Here is a detailed analysis of the amount of time it would take to ride a hypothetical elevator down through the Earth's core and back out the other side of the planet. Apparently, this has something to do with the remake of Total Recall. But it's interesting even if (like me) you have no intention of seeing that movie. (Via Rhett Allain)

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The author misses the best result: it’s 42 minutes regardless of where you want to go, as long as you go in a straight line (and as long as density is constant)

http://en.wikipedia.org/wiki/Gravity_train

This type of travel also plays a big in Jasper Fforde’s Tuesday Next series of novels.

And earlier, in Niven’s “World out of Time”.

http://en.wikipedia.org/wiki/A_World_Out_of_Time

Damon Knight was there first!

http://en.wikipedia.org/wiki/Beyond_the_Barrier

I had no idea that there was so much literature surrounding this idea… I heard about it the first time as a child (40ish years ago). I’ve also heard Joel’s statement about the time being the same no matter where the two points are situated.

What I came by to say, though, was that this is a complex problem algebraically, but very simple with calculus (assuming constant density, of course). Hopefully someone who remembers more calculus than I could enlighten us. Going back to college for monetary gain is a lost cause for me, but I could sure eat up some math and physics and feel like my money was well-spent.

The fine article fails to note that it takes just as long (42 minutes) to go halfway around the world in orbit above all that troublesome dirt as it does to go down a hole though it. This is not a coincidence.

It also fails to mention what to do about that little bit of angular momentum (due to sitting on an earth that’s spinning) you have when you start to fall down the hole. Assuming you manage to make it out the other end with out hitting the side of the bore you’ll end up in atmosphere doing about Mach 2 to Mach 4, sideways.

It’s also trivial to come up with an estimate of the density of air near the core, assuming an adiabatic atmosphere and something like an ideal gas. (hint: it’s nothing near 1.2 kg/m^3)

I had this problem in a homework assignment in my college course in classical dynamics (many years ago). Our text was Marion’s Classical Dynamics. As noted above, the interesting thing about the result is that it takes 42 minutes regardless of where the two endpoints are.

My high school physics teacher pointed out, besides the fact that it’s the same amount of time between any two points, but also it’s the same amount of time (round trip) as an orbit at the same altitude. Though low earth orbit (LEO) is close enough.

So I was confused when taking density into account the time was different. At the extreme, call it a point gravitational source at the center of the earth. I guess a cometary orbit through the center of the earth would be faster.

Yeah, I’m forgetting the math to figure this out. Maybe I’ll have to put down this phone and pick up a pencil and paper…

So, qualitatively, with all the mass at the center of the earth, the rate of acceleration- jerk- increases until you get to the center of the earth, vs. with a uniform density (or even non-uniform with some amount of mass near the surface) the rate of acceleration decreases, so yeah, travel times are obviously going to be different.

By my math, it would take almost eleven minutes to get to the center of the point-gravitational-source earth. Rather t^2=2R^3/(3GM) But I could be completely wrong. I haven’t done calculus in twenty years.

Now I know where the phrase “back of the envelope calculation” came from. That was the handiest piece of paper I could find around the house.