Tube Race: Newsletter Challenge (July 2013)

From what elevation, above the Earth's surface is the, "plastic object dropped"?

There ya go.

42 again.

Not if you are Snowden and you don't have a passport!

The challenge is not to give me a number. Prove it!

Ami, solutions for the Gravity Train are all over the internet.

If you don't like the one from Roger Pink, here is one from Purdue University.

I am sort of surprised to see this as a challenge question...

Way to go, Doorman. (Check the answer - it's been posted.)

P.S. This wasn't Roger's question.

Gravity works. The object is Cargonite, it doesn't ever melt.Elevator passenger rides the theoretical gravity train. It free falls through a section of the planet to arrive at a destination on the other side

If it doesn't melt it still loses momentum and will stop short of the other side...Unlike the theoretical gravity train, this is not stipulated to be in a vacuum....

The OP did stipulate no friction, presumably not from air, either.

The moon would still throw it off....

It is fair to assume that the tube is evacuated of atmosphere, since that would be a necessary condition for drag not to be an issue. It therefore would not burn. The mass wouldn't fail to reach the other side simply because it was no longer solid.

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That said, there is still not enough information provided in the problem.

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The fact that the the Earth is a shape more accurately described as an oblate spheroid than as a sphere (and the problem makes no mention of allowing treatment as anything other than what it is) and the fact that the Earth completes a rotation close to every 24 hours means that the position of the tunnels could have a meaningful effect on the transit.

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The transit from the South pole to somewhere on a shore near the equator would likely be very different than an attempt at the reverse trip.

I was going to say the same thing, except they vaporize, and since frictionless, I assume they are falling in a vacuum, the vapor will pyrolyze to small molecules, and then the arrival time of the vapor becomes rather complicated.

Except for the limiting case of straight through, hypocycloidal paths would be faster.

The path straight through wouldn't qualify as an exception because it is still a hypocycloidal (the radius of the smaller circle being 1/2 of the larger circle within which it is rolling).

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It is an exception because there is no speed improvement.

Fair enough.

Wouldn't the tide get in, adding viscous friction to the equation from all that water, as the continents drift and split the tubes?

And what happens to all the excavated spoil?

Doesn't the object come out of the tube the wrong way up?

What tide?

The puzzle stipulated drilling through the earth not Earth.

The first is soil. The second is a planet.

Soil is only on the top.....or bottom....or outside....

Besides the density is not specified other than to be constant....it could be the density of aerogel....in which case the whole gravity thing kinda falls off a cliff....

The answer could easily be an algebraic expression, not necessarily a numeric answer since R and the two distances, d, are not given.

I thought the gravitational force inside a homogeneous sphere was zero....That means the objects would just float....

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell2.html

From the same site:

Example of falling through the Earthl

Good site Lyn. I remember this as an old math problem and being discussed thoroughly in a classroom setting at the time (1960s). The answer is the same for any tunnel tangent to the centre of the Earth. I always wondered what the minimum depth would be to make the gravity train work. Could you excavate a straight line ( not necessarily a tunnel), tangent to the CoE, from one side of NYC to the other and still make the trip 42 minutes with some additional power? Or could a tube be installed under water, where the curvature of the Earth was able to allow submersion? The time to travel between any two points on the Earth would be reduced but some power would still be required to overcome the issues of friction. I suspect you would not want to slide backwards. Or start rocking in the tunnel without brake power.

The answer is the same for any tunnel tangent to the centre of the Earth.

How can anything be tangent to the center of the earth--a point?

Thanks, I stand corrected. The centre of the tunnel is perpendicular to the CoE.

Same issue: How can something be perpendicular to a point?

Yes, step into it again.

Yes, at the barycenter, but as you move away from the barycenter you have a polarized gravitational field.

Any theoretical mass dropped through a large spherical mass will accelerate toward the center. As it does the gravitational forces pulling it toward the center will steadily reduce. However, the falling mass also has momentum and thus continues past the barycenter much like a pendulum swings past dead center.

In a theoretical situation where all friction is removed the falling mass would oscillate back and forth indefinitely, at least in a Newtonian world.

