What would gravity feel like on a hollow earth?

Jan 30, 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. 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?

I get that if you drilled a hole right through the centre of our earth and out the other side and jumped in, you'd essentially oscillate for a while before coming to rest at the centre point, feeling weightless there. Now do the same with a hollow sphere of identical mass and outer dimensions. Drill a hole through the 1m crust, through the space underneath and out the opposite side. Jump through and what then happens...instant weightlessness? since you're in free space as soon as you leave the outer crust? I guess not. Would you end up in the same centre point as with a solid earth, and if not, do you end up stuck to the inside face of the outer crust, feeling exactly the same gravity as you would if standing on the outside of the crust. That would suggest there's a weightless point 0.5m through the hole you drilled in the 1m crust. If that was the case, the centre of mass of your body would feel weightless, with your legs being pulled by gravity through the hole and your head being pulled in the opposite direction towards the outer surface.

Someone will have worked all this stuff out of course. I can't get a clear picture of what that would look like...
 

Catastrophe

The devil is in the detail
Gravity is proportional to the masses of the items, in this case you and the hollow earth. The latter would be enormously less than that of a solid Earth - depending on its volume (depending on thickness) and its density. The mass of the hollow earth should be considered to reside centrally, so where you are and where go should make no significant difference, just as on our Earth.

At least, that is my view. I stand open to correction if I am in error.
P. S. I know I omitted the distance factor, but I assumed this would be practically constant as the radius.

Cat :)
 
Jun 1, 2020
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Standing on the surface of either, will make no appreciable difference. As Cat mentioned, the mass can be treated as being at the c.g. (center).

However, once you travel inside the Earth, either one, then some of the mass is above you. As you travel closer and closer to the center, this mass begins to make a significant difference to the amount of gravity encountered. Surprisingly, the decrease isn't as much as one might expect because as one gets closer to the center the gravitational increase is by the square of the distance, so although the reduction in mass weakens the gravity, the reduction in distance increases it. The net result is somewhat tricky, especially for the real Earth with all its density variations, but, IIRC, the net gravity will decrease all the way down, just not in a linear way.
 
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Catastrophe

The devil is in the detail
Helio wrote :
"IIRC, the net gravity will decrease all the way down, just not in a linear way. "

I would agree with this. Ooops, am I sure? As you approach the c of g the distance decreases and therefore the gravity increases?
The equation is: force (gravity) = (mass1 x mass2) / d squared.
So as distance decreases (approaches c of g ) force increases (as square of distance).

It is the same reason that it is most difficult to escape Earth gravity, for example, from the surface. The further away from the surface, the lower the escape velocity.
But I agree that we are discussing a hollow sphere here.

What do you think, Helio? At the c of g all force has been attracted to get there, so the same max force (c.f. escape velocity) must be overcome to return to the surface.

Cat :)
 
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Jun 1, 2020
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Helio wrote :
"IIRC, the net gravity will decrease all the way down, just not in a linear way. "

I would agree with this. Ooops, am I sure? As you approach the c of g the distance decreases and therefore the gravity increases?
The equation is: force (gravity) = (mass1 x mass2) / d squared.
So as distance decreases (approaches c of g ) force increases (as square of distance).
Yes, but don't forget that M (mass) is the density x 4/3 x pi x r^3. So r^3 / r^2, in the gravity equation yields a linear result, which means my view that it's not linear was wrong. Oops. :)

Density varies significantly, and that will make the real results much tricker.

What do you think, Helio? At the c of g all force has been attracted to get there, so the same max force (c.f. escape velocity) must be overcome to return to the surface.
The c.g., however, is valid only when the mass is below your feet. When it's not, a FBD might help demonstrate that the c.g. of the mass overhead for an inbound traveler to the core must be taken into account.
 
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Catastrophe

The devil is in the detail
Helio, sorry - I don't see why you are introducing volume. The eqtn is k(M x m)/r^2 so why are you jiggling around with volume, which must be squared since you have M x m which of course you can write as VD x vd (D and d for respective densities) - but WHY?

