If a vacuum tube went through the earth, how long to fall to the other side?

Presume there’s a tunnel that goes straight from the North Pole to the South Pole, and it was a vacuum so wind resistance is not an issue. If an intrepid adventurer were to jump into it, how long would it take to fall to the other side?

You might get an answer similar to this.

Looks like Cecil gave up trying to calculate how long it would take. :smack:

About 45 minutes — or one half the period of an object in low-earth orbit.

You would never reach the other side. Unless past the center of the earth counts as the other side.

Here’s an actual explanation, as opposed to just a bald assertion. And a more accurate figure would be 42 minutes instead of 45.

:smack:

I guess you would, then. I neglected to consider the vacuum.

With no air resistance, I’m assuming you’d accelerate until you reach the center (where your speed would be a little under 7 miles per second).

You would then start slowing down until you stopped at the other side.

Total time: 39 minutes, 56 seconds.

Not exactly what others got, so maybe my constants were off (I used 32 feet per second[sup]2[/sup] for the acceleration due to gravity, and the radius of the earth as 3925 statute miles).

Isn’t that page ignoring the rotation of the earth? I don’t think that a straight hole would work. I beleive that you will hit the side of the shaft.

Not if the hole runs from the NP to the SP…

Unless I need to go back to physics 101 acceleration isn’t a constant. As you approach the center of the earth acceleration approaches 0 because the mass of the earth is equal in all direcctions.

I think you’re right – the gradual decrease of mass under you would slow the acceleration.

So it would take a little longer than 39 minutes, 56 seconds, which makes intuitive sense since my time was shorter than the others’.

I think nowadays you’d use a transistor.

Whatever you use, be sure it’s well grounded.

I got a charge out of that.

IC. But, specifically, a field-effect transistor, no?

I’m incapacitated with laughter. Yet, I solder on. Someday I’ll diode.

Two things are happening:
(1) You are getting closer to the earth’s centre, so attraction is increasing proportionate to r^(-2) – the inverse square law.
(2) The amount of Earth under you is decreasing, so (assuming the Earth has the same density throughout), attraction is decreasing proportionate to r^3 – volume of a sphere is a function of the cube of the radius. (Note that the hollow sphere of the Earth further from the centre than you has a net gravitational attraction of zero).

So by my calculations, acceleration at x% of the earth’s radius is roughly equal to x% of the acceleration at the surface.

IIRC, ignoring the rotation of the earth, if you put a tube through any chord (say NY to Paris) an object’s travel time will be the same as if it went pole to pole.

Isn’t the scenario in the OP just a highly elliptical orbit? Or does the fact that it’s inside the mass of the earth make ordinary formulae for determining orbital periods and the like inaccurate?

Assuming that the Earth is constant density, then the time would be exactly half of the surface orbit time (low Earth orbit is not, of course, a true surface orbit, but it’s pretty close). In fact, though, the density of the Earth increases as you approach the center, and it actually increases in just such a way that for most of the trip, the acceleration really would be more or less constant. So ironically, F. U. Shakespeare’s calculation might actually be the most accurate one in this thread.