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

[Beamin]

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?

[Rerouter]

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,

[newbrain]

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...

[Rerouter]

I was working on the premise that due to large chunks of crust sunken in the mantle, the center of mass of the earth is constantly shifting, as this makes it a non uniform sphere, you would be biased to 1 or more points,

[Electro Detective]

That's the devils hangout and soul storage of trolls, serial sinners, traitors and sell outs     

stay up here as long as you can, it's a trip you don't need, even to curb curiosity   

[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.

[TerraHertz]

I'll go with crushed virtually instantly. Even if somehow there was a cavity at the center, in which you hoped to experience near zero-G, that cavity couldn't be maintained by any conceivable physical barrier. The cavity would implode in nanoseconds. You'd probably be compressed to some pretty impure kind of diamond.

Btw, define 'center'? Do you mean the physical center (averaging radii to Earth's uneven surface), or the gravitational center?(Which if you include the Moon, is a point that orbits around the physical center. Even more complex if you include the Sun and other planets.)

[Rerouter]

I was taking it as a thought experiment, e.g. you find a teleporter to an unobtainium structure near the center.

A similar though would be what would happen in the center of a large asteroid, the net effect is the same.

[Halcyon]

I was taking it as a thought experiment, e.g. you find a teleporter to an unobtainium structure near the center.

A similar though would be what would happen in the center of a large asteroid, the net effect is the same.

Thought experiments are fun. Probably why most of us enjoy Back to the Future. Imagine taking your shiny new smart phone back 50 years (and magically have it work), could you imagine the looks on people's faces? It's fun to think about.

This was also largely the reason why I enjoyed the 'Spellbinder' children's television series as a kid.

[Distelzombie]

It could look like this if you'd survive: 0:29+

[Brumby]

Would you float in the center being pulled from all sides from gravity

This, if you were stationary.

or would you be able to walk around the walls in a circle?

You could only do this if you were walking quickly enough (or running) to create a sufficient "centrifugal" force to counter the walking action.  Gravity from the mass above and below the plane upon which you were standing would, for the most part, cancel out.

[Cyberdragon]

I was taking it as a thought experiment, e.g. you find a teleporter to an unobtainium structure near the center.

A similar though would be what would happen in the center of a large asteroid, the net effect is the same.

You use The Fall from Total Recall.

[calexanian]

+1 for essentially neutral gravity, and so much heat and pressure, you would not want to be there.

[Electro Detective]

I was taking it as a thought experiment, e.g. you find a teleporter to an unobtainium structure near the center.

A similar though would be what would happen in the center of a large asteroid, the net effect is the same.

You use The Fall from Total Recall.

...get the brakes checked and cleaned first !   

[KL27x]

Force of gravity is proportional to the inverse of radius squared. If you're at the center of the earth, the force of gravity is acting on the mass of your body with a huge amount of force. So imagine your belly button is at the center. Your head is barely 1 meter from center. That means your head weighs 10 million pounds, and your spine ain't that strong.

So even if you had a magical chamber in there to make the pressure 1ATM @ 70F, regular atmosphere, you would die instantly. The force of gravity would crush you. And no, you wouldn't be able to walk around the walls of the chamber, even as a zombie. Your mangled corpse would ball up in the gravitational center of earth, which would presumably be at the exact center of the chamber.

[Rerouter]

KL27x, you have any math supporting that?

The premise of this idea is that you are attracted in all directions (more or less) equally, thus the forces cancel,

On the surface, the entire mass of the earth (via vector sums) pulls you in to the ground without crushing you (if your skydiving or piloting an aeroplane you would be the exception due to rapid deceleration)

As to why there is insane pressures in the center of the earth is not due to gravity at that point, but rather gravity acting on the mass further out being pulled towards it,

The reason why the atmospheric pressure drops as you go up is a simple way of seeing this effect, same if you go diving in the ocean, the mass of the water above being accelerated by gravity is what creates the pressure. this continues down to the center, increasing the pressure, but not the gravity

As you head towards the center the vector sum of the acceleration become smaller, due to more of the earths mass being above you and to your sides. rather than beneath you,

https://ux1.eiu.edu/~cfadd/3050/Ch09Gravity/GrvFld.html

The link above gives a good starting point.

[Brumby]

Force of gravity is proportional to the inverse of radius squared. If you're at the center of the earth, the force of gravity is acting on the mass of your body with a huge amount of force. So imagine your belly button is at the center. Your head is barely 1 meter from center. That means your head weighs 10 million pounds, and your spine ain't that strong.

So even if you had a magical chamber in there to make the pressure 1ATM @ 70F, regular atmosphere, you would die instantly. The force of gravity would crush you. And no, you wouldn't be able to walk around the walls of the chamber, even as a zombie. Your mangled corpse would ball up in the gravitational center of earth, which would presumably be at the exact center of the chamber.

Sorry.  That is absolutely incorrect.

The proof comes from the sum of the gravitational effects from all of the components being equal to the gravitational effect of the whole.

Mathematically, this would involve an integral - but we can see the trend of such a calculation by slicing the Earth in 3 orthogonal planes to create 8 segments.  Calculate the gravitational pull of each segment and you will notice the spherically symmetric segment will have the same (or close enough to for this exercise) magnitude but opposite sign.  As a result, the sum of all 4 pairs will be zero - thus the gravitational pull at the centre of the earth will be zero.

[KL27x]

No, you don't need to be crushed by the pressure of the magma. The force of gravity would kill you by itself.

If you landed on the surface of a black dwarf, you would die instantly in the same way. You would weigh so much, your body couldn't support itself.

Pressure would kill you, too. But the force of gravity would be enough to do the job, by itself.

You know what happens if you take an MRI and there's a steel pin in your head, right? Imagine the MRI machine is 1 million x stronger and your whole body is made of iron.

The radius of the earth is nearly 4000 miles. At the surface you weigh 200 lbs. 3 feet away from the center, your body weighs 9.91 x 10^15 lbs.

So your head is gonna weigh about 1x10^15 lbs. And your feet are gonna weigh 1x10^15 lbs in the other direction. And close to your navel, the weigh of your kidneys, for instance, is going to approach infinity. 

Mathematically, this would involve an integral - but we can see the trend of such a calculation by slicing the Earth in 3 orthogonal planes to create 8 segments.  Calculate the gravitational pull of each segment and you will notice the spherically symmetric segment will have the same (or close enough to for this exercise) magnitude but opposite sign.  As a result, the sum of all 4 pairs will be zero - thus the gravitational pull at the centre of the earth will be zero.

Ohh, yeah. I see where I am wrong. That's cool.

The reason why the atmospheric pressure drops as you go up is a simple way of seeing this effect, same if you go diving in the ocean, the mass of the water above being accelerated by gravity is what creates the pressure. this continues down to the center, increasing the pressure, but not the gravity

I contend that if you climb high enough on a pole, you will weigh less due to force of gravity depending on the inverse of radius squared. This is one of the reasons that the atmosphere is less up high. But I see your point. The difference in going down a mineshaft is that you have mass above you, too.

So if you were magically at the exact gravitational center in a special chamber, then yeah. You ought to be fine. I wouldn't be the first volunteer, though.   

[Electro Detective]

Force of gravity is proportional to the inverse of radius squared. If you're at the center of the earth, the force of gravity is acting on the mass of your body with a huge amount of force. So imagine your belly button is at the center. Your head is barely 1 meter from center. That means your head weighs 10 million pounds, and your spine ain't that strong.

So even if you had a magical chamber in there to make the pressure 1ATM @ 70F, regular atmosphere, you would die instantly.

The force of gravity would crush you. And no, you wouldn't be able to walk around the walls of the chamber, even as a zombie.

Your mangled corpse would ball up in the gravitational center of earth, which would presumably be at the exact center of the chamber.

I experienced those same effects after accepting silly advice and joining Facebook   

As soon as I got out of that mindless lemming social media fiasco, the zombie symptoms faded and I could walk, talk and think normally again 


Coincidently, a lot of browser ad junkery went away too, hmmm   

[rbm]

Minute Physics did a video on this falling through the earth, but mentions the physics on your body at the centre of the earth, which should answer the question.

[StuUK]

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?

Not possible the Earth is flat apparently..... 

[Rerouter]

If you landed on the surface of a black drawf, what would kill you is the pressure gradient on the materials you are composed of, due to the acceleration of gravity by that dense mass, your neck cannot support the weight of your head, and causing more problems as you move down your anatomy.

same deal if you put a cup of water on the surface. the approximate acceleration is about 10,000 G's of acceleration near to the surface of an earth sized black drawf, this is overlooking the fact that the surface is not defined as a solid.


So a standard cup of water, lets say filled 10cm high, that would have 10 Bar of pressure between the top surface and the bottom.  (10 atmospheres, or 140 psi, whichever takes your fancy),
Now a humans blood vessels will start bursting somewhere when they experience a pressure of about 2.1 bar (relative to surrounding tissue), so its not hard to see why even lying down it would kill you.

Cool fact I learned while creating this message, the average person would suffer multiple burst blood vessels in there feet, if standing up in a 1.6m pool of metallic mercury, (ignoring the poisoning risks) and start popping arteries if they where a little over 2m tall in a 2m tall pool. (in either case you would not be comfortable, the buoyancy force would nearly rip your feet off if they where anchoring you to the bottom)

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.

Edit: Re-checked my math, in the mercury pool, them feet are coming off, the ankle stops being connected to the leg at around 8KN of force, the pool would have 95... so... fun times ahead for science. and I question the sanity of the researchers who had to put a number to that.

[Rerouter]

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.

[KL27x]

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.

Perfectly stated. And very interesting about the pool of mercury.

Even before reading your comment, I sensed I was missing a piece of the puzzle, because a black dwarf can make a black hole (or something like that), but a star of the same mass doesn't. Difference in density.