You have to put the r's back again in the density (mass / volume)!
P.S. Have you come across dimensional analysis, which is much used in my old subject (Chemical Engineering) and in many other subjects?

Cat :)
 

Catastrophe

The devil is in the detail
OK. Here is the full check by dimensional analysis.

From Wiki or dozens of other places, the Gravitational constant G in the equation

Gravity (Force = mass x acceleration) = G (M x m) / d^2.

is 6.67430 x 10^-11 meter ^3 x kg^-1 x seconds ^-2.

Using m for mass, d for distance and t for time, the equation becomes (with G varying if units are different from SI)

G = m x d x t^-2 x m^-2 x d^2
mass x acceleration by division from RHS

so units of G are m^-1 d^3 t^-2

This corresponds to the dimensions of G kg^-1 metre^3 sec^-2.

When you introduce volume and density, you simply use the substitution:

mass (m) = volume (d^3) x density (m x d ^-3)
mass = volume x mass / volume, reducing to mass = mass.

Obviously there is nothing to be gained dimensionally speaking, from introducing volume and density. Am I missing some point you are making here?

Cat :)









l
 

Catastrophe

The devil is in the detail
Helio,

"When it's not, a FBD might help demonstrate that the c.g. of the mass overhead for an inbound traveler to the core must be taken into account."

For your FBD, you will see the traveller, of course, attracted to the mass overhead, but this will be overcome by the attraction from the rest of the sphere, including the mass on the opposite side of the centre, and the traveller will be attracted to the centre, although there is 'nothing there' because it is the c of g.

Cat :)
 
Jun 1, 2020
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OK. Here is the full check by dimensional analysis.

From Wiki or dozens of other places, the Gravitational constant G in the equation

Gravity (Force = mass x acceleration) = G (M x m) / d^2.

is 6.67430 x 10^-11 meter ^3 x kg^-1 x seconds ^-2.

Using m for mass, d for distance and t for time, the equation becomes (with G varying if units are different from SI)

G = m x d x t^-2 x m^-2 x d^2
mass x acceleration by division from RHS

so units of G are m^-1 d^3 t^-2

This corresponds to the dimensions of G kg^-1 metre^3 sec^-2.

When you introduce volume and density, you simply use the substitution:

mass (m) = volume (d^3) x density (m x d ^-3)
mass = volume x mass / volume, reducing to mass = mass.

Obviously there is nothing to be gained dimensionally speaking, from introducing volume and density. Am I missing some point you are making here?
Yes, that solves the gravitational force at the surface using the known M (mass), but once you head for the center of the Earth M is no longer a constant as only the mass that is below you will be pulling you down (g). It becomes M(r).

IOW, the mass between you and the center becomes less and less. And, the mass overhead becomes more and more.

The only way I now to solve for M(r) is to assume a density and to multiply it by the calculated volume below your feet. This will convert the inverse square term to a linear one.

More here. [scroll up and look at the "Depth" section.]
 
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Helio,

"When it's not, a FBD might help demonstrate that the c.g. of the mass overhead for an inbound traveler to the core must be taken into account."

For your FBD, you will see the traveller, of course, attracted to the mass overhead, but this will be overcome by the attraction from the rest of the sphere, including the mass on the opposite side of the centre, and the traveller will be attracted to the centre, although there is 'nothing there' because it is the c of g.
Yes, but the mass overhead will have its own c.g., hence the need for a FBD.
 

Catastrophe

The devil is in the detail
"Yes, but the mass overhead will have its own c.g "

Sorry? How many c of g'a are there?
Is not the only c of g of the hollow sphere at its centre? (Ignoring minor surface features like mountains, of course.)

Also, are you destroying the dimensional analysis?
I don't understand.
I still don't understand your use of volume. It is simply converted back to mass as you cannot ignore the density element.

Sorry. I just cannot see why you are trying to change the Law of Gravity by introducing volume and ignoring the commensurate density.
Dimensional analysis proves this.

This is just my honest opinion.

