act ?

How would we word the poll?

Edited to better define a vacuum.

Glom

2003-May-14, 11:45 AM

If you ignore friction, I'd expect that it would just oscillate back and forth between the north and south pole. That sounds like one hell of an M3 question. Perhaps AEA?

kilopi

2003-May-14, 02:54 PM

The Straight Dope column on this subject (http://www.straightdope.com/classics/a1_165.html) seems to be short on details. Ignoring friction, you would oscillate back and forth--and the period of the oscillation is the same as the orbital period for an object right at the surface of the Earth. In other words, if you fell through the hole with the space station right over your head, you'd arrive at the other end in time to see it pass over.

ToSeek

2003-May-14, 03:00 PM

The Straight Dope column on this subject (http://www.straightdope.com/classics/a1_165.html) seems to be short on details. Ignoring friction, you would oscillate back and forth--and the period of the oscillation is the same as the orbital period for an object right at the surface of the Earth. In other words, if you fell through the hole with the space station right over your head, you'd arrive at the other end in time to see it pass over.

And if you ignore pesky little considerations like friction, an object dropped in any hole drilled through the Earth, from any point to any point, will take the same amount of time to get to the other end.

And if you ignore pesky little considerations like friction, an object dropped in any hole drilled through the Earth, from any point to any point, will take the same amount of time to get to the other end.

darkhunter

2003-May-14, 07:01 PM

Years ago had that as an extra credit problem in physics. Unfortuanatly I;'m rusty enough at it I can't replicate the math at the moment, but it works out aound 42 (coincidence) minutes. Every one else used the simple harmonic motion formulae, but (me being me) I put it in orbit and proved it would take the same amount of time to as the simple harmonic motion.

Which, of course, ment I had to start over for the extra-extra credit of running it through a chord.....

Which, of course, ment I had to start over for the extra-extra credit of running it through a chord.....

kurtisw

2003-May-14, 08:09 PM

The Straight Dope column on this subject (http://www.straightdope.com/classics/a1_165.html) seems to be short on details. Ignoring friction, you would oscillate back and forth--and the period of the oscillation is the same as the orbital period for an object right at the surface of the Earth. In other words, if you fell through the hole with the space station right over your head, you'd arrive at the other end in time to see it pass over.

This is what would happen if, as the question is worded, the hole runs

from pole to pole. But if the hole runs in any other direction, there are

those nasty Coriolis forces to worry about.

This is what would happen if, as the question is worded, the hole runs

from pole to pole. But if the hole runs in any other direction, there are

those nasty Coriolis forces to worry about.

snowcelt

2003-May-14, 08:54 PM

Wooo! Got into a good argument about this one. Just read Kilopi's hot link and all that we could figure was confirmed. This is where the argument broke down. If we turned the earth into some type of constant earth, would this pendullum like thing actually stop the earth from spinning? I assume that the perpetual motion like action has to get energy from somewhere. Where would this energy come from? Or better yet, if the time span became long enough, would the universe stop?

RichField

2003-May-15, 12:46 AM

You start with the potential energy of the ball above an arbitrary point, in this case taken to be the center of the earth. Assuming a reversible system, the energy just keeps changing from potential, to kinetic and back. You're not really taking it from anywhere.

If it's non conservative, the energy would be dissipated from the ball, not from the rest of the universe.

If it's non conservative, the energy would be dissipated from the ball, not from the rest of the universe.

kilopi

2003-May-15, 02:20 AM

An oscillation like that doesn't need energy--except to overcome friction, as RichField says. If you include friction, the process stops.

This is what would happen if, as the question is worded, the hole runs from pole to pole. But if the hole runs in any other direction, there are those nasty Coriolis forces to worry about.

Not if you ignore friction.

This is what would happen if, as the question is worded, the hole runs from pole to pole. But if the hole runs in any other direction, there are those nasty Coriolis forces to worry about.

Not if you ignore friction.

Pinemarten

2003-May-15, 03:22 AM

I edited an earlier post at this point.

I changed 'a vacuum' to 'Vacuum'.

I changed 'a vacuum' to 'Vacuum'.

Grey

2003-May-15, 03:55 AM

And if you ignore pesky little considerations like friction, an object dropped in any hole drilled through the Earth, from any point to any point, will take the same amount of time to get to the other end.