So... if you put your magic chamber in the center of a black dwarf, would you explode?

[Rerouter]

If you put a magic chamber in the center of a black dwarf, most of the acceleration is cancelled, there will be mass imbalances and as you approach any wall of the sphere you would experience attraction to that wall,

For actual numbers I cannot find easy tools to calculate it, however I am of the opinion you could walk inside it, and any height you can jump to will not break your legs,  (same applies on other planets aswell, just way harder to do a clean landing without practice)

[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 :-)

[Zero999]

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.

That's the problem. There is no material strong enough to resist the extreme pressure and temperature in the centre of the earth. It's also impossible to keep a less dense object in the centre of the earth because it will try to float to the surface. One might think a bubble could remain in the centre, because gravity is close to zero, therefore it should have no buoyancy but in reality the gravity in the middle will be not be perfectly zero and will be unstable, so it would still float to the surface.

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

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.

I agree.

[Brumby]

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 :-)

For this exercise, I will allow you to ignore the void in the centre.

For the record, I'm not saying your outcome is wrong - just that you have not considered both of the critical parameters.

You have more than adequately made mention of the difference in mass to the north and to the south - but you have not given any consideration to the difference in distance to the mass to the north and to the south.

These differences are going to give gravitational variations of opposite sign - so the key to the answer is to find out which variation is the larger.

Edit:  To keep the calculations in line with the original question - the Earth will not have any mass redistribution - just the reference point is to move 1 metre.

[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.

That's the problem. There is no material strong enough to resist the extreme pressure and temperature in the centre of the earth. It's also impossible to keep a less dense object in the centre of the earth because it will try to float to the surface. One might think a bubble could remain in the centre, because gravity is close to zero, therefore it should have no buoyancy but in reality the gravity in the middle will be not be perfectly zero and will be unstable, so it would still float to the surface.

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

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.

I agree.

The material is irrelevant to the thought experiment. And even if not, you could build it out of synthetic diamond. Just make the walls of the sphere how-ever many kilometers thick you need! ... Irrelevant. (As I said before: "If the cavity is stabilized")
The buoyancy on the other hand is not irrelevant: That's an interesting thing to add!     
...
I was just writing stuff about how it would look, when it struck me: There is no bouyancy because there is no gravity. (Relatively. Truly there is always gravity, and if it is just from the Moon, the Sun, the Milkyway or the Great Attractor etc.pp)

[Zero999]

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.

That's the problem. There is no material strong enough to resist the extreme pressure and temperature in the centre of the earth. It's also impossible to keep a less dense object in the centre of the earth because it will try to float to the surface. One might think a bubble could remain in the centre, because gravity is close to zero, therefore it should have no buoyancy but in reality the gravity in the middle will be not be perfectly zero and will be unstable, so it would still float to the surface.

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

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.

I agree.

The material is irrelevant to the thought experiment. And even if not, you could build it out of synthetic diamond. Just make the walls of the sphere how-ever many kilometers thick you need! ... Irrelevant. (As I said before: "If the cavity is stabilized")
The buoyancy on the other hand is not irrelevant: That's an interesting thing to add!     
...
I was just writing stuff about how it would look, when it struck me: There is no bouyancy because there is no gravity. (Relatively. Truly there is always gravity, and if it is just from the Moon, the Sun, the Milkyway or the Great Attractor etc.pp)

I can see your point, but ultimately such a through experiment is pretty silly when the practical applications are considered. It may be seen as educational, but there's very little value in such a thought experiment, without exploring why it isn't possible.

Note that I also didn't say there is no gravity: you've misquoted be. The fact that there is gravity and that it won't be perfectly constant means that any low density object inside the earth will float.

[Distelzombie]

Ok, you're right. It could be beneficial.

BTW, I didn't say that YOU said there is no gravity - so I didn't misquote you.

[Rerouter]

If you cannot see the value in a thought experiment that keeps on reappearing, you just haven't thought outside the box far enough,

reworded "how will I move inside a hollowed out, slowly spinning asteroid"

The earth is not the only example it works for.

[ucanel]

...
I can see your point, but ultimately such a through experiment is pretty silly when the practical applications are considered. It may be seen as educational, but there's very little value in such a thought experiment, without exploring why it isn't possible.
...

What would be the practical application for traveling on a light beam,
someone taught it and we all know what happened next.

As a value, at least such a taught experiments can reduce the risk of Alzheimer.

[Brumby]

I can see your point, but ultimately such a through experiment is pretty silly when the practical applications are considered. It may be seen as educational, but there's very little value in such a thought experiment, without exploring why it isn't possible.

Practical applications are not always the direct objective of a thought experiment - and to limit one's efforts to practical problems in the physical implementation of such experiments utterly destroys the very advantage you gain by doing them.

Without being burdened by the need to come up with all the supporting infrastructure that can achieve the (often impossible) conditions to perform the experiment, the very core concepts which the thought experiment addresses can be assessed and conclusions drawn.  Some of these conclusions may be that there was insufficient information available to come to any definite outcome, while others may give rise to the development of some mathematical processes that can be applied to real world situations.

Some thought experiments can give rise to making observations about a system that cannot be physically constructed - but can describe other qualities that can be tested.  The result is that our understanding of the universe is augmented by taking our thinking from the current physical world, pushing it through an impossible experiment and coming out with better information on our physical world.

Imagination is one of the most powerful traits of the human race - and thought experiments are, IMHO, one of the most powerful exercise of the imagination we can perform.

[hamster_nz]

Imagination is one of the most powerful traits of the human race - and thought experiments are, IMHO, one of the most powerful exercise of the imagination we can perform.

Maybe simulation is the second?   

Code: [Select]

#include <stdio.h>
#include <math.h>

#define R (255)
#define H (150)
static void  do_test(int offset)
{
  int x,y,z;
  double fx = 0;
  double fy = 0;
  double fz = 0;

  for(x = -R; x <= R; x += 2) {
    int dx = x-offset;
    for(y = -R; y <= R; y += 2) {
      int dy = y;
      for(z= -R; z <= R; z += 2) {
        int dz = z;
        int d_sqrd = dx*dx+dy*dy+dz*dz;
        int r_sqrd =  x* x+ y* y+ z* z;
        /* Are we inside the shell? */
        if(r_sqrd <= R*R && r_sqrd > H*H) {
          double f = 1.0/d_sqrd;  // f = G*m1*m2/d^2
          double d = sqrt((double)d_sqrd);
          fx += f * dx / d;
          fy += f * dy / d;
          fz += f * dz / d;
        }
      }
    }
  }
  printf("Offset %5i, force %10.2f %10.2f, %10.2f\n",offset, fx, fy, fz);
}

int main(int c, char *v[])
{
  int i;
  for(i = 0; i < 2*R; i++) {
    do_test(i);
  }
  return 0;
}

So, it seems that if your void is perfectly spherical, centered on the same center as a sphere, then the entire void would be zero-g - I think I've got a little bit of noise/error at the edges, due to the crudeness of my calculations, the finer the grid, the less noise.

As for changes in density in different layers, because 0+0 = 0, as long as any changes in density are also spherical, then the void will still be zero-g.

Humm... interesting!

It seems to be the same reason that if you had a sphere of lightbulbs every point inside the sphere would be evenly illuminated.

I've also added a graph if the sphere was solid and of uniform density. the force falls of linearly.

[CatalinaWOW]

A way of thinking about this appropriate to this forum is to answer thquestion, "What is the voltage gradient inside a conducting sphere charged to a million volts?".  The underlying physics are similar in that both systems involve an inverse square law.  And they answer is testable on the form of a Van de Graf machine.

[Rerouter]

Glad to see it hamster,

[Distelzombie]

You sure the simulation is correct? What did you consider "zero-G"? Let's think this a little further: Without changing the mass or the size/radius of the earth, (make it uniform for good measure) lets blow up the void. ("perfectly spherical, centered on the same center as" the earth) We make it almost as big as the earth itself, say, one centimeter less diameter.
Now, if you stand on the wall of the void-sphere, on the half-centimeter thick wall from the inside, you would feel weightless? Literally 99.9999% of the earth is on the other side of you.
There's just 0.5cm ultra dense material between weightlessness and 1g, right?
Now imagine a hole in the half-centimeter thick earth-shell. You still float inside the shell, as you say. You reach for the hole to lift yourself up on the earth. Now what? I guess it would feel like an impenetrable wall.

I do have to say, there it felt always somehow off, the whole "a little point in the middle of the earth attracts you", but this feels even more off.
Lets look at this picture I drew: The Blue sphere is the earth, light-blue stuff in it is the void and the blue streaks are gravity acting on you, the red blob. Now, there is a significant amount of mass to the left of you, indicated by the Black to Read ratio.
So, there is zero-g? You sure? Can you elaborate why?

[hamster_nz]

You sure the simulation is correct? What did you consider "zero-G"? Let's think this a little further: Without changing the mass or the size/radius of the earth, (make it uniform for good measure) lets blow up the void. ("perfectly spherical, centered on the same center as" the earth) We make it almost as big as the earth itself, say, one centimeter less diameter.
Now, if you stand on the wall of the void-sphere, on the half-centimeter thick wall from the inside, you would feel weightless? Literally 99.9999% of the earth is on the other side of you.
There's just 0.5cm ultra dense material between weightlessness and 1g, right?
Now imagine a hole in the half-centimeter thick earth-shell. You still float inside the shell, as you say. You reach for the hole to lift yourself up on the earth. Now what? I guess it would feel like an impenetrable wall.

