Question Re: Giant Hole Through the Earth

[Mathmagic]

Yes, she will come out the other side, as long as she jumps in, rather than stepping in.

How do you figure?

Assuming no air resistance, you will oscillate between the two ends. If you jump in, you will oscillate between just above the earth, to just above the earth on the other side. This way he can easily kick off the side of the tunnel at the end, and make it onto land.

His center of mass will oscillate back and forth. Since the center of mass is somewhere above the bellybutton, it will be easy to climb out of the hole as he emerges from the other side without needing to jump in.

[roundedge]

Without air resistance, the question has already been answered--by roundedge and calculus.

Change the problem to include calculating the air resistance.... And the ride isn't very fun. Because the pressure goes up as she falls, and the acceleration of gravity goes down, her terminal velocity rapidly decreases.

I'd just like to point out that my solution does include air resistance, and the terminal velocity phenomenon is inherent in the solution, as well as the decrease in the gravitational acceleration. However it models air resistance in a simplistic way, by making it directly proportional to velocity. We could include a position dependency to the air resistance term as well, since pressure is inversely proportional to distance from center, but it requires that air resistance increases noticeably with air pressure, and I'm not entirely certain that's the case. I'd imagine the atmosphere would have to undergo a phase change into a liquid before there would be any noticeable effects. Does anybody know if this is the case?

[EricH]

I'd just like to point out that my solution does include air resistance, and the terminal velocity phenomenon is inherent in the solution, as well as the decrease in the gravitational acceleration.

You're right, my bad.

However it models air resistance in a simplistic way, by making it directly proportional to velocity.

Wikipedia says drag force is proportional to density and v². For an ideal gas, density is proportional to pressure....

We could include a position dependency to the air resistance term as well, since
pressure is inversely proportional to distance from center,

Is it? Despite decreasing gravity? I didn't really check whether the increasing density precisely canceled the decreasing acceleration, but my intuition is that it doesn't....

[gmalivuk]

We could include a position dependency to the air resistance term as well, since
pressure is inversely proportional to distance from center,

Is it? Despite decreasing gravity? I didn't really check whether the increasing density precisely canceled the decreasing acceleration, but my intuition is that it doesn't....

Well, it depends on how compressible air is, really. And any calculations we do assuming it's even close to an ideal gas will be way off, considering how high the pressure at the center would be even if air weren't compressible and the planet were of uniform density. (The fact that it's not uniform means that acceleration due to gravity actually remains nearly constant for awhile, because while more of the earth is above you and thus not contributing to the force on you, what's yet to come is denser and thus contributes more to you now than it did when you were farther away from it, on the surface.)

[Shakleton]

Hey guys!

So, this is my first own topic here and therefore I hope I'm not making a complete idiot out of myself but there is a certain highly hypothetical problem that has bothered me for almost a month now. Neither of my friends or relatives could help me out with it.

Ok, so, any doubts left aside, let's just PRETEND:

1. You could dig a whole from the northpole to the southpole of the earth without

1.1 having the whole earth brake apart or collapse
1.2 causing damage to the earth's core so the earth would blow up for another reason
1.3 dying when you reach the core (because you're digging through fluid metal at almost 7000° C)
1.4 creating a giant volcano at the northpole because fluid metal will shoot to the surface
1.5 your whole hole collapsing for the outer pressure to it.
1.6 anything else ridiculus I'm to lazy to think of

2. You could...

2.1 stabalize it, so it won't go down the tubes
2.2 stuff it's walls with an ultra-thermo-material completely keeping the 7000 ° C out of your giant tube


So let's pretend you did all that. You would have a tube from northpole to southpole, let's say 25 meters (or whatever you like) in diameter. My question is now:

What would happen to you, if you would jump in from northpole?

Ok, you'd start to fall. I figured that out myself. But would you fall past earth's core, almost reach the southpole and then start falling again and almost reaching the northpole again so you would end up floating in the middle of the earth? Would you get slower and slower to stop at the center and float right away?

And either way:
What would happen if you made the whole tube a perfect vacuum and jumped in in a capsule so there would be no air preassure to break?

I hope I hitg the right forum for this. If not, please don't kill me.