I recall hearing someone suggest this for a train system. You just have to overcome trivial engineering details like building stable tunnels through magma (and The Core clearly shows us that this shouldn't be a big problem at all :) ), and put really, really good bearings on the wheels of your train, and you can set up a subway linking any two points on the planet. It doesn't take much power to operate, since you coast from start to finish, and anywhere you go takes 40 minutes!

I recall hearing someone suggest this for a train system. You just have to overcome trivial engineering details like building stable tunnels through magma (and The Core clearly shows us that this shouldn't be a big problem at all :) ), and put really, really good bearings on the wheels of your train, and you can set up a subway linking any two points on the planet. It doesn't take much power to operate, since you coast from start to finish, and anywhere you go takes 40 minutes!

kilopi

2003-May-15, 04:10 AM

It doesn't take much power to operate, since you coast from start to finish, and anywhere you go takes 40 minutes!

How far is the horizon from a really tall building? Let's see, the triangle would have R as one leg and h+R as the hypotenuse, so line-of-sight to horizon would be the square root of the difference of those squared, and then times two, to get to the next building. So, take an elevator to the top of a hundred story building, get on the superconducting track and you can go 80 miles in a little over 40 minutes. No drilling through the core, but no stops. Have to do something about wind resistance too.

Is that right? That's only a one half percent grade.

How far is the horizon from a really tall building? Let's see, the triangle would have R as one leg and h+R as the hypotenuse, so line-of-sight to horizon would be the square root of the difference of those squared, and then times two, to get to the next building. So, take an elevator to the top of a hundred story building, get on the superconducting track and you can go 80 miles in a little over 40 minutes. No drilling through the core, but no stops. Have to do something about wind resistance too.

Is that right? That's only a one half percent grade.

Pinemarten

2003-May-15, 04:25 AM

Is my math right?

My car can peak at 135mi in 60min, and average 78.6mi/hr from Brooks to Salmon Arm. I think I noticed the Rockies on the trip.

My car can peak at 135mi in 60min, and average 78.6mi/hr from Brooks to Salmon Arm. I think I noticed the Rockies on the trip.

Grey

2003-May-15, 05:52 AM

Is that right? That's only a one half percent grade.

You know, back when I heard this, it included the 40 minute figure, and I never checked the math. But now I had to. :) I'm using the Sears Tower, 110 stories, 442 meters high. We imagine a second building, just as high, and your calculation puts it at a distance of 150 km to just skim the planet's surface at the middle. Using conservation of energy to calculate the midpoint velocity, we have:

T = mgh

K = (1/2)mv^2

mgh = (1/2)mv^2

v^2 = 2gh

And we get a peak velocity of 335 km/h. Since the acceleration will be constant, the mean velocity will be just half that, giving us a total transit time of 54 minutes. Not the number I'd heard, but still making pretty good time. :)

I've confirmed that the transit time doesn't vary with changing the height, at least so long as h << R, so if you want to cover more ground, just get taller buildings, or resign yourself to digging. As soon as h starts getting comparable to R, you have to start worrying about the fact the g is no longer constant. This is left as an exercise to the reader. :D A reality check is that if you assume that g does remain constant, the transit time through the planet's center would be 38 minutes.

If you had buildings 10 times as tall (4.4 km! :o ), or are willing to tunnel that deep, you'd cover 475 km in the same 54 minutes, with a mean velocity of 530 km/h. If you had 20 km tall towers, you could space them every 1000 km and you'd get better transit time than flying. Now we just need the engineers to work out the details! :wink:

You know, back when I heard this, it included the 40 minute figure, and I never checked the math. But now I had to. :) I'm using the Sears Tower, 110 stories, 442 meters high. We imagine a second building, just as high, and your calculation puts it at a distance of 150 km to just skim the planet's surface at the middle. Using conservation of energy to calculate the midpoint velocity, we have:

T = mgh

K = (1/2)mv^2

mgh = (1/2)mv^2

v^2 = 2gh

And we get a peak velocity of 335 km/h. Since the acceleration will be constant, the mean velocity will be just half that, giving us a total transit time of 54 minutes. Not the number I'd heard, but still making pretty good time. :)

I've confirmed that the transit time doesn't vary with changing the height, at least so long as h << R, so if you want to cover more ground, just get taller buildings, or resign yourself to digging. As soon as h starts getting comparable to R, you have to start worrying about the fact the g is no longer constant. This is left as an exercise to the reader. :D A reality check is that if you assume that g does remain constant, the transit time through the planet's center would be 38 minutes.