I do have to say, there it felt always somehow off, the whole "a little point in the middle of the earth attracts you", but this feels even more off.
Lets look at this picture I drew: The Blue sphere is the earth, light-blue stuff in it is the void and the blue streaks are gravity acting on you, the red blob. Now, there is a significant amount of mass to the left of you, indicated by the Black to Read ratio.
So, there is zero-g? You sure? Can you elaborate why?

Yes. Strange but true.

Rather than the shell being a solid dense material imagine it being a mesh of tiny light bulbs, with a 1m space between each of them.  When on the inside of the sphere, if you look at somewhere far away (e.g. 10km) each area of vision will have 100x as many light bulbs as the area 1km away. And as light intensity falls off with d^2, and number of bulbs rise with d^2, they cancel each other out.

Anywhere you look you will see a light of constant intensity.

So what happens when you climb out through a gap in that sphere's mesh? Suddenly in on direction you have the same amount of light hitting you as when you were in the sphere, but it is all from the same side - where everything was in balance it no longer is.

Apparently it is the same with gravity... inside the sphere you will be getting pulled equally in all directions, with a net force of zero. Outside the sphere you are only being tugged from one side, so will feel the full force.

[Brumby]

Distelzombie - I understand your scepticism.  I, too, had similar reservations at first, but as I began to look critically at the system, I found myself taking note of a few things that did not indicate a clear answer.

The main consideration is that, if you had a sphere with a thin crust and the rest of the internal structure had no mass to speak of and you were just on the inside of that crust, then certainly, you would have far more mass wanting to exert a gravitational attraction to pull you off that surface - BUT - the distance that it exerts this force is far, far greater than the part of the crust you are actually right next to.

So the question becomes - which is the greater attractive force: the larger mass that is further away or the lesser mass than is right next to you?  When you consider the inverse square of the distance as a factor, I'm not so sure.

It has also been an observation of mine that circular and spherical geometry often have some surprisingly simple formulae drop out of analytical efforts - so the idea of a net zero-G environment throughout the void is not something I could dismiss through casual inspection....

To be happy within myself, I'd have to dust off my calculus skills ... which haven't been exercised for a decade or three......   

[taydin]

When thinking about this scenario, you can't make the calculations with the assumption that the earth's entire mass is at the center. This assumption can be made for other purposes, but here, it would throw you off. You have to think in terms of forces canceling each other when you are perfectly at the center, and a net force when you are not at the center.

[hamster_nz]

When thinking about this scenario, you can't make the calculations with the assumption that the earth's entire mass is at the center. This assumption can be made for other purposes, but here, it would throw you off. You have to think in terms of forces canceling each other when you are perfectly at the center, and a net force when you are not at the center.

In this case you can!

The code above distributes the mass of the hollow shell across 6,911,840 points, and no matter were you are in the inner space all the forces from the shell cancel each other out perfectly.

A bit of research on Wikipedia confirms this - https://en.wikipedia.org/wiki/Shell_theorem :

1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre.
2. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

It seemed intuitively wrong to me too, but Wikipedia says that it is true... and my crude simulation convinces me that it is too (as I've recreated the result using physics and math that I understand).

[Cerebus]

When thinking about this scenario, you can't make the calculations with the assumption that the earth's entire mass is at the center. This assumption can be made for other purposes, but here, it would throw you off. You have to think in terms of forces canceling each other when you are perfectly at the center, and a net force when you are not at the center.

In this case you can!

You're saying there that you can consider all the mass to be at the centre (i.e. treat it as a single point mass).

The code above distributes the mass of the hollow shell across 6,911,840 points, and no matter were you are in the inner space all the forces from the shell cancel each other out perfectly.

And then here you say you had to consider it as 6,911,840 points.

With that obvious a self contradiction, you must surely realise that you're failing to grasp something fundamental here, as you logically can't hold both points of view simultaneously. For what it's worth, taydin is right — in the case of the interior of a hollow shell, you must consider the mass in a distributed fashion to get a correct answer.

[Brumby]

Why are we talking about this again?

That issue was addressed and settled on page 1.

[Distelzombie]

Why are we talking about this again?

That issue was addressed and settled on page 1.

Distelzombie - I understand your scepticism.  I, too, had similar reservations at first, but as I began to look critically at the system, I found myself taking note of a few things that did not indicate a clear answer.

The main consideration is that, if you had a sphere with a thin crust and the rest of the internal structure had no mass to speak of and you were just on the inside of that crust, then certainly, you would have far more mass wanting to exert a gravitational attraction to pull you off that surface - BUT - the distance that it exerts this force is far, far greater than the part of the crust you are actually right next to.

So the question becomes - which is the greater attractive force: the larger mass that is further away or the lesser mass than is right next to you?  When you consider the inverse square of the distance as a factor, I'm not so sure.

It has also been an observation of mine that circular and spherical geometry often have some surprisingly simple formulae drop out of analytical efforts - so the idea of a net zero-G environment throughout the void is not something I could dismiss through causal inspection....

To be happy within myself, I'd have to dust off my calculus skills ... which haven't been exercised for a decade or three......   

What again? You were talking here like it sis not bother you.
I don't understand anymore:
Wikipedia seems to be clear about that: So how come... (CEREBUS)

When thinking about this scenario, you can't make the calculations with the assumption that the earth's entire mass is at the center. This assumption can be made for other purposes, but here, it would throw you off. You have to think in terms of forces canceling each other when you are perfectly at the center, and a net force when you are not at the center.

In this case you can!

You're saying there that you can consider all the mass to be at the centre (i.e. treat it as a single point mass).

The code above distributes the mass of the hollow shell across 6,911,840 points, and no matter were you are in the inner space all the forces from the shell cancel each other out perfectly.

And then here you say you had to consider it as 6,911,840 points.

With that obvious a self contradiction, you must surely realise that you're failing to grasp something fundamental here, as you logically can't hold both points of view simultaneously. For what it's worth, taydin is right — in the case of the interior of a hollow shell, you must consider the mass in a distributed fashion to get a correct answer.

What is wrong here? What is not written in words? Cerebus can you please be precise?

I like problems that show you how wrong common-knowledge can be.
I LOVE problems that are explained.
None of this is true, apparently? Really, Cerebus, I trust your opinion, so please explain (I guess it is just another piece of "english" I didn't get yet)

[hamster_nz]

When thinking about this scenario, you can't make the calculations with the assumption that the earth's entire mass is at the center. This assumption can be made for other purposes, but here, it would throw you off. You have to think in terms of forces canceling each other when you are perfectly at the center, and a net force when you are not at the center.

In this case you can!

You're saying there that you can consider all the mass to be at the centre (i.e. treat it as a single point mass).

When outside the shell, you can treat it as a point mass that is at the center.

Within the void on the interior of the shell the net force is zero, This allows you to ignore the mass of the shell completely.

The code above distributes the mass of the hollow shell across 6,911,840 points, and no matter were you are in the inner space all the forces from the shell cancel each other out perfectly.

And then here you say you had to consider it as 6,911,840 points.

With that obvious a self contradiction, you must surely realise that you're failing to grasp something fundamental here, as you logically can't hold both points of view simultaneously. For what it's worth, taydin is right — in the case of the interior of a hollow shell, you must consider the mass in a distributed fashion to get a correct answer.

No contradiction is present, if analyzed as a shell made up of millions of point masses, you get the same result as if you apply the same two rules:

When outside the shell, you can treat it as a point mass that is at the center.

Within the void on the interior of the shell the net force is zero, This allows you to ignore the mass of the shell completely.

What contradiction to do see? The problem was looked at from three different perspectives (a discrete simulation, through geometric projections, and reading on Wikipedia) and they came to the same answer.

[Distelzombie]

Hm... can someone make a 3D simulation? The graphs from you, hamster, are not labelled correctly and are confusing. (Hm... there's more to it, though. Can you do me a solid and explain it as if i'm a child? I think some meaning is lost in translation.)

[Cerebus]

What contradiction to do see? The problem was looked at from three different perspectives (a discrete simulation, through geometric projections, and reading on Wikipedia) and they came to the same answer.

The contradiction was merely in what you wrote. You said one thing, and then said the opposite, that is pretty much the dictionary definition of contradiction.

[Brumby]

In this case you can!

As I see it, this is the phrase that's caused all the confusion.  Remove it and the 'contradiction' pretty much evaporates.

I wasn't aware of the "Shell theorem" - but my intuition was right to be open to the idea.  The exercise by hamster_nz also led to the same conclusion.

What becomes clear now is that for a point at radius r defined within any given sphere of radius R (where R > r) we have two distinct regions:
1. The sphere that extends from the centre point to radius r, where, for gravitational calculations, all the mass can be considered to be located at a single point at the centre.
and
2. The shell that extends from radius r to radius R, where, for gravitational calculations, there is no net force at the given point.

We can now answer gravitational questions with confidence - and that aspect of the original question with ease....

All the mass that exists between radius r and radius R will have zero net effect.  However, there will be a net gravitational influence from the mass of what lies within the sphere of radius r.  If that were a vacuum, then there would be no gravitational attraction - but if it were filled with air, then there would be gravity from that mass of air ... and the object (person) starting at rest, would fall towards the centre with the appropriate acceleration.

This acceleration, however, would vary.  At a distance ^r/₂ the volume of the shell would now extend from that point out to radius R, leaving only ¹/₈ of the mass to provide gravitational acceleration towards the centre.

We can then see that with any mass in the void, the object will acquire a non zero velocity with acceleration dropping to zero as it passes through the centre and then reversing as it travels away on the other side.  From this we can conclude that the object will slow down until it stops and reverses direction, to repeat the process.  As such, the object will oscillate with the amplitude of those oscillations remaining constant unless acted upon by another force - such as air resistance.

BUT if the void from the centre to radius r contains no mass, then the object will not fall at all.

[Cerebus]

A simple analytical treatment, as Distelzombie pleaded nicely.

