In the past three months, I turned up one discussion each in Comments on Cecil’s Columns and General Questions. Both are digressions from other topics, and I’m sure we’ve had whole threads on the Planck scales, but I didn’t want to strain the servers by searching too deeply.
Here’s the problem I have with this entire discussion: Noone addresses the fact that when one is traveling towards the center of the Earth with gravity pulling one “downward”, there is an oppoosite gravity from the other side of the Earth that is acting on that same falling body. The closer one gets to the center, the more the opposing gravity would cancel the gravity pulling one toward the center. At some point, equilibrium would be reached and one would simply come gently to a stop at the exact center of the gravitational field.
The assumption has always been that since gravity acts to accelerate a body at 32 ft/sec/sec that that gravity remains constant to the center of the Earth. It simply cannot.
It has also always been the assumption that there would be a slingshot effect that would cause one to go past center and then, at some point, stop and return again through center. Each time, that trip would get slightly shorter but, in a perfect vacuum, would continue to infinity never coming to a stop.
I do not believe that to be true. I believe, with no empirical evidence to back me up, that one would come to terminal velocity at about 1/3 of the way to the center whereupon one would then begin to slow and eventually come to a dead stop at the center of the gravitational field where that field is acting equally from all directions.
That’s my story and I’m stickin’ to it.
No, all calculations are done with your assumptions that the acceleration due to gravity decreases gradually to zero at the center of the Earth.
In a perfect vacuum, it would indeed continue on indefinitely, because there would be nothing to slow it down. Just because the acceleration is zero, doesn’t mean the velocity is zero.
Welcome to the board. Better suit up. 
If acceleration ends, then the only thing left to act on the “falling” subject body would be deceleration. Once acceleration falls to zero, the only thing left would be for negative acceleration to begin. That would be the force of the gravity that would be acting on a body “falling” from the other side of the Earth.
If the force of what would be gravitational pull on a body coming the other way is seen by the subject body, that force acting upon the subject body can only be exerted on that body as gravitational push in the physical realm of the subject body.
I believe, again with absolutely no empirical evidence to support my contention, that gravitational equilibrium is reached long before the center of the Earth would be reached, whereupon, opposing forces would begin to act on the “falling” body.
Still stickin’ to it from lovely downtown Kimball, NE.
The h*ck with empirical evidence, you’re mixing up the theoretical evidence. Terminal velocity (which you mention above) is a result of air resistance–but there is no air resistance in a perfect vacuum (which you also bring in to play).
Terminal velocity, in an atmospheric world, would, indeed, be reached due to air resistance.
In my world, in the absence of air, there would still be a terminal velocity reached as the opposing forces began to act upon the “falling” body. Air isn’t the only force out there.
So, in your original post, you meant “maximum velocity” instead of “terminal velocity”? Terminal Velocity is a term used when the force of gravity is matched by the force of air resistance.
Still, there would be no reason to reach maximum velocity 1/3 of the way to the center, especially since the acceleration due to gravity 1/3 of the way down is the same (or a bit greater) than at the surface. That’s because the center of the Earth is so much more dense than the outer layers.
And it doesn’t matter anyway. As long as the net acceleratory force is downward, downward velocity will continue to increase. That’s basic three-laws-of-motion (in fact, first-two-laws-of-motion) territory. Ergo, velocity will increase until the center. Calculating the exact gravitational forces in play involves some fairly awful integrals even without allowing for density changes, but the observation that the big half of the sphere is bigger than the small half is trivial.
If you’re falling through the earth, towards the center, there will continue to be more mass ahead of you than behind you until the time you reach the center. Then the gravitational forces cancel, because there are equal amounts of mass ahead of you and behind you. After the center, there is more mass behind you, and you’ll be dragged back at with the same acceleration profile you experienced during the trip down.
I think jimpeel has a sense of what he’s talking about, but he may have his facts mixed up. There is a point where the Earth’s gravitational acceleration reaches a maximum, and it is slightly more than half the way from the surface towards the center. (Don’t quote me on how far down, it’s just a vague recollection). At that magic point, fifty-some percent the way down, you’re getting close to the center, but have not yet had an appreciable amount of mass pile up behind you. At that location, the resultant force of all the earth’s mass (being the integral product of the mass distribution and the various distances of all that mass) pulls you toward the center stronger than anywhere else on your journey. This is mainly because, as was stated before, the center of the Earth is so dense. If the Earth were a uniform sphere, the point of maximum g would be at the surface.
At the center, you’re done receiving any positive acceleration (assuming positive in is the direction you started moving). You have then reached maximum velocity, and will only have slowing down to look forward to the rest of the way through the planet. That’s because past the center, the pull on you starts going the other (negative) way.
But since you had a positive acceleration the whole way to the center, you kept speeding up. The magnitude of that acceleration changed between surface and center, so you didn’t always speed up at the same rate.
Even if I remember wrong about the point of maximum gravitational acceleration…even if the maximum is at the earth’s surface (as it would be for a sphere of uniform density), that point is still valid. You’ll still reach maximum velocity at the center, and oscillate forever between one surface point and the other, in perfect harmonic motion.
RM Mentock
Point taken, thank you.
What makes the density greater is the very gravity we are addressing. The heavier materials are drawn to the center and compacted by the material above them. One-third was a guess-your-best number but ~1/2 would likely be the better number as the mass of the material “above” the subject body equals the mass of the material “below” the subject body.
More on that below.
audilover
You are correct. I am mere layman who has the thought and process in my cranium but have a bit of a problem with the conveyance of those thoughts. Please bear with me.
