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tracer
2003-Apr-06, 08:34 PM
I just noticed yet another gaffe in The Core.

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Our intrepid heroes, in their nearly-indestructable Earth Ship, tunnel all the way down to the molten outer core of the planet. Yet, while they're walking around on the deck of their ship, they're still feeling the full 9.8 m/s^2 of Earth's surface gravity. (None of them, not even the wacky sidekick characters, exhibits any clumsiness while walking around. Two characters even get pinned underneath 6-foot-tall thermonuclear devices because said devices are "too heavy" for them to lift.)

If you were down that close to the center of the planet, most of the Earth's mass would be "above" you, i.e. farther away from the center of the planet that you were. None of that mass should contribute to the force of gravity you feel. Only the mass beneath you should "count." Why, then, weren't our intrepid heroes running around in 1/2 or even 1/3 Earth normal gravity?

g99
2003-Apr-06, 08:49 PM
I just noticed yet another gaffe in The Core.

Our intrepid heroes, in their nearly-indestructable Earth Ship, tunnel all the way down to the molten outer core of the planet. Yet, while they're walking around on the deck of their ship, they're still feeling the full 9.8 m/s^2 of Earth's surface gravity. (None of them, not even the wacky sidekick characters, exhibits any clumsiness while walking around. Two characters even get pinned underneath 6-foot-tall thermonuclear devices because said devices are "too heavy" for them to lift.)

If you were down that close to the center of the planet, most of the Earth's mass would be "above" you, i.e. farther away from the center of the planet that you were. None of that mass should contribute to the force of gravity you feel. Only the mass beneath you should "count." Why, then, weren't our intrepid heroes running around in 1/2 or even 1/3 Earth normal gravity?

Umm..The unobtanium produces its own localized gravity field?

Colt
2003-Apr-06, 09:32 PM
"Journey to the Center of the Earth" had it more accurately than The Core. They at least realized that gravity would go all wierd near the center of the planet except they should have had to go up through a layer of water to be on the surface of it if I am thinking correctly. -Colt

BigJim
2003-Apr-06, 09:44 PM
Actually, Journey to the Center of the Earth , while one of my favorite movies, had it less accurately than "The Core". In Journey to the Center of the Earth, besides the ridiculousness of an ocean in the core, the only effects that could be considered weird gravity were when they actually reached the center of the Earth. Here, where they should have been weightless, the gravity did not change, but somehow they knew that the center of the Earth was a place that attracted gold, and all the gold on the raft flew off. No explanation was given as to why the core attracts gold- I think this was an old idea like the ether or alchemy.

g99
2003-Apr-06, 09:48 PM
Actually they wont be weightles at the core. Most of the mass of the earth is above them. They would be walking on the ceiling.

skeptED56
2003-Apr-06, 09:59 PM
Actually they wont be weightles at the core. Most of the mass of the earth is above them. They would be walking on the ceiling.
But there is equal mass pulling at them from all directions, so they should be weightless (maybe not absolutely since the Earth isn't a perfect sphere).

g99
2003-Apr-06, 10:03 PM
Actually they wont be weightles at the core. Most of the mass of the earth is above them. They would be walking on the ceiling.
But there is equal mass pulling at them from all directions, so they should be weightless (maybe not absolutely since the Earth isn't a perfect sphere).

Ahhh...I get it. But that would be only if they are in the exact center. What will happen if they are slightly off center? Would they not slowly (very slowly) start to fall towards that part of the planet?

skeptED56
2003-Apr-06, 10:14 PM
Actually they wont be weightles at the core. Most of the mass of the earth is above them. They would be walking on the ceiling.
But there is equal mass pulling at them from all directions, so they should be weightless (maybe not absolutely since the Earth isn't a perfect sphere).

Ahhh...I get it. But that would be only if they are in the exact center. What will happen if they are slightly off center? Would they not slowly (very slowly) start to fall towards that part of the planet?

If they were off center they'd fall in the direction of greatest mass (however slowly). But remeber there is an equal amount of mass on the other "side" of the core, so they'd fall to the center with decreasing weight (anyone correct me if im wrong) and then when they get to the center they would become wieghtless.

tracer
2003-Apr-06, 11:05 PM
In the movie, they never went all the way to the center of the Earth. They stopped while they were in the outer (molten nickel-iron) core.

