# Falling trough the center of the earth? **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [September 25, 2003, 1:08pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730 "2003-09-25T13:08:24Z")\ **Posts on this page:** 20\ **Page:** 4
**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:** [October 21, 2003, 9:41am UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/61 "2003-10-21T09:41:43Z")
> [@](#): > > \*Originally posted by Chronos \* > \*\*These effects are hard to notice with ordinary refrigerator magnets, but with a strong enough magnet you can levitate a frog (or a human, if you had one big enough, but alas, we don’t, yet). \*\* I believe the NFL is working on that. ---
**Author:** ![Gyrate](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/gyrate/32/3588_2.png) [@Gyrate](https://boards.straightdope.com/u/Gyrate)\ **Post date:** [October 21, 2003, 10:09am UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/62 "2003-10-21T10:09:09Z")
Now I’m curious: in a non-vacuum situation, what would be the effect of the increasing air pressure/density (and I’m just assuming that it would increase) on your acceleration as you moved towards the center – would it be significant? > [@](#): > > \*Originally posted by jimpeel \* > \*\*For those who have no patience with people such as myself, a gentle reminder: > > **At some point in your life, you were as ignorant as me.** \*\* It’s worth noting, **jimpeel** , that at least one of those people chastising you has been here even less time than you. Don’t sweat it – there’s an adjustment period for everyone. Anyhoo, welcome to the Straight Dope, and let me just say that’s it’s gratifying to see that the Internet has finally reached Kimball. Don’t worry – I’m sure you’ll have indoor plumbing anyday now. 😃 **Gyrate** (former Lincolnite) ---
**Author:** ![jimpeel](https://avatars.discourse-cdn.com/v4/letter/j/13edae/32.png) [@jimpeel](https://boards.straightdope.com/u/jimpeel)\ **Post date:** [October 21, 2003, 5:00pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/63 "2003-10-21T17:00:06Z")
Thanks for the welcome. > [@](#): > > Now I’m curious: in a non-vacuum situation, what would be the effect of the increasing air pressure/density (and I’m just assuming that it would increase) on your acceleration as you moved towards the center – would it be significant? More significant would be the fact that you were falling toward a big ol’ ball of magma. There would have to be some point where you became the Flying Nun and the rising thermal air/gas currents would stop your progress toward center; and then turn you slowly as they roasted you to golden perfection. 😃 ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [October 22, 2003, 7:31pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/64 "2003-10-22T19:31:58Z")
> [@](#): > > \*Originally posted by Gyrate \* > Now I’m curious: in a non-vacuum situation, what would be the effect of the increasing air pressure/density (and I’m just assuming that it would increase) on your acceleration as you moved towards the center – would it be significant? It’s gotta be enormously significant. In a non-vacuum situation, you’d reach terminal velocity pretty quickly (meaning the force on your body would be balanced by the drag force fom the atmosphere); typical terminal velocity is on the order of [120mph](http://hypertextbook.com/facts/JianHuang.shtml). The drag force is calculated as Drag = (1/2)[symbol]r[/symbol]C[sub]D[/sub]AV[sup]2[/sup]. All other things being equal, an increase of density ([symbol]r[/symbol]) by a factor of four would yield a decrease in velocity (V) by a factor of two. Note that atmosperic pressure and density scale together: doubled pressure implies doubled density (temp being the same). If you look at [a graph of atmosperic pressure](http://web.ics.purdue.edu/~nowack/geos105/lect11-dir/lecture11.htm), you’ll note that it increases by a factor of five in only 10 km or so. My quick estimation says that atmospheric pressure (and thus density) increases approx 20% a mile beneath the Earth’s surface. Actually calculating pressure at the Earth’s center would require an integration that I’m too lazy to do right now, but I’m betting that pressure (and density!) would be hundreds of times larger than at the surface. And, given that, as you fell through the hole, your terminal velocity would deacrease as you fell, eventually getting smaller by some factor of tens (so… down to 5 or 6 mph? In that range?). ---
**Author:** ![aerodave](https://avatars.discourse-cdn.com/v4/letter/a/f08c70/32.png) [@aerodave](https://boards.straightdope.com/u/aerodave)\ **Post date:** [October 22, 2003, 9:12pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/65 "2003-10-22T21:12:55Z")
It’s certainly small compared to the increase in the air density, but remember that you have to factor the varying gravitational pull when calculating the terminal velocity. At locations within the tunnel where gravity pulls stronger, a higher drag force will be required to counteract the increased weight of the body. Therefore, the terminal velocity will be a little higher. The same holds true for areas where _g_ is lower. The terminal velocity will be even slower. Since _g_ approaches zero during the last part of the fall before the center, it leads to an intereting conclusion: If the tunnel is full of air, the object falling will basically come to a dead stop right at the center, without traveling any appreciable distance past it. Overdamped motion at its finest. ---