Assumptions:

1) A planet of radius R with a hollow void at the centre of radius r, R very much greater than r.
Simplifying assumptions: this planet is spherical and the relative sizes of the interior hollow and
outer shell are so great that we can ignore the missing mass caused by the hollow for calculation
purposes.

2) The density of the planet's material is uniform (and for argument's sake is 5500 kgm^-3).

For arguments sake, R = 6300km and r = 15m. This gives a mass of 5.76 x 10²⁴ kg
for the planet and 77.8 x 10⁶ kg for the mass removed by the hollow. The ratio of these
two would be 74.1 x 10¹⁵:1, which justifies the simplifying assumption for R >>> r.

3) A body, initially floating in the centre of the hollow, mass 70kg. Simplifying assumption: the
body is spherical.

4) Newton's general law of gravitation is sufficient here. There are no relativistic or quantum
effects.

5) For the purposes of the above, an object's mass can be treated as a acting at a point as long as
you are topologically outside the object.


The simple case, an argument from symmetry:

Bisect the planet along any great circle, thereby forming two cups. As this is a bisection the two,
equally dense, halves have the same mass. Also their geometric centroids (aka centre of gravity)
are at the same distance from the geometric centroid of the original sphere and thus from the
floating body. The centroid will be at ³/₈ R along a line normal to the
cleaving plane and that passes through the centre of the original sphere. The new centroid's
distance from the centre of the original sphere (and the floating body) = 2,362,500m.

Because we have bisected the sphere we can now rely on the fact that our floating body is now
topologically outside the half sphere and treat the half sphere as a point mass at its centroid.

The force exerted on the body by the gravitational effect of one half sphere will be (by Newton's law):

     G m₁ m₂
     ----------
          r²

       6.674×10⁻¹¹ m³⋅kg⁻¹⋅s⁻² x 5.760  x 10²⁴ kg x 70 kg
=     -------------------------------------------------------------
                              (2,362,500 m)² x 2

The x 2 in the denominator is to cope with half a sphere.

= 1.346 x 10¹⁶ / 2,362,500²

= 2410.924 N towards the centroid of that half sphere.

By symmetry, the other half sphere will exert exactly the same (scalar) force in exactly the
opposite direction, resulting in a zero net force vector.

Repeat this for any great circle and the results will be the same, meaning that, by symmetry, the
net force on any object at the centre of the sphere will be zero.

Now let us try to quantify the effect if you move to one edge of the hollow. You would be 15m
nearer to one centroid and 15m further away from the other. This alters the denominator in the
equation for Newton's law and makes the forces from the two half spheres slightly different.

15m closer to the centroid, force = 2410.955 N
15m further from the centroid, force = 2410.894 N

Difference in forces = 0.061 N, about 6 grammes weight at normal surface gravity.

Conclusion: if you move from the centre of the hollow you will experience a very small net gravitational pull from the side you are nearest to.

[Brumby]

Now let us try to quantify the effect if you move to one edge of the hollow. You would be 15m
nearer to one centroid and 15m further away from the other. This alters the denominator in the
equation for Newton's law and makes the forces from the two half spheres slightly different.

15m closer to the centroid, force = 2410.955 N
15m further from the centroid, force = 2410.894 N

Difference in forces = 0.061 N, about 6 grammes weight at normal surface gravity.

Conclusion: if you move from the centre of the hollow you will experience a very small net gravitational pull from the side you are nearest to.

Incorrect.

You have violated this

Because we have bisected the sphere we can now rely on the fact that our floating body is now
topologically outside the half sphere and treat the half sphere as a point mass at its centroid.

by moving your reference point to a location 15m within the half sphere.

[Brumby]

.... You have to take a 15m slice from that half sphere and calculate the gravitational effect it provides - then subtract that from that half sphere and add it to the other.

I will bet you that will account for your 0.061 N.

[Cerebus]

Now let us try to quantify the effect if you move to one edge of the hollow. You would be 15m
nearer to one centroid and 15m further away from the other. This alters the denominator in the
equation for Newton's law and makes the forces from the two half spheres slightly different.

15m closer to the centroid, force = 2410.955 N
15m further from the centroid, force = 2410.894 N

Difference in forces = 0.061 N, about 6 grammes weight at normal surface gravity.

Conclusion: if you move from the centre of the hollow you will experience a very small net gravitational pull from the side you are nearest to.

Incorrect.

You have violated this

Because we have bisected the sphere we can now rely on the fact that our floating body is now
topologically outside the half sphere and treat the half sphere as a point mass at its centroid.

by moving your reference point to a location 15m within the half sphere.

Somebody does not know what topologically outside means.

[Cerebus]

.... You have to take a 15m slice from that half sphere and calculate the gravitational effect it provides - then subtract that from that half sphere and add it to the other.

I will bet you that will account for your 0.061 N.

You can bet what you like, but if you aren't prepared to do the calculus to prove your thesis then it's a hollow bet. You must, mathematically, overturn the assumption that for gravitational purposes objects act as point masses at their centroids if you are topologically outside the object. If that holds, it already accounts for the 15m of 'overhanging' mass and my solution is correct, if that does not hold I am wrong. To overturn that assumption you must integrate the gravitational effects in three dimensions and show that the resulting vector is different from the vector produced by the centroid approach.

I used the argument from symmetry approach, with the well established principal of objects acting gravitationally at their centroid, precisely because it doesn't involve more than trivial algebra and well established physical rules that anybody here ought to be able to follow. If you say that is incorrect you will have to produce the maths to prove that it is incorrect. I didn't fancy trying to do a triple integral of the gravity equation over a spherical cap/sphere that has a smaller spherical cap/sphere missing from it and I suspect that neither do you.

I'm not even sure that I've the mathematical chops to produce the raw 3D integral let alone the finished analysis. I'm certain that I couldn't confidently get the final analysis - if I could get to the raw integral then I'd let Mathematica or similar do the final heavy lifting, but as I say, I doubt my abilities even to that point.

If you're feeling man enough for a fully analytical solution (I'm certainly not, when it comes to 3D calculus I'm a wimp) then please leave the position of the 'floating body' parametrised and then we'll have an exact analytical solution for any variant on this problem and can argue to our heart's content about all those variants. If there is a proper mathematician in the house please solve this analytically once and for all and put us all out of our misery.

To tempt any proper mathematicians out of the wood, here's a mathematical joke: "What's purple and commutes?" The punchline will be delivered either by a mathematician accidentally revealing themselves, ready to be press-ganged, or in 7 days by me, and none of you will get the joke (which is groan making terrible BTW).

By the way, what we all persist in calling a sphere is more properly called by mathematicians a ball. To a mathematician a ball is solid, the sphere is the outside surface. If you're searching for the appropriate mathematics this distinction may be critical.

[Brumby]

My only error was mistaking your use of the word "topologically" to correctly describe the geometry.

Topological structures have nothing to do with this scenario - the precise geometry is very relevant and your production of incorrect mathematical calculations is enough for me to not even consider any triple integral calculations are necessary.

Simple inspection can show your premise to be completely erroneous.

For starters, we have already addressed the issue where moving your reference point inside the volume of the hemisphere voids your ability to use the equivalent point mass process for gravitational calculations.  We can still use the equivalent point mass process - but we have to define two separate bodies, deal with each of them and then combine the results.

Without having to go into any maths, your premise can be disproven with the inspection of the two bodies required.  The first body is the 15m slice of the half sphere - producing a ring of radius 6300km.  The second body is the remainder of the half sphere.

The equivalent point mass location of the ring is now very, very close to 7.5m on the opposite side of the reference point to the remainder - and the equivalent point mass location of the remainder has moved further away from the original centroid point.

Looking at just these two bodies, the attraction from the remainder is slightly less, because the mass is less.  Also, there is now an attraction force from the ring - which is on the opposite side and will have the opposite effect.


If you like doing your "point mass equivalent" calculations, then do them properly.


My bet still stands.

[Cerebus]

My only error was mistaking your use of the word "topologically" to correctly describe the geometry.

Topological structures have nothing to do with this scenario - the precise geometry is very relevant and your production of incorrect mathematical calculations is enough for me to not even consider any triple integral calculations are necessary.

You make an assertion that I'm incorrect as a justification for not using triple integrals? It's very obvious that one proper solution for this problem (using a Cartesian frame of reference) is integrals of the gravitational forces in the Cartesian x, y, and z, axes - leading to triple integrals. If you really mean that you can't do them, that's OK, neither can I.

Simple inspection can show your premise to be completely erroneous.

For starters, we have already addressed the issue where moving your reference point inside the volume of the hemisphere voids your ability to use the equivalent point mass process for gravitational calculations.  We can still use the equivalent point mass process - but we have to define two separate bodies, deal with each of them and then combine the results.

Without having to go into any maths, your premise can be disproven with the inspection of the two bodies required.  The first body is the 15m slice of the half sphere - producing a ring of radius 6300km.  The second body is the remainder of the half sphere.

The equivalent point mass location of the ring is now very, very close to 7.5m on the opposite side of the reference point to the remainder - and the equivalent point mass location of the remainder has moved further away from the original centroid point.

Looking at just these two bodies, the attraction from the remainder is slightly less, because the mass is less.  Also, there is now an attraction force from the ring - which is on the opposite side and will have the opposite effect.


If you like doing your "point mass equivalent" calculations, then do them properly.


My bet still stands.

It's still a hollow bet as you aren't prepared to undertake the work necessary to win it. It's the same as saying "I bet I could do X if I tried, but I'm not going to try".

It's all still hand waving unless you can put numbers on it or analytically disprove assumption (5) using something resembling mathematics. As I've said, I'm wrong if assumption (5) is wrong and all you need to do to demolish my case is to disprove assumption (5). You haven't disproved it, merely stated a different assumption. I take it you can't disprove it (no small thing, I can't analytically prove it) or won't.

