How long to travel the Earth's diameter?

…by going through the centre?

Based onthis thread, where it was suggested that after digging a hole through the Earth, you could simply jump into the hole, I presume the momentum from your fall being compensated by the effects of your rising after halfway through.

So how long would it take? Since gravity isn’t constant, terminal velocity would change, so this could be quite complicated.

Not an answer to your question (because there’s a lot of variables implicit in that question and some fairly hairy calculus when you include all of them), but a second related question - What would the hole look like in cross-section? Because if the hole was straight through the earth, you would be slammed against the East side of the hole as you fell due to your initial radial velocity of 1000+ mph, and the West side as you shot up from the center back to the surface. I’m guessing some sort of S-shape?

Terminal velocity would apply only if there’s air to fall through, and thus friction. If this is the case, the enormous density at the center of the earth will lead to huge losses - you won’t get past there.

Assuming you can fall in a vaccum with no friction, the period would be the same as half the time to complete a single orbit of the earth at very low altitude.

This site gives the formula as:
P = 2 * pi * sqrt (a[sup]3[/sup]/u)
where
P = orbital period
a = earth’s radius (in this case) = 6378 km
u = gravitational parameter = 3.986*10[sup]5[/sup] kg[sup]3[/sup]/sec[sup]2[/sup]

For me, this yielded a value of 42.2 minutes

Lets not get too complicated. Assume the hole goes from the north pole to the south pole.

That would be the shortest route as well.

It’s always 42 minutes, it doesn’t matter if the tunnel goes through the center of the earth or a chord connecting, say, NYC and London. This is all assuming a frictionless vaccuum in the tunnel.

http://forums.randi.org/archive/index.php/t-16011.html

Sure, if you want to take all the fun out of it. :wink:

So, there are two answers:

1: Not at all.

2: Forty two minutes and 12 seconds.

Tris

And note that 42.2 minutes is also exactly half the time to orbit the earth at minimum altitude, so if you knew that number you didn’t need to do any calculations at all.

We’ve done this thread about 42.2 times in the past as well.

Minimum altitude? You mean, like, hitting trees and buildings all the way?

Hey, if Superman can dodge 'em…

We did this one to death about a month ago. Try searching.

The Earth doesn’t rotate under your feet when you jump in the air, and it doesn’t rotate under your feet if you fall in a hole. Otherwise Olympic divers would see the pool moving out from under them at 1,000 MPH when they leave the board.

You wouldn’t be slammed against the side.

Rotating once a day at the earth’s surface leaves you with a vector of one thousand miles an hour eastward, at the equator, less as your lattitude increases. The rotational rate two thousand miles down is still once a day, but the vector is much less. The earth doesn’t slam into you, you slam into it.

Like I always say in these thread it has to be done with magic, or it cannot be done at all. So, the fact that you don’t die, and you end up in Australia, and it takes 42 minutes but you don’t have to hold your breath are all just aspects of the particular spell you choose to cast. Not hitting the sides is another magical consideration, having nothing to do with physics.

Tris

And as I always say, it’s not magic. Even if such engineering does not exist, one can envision, for instance, a tunnel with sufficiently strong and heatproof walls, with sufficiently low friction that one can slide along the side of it. Or one could examine analogous situations, where the engineering required is not quite so daunting, but the mathematical analysis is the same.

You say engineering, I say magic. When it happens, we’ll see.

Tris

You’re in good company, Arthur C Clarke’s third law states:

“Any sufficiently advanced technology is indistinguishable from magic.”

The point is, there’s a difference between engineering and physics. From an engineering standpoint, this is (currently) impossible. Physically, though, no problem.