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

[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.