If merely calculating, you must be careful that the R >>> r assumption is still safely valid for the reduced system with the annulus - the safety margin of that assumption has just reduced by a factor somewhere near ^R/_r.

By the way, there are now three bodies in the system you've described, not two: one hemisphere (with a spherical cap missing from it), another nearly hemisphere with a 15m slice missing and a 15m x 2 x 6500 km 'almost' annulus with a spherical cap missing from it.

I could probably get away with ignoring the spherical cap as it is tiny compared to the hemisphere it is taken from, a 1:1 x 10¹⁶ ratio, with your 'slice' you've gone from ratios involving cubes to ratios involving squares, a less comfortable proposition at 1:4.69 x 10¹⁰ an effect about 250,000 times bigger. Probably nothing in the grand scheme of things but when making assumptions it is wise to get a feeling for their order of magnitude.

Anyway, I'm now thoroughly bored by this. My recent contribution was done as a favour to Distelzombie, in an attempt to remove some of the fog that seems to have accumulated. To me it seems intuitively obvious that as soon as the system becomes anything other than strictly symmetrical, the general result, that there are zero forces at the centre, fails. My contribution was an attempt to offer some analysis of that. If anybody can do a better job of proving that or disproving that then get to it. Me, I've got to get back to some actual electronics, vis working out if resistor power coefficients are going to bugger up my error budget or not.

[Brumby]

It's all still hand waving unless you can put numbers on it or analytically disprove assumption (5) using something resembling mathematics. As I've said, I'm wrong if assumption (5) is wrong and all you need to do to demolish my case is to disprove assumption (5). You haven't disproved it, merely stated a different assumption. I take it you can't disprove it (no small thing, I can't analytically prove it) or won't.

Hmmm....  Let's see .....
(A)

5) For the purposes of the above, an object's mass can be treated as a acting at a point as long as
you are topologically outside the object.

Let's say we have the original spherical shell as defined by yourself, with a reference point at the centre.  By all accounts (even yours) the net gravitational effect at that point is zero.

Now let me drill a 1m hole through the shell, normal to the surface.  The previously defined reference point is now topologically outside the shell object.  By your assumption, we can now treat the object's mass as acting at a point with respect to the reference point.  The distance between these two points is practically zero.  If we have a 70kg object centred at the reference point, then the gravitational attraction between the two bodies would be practically infinite, because the distance between them is practically zero.

Topologically, I could drill a 1cm hole or even a 1um hole and my argument doesn't change - so using topology in this context is problematic to say the least.  I don't know why you introduced this term into the discussion.  To be honest, it is bizarre, actually.


(B)

Repeat this for any great circle and the results will be the same, meaning that, by symmetry, the
net force on any object at the centre of the sphere will be zero.

Up to this point, I have no quarrel, but then you say:

Now let us try to quantify the effect if you move to one edge of the hollow. You would be 15m
nearer to one centroid and 15m further away from the other.

While this is true, only the centroid you have moved away from is valid to keep using.  The one you have moved towards is no longer relevant because you have departed from the symmetry you so preciously embraced in the previous sentence, in a way that demands a different treatment.

Let's look at this as a 3 bodied system (as you correctly observed) consisting of (A) a hemishphere to the right plus (B) an annulus (or disc, if we want to accept the same approximations you applied using your R>>>r criteria) and (C) a spherical cap which, together, form a hemisphere of equal size and mass to the left.  Let forces acting towards the left be positive.

With the original reference point, the gravitational calculations do, in fact, cancel out.  To put it in some form of mathematical nomenclature:
  -G_A + G_B+C = 0
However, since the gravitational effect is the sum of all component parts - whether these be from a finite number of pieces or from an infinite sum (via an integral) - we can just as accurately state:
  -G_A + G_B + G_C = 0

Now, when you move the reference point, reapplying your calculation logic, the equivalent result is:
  -G'_A + G'_B + G'_C = 0.061 N
The problem with this is, the G'_B component is now on the right side of the reference point, so any influence it has, is now opposite in sign and, as such, must be written:
  -G'_A - G'_B + G'_C = 0.061 N
The proof in showing your calculation for the offset reference point is invalid is clearly shown by the sign error in the G'_B term.  What more can I say?

As I stated earlier (and would be shown by hamster's program) the actual equation should be
  -G'_A - G'_B + G'_C = 0
 ... where the space of the void has no mass.

[hamster_nz]

If you think of gravity as being a field rather than force things become a lot simpler (see https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity)

Make an empty spherical volume in a void. Without a mass inside it total 'sucking force' of gravity  (gravitational flux) across that volumes surface is zero, regardless of size. Mathematically speaking, the surface integral over the spherical surface is zero.

Put a mass at the center of the volume. and the amount of gravitational flux across the surface area of the volume is proportional to the mass. If you make the area 4 times as big (i.e. double the radius) then the sucking force is 1/4th the strength, but over the whole surface area it is the still proportional to mass. (The surface doesn't have to be spherical but it does make analysis nice and easy, as the 'sucking' is always a surface normal vector). Consistent with the 'inverse square' law.

You can have a dense mass in the center, or a less dense mass that fills the volume it doesn't matter as long as the mass is the same. The strength of the flux at radius 'r' is the same - G*m/(r^2) per kilogram. And of course only when the mass (m) is zero, is it zero. Also consistent with the 'inverse square' law.

Now back to the "hollow earth" problem.

As everything is symmetrical, I am sure that everybody agrees that the gravitational flux entering at the edge of the void be of equal magnitude. It might be flowing either at or away from the center (we don't know which yet), but the flux from the shell in one area will be the same as any other area.

As you can't have gravity shadows, and gravity doesn't interact with itself, so the amount of gravitational flux exiting the interior void must also be equal and opposite to that entering it, and is also spread evenly.
 
This proves that that at the edge of the interior void the net gravitational flux from the shell is zero. Any mass put there will feel no force attracting it to the shell.

But you can also this thinking to prove that the flux from the shell through the entire interior void must be zero, hence the gravitational force of the shell has zero effect on anything inside the void.

You can even use this to calculate gravitational flux in the interior of the shell at radius 'r' it is G*m/(r^2) per kilogram, where 'm' is the mass that is within radius R.

[GlennSprigg]

Aaarrgghh... this is frustrating to answer !!!!!!
ONLY because most people can not grasp what is being described !!!!!!  :-)
I've decided to 1st say this initially.....  "The Earth is NOT hollow"......
so get rid of ALL such thoughts in the mean time !!!!!!  xoxoxox
However, IF our 'Earth' was hollow, (or was ANY other object in Space),
and WE were INSIDE such a 'hollow' sphere, then our gravitational attraction
 (internally) to ANY point of the 'Earths' shell, no matter where we are within
that 'void', will be equal, so we would stay in the position we are in.
This is NOT a false statement, so live with it !!!!!!  :-)

[GlennSprigg]

(Not enough room here to 'prove' this... but it is true.....)

[CatalinaWOW]

Just to keep the confusion alive, the Earth is not only not hollow, but it is not a sphere and it is not homogenous.  The gravitational field (due to to the earth, not all the other things in the universe) in a spherical cavity at the "center" of the earth would not be zero.  So though the ideal case has a simple and easy answer, you must go through all of the messy math to get the "real" answer.  Even worse, since the shape and density variations aren't directly measurable, you must go through even messier math to derive them from variations in the orbits of bodies going around the earth.

[Distelzombie]

Aaarrgghh... this is frustrating to answer !!!!!!
ONLY because most people can not grasp what is being described !!!!!!  :-)
I've decided to 1st say this initially.....  "The Earth is NOT hollow"......
so get rid of ALL such thoughts in the mean time !!!!!!  xoxoxox
However, IF our 'Earth' was hollow, (or was ANY other object in Space),
and WE were INSIDE such a 'hollow' sphere, then our gravitational attraction
 (internally) to ANY point of the 'Earths' shell, no matter where we are within
that 'void', will be equal, so we would stay in the position we are in.
This is NOT a false statement, so live with it !!!!!!  :-)

(Not enough room here to 'prove' this... but it is true.....)

And I asked you to prove this. You didn't care to answer for a week or so. So basically it's your fault they're all talking about it - which makes you angry about yourself.
Just saying: "This is NOT a false statement, so live with it !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!" Doesn't make it true, doesn't make your voice heard, doesn't make you seem knowledgeable, doesn't make us stop asking, doesn't make us stop thinking and doesn't make us even just care about what you said. This is no offense. It is critics on your style of argumentation. Always back up your facts or don't even answer, because no one cares otherwise. That's the nature of a scientific discussion.
There is only one case where modern scientific discussions acknowledged and publicised a paper that was not proven correctly: A paper by Steven Hawking about, I think, black holes. I guess it was the one where he said he was wrong about some parts of bh.

I hope you see why that is important.


Thanks everyone for being so detailed about it. I don't lie, I didn't understand it yet. I have to take some time to read all that because ... well I think it's still the withdrawal symptoms from the anti-depri-meds. I can't read straight and I forget so much it's kind of scary actually.

[Brumby]

The very first word of this thread (which appears in the title) is "IF".

How much clearer can that be?

By starting off with that, the rest of the question is clearly a thought experiment - so all the practicalities of the physical possibility of such a void and/or the survivability of the environment are completely, totally and utterly irrelevant.

For those still champing at the bit on that - you have no imagination.  The purpose of a thought experiment is to look past these limitations and focus on some elements of the system being considered in isolation - for the sole purpose of understanding how these components work within themselves.  Even if there is no immediate practical application being pursued, such exercises are still valid as basic research.

So, please, let's give that silliness a rest.


