What would gravity feel like on a hollow earth?

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Nov 17, 2020
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Since gravity starts at the center of an object, and the Earth is hollow with nothing to get in it's way to slow it down, nothing would move.
 
Mar 21, 2021
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If you had 2 earths, one original and the other hollow with a 1m thick crust but with identical mass and dimensions as our earth.
A fun mind bender. I wonder if something got missed, or at least is a reason for confusion. In Dougz's original question, the hollow ball had a 1 meter thick shell but the identical mass as earth. Which would mean the entire 6*10^24 kg mass of the earth is squeezed into this shell. Therefore, to find out how much mass is under each square meter (cubic meter since the crust is 1 meter thick) we divide the mass of the earth by the surface area in m^2. First approximation is ~1.17*10^10 kg of mass for each square/cubic meter.

Now, since the block is 1m^3, the cg of any single block is 0.5 meters from the surface. The acceleration at the surface of the sphere due only to this block would be 3.1 m/sec^2. However, if we have a person 50kG and 2 meters tall (yeah, tall and skinny) and person's cg is 1 meter above the surface, or 1.5 meters above the cg of the single block, the acceleration at belt height would be closer to 0.34 m/sec^2. So, weird. Your feet would "weigh" much more than your head.

Then I thought, what is the effect of the surrounding blocks? if we imagine each block as a separate entity, and placed the blocks in a tic-tac-toe pattern (9 blocks, 1 central). Then for the cg of our tall skinny guy, I come up with a total acc of ~2.27 m/sec^2.at the person's belt line. 0.34 m/sec^2 for the center block and 0.24 m/sec^2 for each surrounding block. (I'm only counting the normal component of the force vector from the ring of blocks) .

So, then take our tic-tac-toe board and add a ring of blocks, for a 5x5 arrangement. 16 more blocks, but a smaller normal force vector per block because of the increased distance from person cg to 2nd ring block cg. Now I get 3.04 m/sec^2. Add blocks to make a 7x7 arrangement and it goes up to 3.78 m/sec^2. A 9x9 arrangement would provide 4.25 m/sec^2 total normal acc. Does this asymptote to 9.8 m/sec^2?

So, my prediction is that walking might be weird, with a serious non-linear gradient very close (cm) to the surface. In fact, I think if you drilled that hole and crawled through, you would not fall but would stick to the inside of the sphere. Maybe if you jumped you would fall. A ball falling from outer space into that hole and through the earth would still see the earth with a cg at it's center the acceleration force and even direction of the force vector would change as you pass through the shell, with gravity pointing "up" very near the under surface.

Of course, I could be seriously wrong. If you can show I'm wrong.. do it!
 
This was the question:
If you had 2 earths, one original and the other hollow with a 1m thick crust but with identical mass and dimensions as our earth. Would you experience gravity identically on the outer surface of both? Also, what would you feel if you stood on the inside surface of the hollow earth? If you stood on the outside of the sphere and jumped through a manhole, where would you end up?

The solution must be based on the centre of gravity (cog) of both being at the centre. My first reaction was that there would be no difference on the outer surface as the cogs are the same. Then I wondered whether the high density crust might cause any differences. But then the lateral forces in each direction cancel each other out and one is left with a resultant towards the centre. So I stand by my first reaction.

The other two are easy. If you stood on the inside of the hollow sphere you would immediately fall towards the centre of gravity i.e., the centre of the sphere. Same if you jumped through a manhole.

I think one way of looking at the problem is to start with Earth, then imaine a spherical hole at the centre, say, 10 metres diameter. Then progressively increase the diameter of this hollow hole until it is within 1 metre of the surface.

I would submit that there is no difference as you go from Earth to the hollow sphere. Surely this is in accord with the whole concept of centre of gravity?

Cat :)
 
A fun mind bender. I wonder if something got missed, or at least is a reason for confusion. In Dougz's original question, the hollow ball had a 1 meter thick shell but the identical mass as earth. Which would mean the entire 6*10^24 kg mass of the earth is squeezed into this shell. Therefore, to find out how much mass is under each square meter (cubic meter since the crust is 1 meter thick) we divide the mass of the earth by the surface area in m^2. First approximation is ~1.17*10^10 kg of mass for each square/cubic meter.

Now, since the block is 1m^3, the cg of any single block is 0.5 meters from the surface. The acceleration at the surface of the sphere due only to this block would be 3.1 m/sec^2. However, if we have a person 50kG and 2 meters tall (yeah, tall and skinny) and person's cg is 1 meter above the surface, or 1.5 meters above the cg of the single block, the acceleration at belt height would be closer to 0.34 m/sec^2. So, weird. Your feet would "weigh" much more than your head.

Then I thought, what is the effect of the surrounding blocks? if we imagine each block as a separate entity, and placed the blocks in a tic-tac-toe pattern (9 blocks, 1 central). Then for the cg of our tall skinny guy, I come up with a total acc of ~2.27 m/sec^2.at the person's belt line. 0.34 m/sec^2 for the center block and 0.24 m/sec^2 for each surrounding block. (I'm only counting the normal component of the force vector from the ring of blocks) .

So, then take our tic-tac-toe board and add a ring of blocks, for a 5x5 arrangement. 16 more blocks, but a smaller normal force vector per block because of the increased distance from person cg to 2nd ring block cg. Now I get 3.04 m/sec^2. Add blocks to make a 7x7 arrangement and it goes up to 3.78 m/sec^2. A 9x9 arrangement would provide 4.25 m/sec^2 total normal acc. Does this asymptote to 9.8 m/sec^2?