**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:** [October 22, 2003, 9:54pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/66 "2003-10-22T21:54:03Z")
It won’t necessarily stop at the centre as there are too many factors affecting it. Also to further complicate things the weight per mass of the air will get less and less approaching the centre where it will equal zero. ---
**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:** [October 23, 2003, 6:52am UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/67 "2003-10-23T06:52:45Z")
**audilover** , the increase of gravity within the tunnel would be small–but it doesn’t decrease, as it would if the density of the Earth were constant. ---
**Author:** ![ras2000](https://avatars.discourse-cdn.com/v4/letter/r/a587f6/32.png) [@ras2000](https://boards.straightdope.com/u/ras2000)\ **Post date:** [October 23, 2003, 3:23pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/68 "2003-10-23T15:23:59Z")
Audilover, I really liked your point, I had originally forgotten about the change in terminal velocity. But just because the terminal speed is falling, doesn’t mean that the body would reach that speed immediately. If that were the case, a thrown ball wouldn’t get far. But the integration involved in figuring out what the speed of the falling body would be at the center of the earth is a bit out of my range. But I’m going to give it a shot over the weekend anyway, with a simplified equation for wind resistance. ---
**Author:** ![aerodave](https://avatars.discourse-cdn.com/v4/letter/a/f08c70/32.png) [@aerodave](https://boards.straightdope.com/u/aerodave)\ **Post date:** [October 23, 2003, 8:28pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/69 "2003-10-23T20:28:02Z")
> [@](#): > > \*Originally posted by RM Mentock \* > \*\ ***audilover** , the increase of gravity within the tunnel would be small–but it doesn’t decrease, as it would if the density of the Earth were constant. \*\* It has to decrease at some point. It’s greater than zero above the center of the earth, but we established a while back that it is exactly zero at the center. to go from _g_\>0 to _g_=0 requires that _g_ decrease. ---
**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:** [October 23, 2003, 9:38pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/70 "2003-10-23T21:38:56Z")
[How’d that comma get in there](#4121628)? 🙂 ---
**Author:** ![Mr.IcaruS](https://avatars.discourse-cdn.com/v4/letter/m/c57346/32.png) [@Mr.IcaruS](https://boards.straightdope.com/u/Mr.IcaruS)\ **Post date:** [October 24, 2003, 9:51am UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/71 "2003-10-24T09:51:33Z")
You just said: for gravity to be more than nothing, gravity needs to increase. why make it so complicated?? ---
**Author:** ![aerodave](https://avatars.discourse-cdn.com/v4/letter/a/f08c70/32.png) [@aerodave](https://boards.straightdope.com/u/aerodave)\ **Post date:** [October 24, 2003, 1:47pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/72 "2003-10-24T13:47:02Z")
> [@](#): > > \*Originally posted by Mr. IcaruS \* > \*\*You just said: for gravity to be more than nothing, gravity needs to increase. why make it so complicated?? \*\* First off, I said the inverse of that: For gravity to be nothing from more than nothing, gravity needs to increase" Secondly, I hardly think I overcomplicated it. Sorry if a variable and a couple comparison signs freaked you out. But in a thread full of ten paragraph diatribes, a three sentence, two line, very clearly laid out reply is hardly complicated. I could have said it in five words, but I thought in its context, I needed to be as clear and explicit as possible. After all, the comment originated from a miscommunication anyway. Welcome to the boards :rolleyes: ---
**Author:** ![aerodave](https://avatars.discourse-cdn.com/v4/letter/a/f08c70/32.png) [@aerodave](https://boards.straightdope.com/u/aerodave)\ **Post date:** [October 24, 2003, 1:50pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/73 "2003-10-24T13:50:25Z")
Oops…I meant “For gravity to be nothing from more than nothing, gravity needs to _decrease_” I can’t even follow my own logic once it’s condensed and compressed that thoroughly. ---
**Author:** ![ras2000](https://avatars.discourse-cdn.com/v4/letter/r/a587f6/32.png) [@ras2000](https://boards.straightdope.com/u/ras2000)\ **Post date:** [October 30, 2003, 4:15pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/74 "2003-10-30T16:15:36Z")
Ok, I gave up on the integral. But another thought occured to me. In a pendulum swinging, when it reaches the bottom, it’s terminal velocity would be 0, since there is no acceleration, and still it keeps going. This brought me to thinking about a mass sliding down a ramp, taking wind resistance into account, but not the friction between the ramp and the mass (kind of the opposite of the simple physics problem). If the body falling through the center of the earth comes to a halt at some point, there would have to be a minimum angle of the ramp, for the mass to start moving. A minimum force required to move the air holding the mass in position. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [October 30, 2003, 8:34pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/75 "2003-10-30T20:34:46Z")