As for the matter of the Earth's geometry - yes, it is an oblate spheroid, not a perfect sphere, but that, in itself, does not make any discussion meaningless.  One source I found stated that the equatorial radius is 6,378 km and the polar radius is 6,356 km - a difference of 22 km or about 0.35%.  While you couldn't really call this totally insignificant, it is certainly not a huge variation - and I would suggest the variation from the spherical is not going to be particularly noticeable.  In fact, for a body at the centre of the Earth, there will be no difference at all.  So long as the Earth maintains point symmetry around that body, all gravitational forces from every constituent part of the Earth will cancel out - and this would still hold true if the Earth was a cube (Please don't start about how the Earth could never be a cube - this is a thought experiment for the purpose of illustrating some maths!)

What happens if we move off-centre within an oblate spheroid?  Now when it comes to the mathematically pure answer to that - I don't know, but for the practical case of the Earth, the variation is not going to be significant, because the eccentricity is so small.  If anyone wants to do the math, please feel free.


Now there has also been a reference to the fact that the Earth does not have a consistent mass distribution.  While this is technically true, so long as each concentric shell around the centre has a consistent density, then the gravity at the centre will still be zero.  It does not matter if one shell layer is ten times the density of another and twenty times as thick, it all works out.

Where this does result in some perturbations of the calculations is where this density is not consistent through a shell - and the Earth's crust is a valid example.  But even here, there is a qualification of one of the basic parameters of the original question that can "save the day" as it were.  That qualification is "Define the 'centre' ".

There are two definitions that immediately come to mind.  The first is the geometric centre - find the middle point of all the points within the volume of the Earth.  The second is the centre of gravity - that magical point towards which the Earth attracts all bodies outside its volume.  In the case of an infinite set of shells having their own consistent density, these two points are the same, but if there is any variation, then these two points may separate.

If they were different, for the purposes of the original question, the centre of gravity would be the point to choose - and the answer to the original question would, again, be zero - for the very reasons that this point was determined to be the centre of gravity.

What happens if you were to move off-centre within such a scenario?  Again, I don't have the mathematically pure answer, but I will stick my neck out and say that, for all intents and purposes there is not going to be much of a variation from the ideal spheroid model.

.... IF our 'Earth' was hollow, (or was ANY other object in Space),
and WE were INSIDE such a 'hollow' sphere, then our gravitational attraction
 (internally) to ANY point of the 'Earths' shell, no matter where we are within
that 'void', will be equal, so we would stay in the position we are in.

This is so close to being perfectly correct (especially for the ideal spherical model), I hesitate to qualify it - but I feel I must.

As I mentioned in a previous post, this observation is correct - so long as the mass of the contents of the void is zero.  Since this is a reasonable assumption for a thought experiment, it does not deserve direct criticism - but my thinking went a little further.....

If there is anything in that void that has any mass whatsoever, then there will be a centrally attracting gravitational force on any body within that void.  For a body at a distance of r from the centre, the gravitational force come from that of the mass of a sphere of radius r from that same centre.  This will result in the body accelerating towards the centre - but, as it does, the radius of the sphere exerting the gravity will decrease - and so will the acceleration.  As the body reaches the centre, the increase in velocity will drop to zero, but the velocity it has gained will carry it through to the other side, where acceleration in the opposite direction from the now growing central sphere will cause it to slow, stop and then reverse it's travel.  The result would be an oscillation.  Air resistance would be the only dampening factor, otherwise this motion would continue indefinitely.

To put this into proper perspective, however, one needs to consider the mass of the air and just how much acceleration that would impart.  From my quick calculations, for a mass of air equivalent to that at standard temperature and pressure within the 15m void mentioned earlier - for a 70kg person starting at rest 10m from the centre, the initial acceleration would be 3.6 nm/_s².  At that rate, it would take more than 77 hours to have moved 1mm.  I don't even want to think of the period of the oscillation.



Just remember ... this is - and always has been - a thought experiment.

[hamster_nz]

Just throwing this out here... As the Earth's mass can be mostly ignored, you are essentially orbiting the sun.

Therefore, if you were to move closer to the sun without a change of orbital speed you will fall towards the sun in a lower elliptical orbit

Likewise if you move away from the center without a change of orbital speed you will find yourself in a higher elllipical orbit, moving away from the sun.

In both cases your obital period willl be different to that of the earth, so you will hit it.

If you are only moved tangentally to the Earth's orbit, you will have the same orbital period (365.25 days) and will slowly oscillate up and down at one cycle per year.

I knew playing  Kerbal Space Program would be of use one day!

[CatalinaWOW]

I agree that spherical layers of unequal density will have no effect.  The density variations I was referring to (called mascons by NASA) are not spherical.  I also agree that the effects of the non-sphericity and the density variations of the actual earth are quite small.  Whether those differences make a difference to the thought experiment depend on the conclusions drawn from the experiment.  If the question is whether this void would be a good zero gravity simulator, the answer is yes.  Better than the "Vomit Comet".   If the question is whether an object at rest with respect to the chamber would remain there forever the answer would be no, even without bringing in Heisenberg or any of the other minutia of post Newtonian physics.

The orbital mechanics comment brings up another factor which depends on the original intent of the thought experiment.

[GlennSprigg]

The very first word of this thread (which appears in the title) is "IF".
How much clearer can that be?
...........................................................
Just remember ... this is - and always has been - a thought experiment.

Absolutely true 'Brumby',including all that you said inbetween !!

I took the original 'question' as a 'Hypothesis', and as such, so much is "out the window" !
Of course I, you, and others, know that the 'Earth' is an 'Oblate Spheroid', and not a perfect
circle/sphere, and as a result all such mathematical 'simplicities' can NOT be truly valid......
I do not agree with or support the original 'question'/'proposition' beyond the hypothesis !

I guess 'I' digressed too, by talking about gravitational forces between an 'object' within a
'void' inside a gravitationally valid 'shell'... The main reason being that I have created real
world 3D computer models of such scenarios... (and much more !! xx)

[Zero999]

I can see your point, but ultimately such a through experiment is pretty silly when the practical applications are considered. It may be seen as educational, but there's very little value in such a thought experiment, without exploring why it isn't possible.

Practical applications are not always the direct objective of a thought experiment - and to limit one's efforts to practical problems in the physical implementation of such experiments utterly destroys the very advantage you gain by doing them.

Without being burdened by the need to come up with all the supporting infrastructure that can achieve the (often impossible) conditions to perform the experiment, the very core concepts which the thought experiment addresses can be assessed and conclusions drawn.  Some of these conclusions may be that there was insufficient information available to come to any definite outcome, while others may give rise to the development of some mathematical processes that can be applied to real world situations.

Some thought experiments can give rise to making observations about a system that cannot be physically constructed - but can describe other qualities that can be tested.  The result is that our understanding of the universe is augmented by taking our thinking from the current physical world, pushing it through an impossible experiment and coming out with better information on our physical world.

Imagination is one of the most powerful traits of the human race - and thought experiments are, IMHO, one of the most powerful exercise of the imagination we can perform.

You missed my point. I didn't say that the thought experiment is totally pointless, just that it's important to explore why it isn't possible.

In any case, I think the question is too narrow for my interest. I find exploring the gravitational anomalies around the earth and why more interesting. The Wikipedia article is very good.
https://en.wikipedia.org/wiki/Gravity_of_Earth

[Distelzombie]

I do not agree with or support the original 'question'/'proposition' beyond the hypothesis !

What does that mean? This is not a case where something is right or wrong.

I have created real world 3D computer models of such scenarios... (and much more !! xx)

Show us. I asked for them before. Please! If you don't show them, your whole argumentation is irrelevant.

Oh, and what does "real world" mean in conjunction with a thought experiment where there is a void at the center of the earth?

[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.

Read post one if that was the question this would be a pretty dull question like what would happen if you put wood into a fire. No.

This is what
I ment the center is hollow  perfectly round 100' and full of air and wouldn't collapse and an enviornment at 15 psi 80%N2 20%O2 at 300 kelvin with no ionizing radiation or heat. Oh and 50% relative humidity and there is light so you can see.

[hamster_nz]

I've done some rough calculations for an oblate spheroid..

If the N/S distance is 258/259ths of the W/E distance (about the same ratio as that of the earth).

and if ~1/20th of the radius is hollow (approx 312km from center to edge, 624kms across)

and you measure the local gravity at 90% of the way to edge (280 km from center)

and the density of the material is of uniform.

Then....

Towards the equator the local gravity is 0.006% of the surface gravity (an acceleration of about 0.6mm/s/s away from the center)

Towards the North/South poles gravity is -0.013% of the surface gravity (an acceleration of about 1.3mm/s/s towards the center)

This would decrease as you get closer to the center. So about a kilometer from the center it would be about single-digit um/s/s acceleration.

[Brumby]

You missed my point. I didn't say that the thought experiment is totally pointless, just that it's important to explore why it isn't possible.

I didn't think much needed to be said on that.

[Brumby]

... and I think hamster needs to back away from the caffeine.

[GlennSprigg]

Distelzombie       please excuse me........
You seem to like pulling apart my comments, whether directed at you or not ??
Moreover, you seem to take almost great 'offense' in what I say ??
I don't know if it is just the 'Language Barrier'.... (BTW i have many German friends).
We all differ in 'opinions', but you seem to enjoy 'Attacking" ?  Not sure why....
I have 'commented' on OTHER posts, too, as you have, and it's the same pattern.

I think I said in some other 'post' something like.... "Never the Twain shall meet"....
I think I was trying to say that (without explaining the saying) that WE may agree to
differ on many things.... but I have only EVER been polite to you my friend......

That being said, I remember a conversation about 'Medication'.....