Would you help me with that problem? I completely don't know what would happen.

Thank you a lot!

Shakleton

P.S: I attached a file to demonstrate what I mean.
P.P.S: I apologize for any mistakes I didn't notice. English isn't my first language so I hope you can forgive me

[SpitValve]

I'm sure we've had this before?

It'll just be a damped oscillation. You'll go back and forth, but you'll reach a lower height each time because of friction until after a very long time you stop.

[arbivark]

Bring a grappling hook, or some electromagnetic boots, or something. Frustrating after 8,000 mile? trip to come up a foot short and fall back down.
Does it matter if the hole is at the poles? I suspect that the experience is the same at any location.
I dimly remember a science fiction story with a similar theme. Guy falls into a parabolic shaped frictionless crater and has to figure out how to get out.

[Shakleton]

I'm sure we've had this before?

Oh, I'm sorry then. I used the forum search and I found nothing and I'm new here for only a couple of days so I didn't get to read much that was posted way ago.

And no, it would not matter if the holes are at the poles. I just used northpole and southpole because they are most likely known as opposite "ends" of the earth.
I read on Wikipedia the earth's radius at the poles was 6357 kilometers. That would be... uhm... where's my calculator? ... stupid American units... when are there going to get the hang of SI-Units (no offense, and yes I know meters are the SI units).... ah, there it is!.... roughly 3950 Miles to the center.

I was just wondering because the acceleration due to gravity would tend to 0 the closer you are to the center. And if you take air friction (does that exist? air friction? I'm not sure!) into account it could slow you down. But what if it was in a vacuum?

Thanks so far anyway!

Shakleton

[yy2bggggs]

I'm sure we've had this before?

viewtopic.php?f=18&t=20308

Not that I think "Physics 'challenge question' ceptnotrly" is a great subject for that thread.

[Shakleton]

Duh! Okay, I didn't see that thread.

So, I think my question is answered. Thread can be closed. Thank you!

Shakleton

[You, sir, name?]

I'm sure we've had this before?

It'll just be a damped oscillation. You'll go back and forth, but you'll reach a lower height each time because of friction until after a very long time you stop.

Are you sure? Wouldn't air density (and subsequently damping coefficient) increase with depth, making it a more complicated differential equation? Furthermore, gravitational force is proportional to the distance from the earth, no? So, you can model it as a spring with the addition of an unpleasant depth dependent drag coefficient.

Something like y''-y'³k(y)+my = C.

[4=5]

if she started falling at the same instant as the hole was created then she wouldn't have a terminal velocity because the air would accelerate right along with her until she smacked right into the air coming the other way and the air hammer effect reversed the giant swirling vortex into a a geyser of air, causing devastation comparable to nuclear explosions in magnatude

a possible model for what might happen
http://oddorama.com/2008/02/25/the-worlds-weirdest-engineering-disaster/

[evilbeanfiend]

I'm sure we've had this before?

It'll just be a damped oscillation. You'll go back and forth, but you'll reach a lower height each time because of friction until after a very long time you stop.

Are you sure? Wouldn't air density (and subsequently damping coefficient) increase with depth, making it a more complicated differential equation? Furthermore, gravitational force is proportional to the distance from the earth, no? So, you can model it as a spring with the addition of an unpleasant depth dependent drag coefficient.

Something like y''-y'³k(y)+my = C.

yes damping and acceleration isn't constant, you are still damped and do still oscillate though.

[John E.]

Are you sure? Wouldn't air density (and subsequently damping coefficient) increase with depth, making it a more complicated differential equation? Furthermore, gravitational force is proportional to the distance from the earth, no? So, you can model it as a spring with the addition of an unpleasant depth dependent drag coefficient.

Something like y''-y'³k(y)+my = C.

I ran into this question regarding digging deep tunnels and the workers getting the bends. I think it was stated that the air pressure increases with depth in a closed hole, such as a tunnel, but when the tunnel was opened to the other side, the air pressure equalized. I think this would be the case here. Still damps, though, just not increasing with depth.