If you had buildings 10 times as tall (4.4 km! :o ), or are willing to tunnel that deep, you'd cover 475 km in the same 54 minutes, with a mean velocity of 530 km/h. If you had 20 km tall towers, you could space them every 1000 km and you'd get better transit time than flying. Now we just need the engineers to work out the details! :wink:

kilopi

2003-May-15, 12:58 PM

And we get a peak velocity of 335 km/h. Since the acceleration will be constant, the mean velocity will be just half that, giving us a total transit time of 54 minutes. Not the number I'd heard, but still making pretty good time.

The reason that you don't get the number you remember is that assumption of constant acceleration. The speed at the center is correct because of conservation of energy, but at the center, the path is flat--that is, no acceleration. The acceleration decreases as you go "down", so you start off with higher acceleration, which mean a higher average speed. The number should be smaller than the shuttle orbital time, so it's close to 43 minutes I believe.

That really illustrates the absurdity of that notion. I said before that the grade at the tower was half a percent, but it's actually two percent (150/6378), isn't it? And it decreases to zero at the center. Can anyone imagine having a track with a two percent grade that can accelerate something by gravity alone up to a speed of over three hundred kilometers per hour? You might as well let them drop down the elevator shaft, and then run level with the ground until they get where they're going.

The reason that you don't get the number you remember is that assumption of constant acceleration. The speed at the center is correct because of conservation of energy, but at the center, the path is flat--that is, no acceleration. The acceleration decreases as you go "down", so you start off with higher acceleration, which mean a higher average speed. The number should be smaller than the shuttle orbital time, so it's close to 43 minutes I believe.

That really illustrates the absurdity of that notion. I said before that the grade at the tower was half a percent, but it's actually two percent (150/6378), isn't it? And it decreases to zero at the center. Can anyone imagine having a track with a two percent grade that can accelerate something by gravity alone up to a speed of over three hundred kilometers per hour? You might as well let them drop down the elevator shaft, and then run level with the ground until they get where they're going.

Grey

2003-May-15, 03:27 PM

The reason that you don't get the number you remember is that assumption of constant acceleration. The speed at the center is correct because of conservation of energy, but at the center, the path is flat--that is, no acceleration. The acceleration decreases as you go "down", so you start off with higher acceleration, which mean a higher average speed. The number should be smaller than the shuttle orbital time, so it's close to 43 minutes I believe.

The shuttle's orbital period is 90 minutes, so we're well under that regardless. Or are you suggesting that it should be under half the orbital time? That would make sense. I suppose that assuming constant g does change the shape of the track, so I'll have to go back and work it out in detail. :-? That will have to wait until I have a bit more time, though.

You might as well let them drop down the elevator shaft, and then run level with the ground until they get where they're going.

Well, you might want a bit of grade, or that sharp turn at the bottom would be a bit tricky to pull out of. Part of the reason for a smooth trajectory is to make sure that you aren't subjecting the passengers to high g-force curves. Othersise, the train arrives nicely if you manage to keep it on the track, but the passengers are all dead. :(

The shuttle's orbital period is 90 minutes, so we're well under that regardless. Or are you suggesting that it should be under half the orbital time? That would make sense. I suppose that assuming constant g does change the shape of the track, so I'll have to go back and work it out in detail. :-? That will have to wait until I have a bit more time, though.

You might as well let them drop down the elevator shaft, and then run level with the ground until they get where they're going.

Well, you might want a bit of grade, or that sharp turn at the bottom would be a bit tricky to pull out of. Part of the reason for a smooth trajectory is to make sure that you aren't subjecting the passengers to high g-force curves. Othersise, the train arrives nicely if you manage to keep it on the track, but the passengers are all dead. :(

kilopi

2003-May-15, 04:16 PM

The shuttle's orbital period is 90 minutes, so we're well under that regardless. Or are you suggesting that it should be under half the orbital time?

Yes, the period of the oscillation is about the same, but in our example, we only go from one side to the other, and not a full oscillation.

Yes, the period of the oscillation is about the same, but in our example, we only go from one side to the other, and not a full oscillation.

