If the center of the earth was hollow what would happen if you stood in it?

[Brumby]

Ok back on track, the radius math applies on the assumption you can treat both objects as point masses of infinite density. When you have one inside the volume of another that math can no lnger be used (its assumptions are broken), and does require falling back to integrals of acceleration for the distrubution of mass around you.

I wasn't quite sure how to make this point, so I stepped past it - but having read the above, it seems the right words were all too easy for Rerouter.   

[Zero999]

Say it was 100ft(32.74833495735m) wide hollow pit and you could go inside it. Would you float in the center being pulled from all sides from gravity or would you be able to walk around the walls in a circle?

Your metric conversion is a little off: 100ft = 30.48m.

This is obviously impossible. It's too hot inside the earth for any human to survive down and the pressure is too high for a hollow cavity to exist in the centre of it. Think of the weight of solid/liquid rock and iron pushing down on the middle of the planet!

It's true that in the earth's centre of gravity, not necessary the geometric centre, the gravitational field will be close to zero, but it's not practical to prove it, in the manner you describe.

[Distelzombie]

Hero999: If you have a tunnel from the surface to this spherical cavity, all you have to worry about is the air pressure and the temperature. (If the cavity is stable in all the molten Iron or worse Elements. [Molten Lead, Uranium ...]) There's no rock pushing on YOU if the cavity is stabilized and with a hole to the surface the only pressure will be the pressure of 6,371 kilometers of air above you = 22.967271909795332 PPa (Peta pascal) or 3331121161161.99 psi
I would say that is enough anyway. But you can reduce it to surface-level pressure by simply adding air-locks with pumps every kilometer.

Also, saying a thought experiment is obviously impossible is pretty useless anyway.


My guess is you can only walk on the walls of this spherical cavity if you have enough speed to overcome the very, very low gravitational pull to the center of the cavity: You need centripetal force or else you'll fall back to the center.
But the gravitational force pulling you towards the center of this relatively small spherical cavity is really small anyways. Imagine: If you're on one side of the 30.48m cavity, there's only a distance of 15.24m to the point of zero-G. Those 15.24m of earth aren't very attractive.

[StuUK]

StuUK, are you saying a flat earth does not have a center point? and that you could not dig down at that center point far enough down to support a 100m radius sphere?

If the earth was disk shaped the same effect would occur using the currently popular mechanics of gravity, If you assume a disk being uniformly accelerated at 10m/s then well your stuck at the bottom of your sphere at 1G.

Fair point... 

[GlennSprigg]

OBVIOUSLY the centre of the Earth is NOT hollow, but IF the Earth WAS a 'Shell',
then the "Laws Of Gravity" are that once inside such a 'Shell', then Gravity is NEUTRAL!!
No matter if you were in the middle, or close to the "Wall", the balanced pull is NEUTRAL.
(I've simulated this in realistic and real-time computer models)

[Beamin]

It would be like being in microgravity, in any case there will be a bias to a certain side, so you would come to rest on the wall of the sphere, technically walk, but difficult,

No

Sorry for the lazy answer, but I spent all afternoon and part of the evening doing integrals with my daughter...

The Earth is not a spherical cow in a vacuum uniform sphere, it is an ovlate spheroid, so there is unequal distribution of mass meaning that you'd experience a slight force in the direction of the equator which, being roughly annular, would tend to centre you in the hollow (if you drift out of the plane of the equator that plane would seem attractive to your mass).

As to what would happen – you'd be crushed and fried. The temperature of the Earth's core is approximately 5700 Kelvin, well in excess of the boiling point of iron at atmospheric pressure (3134 K). The pressure at the core is estimated to be 3.3 - 3.6 million atmospheres. Under any vaguely realistic scenario you'd die instantly there, even if you were unrealistically transported there instantaneously by Mr. Scott in a super-dooper environment suit.

You are not thinking of this creatively. The inside is hollow filled with surface air at a pressure a 1 bar. The walls are 20'C and he structure behind the walls hold back thermal and all em and ionizing radiation. Quite pleasant. But can you jump off the sides to float perfectly in the center or ellipsoid center since is a n ellipsoid plant? Or would you always be stuck to the sides. There is a light in there to so you can see.

[taydin]

If you don't move in that cavity, you would be in a state where your center of mass coincides with the center of mass of the earth.

If you slightly move, so will your center of mass, and you will start a dampening oscillation, and eventually you will reach the above equilibrium.

[Cerebus]

It would be like being in microgravity, in any case there will be a bias to a certain side, so you would come to rest on the wall of the sphere, technically walk, but difficult,

No

Sorry for the lazy answer, but I spent all afternoon and part of the evening doing integrals with my daughter...

The Earth is not a spherical cow in a vacuum uniform sphere, it is an ovlate spheroid, so there is unequal distribution of mass meaning that you'd experience a slight force in the direction of the equator which, being roughly annular, would tend to centre you in the hollow (if you drift out of the plane of the equator that plane would seem attractive to your mass).

