[graham.d]

Kiran, we do know, both from theory and practice, that the field resulting from a spherical shell and inside the shell, for forces obeying the inverse square law, is evrywhere zero. The practical result is obtainable from electrostatics. I doubt there has been any measurement to confirm this with gravity because the apparatus would have to be very large because gravity is such a small force. Nonetheless it has been shown that gravity obeys the inverse square law to high accuracy. It is sufficient to say that it would be very likely to be well behaved and predictable (accurate even with Newtonian mechanics) except in very extreme circumstances. I doubt that even a perfectly hollowed out sphere the size/mass of the earth would show anything significantly different from Newtonian mechanics. Certainly a comet would not and, in any case, where would you get a comet that was perfectly spherical and of perfectly uniform density?

I see no reason to depart from the accepted theories, whether Newtonian or Relativistic, without some reasoned justification for doubting them. I have no doubt that Relativity will get superseded at some point, but I don't think it is likely to be on this basis. 

[Kiran The King Kai]

Kiran, we do know, both from theory and practice, that the field resulting from a spherical shell and inside the shell, for forces obeying the inverse square law, is evrywhere zero. The practical result is obtainable from electrostatics. I doubt there has been any measurement to confirm this with gravity because the apparatus would have to be very large because gravity is such a small force. Nonetheless it has been shown that gravity obeys the inverse square law to high accuracy. It is sufficient to say that it would be very likely to be well behaved and predictable (accurate even with Newtonian mechanics) except in very extreme circumstances. I doubt that even a perfectly hollowed out sphere the size/mass of the earth would show anything significantly different from Newtonian mechanics. Certainly a comet would not and, in any case, where would you get a comet that was perfectly spherical and of perfectly uniform density?

I see no reason to depart from the accepted theories, whether Newtonian or Relativistic, without some reasoned justification for doubting them. I have no doubt that Relativity will get superseded at some point, but I don't think it is likely to be on this basis. 

wow you are cool , are you a Physicist ?

[RAJ1]

I do not think this is correct.  Let me illustrate why field is not zero everywhere inside a hollow earth or ball.

Lets take a lead ball the size of the earth

We hollow it out so its thickness is maybe 1 mile thick, so its mostly hollow.  Like a giant tennis ball made of lead.

If you were on the inside 1 foot from the inside wall (many thousands of miles from the center)

Do you think you would be stationary?  Or would you drift to the center or drift towards the inside wall?

Well remember that gravity weakens the further you are away (inverse square)

So the only place where the field cancels is dead center of the hollow sphere.

If you are placed off center by a few feet you will drift towards the outside walls.   The wall you are closest to will have a stronger gravitational field.  Depending on the material the sphere is made of the acceleration would be low and would not be constant. 

ie.  If you were in space millions of miles from earth you would not fall at 9.8m/sec2  towards earth (assuming no other planets exist)


For those of you who have played the game Halo.  They used a rotating halo to simulate gravity.  But they also could have made the material (the land) denser so it would have enough gravity to hold you to it.   It if were neutron star material it might be a bit too much.  But you get the idea.


=================================================================================================================================================

We are talking about a gravitational field inside a perfect, non-rotating, spherical shell of any thickness but uniform density. The field due to the shell, within the shell, is everywhere zero. There are no fields, no tidal effects. The size of the body being acted upon is not relevent.

[Phractality]

RAJ1, when you study tripple integrals in college, one of your homework assignments will be to integrate Newton's law of gravity for a uniform hollow sphere. When you do so, you will discover for yourself that the field inside the sphere is zero, and outside it is the same as if all the mass were concentrated at the center. Until you discover it for yourself, you'll just have to take the word of your seniors.

As for general relativity, it yields exactly the same results as Newtonian physics except in extreme cases, like black holes and the whole universe. A hollow Earth is not such an extreme case.

The reason the Earth can't be hollow is because two halves of the hollow sphere would be attracted to one another. The pressure where two halves meet would be greater than the strength of the material. So pieces would break off and float inside, where they would be attracted to each other's gravity and form a solid ball. That ball would attract other chuncks from the near side of the hollow sphere, and very quickly the whole thing would collapse into one solid ball.