O_Desfibrador

2003-May-19, 05:09 PM

Grey wrote

Using conservation of energy to calculate the midpoint velocity, we have:

T = mgh

K = (1/2)mv^2

mgh = (1/2)mv^2

v^2 = 2gh

As Kilopi wrote, acceleration is not constant. (The reason for this is that you only feel the pull of the mass thatīs closer to the centre that you). But you are also counting the surface of the Earth as the zero level of the potential energy, which means you must have negative potential energy inside the Earth. That means that your kinetic energy must be higher. If we use the formula for the harmonic oscilator to calculate the speed at the centre (maximum speed, zero acceleration) we get Vmax=w*R, where w is the angular frecuency and R the amplitude of the oscillation (Earth radius). That translates in a top speed of about 8 km/s. Now, thatīs fast. (Remember that the acceleration is never higher than that at the surface)[/b]

Using conservation of energy to calculate the midpoint velocity, we have:

T = mgh

K = (1/2)mv^2

mgh = (1/2)mv^2

v^2 = 2gh

As Kilopi wrote, acceleration is not constant. (The reason for this is that you only feel the pull of the mass thatīs closer to the centre that you). But you are also counting the surface of the Earth as the zero level of the potential energy, which means you must have negative potential energy inside the Earth. That means that your kinetic energy must be higher. If we use the formula for the harmonic oscilator to calculate the speed at the centre (maximum speed, zero acceleration) we get Vmax=w*R, where w is the angular frecuency and R the amplitude of the oscillation (Earth radius). That translates in a top speed of about 8 km/s. Now, thatīs fast. (Remember that the acceleration is never higher than that at the surface)[/b]

kilopi

2003-May-19, 05:17 PM

As Kilopi wrote, acceleration is not constant. (The reason for this is that you only feel the pull of the mass thatīs closer to the centre that you).

No. In this example, the path is never inside the Earth.

The reason that the acceleration is not constant is that the angle of the path changes as the path is traversed and, to a limited extent, because the distance to the center changes. And that would be true even for a path that went through the Earth, although there would be the other effect as well (EXcept, as we've talked about before, the acceleration due to gravity doesn't really decrease substantially in the real Earth until you get past the core-mantle boundary, half way to the center of the Earth. I think the last time we discussed this was in the movie The Core thread.)

No. In this example, the path is never inside the Earth.

The reason that the acceleration is not constant is that the angle of the path changes as the path is traversed and, to a limited extent, because the distance to the center changes. And that would be true even for a path that went through the Earth, although there would be the other effect as well (EXcept, as we've talked about before, the acceleration due to gravity doesn't really decrease substantially in the real Earth until you get past the core-mantle boundary, half way to the center of the Earth. I think the last time we discussed this was in the movie The Core thread.)

O_Desfibrador

2003-May-19, 05:26 PM

Sorry, I misunderstood the example. Should have slept last night, instead of going to that stupid party. And now I must go again to the University to do some programming. I wonder who made my timetable. I generally donīt like classes later than 19:30, but thatīs Spanish organisation for you.

AstroSmurf

2003-May-20, 09:08 AM

It would be a quite complicated motion, actually, due to two things. This is even if we ignore friction.

First of all, the gravitational forces would vary depending on how far the bowling ball has gotten. This has nothing to do with being "inside" the Earth - for instance, the gravity inside a uniform spherical shell is exactly zero. For a uniform sphere, the force would vary linearly with the distance from the centre, but as the Earth isn't uniform, some integrating would be needed.

Secondly, there are the coriolis forces. We've disregarded friction, but even without it the ball won't be falling freely anywhere but at the poles. Anywhere else, and it'll be skating downwards along one side of the hole until it reaches the centre, and then switch to skating upwards along the opposite side. Even disregarding friction, this would affect the fall, though I can't say how without digging into the math. My guess is that the normal forces wouldn't actually affect the oscillation movement, but I can't swear to it.

// The Astro Smurf

First of all, the gravitational forces would vary depending on how far the bowling ball has gotten. This has nothing to do with being "inside" the Earth - for instance, the gravity inside a uniform spherical shell is exactly zero. For a uniform sphere, the force would vary linearly with the distance from the centre, but as the Earth isn't uniform, some integrating would be needed.

Secondly, there are the coriolis forces. We've disregarded friction, but even without it the ball won't be falling freely anywhere but at the poles. Anywhere else, and it'll be skating downwards along one side of the hole until it reaches the centre, and then switch to skating upwards along the opposite side. Even disregarding friction, this would affect the fall, though I can't say how without digging into the math. My guess is that the normal forces wouldn't actually affect the oscillation movement, but I can't swear to it.