As to what would happen – you'd be crushed and fried. The temperature of the Earth's core is approximately 5700 Kelvin, well in excess of the boiling point of iron at atmospheric pressure (3134 K). The pressure at the core is estimated to be 3.3 - 3.6 million atmospheres. Under any vaguely realistic scenario you'd die instantly there, even if you were unrealistically transported there instantaneously by Mr. Scott in a super-dooper environment suit.

You are not thinking of this creatively. The inside is hollow filled with surface air at a pressure a 1 bar. The walls are 20'C and he structure behind the walls hold back thermal and all em and ionizing radiation. Quite pleasant. But can you jump off the sides to float perfectly in the center or ellipsoid center since is a n ellipsoid plant? Or would you always be stuck to the sides. There is a light in there to so you can see.

If you want the answer to the 'what would gravity do' then you already have that in spades, described by several people in several different ways. The only arguments I haven't seen are (1) The simplest, the argument from Newton's law of gravitation with a symmetry argument (the absolute minimum maths required), (2) The most complex, an analytic integration of the Earth's mass based on a well established geoid model such as WGS-84 (much maths).

As to the rest; I never understand why people ask a 'physics what if' on here, and the second any actual physics intervenes they want to change the rules so that physics doesn't apply. If you want a "creative" answer you'd be better served by asking on a soft Sci-Fi forum or a creative writing forum instead of an engineering one. Ask on an engineering forum and people will put pesky science and physics in the way of your creative flights of fancy.

[Beamin]

So some say you get stuck to the walls
Some say you float in the center
Somesay you just float all around.

What did the simulations say? Pictures?

[Rerouter]

Still not backed by simulation, but in any case it would come down to an accurate descript of the earths mass and density at a wide array of sample points

So lets break the problem down in to 2, the spherical cow model (world is uniform density and spherical), and the more realistic at the center of mass.

1. In this model, its is perfect "zero-g" anywhere in the sphere has no restoring force, the math behind it is shell theorem being as you move further away from the center, the forces towards the wall your heading too get a little stronger, but the weaker forces of the walls your moving away from become angles that tug you more in the opposite direction.

Good link: https://www.quora.com/Why-is-gravity-inside-a-spherical-shell-considered-to-be-zero

2. Center of mass, in most cases would be viewed as the above problem, however the density surrounding the magic room is not exactly even, so there will be imbalances that pull you to certain points, equally the world does spin (360 degrees per 24 hours), its rotation would very weakly make you stick to the walls about the axis of rotation, Then you have fun stuff like the moons contribution tugging you towards it (this model only cancels out the effect of the planets gravity).

number 1, you float, if you get to a wall, you could walk on the wall, relying on your forward speed to keep you there
number 2, you would be tugged to a wall, could walk anywhere, but you cannot stop and expect to stay stuck everywhere.

[Distelzombie]

OBVIOUSLY the centre of the Earth is NOT hollow, but IF the Earth WAS a 'Shell',
then the "Laws Of Gravity" are that once inside such a 'Shell', then Gravity is NEUTRAL!!
No matter if you were in the middle, or close to the "Wall", the balanced pull is NEUTRAL.
(I've simulated this in realistic and real-time computer models)

I don't think your simulations are accurate in such low levels of gravitation. If you're on one side, there's a little more matter on the other side. As simple as that. That's ... just simple logic. I mean that's the reason that there even is a center.
Of course, as I already said, the pull you'd feel is very VERY tiny. And that's why I think the program you did the simulations in is not accurate enough for that.
Can you show us? Maybe that would sort out the issues.

On another note, I guess you can swing yourself around if you end up in the zero-G point. Much like on swing. Because there is air you have something to push against. And the forces are very weak, that pull you back to the center.

[Electro Detective]

Too much speculation and conflictiing theory here    

Time to hit on Youtube perhaps ?   

[Distelzombie]

Too much speculation and conflictiing theory here    

Time to hit on Youtube perhaps ?   

Just go and do your own stuff. Leave us alone, OK? With all the shit you wrote everywhere you just simply have to be drunk. That's a prerequisite.

[Cerebus]

Too much speculation and conflictiing theory here    

Time to hit on Youtube perhaps ?   

Really?

With the discussion above anybody who has done secondary school physics ought to be able to work out with a few minutes thought which answers are correct (for their given assumptions) and which are speculation.

And you tube as the arbiter? (Normally your smiley would indicate humour, but your tendency to sprinkle smileys about like punctuation rather devalues it as a reliable indicator).

[Cerebus]

Too much speculation and conflictiing theory here    

Time to hit on Youtube perhaps ?   

Just go and do your own stuff. Leave us alone, OK? With all the shit you wrote everywhere you just simply have to be drunk. That's a prerequisite.

A bit unlikely. It's a bit early (Aus time) to be drunk, even on a Saturday, even for an Australian. [Ducks as a dozen empty EMU bottles fly his way.]

[ucanel]

At first glance I thought @KL27x is wrong but know I think similar,
if the density of the near outer part the 100ft chamber is would be very high as if it is,
then only the very center of the chamber would be at zero gravity and
the gravity difference between even the 1 feet up and down from the center
would be very very high and would be opposite force vector,

so anything bigger than may be an atom could fall apart to the walls of this chamber.