[evilbeanfiend]

I ran into this question regarding digging deep tunnels and the workers getting the bends. I think it was stated that the air pressure increases with depth in a closed hole, such as a tunnel, but when the tunnel was opened to the other side, the air pressure equalized. I think this would be the case here. Still damps, though, just not increasing with depth.

no, proof: the outside is obviously open but if you go high enough the pressure will be different. the closed tunnel thing is a different effect, the gravitational effect would be negligible on the scale of most human made tunnels.

[zenten]

if she started falling at the same instant as the hole was created then she wouldn't have a terminal velocity because the air would accelerate right along with her until she smacked right into the air coming the other way and the air hammer effect reversed the giant swirling vortex into a a geyser of air, causing devastation comparable to nuclear explosions in magnatude

a possible model for what might happen
http://oddorama.com/2008/02/25/the-worlds-weirdest-engineering-disaster/

It would probably be worse than that, if enough air could fit in the tunnel to cause a significant drop in global atmospheric pressure.

[Mathmagic]

Okay, this is inspired by the thread over in General regarding the same thing. I made a post in there, but I'm looking for some help in finishing my calculations:

But the air resistance would slow you down, as at lower forces of gravity, terminal velocity is lower.

Okay, so using conservation of energy, I got the following expression (this is ignoring Coriolis forces, because I don't feel like getting too complicated) :

E_total = E_potential + E_kinetic + Work_{air resistance}

Our total energy is just the energy we started out with (the total gravitational potential energy at t = 0), which is G*m*m_earth/r_earth

Our kinetic energy is simply the usual (1/2)m*v(t)².

The work done by air resistance is simply the line integral of the force exerted by the air resistance taken along the path, which is just along the radius of the earth which we are falling along. The force of air resistance is given by (1/2)*ρ*A*C_d*v(t)², which is exerted in the direction OPPOSITE of motion.

The potential energy is simply the gravitation potential energy, G*m*m_earth(t)/r(t).

The mass of the earth at time t is just the "effective mass", which is (4/3)π*r(t)³*d, where d is the density of the earth. We can assume uniform density for our purposes here.

Substituting all these expressions into the original equation, we get:

G*m*m_earth/r_earth = (4/3)π*G*m*d*r(t)² + (1/2)m*v(t)² + (1/2)ρ*A*C_d*Integral[v(t)²dr, 0..r_earth]

So I'm trying to do the change of variables (get v in terms of r), but I'm kind of stuck right now.

I wrote out the law of motion for this scenario:

F_net = G*m*m_earth(t)/r(t)² - (1/2)ρ*A*C_d*v(t)²

Substituting for m_earth(t) and for F_net, we get

m*dv/dt = G*m*(4/3)π*d*r(t) - (1/2)ρ*A*C_d*v(t)²

=> (dr/dt)*(dv/dr) = G*(4/3)π*d*r(t) - (1/2)ρ/m*A*C_d*v(t)²
=> v(t)*(dv/dr) = G*(4/3)π*d*r(t) - (1/2)ρ/m*A*C_d*v(t)²

=> dv/dr = G*(4/3)π*d*r(t)/v(t) - (1/2)ρ/m*A*C_d*v(t)

Normally, you could do separation of variables and integrate to get v in terms of r, but we have the r(t)/v(t) term which is causing problems. Anyone have any ideas?

[J Spade]

I'm thinking oscillation until she smashes into the magma around her due to planetary rotation.

If the class was at the north or south poles, air resistance will slow her down after a long time, but she'll likely starve or die of dehydration first as she repeatedly flies through her own excretions.

If someone is available at the other end, they can hold down a rope for her to grab at the peak of her oscillation.


Also, there would not be a hole, as all dirt removed by the boy would settle in the core. It would actually be very hard to 'dig' through the core by conventional means.

[cascabel]

I've asked around about this little "thought experiment" I encountered at some random blog, and I'm not too sure if it's been talked about anywhere else, so...

You of course do know that if you're, for instance, buttering a slice of bread and you lose your hold on the slice, it falls. It could be on the tabletop, it could be on your lap, or it could be on the ground. But, it will fall.

And we all of course explain it through the deep concept of gravity. How big bodies (e.g. the Earth) attract small bodies (which would be the bread slice) to themselves.