// The Astro Smurf

informant

2003-May-20, 12:28 PM

This is what would happen if, as the question is worded, the hole runs

from pole to pole. But if the hole runs in any other direction, there are

those nasty Coriolis forces to worry about.

And that nasty rotation of Earth too, I think. The hole would rotate while you were falling, and you would clash into its walls.

from pole to pole. But if the hole runs in any other direction, there are

those nasty Coriolis forces to worry about.

And that nasty rotation of Earth too, I think. The hole would rotate while you were falling, and you would clash into its walls.

RichField

2003-May-20, 02:21 PM

Secondly, there are the coriolis forces. We've disregarded friction, but even without it the ball won't be falling freely anywhere but at the poles. Anywhere else, and it'll be skating downwards along one side of the hole until it reaches the centre, and then switch to skating upwards along the opposite side. Even disregarding friction, this would affect the fall, though I can't say how without digging into the math. My guess is that the normal forces wouldn't actually affect the oscillation movement, but I can't swear to it.

// The Astro Smurf

I agree it shouldn't affect the oscillation without friction. Like you said, I'm not really digging into the math either, but if you turn it on it's 'side', apply a forcing function to simulate the variable gravitational force and vertical (positive or negative) forces to represent coriolis forces, without friction, the ball just slides in a channel. The normal forces prevent motion in the (now) vertical direction, but without friction can't slow the motion or induce rotation in the ball.

The only thing I'm not sure about is whether we're talking about the original problem (that's how I interpreted it). Or kilopi's similar topic to which I'm not sure my above statements are applicable.

Rich

// The Astro Smurf

I agree it shouldn't affect the oscillation without friction. Like you said, I'm not really digging into the math either, but if you turn it on it's 'side', apply a forcing function to simulate the variable gravitational force and vertical (positive or negative) forces to represent coriolis forces, without friction, the ball just slides in a channel. The normal forces prevent motion in the (now) vertical direction, but without friction can't slow the motion or induce rotation in the ball.

The only thing I'm not sure about is whether we're talking about the original problem (that's how I interpreted it). Or kilopi's similar topic to which I'm not sure my above statements are applicable.

Rich

Pinemarten

2003-May-20, 02:45 PM

I disagree. Without doing the 'math'. The ball would follow the oscillation pattern and eventually spin to coincide with the coriolis force.

No energy added , no energy taken; and no energy to allow it to 'rub' the sides of the hole.

If more precise definitions are needed of the OP, I will be happy to help.

On a humourous note; how would we word the poll?

No energy added , no energy taken; and no energy to allow it to 'rub' the sides of the hole.

If more precise definitions are needed of the OP, I will be happy to help.

On a humourous note; how would we word the poll?

AstroSmurf

2003-May-20, 03:37 PM

Physics is cool.

According to the original scenario, what you get is a really cool way to tell the time. The ball will keep going more or less forever, oscillating from pole to pole.

If you drill the hole anywhere but at the poles, you'll get the same effect, only the ball will go skating along the sides. There is no way for the ball to avoid skating along the side.

Very simple explanation: If you think about the top end of the hole, and a point halfway down to the centre, the uppermost point is spinning twice as fast around the earth's axis than the the point halfway down. So, when the ball is falling, it needs to slow its "sideways" motion relative to the centre. After it's passed the centre and starts rising, it needs to accelerate the sideways motion again, to keep up with the sides of the hole. After it reaches the other end and starts falling again, it needs to slow down, and so on.

This is what people call coriolis forces, but they're simply an effect of inertia, similar to the centripetal or centrifictional force. The same forces can be felt if you try to climb towards the midpoint of a spinning merry-go-round, and astronauts who lived in a rotating space station would have to consider these effects as well.

// The Astrosmurf

According to the original scenario, what you get is a really cool way to tell the time. The ball will keep going more or less forever, oscillating from pole to pole.

If you drill the hole anywhere but at the poles, you'll get the same effect, only the ball will go skating along the sides. There is no way for the ball to avoid skating along the side.

Very simple explanation: If you think about the top end of the hole, and a point halfway down to the centre, the uppermost point is spinning twice as fast around the earth's axis than the the point halfway down. So, when the ball is falling, it needs to slow its "sideways" motion relative to the centre. After it's passed the centre and starts rising, it needs to accelerate the sideways motion again, to keep up with the sides of the hole. After it reaches the other end and starts falling again, it needs to slow down, and so on.