A sphere is by actual definition hollow, and the gravity inside is zero. Not so with a solid ball.

A "ball" cannot be solid. It is the hollow, inside of a sphere.

Or, is it the other way around?

Me thinks we are over-thinking this.

I call time out for shots.

shots??

Actually, the definition extends to both hollow and solid.

Either case will have a gravity null at its very center of mass. Moving off the mass center will induce a gravitational imbalance.

A sphere, being defined as the set of points equidistant from its center, does not include the whole solid. (Unless one actually says "spherical solid" or, informally, "ball".)
I understood that inside a hollow sphere, gravity is zero throughout.

That is also my understanding... isomorphic to the electric field inside a charged conducting sphere.

A true sphere, by that definition, would be massless as there would be no thickness to the shell.

However, if the shell has a specific thickness and it is uniform, then the net gravitational field would be zero as explained Here

Do the paths projected appear to be off center to you? They do to me.

Try to find this expression!

We are allowed to use a sort of reverse logic here. The wording of the question, with no dimensions tells us that it must be a special case with the same time for both tubes. That being true, it must be the same time for a tube running through the center and that is the aforementioned 42.2 minutes.

Time out for shots, again.

I may have to have a nap if this thread keeps going.

Trick question.

'...Each object is a replica of the other....'

I don't think that is possible.

If we take variable x to denote the distance covered by the object in the tube, and let 0 be in the middle of the tube, right above the center of earth. (i.e. the object starts its trip at -√R²-d² and pops out at +√R²-d²)

The acceleration a(x) sensed by the object at each point x is (... some school maths here...)

a(x) = g (r(x) / R) where r(x) is the distance from the center of the earth. I took into account that the acceleration at any point inside earth is due to the mass inside the sphere that lies right below the object, as all mass over it cancels its effect out.

Now, the acceleration that is along the length of the tube is (.... simple trigonometry here...) x / r(x) of the amount above, so eventually, a_x(x) = g . r(x) / R . x / r(x) = g . x / R

Amazingly how, it is equivalent to the gravity elevator passing through the center of the earth, as Doorman and lyn alrady mentioned (Example of falling through the Earthl) Refer to this site for the rest of the solution.

The final conclusion is no matter what's the distance of the tube from the center of the earth (i.e. no matter the d₁ and d₂ in the figure) the physics are the same, so the time of the travel is the same.

Before looking at other replies, my guess is time is same for both. Each object moves in simple harmonic motion and the period is independent of amplitude.

If we apply a bubble level to the two lines, and they are level,as drawn,the balls will not roll forward,they will remain stationary,as on a flat surface.

However, if the lines are straight, as a photon would travel,the balls would not roll from one end to the other,but would tend to roll back to the end from which they were dropped.

At least, that is a carpenter's take on the problem.

Consider a very long highway, laid out with a laser as a guide.Periodic corrections have to be made to prevent building of a ramp,because the laser travels in a straight line, and if followed infinitely,would go off into space,but level is in reference to the curvature and center of the Earth.

A straight line is not the same as a level line.

Imagine you are standing on the object, your body would be in line with the radius and the tube would be a hole, slope down and sideways. The ball would slide down since there is no friction.

down is up when you pass the middle

I don't think it ever would reach the other side. If the force of gravity is the only force involved then as the object travels on a path tangent to the center of the gravitational mass it will continue to move until it reaches the point closest to that center. Allowing for some momentum, it would pass that point and begin slowing down and eventually stop, reverse direction and the process would repeat until it came to rest at the point nearest the center of mass within the constrains of the tube.

Because each of the tubes is almost horizontal to the surface of the earth, you would hear a resounding thud as the plastic object landed on the the wall of the frictionless tube. (I want one of those for my potato cannon) Neither object will move or will only move gradually as lunar cycles nudge it back and forth, as gravity will eventually locate it at the center of that tube. It will take many moons.

Who cares?

a) NONE

b) FOREVER. At midway they reach horizontal position.

Neither will come out as after passing the centre line gravity will not permit it to climb out

Whichever object is dropped first, as it hasn't been stated if the objects are being dropped at the same time.

Also, wouldn't the "plastic" object melt as it approached the center of the tube because of rediant heat from the Earth's core?