[GlennSprigg]

Distelzombie        Oh... and YES I have created realistic computer models of such effects,
and although I would normally GLADLY share it, my 'interest' in compiling and sharing with
YOU is somewhere around zero now, mate.   (It was/is a fascinating subject/project !!!)

[Beamin]

OK the short answer is you could walk around on the walls but the gravity pulling you toward your feet would be very low because you would have all the mass above your head pulling you up.

If you jumped you could reach the center and float  because the gravity would be equal in all directions?

[GlennSprigg]

...... I have re-read this in it's 'entirety', and out of personal interest, as well as the
interest of the 'Majority' here....   I WILL piece together an answer verbally & graphically.
(Much to the 'appeasement' of 'Distelzombie'....   argh...   :-)
(Please give me a few days !!!!!)

[Brumby]

OK the short answer is you could walk around on the walls but the gravity pulling you toward your feet would be very low because you would have all the mass above your head pulling you up.

If we use the ideal spherical model, my understanding is that this is incorrect - as per the shell theorem.  If there is no mass in the void, the gravity at all points of the void will be zero.

If you jumped you could reach the center and float  because the gravity would be equal in all directions?

If you jumped, you would have a velocity that would carry you to an opposite wall.  To float, you would need to be able to stop yourself - and wherever you stopped is where you would stay.

[Beamin]

OK the short answer is you could walk around on the walls but the gravity pulling you toward your feet would be very low because you would have all the mass above your head pulling you up.

If we use the ideal spherical model, my understanding is that this is incorrect - as per the shell theorem.  If there is no mass in the void, the gravity at all points of the void will be zero.

If you jumped you could reach the center and float  because the gravity would be equal in all directions?

If you jumped, you would have a velocity that would carry you to an opposite wall.  To float, you would need to be able to stop yourself - and wherever you stopped is where you would stay.

But you have a huge amount of mass above you that doesn't just go away or are you saying it cancels out?

Plus I think the core has huge iron crystals in it that are some 30 miles long according to science.

[Zero999]

OK the short answer is you could walk around on the walls but the gravity pulling you toward your feet would be very low because you would have all the mass above your head pulling you up.

If we use the ideal spherical model, my understanding is that this is incorrect - as per the shell theorem.  If there is no mass in the void, the gravity at all points of the void will be zero.

If you jumped you could reach the center and float  because the gravity would be equal in all directions?

If you jumped, you would have a velocity that would carry you to an opposite wall.  To float, you would need to be able to stop yourself - and wherever you stopped is where you would stay.

But you have a huge amount of mass above you that doesn't just go away or are you saying it cancels out?

Well if you're in the gravitational centre, there's an equal amount of mass above you, as below you, pulling you in opposite directions, thus cancelling eacth other out.

Plus I think the core has huge iron crystals in it that are some 30 miles long according to science.

I thought the centre of the earth was liquid iron? In any case, it's impossible to build a capsule capable of sustaining human life in the centre of the earth.

[rhb]

I'd like to nominate this for "stupidest thread" although @beamin has started a number of close competitors.   The gravitational forces on a spherical particle contained inside a spherical shell is an advanced undergraduate physics problem.  And not an especially hard one at that.  Real world analogies are meaningless because the real world is far too complex and variable.

Aside from the earth not being a sphere, the gravity varies from point to point throughout the entire earth because the composition is not uniform.   Moreover, it is spinning.  So the gravitational forces in any direction are constantly changing.  That has never been measured and never will be measured because no instrument would survive the conditions even 10 miles below your feet.

How many angels can dance on the head of a pin?  That's as sensible a question.  But neither have anything remotely to do with electronics or even science.

A sensible question would address the forces on the gyros of the Gravity B probe.  But that would require actual work.

[Zero999]

I'd like to nominate this for "stupidest thread" although @beamin has started a number of close competitors.   The gravitational forces on a spherical particle contained inside a spherical shell is an advanced undergraduate physics problem.  And not an especially hard one at that.  Real world analogies are meaningless because the real world is far too complex and variable.

Aside from the earth not being a sphere, the gravity varies from point to point throughout the entire earth because the composition is not uniform.   Moreover, it is spinning.  So the gravitational forces in any direction are constantly changing.  That has never been measured and never will be measured because no instrument would survive the conditions even 10 miles below your feet.

How many angels can dance on the head of a pin?  That's as sensible a question.  But neither have anything remotely to do with electronics or even science.

A sensible question would address the forces on the gyros of the Gravity B probe.  But that would require actual work.

I agree with you, this is a silly question, but don't think there is any ill intent.

[Brumby]

It is not a silly question at all.  It is a thought experiment - and has been stated as such many times.

It has never been intended as a practical problem in itself, but more to evoke an understanding of the physics behind it.  That understanding allows further thinking to expand on the principles identified with the possibility of an analogous situation being understood.  It also holds true for considering the gravitational effects on the material that does exist at the centre of the Earth - for whatever reason you might want to know that.

As for the minutiae of irregular mass distribution, that is a distraction from the fundamental concept being discussed.  Certainly is is not an invalid component, but it's effect needs to be referenced against the whole body being assessed, based on some identifiable properties.  Such properties can only be determined by the use of assumptions which give us a starting point.  Then each assumption is addressed in turn and the model refined.  For example, we may start out with the assumption that the density throughout the whole body is the same.  After investigating this, we refine the model to allow a spherically symmetric distribution of mass.  We can then go on and look at other details.

The important point, though, is that we had to start somewhere - which is where this thought experiment holds perfect validity.

[GlennSprigg]

One thing that seems to be 'assumed' in  regards to this concept of 'Walking' inside such a Sphere...
is the relative 'scale', which I do not think was originally considered/spoken-of in the original Hypothesis.
If a 'Man' was 6' tall, walking inside a "very massive" 20' shell, then obviously we would be talking
about the obvious difference between "Head & Shoulders" attraction above, compared to "Feet" attraction
below....  as the 'Man' has no specified 'Centre' for calculations or in thought of the 'Real World' hypothesis.

I would LIKE TO THINK that the 'Man' is but a DOT in 'space', and that the 'Hollow Sphere' was the size of
Earth !!!, so that RELATIVELY speaking, his 'head' & 'feet' are considered the same, in relative scale.
That being the case, there is COUNTLESS thousands of sites/refs that agree that INSIDE of a hollow 'Mass',
(and interestingly it does NOT have to be Spherical !!) there is a 100% balance of equilibrium within !!

[CatalinaWOW]

One thing that seems to be 'assumed' in  regards to this concept of 'Walking' inside such a Sphere...
is the relative 'scale', which I do not think was originally considered/spoken-of in the original Hypothesis.
If a 'Man' was 6' tall, walking inside a "very massive" 20' shell, then obviously we would be talking
about the obvious difference between "Head & Shoulders" attraction above, compared to "Feet" attraction
below....  as the 'Man' has no specified 'Centre' for calculations or in thought of the 'Real World' hypothesis.

I would LIKE TO THINK that the 'Man' is but a DOT in 'space', and that the 'Hollow Sphere' was the size of
Earth !!!, so that RELATIVELY speaking, his 'head' & 'feet' are considered the same, in relative scale.
That being the case, there is COUNTLESS thousands of sites/refs that agree that INSIDE of a hollow 'Mass',
(and interestingly it does NOT have to be Spherical !!) there is a 100% balance of equilibrium within !!

A little thought shows that the last statement is not generically true.  As I sit in my totally enclosed region of space (my house) I am not floating off of my chair.  The symmetry of the matter outside the hollow does matter.

[ez24]

Plus I think the core has huge iron crystals in it that are some 30 miles long according to science.

But I thought you carved out a hollow area which would include the crystals.   My theory is that it would look like the inside of the space station.  With little energy you could float to the other side.

[Brumby]

(and interestingly it does NOT have to be Spherical !!)

Be careful with that.  I have not looked closely at the maths, but I would consider it essential to at least qualify it with the mass of the body having point symmetry about its centre.

One thing that seems to be 'assumed' in  regards to this concept of 'Walking' inside such a Sphere...
is the relative 'scale', which I do not think was originally considered/spoken-of in the original Hypothesis.
If a 'Man' was 6' tall, walking inside a "very massive" 20' shell, then obviously we would be talking
about the obvious difference between "Head & Shoulders" attraction above, compared to "Feet" attraction
below....  as the 'Man' has no specified 'Centre' for calculations or in thought of the 'Real World' hypothesis.

I would LIKE TO THINK that the 'Man' is but a DOT in 'space', and that the 'Hollow Sphere' was the size of
Earth !!!, so that RELATIVELY speaking, his 'head' & 'feet' are considered the same, in relative scale.

That, actually, is irrelevant.

Consider each part of a person's body in isolation.  Each of those will be subject to the same mathematical process to assess the gravitational effects - and each of those calculations will result in a zero net result.  You can continue this analysis to infinitely small divisions, so the scale of the human isn't really a factor at all.

[CatalinaWOW]

Some of the confusion about this may come from comparison with electrostatic fields - which I contributed to with my earlier statement on the proper approach to this problem on this forum.  The fundamental difference is that the electric field parallel to a conductor is nominally zero, and is indeed zero at the surface of a perfect conductor.  There is no such thing as a gravitational conductor (as far as we know today) which makes it a fundamentally different situation.

While I am still not sure what the fundamental idea Brumby is aiming for in his thought experiment, the basic answer is that the net gravitational pull at the center of an ideal spherical assembly of mass is zero.  Once the symmetry of that ideal case is lost the devil is in the details.  There will be a point near the center of any assembly of mass where the net gravitational pull from that mass is zero.  The exact location becomes computationally difficult for anything other than homogeneous simple geometric solids (a cube for example is also easy) but exists none the less.  As another aside, you can mix shapes.  Put a cubical cavity in a spherical mass.  As long as their centers match the pull at the center will be zero.