This is what people call coriolis forces, but they're simply an effect of inertia, similar to the centripetal or centrifictional force. The same forces can be felt if you try to climb towards the midpoint of a spinning merry-go-round, and astronauts who lived in a rotating space station would have to consider these effects as well.

// The Astrosmurf

RichField

2003-May-20, 05:00 PM

I disagree. Without doing the 'math'. The ball would follow the oscillation pattern and eventually spin to coincide with the coriolis force.

No energy added , no energy taken; and no energy to allow it to 'rub' the sides of the hole.

If more precise definitions are needed of the OP, I will be happy to help.

On a humourous note; how would we word the poll?

Follow which oscillation pattern? The one from pole to pole or the one back and forth across the hole. Unless you can show where off center forces would be generated on the ball, I don't see how it can start to spin.

I think I may need those more precise definitions.

AstroSmurf:

I don't disagree with your above post and my previous post was agreeing with your assessment that the oscillation movement wouldn't be affected. I'm not sure whether you were responding to something I said or were just making a general statement.

Rich

No energy added , no energy taken; and no energy to allow it to 'rub' the sides of the hole.

If more precise definitions are needed of the OP, I will be happy to help.

On a humourous note; how would we word the poll?

Follow which oscillation pattern? The one from pole to pole or the one back and forth across the hole. Unless you can show where off center forces would be generated on the ball, I don't see how it can start to spin.

I think I may need those more precise definitions.

AstroSmurf:

I don't disagree with your above post and my previous post was agreeing with your assessment that the oscillation movement wouldn't be affected. I'm not sure whether you were responding to something I said or were just making a general statement.

Rich

Zathras

2003-May-20, 07:32 PM

I disagree. Without doing the 'math'. The ball would follow the oscillation pattern and eventually spin to coincide with the coriolis force.

. . .

No spinning should take place, unless there is friction or the edges are jagged. If the tunnel walls are smooth, the force from the wall on the sphere would be directed normal to the wall, and therefore normal to the surface of the sphere, since the wall should be tangent to the sphere. If the force is tangent to the sphere, it will be directed through the center of the sphere. Since the radius and force are parallel, there will be no "lever arm," and therefore no torque, and therefore no spinning.

. . .

No spinning should take place, unless there is friction or the edges are jagged. If the tunnel walls are smooth, the force from the wall on the sphere would be directed normal to the wall, and therefore normal to the surface of the sphere, since the wall should be tangent to the sphere. If the force is tangent to the sphere, it will be directed through the center of the sphere. Since the radius and force are parallel, there will be no "lever arm," and therefore no torque, and therefore no spinning.

Pinemarten

2003-May-22, 08:45 AM

I stand corrected. The coriolis force would have no affect on the ball, since it is at the centre of the axis of the greater rotational mass.

The oscillation pattern I was refering to is the popular one; that states that the ball would arrive at an equal distance from the centre of the earth from where it started, and then return to where it started. This should repeat forever.

The oscillation pattern I was refering to is the popular one; that states that the ball would arrive at an equal distance from the centre of the earth from where it started, and then return to where it started. This should repeat forever.

AstroSmurf

2003-May-22, 09:24 AM

RichField: I was disagreeing with Pinemarten, but I may have misunderstood his note. As I read it, he argued that the ball would eventually fall freely through the hole no matter where it was drilled - my point was that anywhere but at the poles, it would have to skate along the sides.

The effects of that skating are purely frictional, however, so disregarding friction we'd still get more or less the same oscillation.

I'm not sure what the poll would be about, but I'll give it a go:

Let's say you drill a hole through the Earth, from pole to pole, and drop a bowling ball through that hole. Disregarding friction, the ball would oscillate back and forth between the poles.

How would you use this?

- Practical: Sending mail to the south pole station

- Prank: "Take a look down this hole and tell me what you see"

- Pastime: Can you say "Bungee jump"?

// The Astrosmurf

The effects of that skating are purely frictional, however, so disregarding friction we'd still get more or less the same oscillation.

I'm not sure what the poll would be about, but I'll give it a go:

Let's say you drill a hole through the Earth, from pole to pole, and drop a bowling ball through that hole. Disregarding friction, the ball would oscillate back and forth between the poles.

How would you use this?

- Practical: Sending mail to the south pole station

- Prank: "Take a look down this hole and tell me what you see"

- Pastime: Can you say "Bungee jump"?

// The Astrosmurf

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