All orthodox up to this point, yes?

Allow me now to present the conundrum.

It is said that gravitational force tends to attract smaller bodies to as near as they can be to the bigger body's center of mass.

Now, suppose someone did succeed in boring a hole straight through the Earth and passing by right its center. (There's a little ambiguity in where the hole would be placed, like whether it's at the Poles or somewhere on the equator, but let's relegate that argument to the replies... ) So you get a hole that passes through the crust, the mantle, the core... and up until you pass through the crust again at the opposite end. (I'd maybe draw a pic later for easier visualization.)

Now, remember that bread slice I mentioned earlier? Suppose that while buttering it, it accidentally falls into that hole.

What happens to it? Remember that we bored a hole such that there's "nothing" in the center. Will it fall up to a certain point, or would it do something else?

(Assume for the purpose of this thought exercise that this particular slice of buttered bread will not be toasted by the high temperatures inside the Earth, and will thus always remain intact.)

Thoughts?

[Ann_on_a_mouse]

I've heard about these before, they're called gravity sleds. While I don't know much about the fate of your toast, I can speak with some certainty about the mechanics of the sled. The toast would gravitate towards the center of the hole just as if dropped in any other scenario. It can be assumed that the toast would achieve terminal velocity within this time due to the length of the trip. Once it reached the center of the tube, it would briefly reach a point where it had no forces acting on it. Inertia would carry it further. How much further, I can not say. As the link says, though, a human can successfully get out of the hole on the other side.

[Mathmagic]

How is this a "conundrum"?

[thoughtfully]

The main problem is the Coriolis force due to the Earth's rotation, which will cause the falling body to move towards one side of the tunnel, unless the tunnel is between the poles. If the air was evacuated from the tunnel, the falling body would be in Simple Harmonic Motion, just like a pendulum.

It turns out that the natural period (actually, a half-period) of this motion (for a tunnel between any two points on the surface, not only antipodes) is a little over forty-two minutes
http://home.att.net/~numericana/answer/ ... htm#subway

[Bright Shadows]

On a side note, while the net forces acting on the "toast" would be zero, were the toast non-uniform, it would eventually be torn to bits until it was, methinks. What with the sides of the earth having gravity. So we could get spheres like that, for no specified purpose.

Having not taken an applicable physics class, I'm basing this post purely on my sub-par opinion. Ignore at will.

[gmalivuk]

No, because the gravitational forces on anything in the center of the Earth are zero. Nothing to tear it apart, because there's no force differential across the length of the toast (as there would be, for instance, were it falling into a black hole).

[Mathmagic]

We've already had quite an extensive discussion about a "hole through the center of the earth" question before. It would be quite beneficial to read/post in that thread before thinking of posting here about it.

Hence why I merged the threads while you were posting that.

[Willis888]

I'd hate to be the one to have to tell her this, but after her boyfriend dug a few meters down he was eaten by turtles.

[meat.paste]

I recall reading something about this one time...don't remember the source.

Basically, neglecting air resistance and contact friction with the walls (in other words, anything that would slow you down in any way), a trip from one end to the other purely under the force of gravity would take about 45 minutes.

The interesting thing is that if you dig a hole that forms a straight line between any two points on the earth, e.g., New York to London, and travel this hole purely under the force of gravity, the time it would take to traverse the distance would be the same, regardless of the relative location of the two points.

As I recall, when I calculated the oscillation time, it was 31 minutes. But, it's been a few decades.

[Senefen]

She'll accelerate to the core, then decelerate leaving it. With no air resistance she'll end up the same distance form the core that she jumped in from but on the opposite side of the planet. Her velocity will reach zero and she'll start accelerating back in, back and forth back and forth
She doesn't stop at the core, she'll start decellerating there.
However with air resistance she'll slow down more and more and end up stuck at the core.

[thornahawk]

Only slightly related, but oh well... how would a large enough toroidal (doughnut-shaped) body exert gravity?

