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.

]]>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…

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

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)

]]>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.

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