Most, but by no means all, of the Earth's mass would have been "above" their depth at that point. "Down" would still have been toward the center of the Earth, but their weight should have been a lot lower than it was at the surface.

skeptED56
2003-Apr-07, 12:07 AM
For an example say a group of scientists are drilling to the core of a planet (this planet happens to be a perfect sphere and alll mass and densities are all uniform throughout the entire thing) 1000 km in diameter making its radius 500 (duh). If you were at the center all the mass in the 500 km to the surface in every direction will be pulling you each way so you would be weightless. If you drill from the surface 480 km into the planet with 20 km left to go to the center you will feel the affects of 40 km pulling you towards the center because the 480 km you have traveled (which would be "up) cancels out 480 km of the 520 km depth directly ahead of you from you to the opposite point on the surface from where you started out( "down"). So the 20 km of mass between you and the center would be added to 20 km of another radius (or the second half of the diameter) from the center to the opposite point on the planet's surface form which you started drilling that didn't get canceled out by the 480 km behind you. The point is even if you've traveled almost all of the radius you've traveled a a little less than half of the diameter meaning that you can never fall "up".

Durandal
2003-Apr-07, 01:35 AM
I just noticed yet another gaffe in The Core.

But the writer said the science was dead-on! I can't believe Hollywood would lie to me like that!

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Our intrepid heroes, in their nearly-indestructable Earth Ship, tunnel all the way down to the molten outer core of the planet. Yet, while they're walking around on the deck of their ship, they're still feeling the full 9.8 m/s^2 of Earth's surface gravity. (None of them, not even the wacky sidekick characters, exhibits any clumsiness while walking around. Two characters even get pinned underneath 6-foot-tall thermonuclear devices because said devices are "too heavy" for them to lift.)

If you were down that close to the center of the planet, most of the Earth's mass would be "above" you, i.e. farther away from the center of the planet that you were. None of that mass should contribute to the force of gravity you feel. Only the mass beneath you should "count." Why, then, weren't our intrepid heroes running around in 1/2 or even 1/3 Earth normal gravity?

Pretty much, yeah. However, attractive force increases according to the inverse square law. So, since they're closer to the core of the Earth, they'd be experiencing more attraction in that direction. I'm too lazy right now to work out the numbers of falling-off radius versus falling-off mass, but since the mass distribution of the Earth isn't uniform, it could make for some whacky gravitational effects.

tracer
2003-Apr-07, 06:19 AM
I doubt that the interior of the Earth would be so nonuniform that it would make gravitational effects that were all that wacky -- until you started tunnelling into the inner (solid iron-nickel) core and got really close to the geometric center of the planet, which they didn't do in the movie.

The Earth started as a big molten ball of metal and rock and cooled over hundreds of millions of years before it had a comfy solid surface on it. That was plenty of time for all the dense stuff to settle down at the center and all the less-dense stuff to "float" to the top, and the distribution is probably pretty uniform. [I'm guessing it was this "densest stuff sinks to the center" tendency that inspired what's-his-face to hypothesize a giant nuclear fission reactor in the Earth's inner core -- solid Uranium under normal conditions of temperature and pressure has a specific gravity of 19.05, which puts its density only slightly behind tungsten (19.25) and gold (19.3), and a weensy bit farther behind rhenium (21.02), platinum (21.09), osmium (22.61), and iridium (22.65). If the densest stuff really does sink to the center as a uniform, compacted layer, then there should be a thin, uniform spherical shell of uranium near the center of the inner core surrounding a (perhaps thicker) spherical shell of tungsten, and so on in concentric shells of greater density as you go downward, until you hit iridium at the dead center.]

kilopi
2003-Apr-07, 10:53 AM
I doubt that the interior of the Earth would be so nonuniform that it would make gravitational effects that were all that wacky
Nope. It's true, they are exactly that wacky.

As much as it pains me to say this, The Core got that science right. Of course, it could have been by accident, since the force of gravity at the top of the core is very nearly the same as the force of gravity on the surface of the Earth. In fact, it's pretty much constant all the way down to the top of the core--increases a little bit in fact--so they didn't have to do anything to get it right. (That seems to be in keeping with the rest of the movie. :) )

The core is more or less twice as dense as the entire Earth, and has a radius of about half the Earth. The force of gravity is proportional to the mass divided by radius squared, but mass is proportional to density times the cube of radius. As has been pointed out, the mass above your radius can more or less be ignored. So the force is proportional to density times radius, and since one is twice and the other half, the result is that the gravity at the core is about the same as at the surface.