So, my prediction is that walking might be weird, with a serious non-linear gradient very close (cm) to the surface. In fact, I think if you drilled that hole and crawled through, you would not fall but would stick to the inside of the sphere. Maybe if you jumped you would fall. A ball falling from outer space into that hole and through the earth would still see the earth with a cg at it's center the acceleration force and even direction of the force vector would change as you pass through the shell, with gravity pointing "up" very near the under surface.

Of course, I could be seriously wrong. If you can show I'm wrong.. do it!
Do you not accept that the cogs must both be at the centre? Thus the forces on head and toe are divided by distance squared, where distances are r and (r+h) where r is Earth radius and h is height of the man. Thus h would be immeasurably small in relation to r. So h is irrelevant. I had my misgivings, but surely centre of gravity must remain at the root of all (Newtonian) gravitational calculations?



Cat J
 
Mar 21, 2021
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This was the question:
If you had 2 earths, one original and the other hollow with a 1m thick crust but with identical mass and dimensions as our earth. Would you experience gravity identically on the outer surface of both? Also, what would you feel if you stood on the inside surface of the hollow earth? If you stood on the outside of the sphere and jumped through a manhole, where would you end up?

The solution must be based on the centre of gravity (cog) of both being at the centre. My first reaction was that there would be no difference on the outer surface as the cogs are the same. Then I wondered whether the high density crust might cause any differences. But then the lateral forces in each direction cancel each other out and one is left with a resultant towards the centre. So I stand by my first reaction.

The other two are easy. If you stood on the inside of the hollow sphere you would immediately fall towards the centre of gravity i.e., the centre of the sphere. Same if you jumped through a manhole.

I think one way of looking at the problem is to start with Earth, then imaine a spherical hole at the centre, say, 10 metres diameter. Then progressively increase the diameter of this hollow hole until it is within 1 metre of the surface.

I would submit that there is no difference as you go from Earth to the hollow sphere. Surely this is in accord with the whole concept of centre of gravity?

Cat :)
I hear you, Cat, and you may be right. However, I can't discount short distance near field effects. I think the high density matter (and we are talking like white dwarf type density) of the shell is relevant. If this mythical 1 meter cube I've invented was sitting here on my table, right at the surface of the cube will be an acceleration of ~3.1 m/sec^2, (not including earths gravity, of course) toward the center of mass of the cube. This falls off very quickly with distance, given the extremely small radii we are working with. If you (earth gravity back on) had a pendulum, and swung it very close to the cube, the resulting force near the cube would be a vector combo of earth's gravity and the gravity of the cube itself.

There are some differences between solid spheres and hollow spheres of the same mass and outer dimensions. Like, if you take a pair of such balls and roll them down a ramp, will they go down identically? No, because the hollow sphere has a larger moment of inertia, and so the solid sphere finishes first, because it had less resistance to the motion.

Maybe you will fall through the hole, but as our little bat is clinging to the underside of the hollow sphere, looking at a 4000 mile swan dive, won't he feel a total acceleration vector that is less than 9.8 m/sec^2? The block has significant mass, and there will be a gravitational force upward (along with a gravity vector downward) for that little hanging bat per the gravity equation (maybe not enough to cancel the direction of the force vector, but not 9.8 m/sec^2) . I don't see how we get around it. If you moved an object slowly through a small hole in our sphere/block, do you see it having exactly 9.8 m/sec^2 during it's travel through the 1 meter hole?

Another thought experiment: Instead of the hollow sphere, what if you had a black hole with the exact mass of the earth 4000 miles away (like earth radii), and our mountain-mass 1 meter cubic block 1 cm above you (you are extremely tiny), wouldn't the resulting force vector be less than 9.8 m/sec^2. Ok, having thought it through I agree that the guy will probably fall through the hole, and walking would be mostly normal, but I do see a near field effect over very short distances that either increases or decreases the total acceleration vector just above and just below our hollow sphere. Thanks for responding, I need a mental work out while bored to tears waiting for my next COVID vaccine shot!
 
Epiphany
Thank you for that. As I said, I did have second thoughts, but I find it difficult to go against cog in Newtonian gravitation.

OK, you have your little but very dense cube (are we not here discussing an impossibility which, in itself, may invalidate any argument?) but you have zillions of little cubes, albeit further away, pulling against it. Is this not the essence of the cog concept?

Cat :)

P.S. Do you have any refutation of my solid sphere with ever increasing central voids?
 
Jun 1, 2020
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The solution must be based on the centre of gravity (cog) of both being at the centre. My first reaction was that there would be no difference on the outer surface as the cogs are the same. Then I wondered whether the high density crust might cause any differences. But then the lateral forces in each direction cancel each other out and one is left with a resultant towards the centre. So I stand by my first reaction.
Yes, that makes sense. The lateral forces become more and more canceling as the the shell thickness decreases.

If the 0.5m CoG distance is cut in half, for example, then the inverse square law would increase g by 4x, so the 1/3 normal g would be 1-1/3 normal g, exceeding the Newton principle. But that only works if the lateral component of those force vectors are completely ignored. As the distance to a cog decreases, the more the lateral vector components increase, so I can't see how it would ever exceed 1 g, even for an ant.
 
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