Cat :)
 
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Catastrophe

The devil is in the detail
Yes, that solves the gravitational force at the surface using the known M (mass), but once you head for the center of the Earth M is no longer a constant as only the mass that is below you will be pulling you down (g). It becomes M(r).

IOW, the mass between you and the center becomes less and less. And, the mass overhead becomes more and more.

The only way I now to solve for M(r) is to assume a density and to multiply it by the calculated volume below your feet. This will convert the inverse square term to a linear one.

More here. [scroll up and look at the "Depth" section.]
"Yes, that solves the gravitational force at the surface using the known M (mass), but once you head for the center of the Earth M is no longer a constant as only the mass that is below you will be pulling you down (g). It becomes M(r). "

"IOW, the mass between you and the center becomes less and less. And, the mass overhead becomes more and more."
Are you not accepting that the centre of gravity is at the centre?
"A free body diagram (FBD) is a visual representation of the forces acting on an object."
I have done the free body diagram, and I do not see the problem.

See my post #8:
"For your FBD, you will see the traveller, of course, attracted to the mass overhead, but this will be overcome by the attraction from the rest of the sphere, including the mass on the opposite side of the centre, and the traveller will be attracted to the centre, although there is 'nothing there' because it is the c of g."

Cat :)
 
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Catastrophe

The devil is in the detail
Here is a new idea.

Imagine there is a 1% hole at the centre of the Earth. Where is the c of g?

Now make it a 5%, 10%, 99% hole. When does the c of g stop being at the centre?

I am, of course, ignoring quibbles like mountains on the surface causing tiny variations.

Cat :)
 
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"Yes, but the mass overhead will have its own c.g "

Sorry? How many c of g'a are there?
Is not the only c of g of the hollow sphere at its centre? (Ignoring minor surface features like mountains, of course.)
But, as in a FBD, one must do the summation of every relevant sectioned area's c.g.

Consider 4 people on a see-saw in perfect balance. Then have one person slide to the center. To determine the net torque, the single person remaining on the end must still be included in the FBD, especially if the weights are all different. Right?

Similarly, any mass overhead is gravitationally attracting the inward bound traveler. This overhead mass is similar to the non-moving, single person on the end of the see saw.

I still don't understand your use of volume. It is simply converted back to mass as you cannot ignore the density element.
If we're given the depth of a person traveling to the center, say 1000 km, then what is needed to calculate what the mass is underfoot? This must be done geometrically, hence we would pull-up all those volumetric formulae so that we could determine that volume, then we multiply that by the density and we now have the mass below us, hence the value we need for calculating the gravity below us. The volume difference between the whole and this lower volume can be subtracted to give us the overhead volume, allowing the amount of gravitational attraction overhead. The net g is what we are after.

[I looked at some of those forumale, BTW, and they are, not surprisingly, different for an annulus (i.e. hollow Earth). So there would be difference in gravity as one travels to the center compared to a homogenous sphere.]

Sorry. I just cannot see why you are trying to change the Law of Gravity by introducing volume and ignoring the commensurate density. Dimensional analysis proves this.
I don't understand this use of dim. analysis. Isn't it just...

kg/km^3 (density) * km^3 (Vol.) = kg (Mass)
 
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Catastrophe

The devil is in the detail
At the risk of repeating myself, we could save pages of discussion if you will please just answer this?

Imagine there is a 1% hole at the centre of the Earth. Where is the c of g?

Now make it a 5%, 10%, 99% hole. When does the c of g stop being at the centre?

I cannot see why you do not seem to agree with Newton's Law? At the centre of a hollow sphere one would be pulled equally in all directions. It is the cog even if there is nothing there. If one moves away from it there will be a net force to re6urn to the centre.

If we cannot agree on this, I think we will just have to agree to differ. This is just one point on which we disagree. Doesn't stop us discussing other things as friendly as ever.

Cat :)
 

Catastrophe

The devil is in the detail
Sorry, I will add one point:
"I don't understand this use of dim. analysis. Isn't it just... "

Dimensional analysis is a way of checking things are in order
That is why it is clear that you are doing something strange (IMHO) by ignoring mass in the equation and substituting volume and ignoring density which, of course, turns the mass equation straight back into a mass equation. So why involve volume except to get some incorrect result, which is shown by the dimensional analysis?

