# Digging a hole to through the Earth **URL:** \ **Category:** Factual Questions\ **Created:** [June 10, 2003, 1:22pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849 "2003-06-10T13:22:15Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![Sock\_Munkey](https://avatars.discourse-cdn.com/v4/letter/s/aeb1de/32.png) [@Sock\_Munkey](https://boards.straightdope.com/u/Sock_Munkey)\ **Post date:** [June 11, 2003, 12:17pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/21 "2003-06-11T12:17:55Z")
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? ---
**Author:** ![newcrasher](https://avatars.discourse-cdn.com/v4/letter/n/b3f665/32.png) [@newcrasher](https://boards.straightdope.com/u/newcrasher)\ **Post date:** [June 11, 2003, 12:21pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/22 "2003-06-11T12:21:43Z")
> [@](#): > > \*Originally posted by RM Mentock \* > \*\*The Earth is known to have a liquid core about half way down–but sometime after that was discovered, we found that inside that liquid core, there seemed to be a solid inner core. So now we say that it has a liquid outer core, and a solid inner core. Beneath the solid crust, which has cooled, is a solid mantle, which is nearly liquid in a zone just under the crust. \*\* mmmmmmmmmmmm… sounds yummy… ---
**Author:** ![MC\_Master\_of\_Ceremonies](https://avatars.discourse-cdn.com/v4/letter/m/bb73d2/32.png) [@MC\_Master\_of\_Ceremonies](https://boards.straightdope.com/u/MC_Master_of_Ceremonies)\ **Post date:** [June 11, 2003, 12:31pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/23 "2003-06-11T12:31:05Z")
> [@](#): > > \*Originally posted by nocturnal\_tick \* > \*\*but how long would it take for the ball to reach the end? that is, of course, it was a perfect situation (vacuum, straight tunnel etc.) \*\* 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. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [June 11, 2003, 12:34pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/24 "2003-06-11T12:34:52Z")
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. ---
**Author:** ![MC\_Master\_of\_Ceremonies](https://avatars.discourse-cdn.com/v4/letter/m/bb73d2/32.png) [@MC\_Master\_of\_Ceremonies](https://boards.straightdope.com/u/MC_Master_of_Ceremonies)\ **Post date:** [June 11, 2003, 12:41pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/25 "2003-06-11T12:41:11Z")
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. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [June 11, 2003, 12:45pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/26 "2003-06-11T12:45:58Z")
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. ---
**Author:** ![MC\_Master\_of\_Ceremonies](https://avatars.discourse-cdn.com/v4/letter/m/bb73d2/32.png) [@MC\_Master\_of\_Ceremonies](https://boards.straightdope.com/u/MC_Master_of_Ceremonies)\ **Post date:** [June 11, 2003, 1:07pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/27 "2003-06-11T13:07:03Z")
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. ---
**Author:** ![netscape\_6](https://avatars.discourse-cdn.com/v4/letter/n/8c91f0/32.png) [@netscape\_6](https://boards.straightdope.com/u/netscape_6)\ **Post date:** [June 11, 2003, 1:19pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/28 "2003-06-11T13:19:08Z")
Would tidel forces have any affect? ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [June 11, 2003, 1:20pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/29 "2003-06-11T13:20:12Z")
Would it be possible to put this question in a sticky? ---
**Author:** ![MC\_Master\_of\_Ceremonies](https://avatars.discourse-cdn.com/v4/letter/m/bb73d2/32.png) [@MC\_Master\_of\_Ceremonies](https://boards.straightdope.com/u/MC_Master_of_Ceremonies)\ **Post date:** [June 11, 2003, 1:29pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/30 "2003-06-11T13:29:20Z")
Tidal force swoukldn’t have any effect as we’re considering the ball as a point mass. ---
**Author:** ![Kalhoun](https://avatars.discourse-cdn.com/v4/letter/k/3bc359/32.png) [@Kalhoun](https://boards.straightdope.com/u/Kalhoun)\ **Post date:** [June 11, 2003, 5:27pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/31 "2003-06-11T17:27:42Z")
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! ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [June 11, 2003, 5:50pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/32 "2003-06-11T17:50:10Z")
Heh. **Kalhoun** thinks this math problem is massive. 🙂 ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [June 11, 2003, 6:07pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/33 "2003-06-11T18:07:42Z")
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. ---
**Author:** ![Max\_Torque](https://avatars.discourse-cdn.com/v4/letter/m/e9bcb4/32.png) [@Max\_Torque](https://boards.straightdope.com/u/Max_Torque)\ **Post date:** [June 11, 2003, 8:58pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/34 "2003-06-11T20:58:41Z")
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? ---
**Author:** ![ltfire](https://avatars.discourse-cdn.com/v4/letter/l/7ea924/32.png) [@ltfire](https://boards.straightdope.com/u/ltfire)\ **Post date:** [June 11, 2003, 9:10pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/35 "2003-06-11T21:10:27Z")
FLAT dammit! The earth is FLAT. My ancestors told me so…😃 ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [June 11, 2003, 9:15pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/36 "2003-06-11T21:15:24Z")
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. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [June 11, 2003, 9:20pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/37 "2003-06-11T21:20:33Z")
**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**](http://boards.straightdope.com/sdmb/showthread.php?s=&postid=3113283#post3113283) ---
**Author:** ![Aceospades](https://avatars.discourse-cdn.com/v4/letter/a/848f3c/32.png) [@Aceospades](https://boards.straightdope.com/u/Aceospades)\ **Post date:** [June 11, 2003, 10:59pm UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/38 "2003-06-11T22:59:01Z")
so how far down is .001%? ---
**Author:** ![KeithT](https://avatars.discourse-cdn.com/v4/letter/k/e47774/32.png) [@KeithT](https://boards.straightdope.com/u/KeithT)\ **Post date:** [June 12, 2003, 2:52am UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/39 "2003-06-12T02:52:38Z")
> [@](#): > > \*Originally posted by Aceospades \* > \*\*so how far down is .001%? \*\* 3963 miles (radius) \* .001 \* .01 \* 5280 feet/mile = 209 feet ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [June 12, 2003, 3:11am UTC](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849/40 "2003-06-12T03:11:43Z")
I agree, but if it’s 0.01% of the way through, then it’s twice that, or 417ft 1.3in. [Previous page](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849.md?page=1) [Next page](https://boards.straightdope.com/t/digging-a-hole-to-through-the-earth/180849.md?page=3)