If the densest stuff really does sink to the center
The dense stuff does not sink to the bottom. The upper layer of the Earth has a much higher concentration of radioactive elements than the lower layers.

JS Princeton
2003-Apr-07, 01:37 PM
The definitive word:

You can basically consider the Earth to be of uniform density for our purposes. As you headed toward the center of the Earth, the acceleration due to gravity would decrease linearly according to the equation:

a = G*4/3*pi*r*rho

Where rho is the density and r is the radius.

If they were halfway to the core then they should experience half the gravity. At the core (r=0) there would be no acceleration due to gravity at all.

kilopi
2003-Apr-07, 01:47 PM
The definitive word:

You can basically consider the Earth to be of uniform density for our purposes. As you headed toward the center of the Earth, the acceleration due to gravity would decrease linearly according to the equation:

a = G*4/3*pi*r*rho

Where rho is the density and r is the radius.

If they were halfway to the core then they should experience half the gravity. At the core (r=0) there would be no acceleration due to gravity at all.
Three points. If you get 12 points you lose your license. :)

tracer
2003-Apr-07, 03:41 PM
Sorry, but I gotta agree with kilopi on this one.

Your formula is correct, JS Princeton, but only if rho (the density) is constant all the way down. It clearly is not -- the Earth's average density is some 5.5 grams per cubic centimeter, but according to this webpage (http://www.clas.ufl.edu/users/emartin/OCE1005f00/lecturenotes/structpt.htm), the average density of the entire core is 10-12 grams per cubic centimeter, or about twice the average density of the Earth as a whole.

According to this U.S. Geological Survey page (http://pubs.usgs.gov/publications/text/inside.html), the top of the outer core is about 2900 km below the surface of the Earth. Since the Earth is 6378 km in radius, this makes the core's radius equal to 0.55 times the radius of the Earth, which would make the core's volume equal to 0.17 times the volume of the Earth. (The volume of a sphere is directly proportional to its radius cubed.)

Now then, the general formula for gravitational acceleration in Earth g's is M/r^2, where M is the mass beneath you (in Earth masses) and r is your distance from the center (in Earth radii). Assuming a volume of 0.17 Earth volumes and a density of 2 Earth densities :) , the core would have a mass M of 0.34 Earth masses.

Therefore, when our intrepid heroes in the movie were at the tippy-top edge of the outer core, they should have been experiencing gravity of 0.34 / 0.55^2 = ... holy cats ... 1.12 times normal Earth surface gravity! :o :o :o

I take back everything I said at the beginning of this thread. They would weight slightly MORE at the surface of the core, not less!!

Mea culpa, mea culpa, mea maxima culpa!

JS Princeton
2003-Apr-07, 04:54 PM
So therein lies the point. If you multiply rho by 2 and divide r by less than 2 you end up with more gravity.

Us fly-by-night approximators need to be clear about our assumptions. My assumption that the average density was good enough was wrong, dead wrong.

However, what I don't get is that the Cavendish experiment measures a mean density of 5 g/cc. Where on "Earth" does this number come from? After all, with a radius of 6300 km and a mass of 6.E27 grams, you get a much higher mean density.

Thanks for the 3 points, kilopi, I'll store them in my pocket.

kilopi
2003-Apr-07, 05:32 PM
However, what I don't get is that the Cavendish experiment measures a mean density of 5 g/cc. Where on "Earth" does this number come from? After all, with a radius of 6300 km and a mass of 6.E27 grams, you get a much higher mean density.
Nine Planets (http://seds.lpl.arizona.edu/nineplanets/nineplanets/earth.html) gives those figures as 6378.2 km and 5.972E27 grams. That's close to what you got, and using them I get 5.495 gr/cm^3, which is very close to the 5.5 that I've always used--and is the one tracer used.

ToSeek
2003-Apr-07, 07:33 PM
Us fly-by-night approximators need to be clear about our assumptions. My assumption that the average density was good enough was wrong, dead wrong.

Let us first assume that the horse is a massless, frictionless sphere....

(From some joke about theoretical physicists going into horse racing.)

JS Princeton
2003-Apr-09, 04:59 PM
However, what I don't get is that the Cavendish experiment measures a mean density of 5 g/cc. Where on "Earth" does this number come from? After all, with a radius of 6300 km and a mass of 6.E27 grams, you get a much higher mean density.
Nine Planets (http://seds.lpl.arizona.edu/nineplanets/nineplanets/earth.html) gives those figures as 6378.2 km and 5.972E27 grams. That's close to what you got, and using them I get 5.495 gr/cm^3, which is very close to the 5.5 that I've always used--and is the one tracer used.

