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

[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 !!!!!)