---Just saying...

The law of gravity says both will stop at the shortest radius distance from the center off the earth at the same time, which is the center of each objects tube. No object without its own power source can defy gravity and rise against it.

Think again.

Kinetic energy - Wikipedia, the free encyclopedia

It won't stop at the center.

Welcome.

Nope.

For a long discussion that we've already had about how this really works, have a look at this thread, posted in Roger's Equations blog.

The object in tube "B" would reach the end sooner, not because of the length, but because of it's distance from the earths center. Both objects would arrive at the center at the same time, but because of harmonic motion is greater on object "A" it will slow down sooner and longer.

Nope! Object B never sees nearly the gravitational pull on it that A does. Thus travel times are very close.

I've unsubscribed twice from this, but always come back.

It's sort of entertaining to watch people not read stuff, isn't it?

It makes me crazy. Adds to the confusion and misleads the poor OP.

Yep! In addition to the ones that are "intentionally" quips, some seem like they have no idea at all.

32/sec per sec on both.

Only above the surface. Once below, the attractions change.

It'll be zero on the other side, till it starts back "down".

this is a theoretical excercize that bears no relationship to the given problem. So you'd like to keep adding qualifications till it works out that it wall fall right through, and bounce out the other side. Playing with yourself may be fun, but it doesn't really get the job done.

If the "tube" is evacuated, could you not then open one end and use gravity and atmospheric pressure to drive the plastic object through to the other side? This would of course require a minimum "seal" around the plastic object. Friction should be minimal and could be compensated for with lubrication. Simply extend the "tube" through the surface, with an airlock, to accommodate a capsule for transportation and you have a reliable method for getting stuff back and forth quickly with little cost. Evacuation of the "tube" would not have to be complete vacuum but would have to be less than atmospheric. What do ya think?

Once again, what we have here is, failure to communicate", coupled with the burning desire to "be right" at all costs.

Bye.

Ignoring everything that matters except gravity B wins, and does so because it takes less time to fall through a shorter distance. Not gonna do the math, I want to go home.

No, B doesn't win. It's a tie.

I agree with you (for once). There is obviously a closed form solution to this problem, even though many ignored the link to it. I was making a poor attempt at being flippant.

One thing the OP did not point out, are we looking downward from the north pole when we see the diagram, if so, then the rotation of the planet should be factored in.

If we are looking at "earth" from the equator, how does that factor in as well? Nevermind, it is a ball of dirt, not rotating.

As to the poster who asked what we think of his elevator from hell ride with airlocks I say " You take the first ride ", and also there needs to be a sign posted "Abandon ye hope all who enter herein".

a) the shortest way needs the shortest tiime! (acceleration in tube A decreases faster than the acceleration in tube B)

b) the gravitational acceleration inside/at the centerline of the tube is

g_tube=4/3*pi*rho*f*sqr(d²+x²)*cos alpha

where abs(x)<=sqrt(R²-d²) and tan alpha=x/d!

v(t)=g_tube*t

x(t)=g_tube/2*t²

there are small differences to the real gravitational acceleration by reason of the not constant mass environment!

Tube race is sin - not cos!

c) if r=R then the racing object stays at the radius of the earth, if r=R there can be an oscillation - but the oscilaltion time cannot be constant depending from the distance to the centerline

I'll leave the debate about what shape we are really dealing with up to you experts.

However, all depictions of the object in question show a perfect circle.

I'll have to take your word that the circle is perfect. My monitor isn't capable of displaying perfection.

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But even now that you have alerted me to the the perfection of this circle, it still might be a cross section of an oblate spheroid and not necessarily that of a perfect sphere. (I'm assuming you are suggesting the circle represents a sphere and not that two tunnels were somehow drilled through a two dimensional circle.)

OK, it's an imperfect circle. That wasn't the point.

My point is that this anal bickering over such small details is pointless.

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Oh! Were you 'bickering' anally?

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I missed it.

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Listen, if I have to more explicitly announce when I am joking, it is only fair if others announce when they are bickering anally (pointlessly or otherwise).

Maybe I'm just not smart enough to differentiate between the two, after 80 something exchanges.