Right, so the number is right, for the average, does that mean that the mantle in the crust have a lower density than 5 g/cc?

Weighting the volume with 17%, and considering the core to have a mean density twice that of the average, that means the outer density of the mantle and the crust is closer to 4 g/cc.

Ah the places you'll go.

tracer
2003-Apr-09, 05:26 PM

Of course, in the movie, when they finally reched the outer core, they discovered that its density was less than we had previously estimated. If the cores average density was only 8 g/cm^3, our intrepid heroes should have been feeling a gravitational force of only about 0.8 times Earth normal surface gravity.

kilopi
2003-Apr-09, 05:41 PM
Density of granite is less than 3g/cc

tracer
2003-Apr-09, 08:46 PM
And because there mantle contains more iron, magnesium, and calcium than the crust, and is also denser due to being compressed (pressure inside the Earth increases with depth), a mantle density above that of granite is to be expected.

kilopi
2003-Apr-10, 03:44 AM
The crust is mostly silicate, SO2 or SO4 etc.

tracer
2003-Apr-10, 10:29 PM
I hope you mean SiO2, not SO2. I'd hate to have to worry about breathing sulfur dioxide every time I walked around a patch of silicate!

copabera
2003-Apr-10, 10:58 PM
So, given that gravity decreases to (very close to) zero as we approach the center of the Earth, wouldn't that mean that pressure also decreases? Wouldn't heat be reduced as a result? If we could get through the initial 3900 miles, wouldn't the last leg be a cake walk?

I'll bet the center of the Earth (or that of Jupiter for that matter!) could be a very cool place!

tracer
2003-Apr-10, 11:55 PM
Heh. No. The lack of "surface gravity" if you "stand" at a particular point within an object doesn't do a whit to reduce the pressure from all the material above you, which will all be experiencing higher gravity.

The center of the sun is at zero-gee, but it's hot enough and high-pressure enough to fuse protons into alpha particles.

kilopi
2003-Apr-11, 05:30 AM
I hope you mean SiO2, not SO2. I'd hate to have to worry about breathing sulfur dioxide every time I walked around a patch of silicate!LOL. Yep, I guess I was too focussed on not being able to subscript (http://www.badastronomy.com/phpBB/viewtopic.php?t=4584). :)

tracer
2003-Apr-11, 06:49 AM
You'd think on a message board devoted to astronomy, you'd have formatting tags to do proper exponents and chemical formulas. ;) Heck, vBulletin and UBB both support [sub] and [super] tags.

tracer
2003-Apr-12, 10:44 PM
Okay, I just saw The Core for a second time. (I must have a masochistic streak in me.)

This time, I noticed something I missed before. When they were bench-pressing the nuclear (excuse me, nuke-you-lar) warheads to move them into different compartments, the Virgil was a lot farther down than I previously thought. They were skimming along the top of the inner core.

That changes the gravity equation. The radius of the inner core is only 0.2 times the radius of the Earth, giving it a volume equal to a mere 0.0080 times the Earth's volume. Assuming that the inner core is twice as dense as the Earth as a whole, this would make its mass equal to 0.016 Earth masses.

Therefore, when our intrepid heroes in the movie were at the tippy-top edge of the inner core, they should have been experiencing gravity of 0.016 / 0.2^2 = 0.4 times the surface gravity of the Earth!

This is about the same gravity as you'd experience standing on the surface of Mars. So, like I asked in the first post in this thread, why weren't they walking funny or having an easier time picking up those nuclear bombs?

Betenoire
2003-Apr-14, 03:59 AM
On the question of gravity, the calculators are going by the standing estimates for the density of the core. But remember, in the movie the outer core is not as dense as we think.
My question is how the outer core could magically be less dense than previously thought. Wouldn't that mean a lot different composition for the inner core? Say, a medium size primordial black hole or some such to compensate for the lack of mass of the outer core?

tracer
2003-Apr-14, 05:19 AM
Good point. If we currently know the total mass of the core (inner and outer combined) and are only guessing that the outer core is the same density as the inner core -- then if the outer core turned out to be less dense than we had previously assumed (like it did in the movie), that would mean the inner core would have to be more dense than we had previously assumed so that the total mass of the outer and inner core would equal our previously-calculated values.

