# Tube through the Earth

I performed this calculation about 10 years ago. If I recall correctly, it was a surprisingly short trip - 13 minutes or 23 or 32 or something like that. The velocity at the center was incredible. The other assumption you have to make to do this (easily) is that the Earth has a constant density, which of course it doesn’t.

The important thing to realize is that any point on your journey down the tube, the gravitational attraction from the sphere of earth above you is equalized and can be ignored. In fact, if the earth were a hollow sphere, you would experience equal pull from all directions immediately upon stepping inside (Sorry, Dyson Sphere nerds.) Anyway, the gravitational pull that accelerates you comes from only the mass below, which since it is a sphere, can be assumed to be a point mass at the center of the earth. Now you just have a two body problem where one of the masses changes over time - a lot simpler.

Now, if you want an even more interesting problem, try this: Since we have already assumed frictionless tubes, how about tubes that do not go through the center of the earth, but along a chord between any two surface points? You could slide (gravity would pull you along the vector) from New York to Moscow, for instance. How long would it take for gravity to pull you through that tube? The mechanics are the same - acceleration to the midpoint, deceleration to the destination. Interestingly enough, so is the travel time. One could, in theory, select a central hub location, build a network of tubes, and offer travel or shipping – anywhere in the world to anywhere else in less than an hour! How about that?

No one bothered to mention that in the model that contains air, the atmospheric pressure would make breathing pretty darned hard near the center.

I’m pretty sure it would alter terminal velocity as well. I’m not sure if that would make the trip longer or shorter or if the factors would cancel out.

The time it takes to go from one side to the other in a frictionless tube is actually pretty trivial to calculate, it’s the exact same time a circular orbit would take at the radius of the Earth’s surface. Because, guess what, you’re in orbit, i.e. free fall, in both cases. Falling through the tube is just a degenerate case of an orbit with zero angular momentum.

Since a complete orbit takes about 90 minutes, it would take about 45 minutes to get to the other end of the tube, or 90 minutes round trip.

Atmospheric pressure would indeed get pretty high at the bottom of the hole, but this should not affect breathing at all, since the air pressure inside your lungs would be the same as outside. Scuba divers have no problem breathing at depths of 100+ feet, where the water pressure on their bodies (and the air pressure inside their lungs) is around 60 psi.

Ummmm, since drag reduces one’s speed, one might reasonably conclude that it would take longer. In fact, you wouldn’t make it all the way up to the far surface of the earth without some kind of propulsion to get you up to the top of the hole.

Now I did neglect to mention the complication of the fact that the Earth isn’t a point mass, so as you’re falling the mass of the Earth outside your current radius make no contribution to the gravity since it all exactly cancels out. But it all still works out the same if I remember correctly.

Actually, a spreadsheet calc shows that the pressure runs away pretty quickly as you get any distance below the surface of the earth. The rate-of-change with respect to Z increases as you move toward the core; at 50,000 feet (just ten miles) below the surface, atmospheric pressure (assuming a constant 68 degrees F) is around 67 psi. At 100K feet (20 miles), it’s 308 psi. I ran out of space on my spreadsheet, but at 3.9 million feet (just 739 miles) below the surface, the atmospheric pressure is 2.2X10\$^46 psi. In other words, you’d have a supercritical fluid.

I suspect that if a hole of any significant diameter were created that penetrated through the earth, the mass of air required to pressurize it might seriously deplete the atmosphere above the surface.

That 10^46 psi result seemed implausible, so I went Googling. From a NASA site on the Galileo probe:

Might want to check your calculations.

Hey, gork, it’s a good idea to provide a link to the column/article you are talking about, maybe this one:

…since this thread has an infinite lifetime and that article won’t be quite as prominent after a week.

And welcome to the SDMB. Stick around. Who knows, you might learn something, whether you like it or not. Not that I doubt you, but why, and is there a geometric or mathematical proof of that?

It’s straightforward for the tube through the center: First, realize that it’s a simple harmonic oscillator: The force on you is proportional to your distance from the center. Then, you just have to realize that if you take a single component of a circular motion, that’s a simple harmonic oscillator, too, and the acceleration at the endpoint is the same in both cases. So the Y component of position is always the same for the guy in the hole and the guy in orbit.

I don’t know if there’s a similar simple calculation for the chord, or if that one requires calculus.

Good question. This can be seen in several steps:

1. As others have mentioned, you can ignore all the mass that is further away from the center than you are, since the net force inside a hollow spherical shell is zero. (I think Newton proved this geometrically in his Principia). It is a property of an inverse square law centripetal force.