> [@](#): > > In a pendulum swinging, when it reaches the bottom, it’s terminal velocity would be 0, since there is no acceleration, and still it keeps going. Would you mind rephrasing this? I can’t see what you’re trying to say, here. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [October 30, 2003, 10:06pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/76 "2003-10-30T22:06:49Z")
I do believe **ras2000** is stating something like, “1. Terminal velocity is the velocity where aerodynamic drag forces balance inertial acceleration forces. 2. A pendulum at the bottom of its swing has zero tangential acceleration. 3. Therefore, a pendulum at the bottom of its swing has no terminal velocity.” However, that line of thought is not quite correct because there _is_ a normal acceleration of the pendulum bob, so all the forces are not balanced out. However, I’m not quite sure how that relates to earlier discussion. > [@](#): > > A minimum force required to move the air holding the mass in position. So far as I know, standard fluid dynamics has the drag force proportional to some power of hte velocity. For a body on a frictionless ramp, any ramp angle would produce some gravitational component along the ramp; at zero velocity, no drag force would oppose it, and the body would accelerate. ---
**Author:** ![ras2000](https://avatars.discourse-cdn.com/v4/letter/r/a587f6/32.png) [@ras2000](https://boards.straightdope.com/u/ras2000)\ **Post date:** [October 31, 2003, 8:05am UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/77 "2003-10-31T08:05:25Z")
Zut: Well if that is true, then I’ve been correct all along. But what do you mean all the forces don’t cancel out at the bottom? Are you talking about the rotation, the masse changing directions? With a long string, that force would be very close to zero, and not in the same direction as the drag, actually perpendicular to it. ---
**Author:** ![ras2000](https://avatars.discourse-cdn.com/v4/letter/r/a587f6/32.png) [@ras2000](https://boards.straightdope.com/u/ras2000)\ **Post date:** [October 31, 2003, 8:20am UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/78 "2003-10-31T08:20:47Z")
Chronos, zut explained it pretty well, with terminal velocity, the wind drag and the gravitational pull cancel each other out, Newtons 2. I think it is, “a partical that isn’t affected by any forces or is affected by forces that cancel each other out will either be still, or move with constant speed”. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [October 31, 2003, 12:35pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/79 "2003-10-31T12:35:56Z")
> [@](#): > > \*Originally posted by ras2000 \* > Zut: Well if that is true, then I’ve been correct all along. OK, now _I’m_ confused. If _what_ is true, then you’ve been right about _what_ all along? > [@](#): > > But what do you mean all the forces don’t cancel out at the bottom? Are you talking about the rotation, the masse changing directions? With a long string, that force would be very close to zero, and not in the same direction as the drag, actually perpendicular to it. Yes, all that’s true, but the terminal velocity is defined as “the velocity with which a body moves relative to a fluid when the resultant force acting on it (due to friction, gravity, and so forth) is zero” (just looked it up in my ME dictionary). Resultant forces on a pendulum are _not_ zero, as you point out. The fact that the resultant force is perpendicular to the velocity is irrelevent. For example, suppose you throw a ball to a friend. The same conditions (acceleration perpendicular to velocity) are true for the ball at the top of its arc. Would you suggest that the ball, at that instant, has a terminal velocity of zero? Clearly this is not the case. ---
**Author:** ![ras2000](https://avatars.discourse-cdn.com/v4/letter/r/a587f6/32.png) [@ras2000](https://boards.straightdope.com/u/ras2000)\ **Post date:** [October 31, 2003, 12:57pm UTC](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730/80 "2003-10-31T12:57:36Z")
Zut, first point, when I started this thread, I asked whether or not the body falling would ever stop. I felt it wouldn’t because the closer the speed got to zero, the closer the drag would get to zero, making the max speed near zero asymptotic, never reaching zero. Second point, the forces are vectors. The resulting vector from the swinging pendulum would be the normal added to the wind drag, pointing almost in the opposite direction of the motion. It is way above terminal velocity. And I would state that the ball has a terminal velocity of zero at the height of the arc. I guess I disagree with your book. It is, as I said earlier, possible for something to move at a higher speed than the terminal velocity. [Previous page](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730.md?page=3) [Next page](https://boards.straightdope.com/t/falling-trough-the-center-of-the-earth/203730.md?page=5)