Unfortunately, in the movie, they didn't say how much less dense the outer core was than the inner core, merely that it was sufficiently rarefied enough for the Virgil to drop through it like a rock through a pond. However, we can assume that if the outer core is primarily iron and nickel, which it would pretty much have to be in order to generate a magnetic field while rotating, then it must have a density at least as high as the specific gravity of iron at "normal" temperatures and pressures, i.e. 7.87 grams per cubic centimeter.

SO ... we know that the volume of the whole core is 0.17 times the volume of the Earth, and that the volume of the inner core alone is 0.008 times the volume of the Earth. This means the inner core's volume is about 0.05 times the volume of the whole core. We also know that the mass of the whole core is 0.34 times the mass of the Earth. Now, if the outer core turned out to have a density of only 7.87 g/cm^3, or 1.43 times the average density of the Earth, that would mean that the mass of the outer core would be (0.17 - 0.008) * 1.43 = 0.23 times the mass of the Earth. This would make the inner core a whopping 0.11 times the mass of the Earth, which would make its density 13.8 times the average density of the Earth, or over 6.8 times our current scientific estimates.

And the surface gravity at the top of this super-dense inner core? It would be 0.11 / 0.2^2 = 2.8 times the surface gravity of the Earth.

So how were they able to stand up at all? ;)

kilopi
2003-Apr-14, 12:36 PM
And the surface gravity at the top of this super-dense inner core? It would be 0.11 / 0.2^2 = 2.8 times the surface gravity of the Earth.

So how were they able to stand up at all?)
Your calculation is based upon the lowest outer core figure you could justify. Just increase it a bit to a more conventional figure, and you can get any value you want between the 2.8 times surface gravity and the 0.4 times you calculated earlier (and, actually, based upon conventional values, that should be more like 0.5 I think).

So, those screenwriters were way ahead of you!

Every so often I hear a comment in a movie that seems to be unrelated to the action, and I wonder why the character said it, only to find later that it plugs a plot hole that would have been created by subsequent developments.

myoffe
2003-Apr-14, 05:32 PM
..... when our intrepid heroes in the movie were at the tippy-top edge of the outer core, they should have been experiencing gravity of 0.34 / 0.55^2 = ... holy cats ... 1.12 times normal Earth surface gravity! :o :o :o

I take back everything I said at the beginning of this thread. They would weight slightly MORE at the surface of the core, not less!!

Mea culpa, mea culpa, mea maxima culpa!

Not quite culpa. The fact is that the "heavier" or denser elements have a tendency to concentrate toward the center of the planet, for some Archimedean principle or another. (Otherwise we'd have an atmosphere of solid iron and a core made up of low pressure gases – a physical absurdity). Thus, less dense entities, human bodies, relatively hollow ships, etc., tend to be *repelled* outward toward the surface. The equation cited above assumes that there is nothing denser competing for the space at the center. If there is, the denser object “win” the gravitational tug of war causing the less dense bodies to be repelled outward!

David Hall
2003-Apr-14, 06:06 PM
Hmm, just thinking. if the material this craft was travelling through was that dense, then the ship would have been incredibly buoyant, and as such would have been pretty difficult to keep moving in the down direction. It would constantly want to float up.

Same for the geode, I would think.

Edit: D'oh! Myoffe pretty much just said the same thing in the last post. I should have read it more carefully first. :-)

tracer
2003-Apr-14, 06:40 PM
True, but we can assume that the Virgil had some kind of magical buoyancy counterweights. Maybe it had ballast tanks full of white dwarf matter.

In any case, if the Virgil can maintain a constant depth, the people inside the Virgil should be feeling a force of gravity equal only to their weight at that depth. Thus, they'd feel 1.12 G's at the top of the outer core, and anywhere from 0.4 to 2.8 G's at the top of the inner core depending on how badly "science" got the outer core's density wrong.

myoffe
2003-Apr-14, 07:09 PM
True, but we can assume that the Virgil had some kind of magical buoyancy counterweights. Maybe it had ballast tanks full of white dwarf matter.

In any case, if the Virgil can maintain a constant depth, the people inside the Virgil should be feeling a force of gravity equal only to their weight at that depth. Thus, they'd feel 1.12 G's at the top of the outer core, and anywhere from 0.4 to 2.8 G's at the top of the inner core depending on how badly "science" got the outer core's density wrong.

A few points.

1. I don't know about the "dwarf" stuff. If we assume magic, then the whole concept of "bad" vs. "good" astronomy becomes meaningless. This would be “badmagic.com”....