I'm not adding anything here, so I'm gone.

Well if there is no friction in the tube, then we would have to assume a state of perfect vacuum existed,,,,,theoretically possible or not....and since the true nature of gravity is not known, as far as I know, then it may be possible that gravity is some sort of mini-particle, and thus would not exist in this 'perfect vacuum'.....Therefore the plastic objects would exist in both states, gravitational and non-gravitational, much the same as that poor, but famous kitty, of schrodinger's....so it is both floating and falling at the same time....

Dang, this was the basis of my perpetual motion machine, now it is out.

It would be equal. No friction means no terminal velocity. So object in tube A is traveling much faster at the mid point of the tube where it starts to encounter gravity to slow it down. It slows at the same rate it accelerated. Tube B situation is exactly the same. You don't need to discuss overcoming friction, there isn't any. The question is not how do we build this? So after 1 hour 24 minutes and 24 seconds when the objects return to their respective starting points, we could possibly return to the original problem with an answer. But nobody will like it.

Intuitively, in frictionless tubes through the ''world'', both identical plastic objects would accelerate to the midpoint of their respective tubes, and then decelerate to zero speed when reaching the surface level at the opposite ends, and then fall back, repeating the cycle, endlessly, under uniform gravity. Both objects would reach the other end of their tube simultaneously, if there is absolutely no change, what-so- ever, in the entire gravitational field.

what if d2=R and d1=0?

From the given figure, if d1 were equal to zero, then plastic-ball-one would be at the center of the "earth", and would not move at all...

Likewise, if d2 were equated to R, then plastic-ball-two would be at the south pole, and not moving either...

no comment

Let me correct myself by saying that neither object-one, nor object-two, would be moving at all...

In this case where d1=0, then the ball path is through the exact center of mass of "earth", the object is accelerated through the path and oscillates in its trajectory.

Where d2=R, then this ball's pathlength (tunnel) is zero length, as the ball rests on the pole (presumtively this is the pole) and travels not at all.

This is very interesting, and correct. It complies with all of the data supplied and required in the OP. I can't wait to see the indignation of all of those previous posters who replied time after time that this riddle was previously solved and unworthy of debate. Excellent job, James.

There are more answers, I'm sure. The OP expresses no restriction on lunar gravitational impact, terrestrial rotational impact, seasonal gravitational impact deviation. He does say "the earth" which leaves no doubt as to the intended planet, he does not identify latitude or longitude of these penetrations.

I work with a brilliant software engineer who makes fantastic jumps of logic between relatively unrelated political events. It's really hilarious, and so entertaining, but only occasionally relevant or even informative. It's usually just wrong, driven by conviction, and belief in the sanctity of experiential logic. This posted solution that we've all read by now is completely theoretical, and requires the reader to fill in the blanks with rigid assumptions in order to create a solvable problem. Then just dig in your heels.

Let me append my self-correction by observing that all 4 of the original starting points are located at the end of a radius, each starting from the center of the (world) and pointing outward to the (world surface). When the radius is specified as zero, then the object would start where the radius starts, which is, in all four cases, at the center of the (world), and then follow the (arrowhead) of said radius, which would move zero units of measure from the center of the (world), which would place the object in question exactly at the center of said (world)...

Don't yah just luv semantics?...

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'....Let me append my self-correction by observing that all 4 of the original starting points are located at the end of a radius, each starting from the center of the (world) and pointing outward to the (world surface).....'

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um, what?!? Are you observing the same drawing? There are only two objects, so only two starting points. While there is one ray depicting the radius, it does not align with any of the possible depicted starting points.

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d₁ and d₂ were not used to specify radius. But even if it had been specified that d₁=r=0, then the object would not be at the center of said world, because said world would not have a center, as it would have no volume (and assuming the object, being plastic, does have volume, it couldn't be in the center of something with no volume).

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The paths (A and B) are chords. d₁ and d₂ are the shortest distance from center of the circle to any point on the chord.

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The starting points were understood to be from the surface as can be inferred from the description of 'dropping' the objects into the holes.

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I am quite enamored with semantics, but I get the distinct feeling you haven't really had the pleasure of her acquaintance.