[hamster_nz]

If you are at the 'gravitational center' of the Earth, in a void you would be in zero G. The gravitational pull from above matches the gravitational pull from below. If you are one side, there would be a restoring force, pushing you back to the center.  The easy way I know to visualize this is....

When you move 'north' by 1m, the gravitational pull from mass that north of you is no longer in balance with the pull to the south of you - there is now an extra 1m slice of the earth that is now below you.

If you were to subtract what is north of you from what is south of you, you will have a dome that is paper-thin around the equator, and 2m thick at the South pole. The gravity from the mass of this dome will be the net gravitational force acting on you, pulling you south.

A dome of earth, 6,900 away is officially 'not very much' compared to an entire planet, so the force will be very, very feeble. It would be in the order of 1/10,000,000th of G I am guessing (only because it has to be 1G at the surface, 6,371,000 meters above.

[Brumby]

When you move 'north' by 1m, the gravitational pull from mass that north of you is no longer in balance with the pull to the south of you - there is now an extra 1m slice of the earth that is now below you.

Yes, but you are now 1m further away from the centre of mass to the south and 1m closer to the centre of mass to the north.  This would result in a smaller attractive force from the south and a larger one to the north.

The question then becomes: Which of these effects will be greater: The gravitational difference due to mass differences (your observation) or the gravitational difference due to distance differences (my observation)?



Who wants to do the math?

[hamster_nz]

No math needed.

Look at it this way Start with everything in equilibrium

Rather than moving the person from the center of gravity, scrape off one meter off of the north  hemisphere and spread it over the southern hemisphere (without moving the person or the rest of the planet).

Gravity will now pull the person south.

Any other answer has to be wrong.

[Brumby]

No math needed.

Look at it this way Start with everything in equilibrium

Rather than moving the person from the center of gravity, scrape off one meter off of the north  hemisphere and spread it over the southern hemisphere (without moving the person or the rest of the planet).

Gravity will now pull the person south.

Not necessarily.  The distribution of that mass is very important in determining the net force.

Any other answer has to be wrong.

Not necessarily.  It depends on the distribution.

[taydin]

At first glance I thought @KL27x is wrong but know I think similar,

@KL27x would right if the hypothetical question was about a black hole's center being hollow (whatever that would be like ), or a neutron star, or a star. In other words, an object much more massive than the earth. In that case, a few meters of distance would cause very large gravitational force differential and those forces would tear you apart. But that won't happen with the earth, and we all are living proof

[hamster_nz]

No math needed.

Look at it this way Start with everything in equilibrium

Rather than moving the person from the center of gravity, scrape off one meter off of the north  hemisphere and spread it over the southern hemisphere (without moving the person or the rest of the planet).

Gravity will now pull the person south.

Not necessarily.  The distribution of that mass is very important in determining the net force.

Any other answer has to be wrong.

Not necessarily.  It depends on the distribution.

Hummm.. so let's thing about that. What distribution makes sense? Dense material on the outside, less dense material on the inside? No. Can't be. We would all sink because it would be unstable.

How about less dense material to the outside, denser materials to the inside? Sure. Like an iron core. That makes sense and unlike other option is inherently stable.

So as you move 1m north, proportionally even more of the densest core material is now south of you.  Were you able to move to the edge of the core, then all of the dense core would be south of you, but you would still have the thickness of the less dense mantel and crust above you.

On the other hand, i guess it could always be hollow and filled with living dinosaurs...

Am I going to have to write code to prove it?

[Rerouter]

writing code would be nice, Its always interesting to see the math backed up. (Shell theorem by the way)

The step you would be proofing is that there is a difference in the sum of gravity vectors as you move in the sphere,

To make it even simpler, you could start with 6 fixed point masses, located at X+ X- Y+ Y- Z+ and Z- and compute the components from gravity of each to your "person" acting as a point mass to simplify the math

If it works there, you can distribute points more densely about a spherical shape and see if it still holds up.

[Electro Detective]

Too much speculation and conflictiing theory here    

Time to hit on Youtube perhaps ?   

Just go and do your own stuff. Leave us alone, OK? With all the shit you wrote everywhere you just simply have to be drunk. That's a prerequisite.

NO ONE HERE can back up the fancy entertainment come math BS: it's  straight up ---T H E O R Y---  mashed with 'my education is better than yours' oneupmanship,  end of story   

I'll still take my chances on Youtube for info and disinfo, and maybe get lucky like I found EEVblog   

Oh, and there's internet 'Search' too 


btw Distelzombie, you're not short on sh!t yourself, so go easy on the stone throwing and drinking too 

[hamster_nz]

If I get some time I will give it a vrack. But I have to be careful...

If the object is a giant dumbbell (two large masses at either end of a massless rod) then the gradients are away from the center of mass. However if the object is a  continuous rod of uniform density, then the graident is towards the centre.

But a small hall-sized void in the center of a planet isn't that special a case :-)