~ Werner

[Rohit507]

i suspect as the reached the center of the earth she would become a beautitful diamond

the pressure and temp of the air would be really high.

the pressure would be twice the mass of air above her on one side of the earth.

you probably want a good formula for this, i guess you ought to find the total mass of air above her (on one side) divide by her cross sectional area.

you probably need an equation for pressure at a height r taking into to account the gravitational pull of the earth and things.

damn that equation is confusing, can anyone figure it out?

is there an equation for density of air at a pressure?

[Matterwave1]

First order simple approximation has simple harmonic motion between the 2 end points forever.

Adding in air resistance has damped harmonic motion.

Adding in the rotation of the earth has more effects. Would make her motion no longer in a straight line, assuming the hole is larger than her body. If she fits very snuggly inside the hole, I would suspect her velocity would vary.

Then you can add in the motion of the earth around the sun, the motion of the sun around the galactic center, and the motion of the galaxy.

You could also then add in the gravity from the sun/moon/other celestial bodies.

[seladore]

I recall reading something about this one time...don't remember the source.

Basically, neglecting air resistance and contact friction with the walls (in other words, anything that would slow you down in any way), a trip from one end to the other purely under the force of gravity would take about 45 minutes.

The interesting thing is that if you dig a hole that forms a straight line between any two points on the earth, e.g., New York to London, and travel this hole purely under the force of gravity, the time it would take to traverse the distance would be the same, regardless of the relative location of the two points.

As I recall, when I calculated the oscillation time, it was 31 minutes. But, it's been a few decades.

It's actually about 84 minutes.

Interestingly, ignoring friction, this is identical to the period of a low earth orbit.

[Vieto]

Then you can add in the motion of the earth around the sun, the motion of the sun around the galactic center, and the motion of the galaxy.

Actually, Einstien's theory of relativity proved that this would have no effect, as the person would be moving the same as Earth relative to the sun.

[Senefen]

Only slightly related, but oh well... how would a large enough toroidal (doughnut-shaped) body exert gravity?

~ Werner

I assume it's focused at the centre. If you have a big hollow sphere the gravity is focuses at the centre and it behaves as a point (like a non-hollow sphere). I imagine a toroidal body would act the same. Read this maybe. (not sure if there's anything helpful there)

is there an equation for density of air at a pressure?

I know there is one, it's a derivitive of
PV=nRT (pressure[kPa]*volume[L]=Number of moles[mol]* 8.31*Temperature[K])
R = Ryberg constant
I think.
You have volume and moles there, you should be able to get density from that.

[Vieto]

PV=nRT (pressure[kPa]*volume[L]=Number of moles[mol]* 8.31*Temperature[K])
R = Ryberg constant
I think.
You have volume and moles there, you should be able to get density from that.

gah! I'm working on an assignment for that equation! Get out of my head!

[thornahawk]

I know there is one, it's a derivative of
PV=nRT (pressure[kPa]*volume[L]=Number of moles[mol]* 8.31*Temperature[K])
R = Ryberg constant

My physics ain't überly good, but the Rydberg constant and the molar gas constant have different units, much less values.

~ Werner

[Senefen]

I know there is one, it's a derivative of
PV=nRT (pressure[kPa]*volume[L]=Number of moles[mol]* 8.31*Temperature[K])
R = Ryberg constant

My physics ain't überly good, but the Rydberg constant and the molar gas constant have different units, much less values.

~ Werner

Ya you're right, I'm either crossing wires or have been taught the wrong name, with my uni both are equally possible

[gmalivuk]

That equation is actually pretty useless here, considering that you're not talking about gas in a small closed container...

[jmorgan3]

could you analyze a column of air, starting at sea level and then working down inductively, since you know the air above a given height, therefore you know it's pressure, therefore you know its density, so now you can calculate the air below it?

Yes, using the procedure in this pdf (created by Dr. Eric Johnson, GIT; I think I'm allowed to link to this). Note: some of the slides make the assumption that g is constant, which will not hold for this situation. I doubt you'll be able to find an analytical solution, but numerical methods should get you close enough.

[gmalivuk]

Yeah, you'd pretty much have to use numerical methods, even if you make simplifying assumptions like that Earth has a constant density, and also suppose that nothing in the air is going to liquefy or solidify at the kinds of pressures we're talking about (which is in effect what you're doing if you use the gas law all the way down).