The same kind of symmetry applies to objects placed at the center of the cavity.  A sphere centered in the cavity would stay put.  Any shape with it's gravitational center placed at the spheres center would stay put.  But a human whose center of gravity was placed at the center would have to remain absolutely motionless to maintain that point.  Any position changes that changed the persons center of gravity would spoil the balance.

Finally, the equilibrium is unstable, in the same way that a pin balanced on it's point is unstable.  Any displacement results in further motion away from the balance point.   The forces will be quite small for small displacements, but they are there.

[Brumby]

While I am still not sure what the fundamental idea Brumby is aiming for in his thought experiment

Keep reading....

the basic answer is that the net gravitational pull at the center of an ideal spherical assembly of mass is zero.

That is the obvious - but there's more ...

  Once the symmetry of that ideal case is lost the devil is in the details.

The only symmetry that is important is that of the spherical body (which goes a bit deeper as well)

The exact location becomes computationally difficult for anything other than homogeneous simple geometric solids (a cube for example is also easy) but exists none the less.

You would think.  However, if we stick with the spherical, the answers are surprisingly simple.

The thing that seems mysterious is covered by the Shell Theorem - specifically this point: If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.
It's that last phrase that is not intuitive - but I had reservations that such might be possible, which is why I said this earlier:

It has also been an observation of mine that circular and spherical geometry often have some surprisingly simple formulae drop out of analytical efforts - so the idea of a net zero-G environment throughout the void is not something I could dismiss through casual inspection....

[CatalinaWOW]

OK, it finally broke through to me - sorry about my slow mind, but need to keep the details of the theorem in mind.  Easy to go astray as above with the idea that gravity is zero inside any spherical void.  Only true if the mass is spherically arranged around the void, which is only true for the special case of a void centered in the earth.  Also the tidal effects due to orbiting the sun still apply.

[Brumby]

Not to worry - we've all had moments like that ... and, yes, I agree with your other comments.

What I like to take from this is the extension of this into more realistic scenarios...

For starters, the requirements for a shell are that it is spherical, of constant thickness and constant density.  There are no constraints on the actual diameter nor the actual thickness.  If you had a shell with an outer diameter of 12,000km and a thickness of 1,000km, then the 10,000km diameter void within would present zero gravity at any point within.  You could also take another shell with an outer diameter of 10,000km and a thickness of 500km, then the 9,000km diameter void within would also present zero gravity at any point within.  The density of these two shells does not have to be the same for this to hold true.  What's more, you could place the smaller shell inside the larger one and the 9,000km diameter void within would still present zero gravity.

Extending this idea even further and you could have thousands of concentric shells and the void within the smallest of them would still present zero gravity.  This provides us with a model which offers characteristics that are more closely related to a real world example.  Yes, the concept of a real, physical void is absurd, but the layered structure will better reflect planetary construction.

At this point, I would like to address the "finer details" that have been mentioned such as irregular mass distribution.  While I don't argue against their existence, I have to question their relevance to the discussion.  It is like debating relativistic effects in a discussion on Newtonian mechanics where those effects can be ignored and the discussion can still yield functionally valid results.  I would say: Let's make a note of these things as they may be useful if we want to delve deeper, but let's not try and build such a detailed model at the outset where the basic observations are being explored.

As a last point, let us consider the above example with multiple shells and a 9,000km diameter (4,500km radius) void - and we set our observation point on the inner surface of the smallest shell, some 4,500km from the centre of our body.  As I have said before, if that void has no mass then the gravity at our observation point will be zero ... but what if that void did have mass?  The answer is, there would be gravity.  In my previous mention of this, I used an example where the void was rather small in diameter and was filled with air.  This resulted in very low gravity, but it was non-zero.

Now, instead of air, what would happen if we placed the Moon within the void?  It could float about in an essentially weightless state - yet it would exert it's own gravity.  The gravitational effect on the observation point set above would be a simple calculation based on Newton's Law of Gravity in relation to the Moon and the observation point ... the shell structure they are within does not even come into it.

Up to this point, we have explored "impossible" scenarios - just to explore the maths - but here we can step into a real world example that uses these explorations to give a useful start.....

Let us revisit the above example with the 9,000km diameter void, remove the Moon and fill the void with a 9,000km diameter ball of matter (the sort of thing you might find at the centre of the Earth) and for whatever reason, you need the calculation for the gravitational effects at a point 4,500km from the centre.  You can now do it and the answer is delightfully simple - it is Newton's Law of Gravity applied to the 9,000km diameter ball of matter.  Everything outside that distance can be considered as having zero effect (from the Shell Theorem) and everything inside it can be considered just the same as any other planet, with the observation point being on its surface.

Simples.

[Rerouter]

As someone who enjoys physics, There is fun in starting with the spherical cow in a vacuum, then adding back in real world effects to build a better understanding piece by piece, And those small effects are what make the difference between no net acceleration and more interesting interactions,

To put some numbers to the non uniformity of the distribution of mass around the earth, we have gravitational anomaly maps, this one also gives number for the acceleration due to the sun and moon.

https://earthscience.stackexchange.com/questions/7449/does-gravity-differ-from-place-to-place-on-earth

The real fun in this situation happens when you ask what happens when you jump off the inside the shell, the larger the shell, the more the anomalies and the sun and moons influence add up, leading to weirder and weirder paths as the shell radius gets bigger, not to mention the earth is spinning, so unless the shell is very small, your interaction may not be very pleasent when you reach another side.

[Brumby]

As someone who enjoys physics, There is fun in starting with the spherical cow in a vacuum, then adding back in real world effects to build a better understanding piece by piece, And those small effects are what make the difference between no net acceleration and more interesting interactions,

Exactly why starting off with an ideal situation allows such progression and thought experiments on impossible scenarios can deliver useful models - which, without such simplistic origins, may never have been developed.

It is fun.

[Beamin]

It is not a silly question at all.  It is a thought experiment - and has been stated as such many times.

It has never been intended as a practical problem in itself, but more to evoke an understanding of the physics behind it.  That understanding allows further thinking to expand on the principles identified with the possibility of an analogous situation being understood.  It also holds true for considering the gravitational effects on the material that does exist at the centre of the Earth - for whatever reason you might want to know that.

As for the minutiae of irregular mass distribution, that is a distraction from the fundamental concept being discussed.  Certainly is is not an invalid component, but it's effect needs to be referenced against the whole body being assessed, based on some identifiable properties.  Such properties can only be determined by the use of assumptions which give us a starting point.  Then each assumption is addressed in turn and the model refined.  For example, we may start out with the assumption that the density throughout the whole body is the same.  After investigating this, we refine the model to allow a spherically symmetric distribution of mass.  We can then go on and look at other details.

The important point, though, is that we had to start somewhere - which is where this thought experiment holds perfect validity.

100% spot on. My only fault was making this thread before I was warned not to make off topic science threads which I have stopped and stopped replying to except for this one since people seem to like answering it. The purpose of many of my questions is invoke deeper thought and discussion of that which you can't get by "googling" it, plus there are some very smart people on this board who have got me to look at things differently. I should have called this thread "what kind of technology would be needed to measure this effect and how to build it?" But saying this is a "stupid question" is like saying Schrodinger's cat is just dead or it doesn't matter. His cat may have seemed stupid to those who couldn't think outside the radio active cat containing box. Most people in life who told me I had stupid simple questions often didn't understand what I was talking about and I rarely clue them in rather having them come up with it on their own. I have figured out many hard problems with thought experiments.

I wonder if you could measure this by sending a perfect crystal down and recording its distortion through EM disturbances like a harmonic oscillator in RFID cards.

[GlennSprigg]

Although I 'Think' I know, I will quote nothing more, before producing a calculable 'model'.
"I'll be back" !!!!  xox

[Beamin]

Although I 'Think' I know, I will quote nothing more, before producing a calculable 'model'.
"I'll be back" !!!!  xox

Please hurry I'm setting up a company that gives tours of the center of the earth and need to see if we need to bring chairs or ropes to keep people from floating away. Some tritanium alloy should work.

[GlennSprigg]

Ha !!  ,   It's still in the 'pipeline'.... 
My 'old' work on this was on another old laptop, using real-time 2D physics, which I just recently found !
I had coded the Java/Javascript to take into account various physical size/shape & Mass bodies in space,
with both Gravitational &/or Magnetic properties, involving real time forces both from surface to surface,
as well as between centres of such masses.  (Graphically demonstrated in real time).
This also applies for a shape 'inside' of a shape. However...

Awhile back, 'CatalinaWOW' said in answer to me....
A little thought shows that the last statement is not generically true.  As I sit in my totally enclosed region of space (my house) I am not floating off of my chair.  The symmetry of the matter outside the hollow does matter.
Well...Obviously the scale/masses of a person in a chair reacting to his 'house', and while 'ON' the Earth's
surface, has NO bearing on the original question what so ever xxx, but one thing I realized I should clarify,
is that MY comments about being inside a 'hollow shell', should state, (as that's what I mean), that the WALLS of
such a 'shell' should be of EVEN thickness/mass, be it a circle or an ellipse etc !!!
(NOT obviously (for example) like an 'egg' shaped Hollow, inside a 'cube' outer shape !!)

That put me off, but if interested, I will re-create the scenarios/info, with a real time video-clip of the action.
P.S.
I know you are 'stirring', however in regard to your proposed new 'travel Company'.....  ....
The Earth, as it is not hollow, is 'different' than the scenarios talked about, and so the 'Gravitational-Pull'
on you as you 'descended', would only 'gradually' decrease, until at the 'centre' where the 'balance' of the
gravitational 'pull', is now 'equal' in all directions !, (although 'effectively' now zero) .
Have a great day !