Digging it between the poles would be best.
On a side note, with our current technology how deep a hole are we capable of making?
mmmmmmmmmmmm…
sounds yummy…
If we consider the idealized case where there is no energy loss to the ball and take the Earths radius to be 6371000m (an approximation of the mean radius) and drop the ball from rest.
s = 6371000
u = 0
a = 9.81
s = ut+(1/2)a(t)^2
63721000 = 0t + (1/2 * 9.81 * t^2) =>
t^2 = 63721000/ (1/2*9.81) = 12991030 => t = 3604 seconds to reach the centre
by symetry the time for it to reach the other side will be the same, therefore the total time will be 7208 seconds, which is a 2hrs and 8 seconds.
Somewhere in your analysis 6371000 became 63721000. That could make a big difference.
But also, you’re assuming uniform acceleration, which would not be true if the Earth is uniform density. For uniform density, the time is around 42 minutes, IIRC.
d’oh I did have a typo that worked itself into th equation. The actual answer would of been 38 mins but as ypu said that’s for uniform acceleration only ehich would be wrong in this case.
Actually, I think that uniform acceleration is good for part of the journey. Because the core is denser than the outer layers, it kind of compensates. I once saw a link that said that the acceleration is uniform down to the edge of the core, and thereafter, it was roughly uniform density (which would imply linearly-decreasing acceleration). So, it would be somewhere between 38 and 42 minutes.
Surely, for unifom density it would be simple harmonic motion. Which would give:
at x=a accelration =(w^2)a => 9.81 = 6371000* w^2 => w= 0.00124
T= (2*pi)/w = 84 mins. As T is the period for the oscillation, the time taken to get from top to bottom would be 42 mins.
Would tidel forces have any affect?
Would it be possible to put this question in a sticky?
Tidal force swoukldn’t have any effect as we’re considering the ball as a point mass.
C’mon you guys! I was a little kid! My brother and I did our best, but I didn’t think our inability to dig would turn it into a massive math problem for you folks! Now put your slide rules down (if that’s what you use to figure this stuff out) and get back to goofin’ off. Seriously!
Heh. Kalhoun thinks this math problem is massive. 
Unless you are at the equator or the poles, at any other mid-latitude, the vertical does not go through the center of the earth and will come out on the other side at a higher latitude. AND if you dig vertically from the point you reach you will not return to your original point. I hope you guys are taking this into account in your projects.
One question I’ve always had about this, maybe you physics-minded folks can tell me:
It’s always seemed to me that, if you did dig such a hole and dropped something down into it, it would never get farther than the center. The reason for this, I figure, is a combination of terminal velocity and gravity.
See, by the time the ball (or whatever) has passed, say, 1/4 of the way through the Earth, there’s a significant chunk of matter formerly below it that is now above it. As a result, the mass that tugs inward on the ball is smaller. And the terminal velocity of our falling ball should likewise be reduced, as a large gravitational mass (though smaller than the mass in the ball’s path) begins tugging it in the opposite direction.
As the ball falls, this continues, as more and more Earthly mass passes more and more quickly to the other side of the ball. In effect, this should gradually “brake” the ball until it comes gently to rest, motionless, at the Earth’s core.
So, am I right about this? I suppose the fundamental question is, on a planet with an atmosphere like Earth’s, but with a smaller planetary mass, is terminal velocity the same?
FLAT dammit! The earth is FLAT.
My ancestors told me so…
Until the pbject gets to the center it has more mass pulling down than mass pulling up so it would be accelerating (although at a decreasing rate of acceleration) until it passed the center and only after this point the acceleration would become negative.
Max Torque: Terminal Velocity only comes into play when you have air resistance. In that case, what you have is effectively a damped harmonic oscillator. A DHO may or may not come to rest, depending on the nature of the damping force. I believe that the most realistic damping force will be proportional to the velocity of the object. In this case it would come to rest, but the behavior still varies based on the damping constant. cf. this previous post by Elemental
so how far down is .001%?
3963 miles (radius) * .001 * .01 * 5280 feet/mile = 209 feet
I agree, but if it’s 0.01% of the way through, then it’s twice that, or 417ft 1.3in.