Let me put my opinion (as we both know, no one has a monopoly on absolute truth) succinctly.
I can see no reason to involve volume except to obfuscate matters in arriving at an erroneous conclusion. Helio, of course I am not suggesting that this is intentional. I really think that you believe you have found something. Unfortunately it contradicts dimensional analysis.

I think what is happening is that you are converting mass to volume and acting on volume without retaining the inseparable concomitant of density which, by dimensional analysis, would otherwise convert it straight back to mass. In other words, involving volume is only going to introduce the possibility of error. If done without remembering concomitant density it will absolutely ensure error.

I suggest we either both just agree to differ or, if you prefer, agree to leave it a week during which time we both agree to go over each other's posts and begin discussion again.

I hope you will see that I am trying my best to arrive at a friendly resolution.

Best wishes,

Cat :)
 
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At the risk of repeating myself, we could save pages of discussion if you will please just answer this?

Imagine there is a 1% hole at the centre of the Earth. Where is the c of g?

Now make it a 5%, 10%, 99% hole. When does the c of g stop being at the centre?

I cannot see why you do not seem to agree with Newton's Law? At the centre of a hollow sphere one would be pulled equally in all directions. It is the cog even if there is nothing there. If one moves away from it there will be a net force to re6urn to the centre.

If we cannot agree on this, I think we will just have to agree to differ. This is just one point on which we disagree. Doesn't stop us discussing other things as friendly as ever.

Cat :)
Ah, I see where we are off the page with one another. You’re certainly correct that the Earth’s c.g. is at the center and it doesn’t move around. But once you begin to travel toward that c.g. then some mass will be above you. Gravity‘s force comes from mass, including mass above. This forces us to determine the variables in Newton’s equation. We can no longer us the total mass of Earth for M, but if we know the mass values from around us, and their respective c.g,s, we can determine the gravity value for any depth.
 

Catastrophe

The devil is in the detail
Helio,

"But once you begin to travel toward that c.g. then some mass will be above you. Gravity‘s force comes from mass, including mass above." My emphasis.

NO, sorry. There is mass "above", of course, but much more mass "below". That is on the other side of the cog. Please beware of using "above", "below", "on", and similar words.

Please answer this:

Imagine there is a 1% hole at the centre of the Earth. Where is the c of g?

Now make it a 5%, 10%, 99% hole. When does the c of g stop being at the centre?

Cat :)
 
Jun 1, 2020
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Please answer this:

Imagine there is a 1% hole at the centre of the Earth. Where is the c of g?

Now make it a 5%, 10%, 99% hole. When does the c of g stop being at the centre?
Again, the c.g. of the Earth will always be the same for the planet as a whole. It's been demonstrated, by Newton IIRC, that the mass of a body can be treated as if it were concentrated at the c.g. This is true whether it is solid or hollow. We agree on this point.

But, this changes once you are surrounded by other influential masses, which is what happens when one is traveling inside the Earth toward the center.

If someone were between two binary stars, both masses would have to taken into account to determine the net gravitational force, right? Similarly, what if a person traveled, say, 1000 km inside the Earth, then someone sliced-off all the Earth above the traveler and moved this mass upward, say, 2000 km. Now the Earth no longer is a sphere, so its c.g. is changed, and the mass above has its own c.g. Does this help?
 

Catastrophe

The devil is in the detail
Again, the c.g. of the Earth will always be the same for the planet as a whole. It's been demonstrated, by Newton IIRC, that the mass of a body can be treated as if it were concentrated at the c.g. This is true whether it is solid or hollow. We agree on this point.

But, this changes once you are surrounded by other influential masses, which is what happens when one is traveling inside the Earth toward the center.