2. Next you can easily see that the force inside the tube is just like an ideal spring, i.e. the gravitational force is directly proportional to the distance. That’s because the mass inside your position is proportional the volume, which is proportial to the radius cubed. The force is the same as if that mass were all concentrated at the center (also proved geometrically by Newton and also a consquence of the inverse square law). Since that force goes as the inverse square, the net result is a force proportional to distance, and with a magnitude that has to agree at the surface of the Earth.

3. Since the force is that of an ideal spring, the motion is is exactly sinusoidal, the so-called harmonic oscillator. Now imagine you had a particle traveling in a two dimensional harmonical oscillator potential, placed at a distance from the center equal to the Earth’s radius and traveling perpendicular to the radius at exactly the right speed to give a circular orbit. Since the motion in X and Y are independent (a circle is the sum of two sinusoids at 90 degree phase offset), it has to take exactly the same length of time to complete an orbit as to complete one oscillation along the X or Y direction. Since the force on a satellite in circular orbit is constant, it has to give the same result as the two dimensional harmonic oscillator potential.

There may be an easier way to see this, but this works for me without having to write down any equations (other than R^3/R^2 = R).

Oops, Chronos beat me to it and used less words.

Yeah, but you’re paid by the word, right?

I can’t brain today, I have the dumb. :smack:

Thanks for performing the sanity check for me. I went back and revised my calcs. Atmospheric density increases with depth, and if left unchecked it quickly runs away to neutron-star levels. This time I put a cap on it at the density of liquid oxygen, about 1.14 grams per cc; that brought things nicely under control. I also accounted for the decrease in gravity as one moves toward the center, and I rearranged the spreadsheet so that I could run it all the way to that depth. Now the atmospheric pressure at the center of the earth appears to be a much more comfortable 5.03 million psi.

Which still makes me wonder if it wouldn’t drain most of the air off of the surface of the earth to fill it. (someone else can do the math on that one.)

I suspect the rest of the parts that are involved in keeping a body breathing would have slightly less efficient funtionality at 5 million psi.

Any idiot understands that you don’t go all the way through. Cecil even pointed it out in his article. Something about waving at bunnies.
My thoughts were on the pendulum effect - while the trip would take longer, the pendulum swing wouldn’t take you as far to the other side and you’d have to deal with all that extra resistance on each swing to and from center. The question wasn’t how long it takes to get to the other side but how long until motion ceases at the center.
IE - in a vaccuum it takes X time to come to a complete stop at the center
given a realisitic atmosphere it takes Y time to come to a complete stop at the center - which is greater, x or y?

The diameter of the Earth is ~ 12750km. Assuming a nicely-sized two-meter diameter hole (which number I just made up, but is reasonably human-scale) gives a hole volume of 40X10[sup]6[/sup] cubic meters. As an upper bound, the entire hole could be filled with 1.14 grams per cc liquid atmosphere, which would have a total mass of **4.5X10[sup]10[/sup] kg **(40X10[sup]6[/sup]m[sup]3[/sup]*1000000cc/m[sup]3[/sup]1.14 g/cc(1/1000)kg/g). The entire mass of the atmosphere is around 5 × 10[sup]18[/sup] kg, so the hole would suck away less than a one-hundred-millionth of the atmosphere.

Catastrophe averted… through science!

In a vacuum with no other frictional effects present, you’d never stop, so clearly you’d stop sooner with atmosphere present. Even factoring in other frictional effects (bouncing off the walls of your tunnel, I suppose), the existance of an atmosphere of any type will only increase the amount of energy your yo-yoing body sheds. Less energy = less speed at any given point = stopping sooner.

Ah. Your initial comment was about whether air drag would make the trip longer or shorter, which is an entirely different matter from how long until the oscillatory motions cease.

In a vacuum you will never come to a stop at the center; you’ll keep oscillating back and forth indefinitely, from one side of the earth to the other.

In an atmosphere you’ll never come to a complete stop at center, either; the amplitude of your oscillations will asymptotically approach zero over time.

It’s a standard on freshman Physics homeworks to have 3 questions in a row asking the time period for 1. A surface level orbit. 2. A hole thru the center of the Earth. 3. A chord/tunnel thru part of the Earth. (All frictionless, etc.) Students of course want to know how long the chord is. We tell them to just work it out. There’s a reason all 3 are asked together.

Well, yes, I know the answer to the chord question, and I can get it via calculus, but I just don’t know if there’s a non-calculus answer.