2. Ignoring buoyancy for a moment, tracer's calculations seem to ignore the fact that besides the core there is also the mass of mantle and crust "below" and to the sides of the surface of the core. This is decidedly *not* canceled out by the mass directly above and to the sides because the mass between the planes tangent to the surfaces of the core tend to pull the ship down, and there is no canceling force for that. I don't think this issue can be solved algebraically - we need calculus for this.

3. even assuming that there was this magical stuff counteracting the buoyancy of the ship, that magic would affect the ship presumable. There is another buoyancy factor created by the relationship between the inhabitants and atmosphere of the ship. . But let's assume that that is negligible.....

tracer
2003-Apr-14, 08:42 PM
2. Ignoring buoyancy for a moment, tracer's calculations seem to ignore the fact that besides the core there is also the mass of mantle and crust "below" and to the sides of the surface of the core. This is decidedly *not* canceled out by the mass directly above and to the sides because the mass between the planes tangent to the surfaces of the core tend to pull the ship down, and there is no canceling force for that. I don't think this issue can be solved algebraically - we need calculus for this.
People have already done the calculus centuries ago. And the verdict is that if you are anywhere inside a uniform spherical shell of material, the gravitational contribution from all the material on one side of you exactly cancels out the gravitational contribution from the material on the other side of you, regardless of where inside this sphere you are.

Net result: Assuming each spherical shell of material above you is uniform (constant density all along the sphere at any one given depth), their gravitational contributions completely cancel each other out, so all you are left with is the gravitational contribution from material in spherical shells that you're still outside of, i.e. everything at your depth and lower.

myoffe
2003-Apr-14, 09:36 PM
People have already done the calculus centuries ago. And the verdict is that if you are anywhere inside a uniform spherical shell of material, the gravitational contribution from all the material on one side of you exactly cancels out the gravitational contribution from the material on the other side of you, regardless of where inside this sphere you are.

If by "side" Tracer means to the left and to the right, it is correct - intuitively. If Tracer means down (toward the center) and up (away from the center), which is the real problem here, I don't think it is correct. Otherwise people would be floating around whenever they went into an undeground cave. Am I missing something here?

Glom
2003-Apr-14, 11:06 PM
Otherwise people would be floating around whenever they went into an undeground cave. Am I missing something here?

The average underground cave is insignificantly deep in the sphere. The peach comparison in the movie, believe it or not, was good science. The skin is very thin compared to the juicy bit and therefore does analogise very well with the crust. You'd start to float around once you got near the centre.

tracer
2003-Apr-14, 11:07 PM
Yep -- you're missing the fact that it's only the material in the spherical shell above your altitude/depth that cancels itself out. All material at your depth or lower -- not just right below your feet, but anywhere on the globe -- contributes to your weight.

Draw two concentric circles. Label the inner circle "core" and the outer circle "mantle/crust". Pretend that these circles are just a 2-dimensional slice of a 3-dimensional spherical globe. Now, put an X at the top of the inner circle, to represent "you are here." All the material in the inner (core) circle contributes to how much the X weighs. None of the material in the outer (mantle/crust) circle contributes to how much the X weighs. That's all I'm trying to say.

And, yes, when you go into an underground cave, your weight can decrease. Very very very slightly. And only if the cave goes below sea level, like a mine shaft.

kilopi
2003-Apr-15, 12:32 PM
If by "side" Tracer means to the left and to the right, it is correct - intuitively. If Tracer means down (toward the center) and up (away from the center), which is the real problem here, I don't think it is correct.
It is correct. An early problem in a calculus course.

Otherwise people would be floating around whenever they went into an undeground cave. Am I missing something here?
Yes, I'm not sure why you'd think they'd be floating around in a cave. The solution mentioned by Tracer says the effect in a cave would be by all the material contained within a radius from yourself to the center of the Earth--that's a considerable amount of material.

And, yes, when you go into an underground cave, your weight can decrease. Very very very slightly. And only if the cave goes below sea level, like a mine shaft.
It can decrease, or it can increase. Remember, the force of gravity at the core-mantle boundary is greater than at the surface. In other words, gravity increases slightly as you go down.

The bouyancy thing is almost irrelevant. In order for bouyancy to play a part, the material has to flow on the order of the time scale involved--the quick trip to the Earth's core. Even magma is too viscous for bouyancy effects to make much of a difference, probably.