If someone were between two binary stars, both masses would have to taken into account to determine the net gravitational force, right? Similarly, what if a person traveled, say, 1000 km inside the Earth, then someone sliced-off all the Earth above the traveler and moved this mass upward, say, 2000 km. Now the Earth no longer is a sphere, so its c.g. is changed, and the mass above has its own c.g. Does this help?
"
Again, the c.g. of the Earth will always be the same for the planet as a whole. It's been demonstrated, by Newton IIRC, that the mass of a body can be treated as if it were concentrated at the c.g. This is true whether it is solid or hollow. We agree on this point." Totally.

"But, this changes once you are surrounded by other influential masses, which is what happens when one is traveling inside the Earth toward the center." Where has this come from? What are these substantial masses which are near enough to be influential? When one is travelling towards the centre? Is one (the traveller) a substantial gravitational influence? I did not realise that our hollow sphere was paper thin and 1000 feet across. Clearly I should have asked for definition of this. My understanding was clear, I think, when I spoke of mountains on the surface. If we are to continue, can we please agree what we are talking about, including its newly acquired neighbours?

" Now the Earth no longer is a sphere, so its c.g. is changed, and the mass above has its own c.g. Does this help?". I am in total agreement with this. The slice removed will have its own cog and, being removed, will have less gravitational influence. The cog of the remaining section will have its own cog which will be "inwards" of its former cog. The traveller will now be drawn to this new cog. This is in accordance with the first point we agreed above.

Are we making progress?

Cat :)
 
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Catastrophe

The devil is in the detail
Helio, my friend, I have just checked the title of this frame, which is:

What would gravity feel like on a hollow earth? (My emphasis).

Do you think, in view of this, that we should continue our discussion in a conversation?
It also mentions hollow "Earth", not a paper thin sphere, so we should agree what we are talking about.

Best wishes

Cat :)
 
Jun 1, 2020
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Helio, my friend, I have just checked the title of this frame, which is:

What would gravity feel like on a hollow earth? (My emphasis).
Yes, you addressed that question nicely in your response to the OP.

Once we elected to go deeper (pun intended :)) then it go more interesting, at least for me. But I have nothing new to add at this point.
 
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Catastrophe

The devil is in the detail
"Once we elected to go deeper (pun intended :)) then it go more interesting, at least for me. But I have nothing new to add at this point. "

OK, if, on reflection you would like to continue, I suggest we go to conversation. I did not understand what you were getting at as I quoted in my second para of #20, but I won't lose any sleep over it.

Cat :)
 
Mar 15, 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. 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?

I get that if you drilled a hole right through the centre of our earth and out the other side and jumped in, you'd essentially oscillate for a while before coming to rest at the centre point, feeling weightless there. Now do the same with a hollow sphere of identical mass and outer dimensions. Drill a hole through the 1m crust, through the space underneath and out the opposite side. Jump through and what then happens...instant weightlessness? since you're in free space as soon as you leave the outer crust? I guess not. Would you end up in the same centre point as with a solid earth, and if not, do you end up stuck to the inside face of the outer crust, feeling exactly the same gravity as you would if standing on the outside of the crust. That would suggest there's a weightless point 0.5m through the hole you drilled in the 1m crust. If that was the case, the centre of mass of your body would feel weightless, with your legs being pulled by gravity through the hole and your head being pulled in the opposite direction towards the outer surface.

Someone will have worked all this stuff out of course. I can't get a clear picture of what that would look like...
First question I would ask: Is it spinning and at what rate? 2nd Is there a moon? 3rd a Sun?
 

Catastrophe

The devil is in the detail
My answers, not totally guaranteed due to unstated factors, are: See post #24.
QUOTE
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.
QUOTE

QUOTE Question 1
Would you experience gravity identically on the outer surface of both?
QUOTE
Answer 1: Yes since the cog would be the same - at the centre.

QUOTE Question 2
Also, what would you feel if you stood on the inside surface of the hollow earth?
QUOTE
Answer 2 Pulled towards the centre.

QUOTE Question 3
If you stood on the outside of the sphere and jumped through a manhole, where would you end up?
QUOTE
Answer 3 You would accelerate towards the centre.

Cat :)
 
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