As a simple calculation of the volume (totally disregarding mass) of a perfect sphere, a sphere of r=2 would have a volume of ~33.5. The direct division of that sphere to half its radius, or r=1, however, reveals that volume to be ~4.2.
So, as the math books always say “as can readily be seen”, halving the sphere produces a volume that is far and away from the figure of ~16.525 to get a figure that is half the volume of the first sphere. That figure would be r=~1.58 or “as can readily be seen” ~79%.
This means that when the subject body reaches a point that is ~21% of the way to the center of the Earth (disregarding the mass of the various materials which make up the Earth) he would have as much material, by volume, “above” him as “below” him.
So maybe my ~1/3 guesstimate wasn’t so far off; but all things considered, and all of those things being infinitely variable, ~1/2 probably is closer.
I had trouble following this at first, but I think I figured it out. Did you mean 16.75, instead of 16.525?
Weirdly enough, the effect of all the mass at higher depth cancels out. Even though there’s a lot more of it farther on, the mass directly above is much closer–and its gravitational effect is much stronger. So, at depth, the acceleration is pretty much just as if the only mass left were the mass in the entire sphere below.
Because of density considerations, the acceleration increases a bit but mostly stays the same from the surface to the core-mantle boundary, about half way down.
I’m sorry, but this recent line of discussion, not to put too fine a point on it, is pretty stupid.
The equilibrium point is reached at the center of the earth. Period.
>>>> This means that when the subject body reaches a point that is ~21% of the way to the center of the Earth (disregarding the mass of the various materials which make up the Earth) he would have as much material, by volume, “above” him as “below” him.
<<<<
Uh, no. Apparently you are attempting to divide the earth into an inner core “below” you and a hollow shell “above” you. The problem is that the hollow shell isn’t all above you. Most of it, in fact, until you reach the center, is below you, pulling you toward the center.
RM Mentock
:eek:
Yep! Screwed the pooch on that one, didn’t I? :smack:
markci
I guess tact isn’t one of your finer points, now, is it? I apologize for being too lowbrow for your taste.
The fact is that none of us has ever actually tried this experiment so what would actually happen is up for discussion. I thought that was the purpose of discussion boards. I threw something into the mix that is heretofore unconsidered (and never will be if you have any say about it) to glean discussion on the plausibility, not possibility, that it may or may not have merit.
That may seem “stupid” to you but the Theory of Relativity is still called the “Theory” of Relativity even though it has been proven to be correct on numerous occasions and levels. It should rightly be called the Law of Relativity but the debate goes on unabated nonetheless.
Thanks for the warm welcome.
It is below me if you make a cutting plane straight through the planet perpendicular to the path of the subject body. The truth is that it is all around the subject body and should act upon it from all sides.
A round ball under water has pressure acting on it from all sides equally regardless of how much water is above or below it. In the case of a mass, the mass exerts gravity on all sides but in varying degrees dependent upon the mass location relative to the subject body. It seems to me that the gravity acting upon the subject body has to change relative to the mass surrounding it and where the majority of the mass resides relative to the subject body.
It would be true that the majority of the mass would be “below” the subject body pulling it “downward”; but at the same time the thickness of the mass to the sides, prependicular to the path of the subject body is increasing and must exert some influence on the subject body. Even with a clear path straight through the exact center of the Earth, gravity still has to act from all sides of the subject body and cannot be ignored.
Any thoughts – ad homs excluded?
No, the mass to the sides exerts no net graviational force at all, because it’s symmetrical. The only net force is toward the center.
And “Theory of Relativity” doesn’t mean “Hypothesis of Relativity” any more than “Set Theory” means “Set Hypothesis”.
For fuck’s sake, by symmetry, the point where the forces balance out is the center of the sphere. Period.
Cripes, why am I debating physics with people who obviously couldn’t pass PHYS 101 if their life depended on it?
By the way, the gravitational force inside a hollow spherical shell is zero for ANY point inside the shell. This is a pretty standard mechanics problem in sophomore-level physics. So you can treat the outer shell as if it didn’t exist.
I made the same point earlier, in the post immediately above where you said the recent line of discussion was pretty stupid.
But don’t worry about me, I didn’t take it personally.
I see gravity as a three dimensional entity while those espousing otherwise see it as linearly two dimensional. I don’t believe gravity is linear. I believe that the three dimensional aspect of gravity, those forces from other than “down”, will act upon the subject body; and it will not act in the classic pendulum theory – note theory – that has always been thought to be correct.
I really don’t care if I irritate some on this board and drive them to the posting of ad homs and expletives. That is not my purpose here. I would simply like to discuss, reasonably, why my hypothesis is flawed; and why non-axial forces perpendicular to the subject body would have no effect on it.
By the way, my model is the Earth, not a hollow shell, so we can drop that argument. Of course hollow has no gravitational force. It has no mass – that’s why they call it hollow.
By the way, there was mention of the Earth’s rotation and, as such, to successfully traverse through the Earth one would have to go directly down through the poles. This ignores the fact that the magnetic poles of the Earth are not coincident with the rotational poles.
What effect would that have on the subject body; or would it be something like Sound File as the subject body slid down the wall of the hole between the poles?
You have yet to debate anything. Your entire premise is based on epithets, ad homs, and “I’m right, you’re wrong.”
Uh huh!
Nuh uh!
Uh huh!
Nuh uh!
is NOT a debate.
Know the difference.
If you have nothing to add, then nothing is what you should say.