A small sphere can be made of just about any solid, but when you get to the radius of Earth, it would need to be quite strong. You can do the math yourself; it's pretty simple. Consider two hemispherical shells of a given thickness and radius. The center of mass of a thin hemisphere is located midway between the center and the surface. Calculate the mass of each hemisphere (m = 2πρr²δr), use Newton's equation (F = G(m²/r²)) to get the force of gravity and divide by the area (a = 2πrδr) where the two hemispheres meet. That will give you the compressive pressure. Compare that to a table of the compressive strenghts of various materials.

[RAJ1]

Phractality,  thanks for the response.   

I see the error in the logic and looked at some equations.   I'd seen this  question posted a few places and never saw anyone give a detailed explanation.  I would like you to take a look at this site and see what ya think. 

newbielink:http://cseligman.com/text/planets/internalpressure.htm [nonactive]

My error was not factoring in that while you could be closer to the inside wall of the sphere the area or mass to the side furthest from you would be weaker but the amount of mass is larger. (ie. weaker but more numerous gravity vectors which as a whole will cancel out the force to your right.  I'm not sure that made sense but there is a good diagram and explanation as to why.  And of course newtons equations also fit in with that.

Yes I know there would be a maximum size before the planet or object would break apart.  Wonder how big it would be for say a iron ball. 

I was a physics major in undergrad, had a bout of encephalitis recently and lost memory, its still coming back so I have to relearn many things.  I still like it but obviously my logic is not what it use to be.   

Thanks for the help.

Let me know if that is a good page in your opinion.

newbielink:http://cseligman.com/text/planets/internalpressure.htm [nonactive]

RAJ1, when you study tripple integrals in college, one of your homework assignments will be to integrate Newton's law of gravity for a uniform hollow sphere. When you do so, you will discover for yourself that the field inside the sphere is zero, and outside it is the same as if all the mass were concentrated at the center. Until you discover it for yourself, you'll just have to take the word of your seniors.

As for general relativity, it yields exactly the same results as Newtonian physics except in extreme cases, like black holes and the whole universe. A hollow Earth is not such an extreme case.

The reason the Earth can't be hollow is because two halves of the hollow sphere would be attracted to one another. The pressure where two halves meet would be greater than the strength of the material. So pieces would break off and float inside, where they would be attracted to each other's gravity and form a solid ball. That ball would attract other chuncks from the near side of the hollow sphere, and very quickly the whole thing would collapse into one solid ball.

A small sphere can be made of just about any solid, but when you get to the radius of Earth, it would need to be quite strong. You can do the math yourself; it's pretty simple. Consider two hemispherical shells of a given thickness and radius. The newbielink:http://thesaurus.maths.org/mmkb/entry.html;jsessionid=D9BBC8A1DCD98DE74989381B0B0A23AC?action=entryByConcept&id=3610&langcode=en [nonactive] is located midway between the center and the surface. Calculate the mass of each hemisphere (m = 2πρr²δr), use Newton's equation (F = G(m²/r²)) to get the force of gravity and divide by the area (a = 2πrδr) where the two hemispheres meet. That will give you the compressive pressure. Compare that to a table of the compressive strenghts of various materials.

[yor_on]

So stay with us and argue Raj, your memory will come back as you find yourself defending your views Or you will create new and even more advanced ideas/memories as you find yourself forced to explain your thoughts in detail. It's good for the soul to argue, well, in moderation at least

[oatman]

I read somewhere this exact scenario was put forward as a means of time travel, hypothetically.

Maybe if you could position yourself right in the centre of some huge object say Jupiter, time would be sufficiently affected by the gravity to make it pass extremely slowly for you.

getting in and out could be a bit tricky though!

[RAJ1]

Sorry to beat a dead horse here.

But could someone try to explain (simplify) this scenario.

Its hypothetical so for now just ignore what i am about to describe is impossible to do.  Pretend the material is of infinite strength and will not bend or break.

Ok

Lets make a giant ping pong ball with radius of the milkyway.  The thickness will be say 10 feet thick, the center is void of air, just empty space.
The material its made of is the density of a neutron star.

The curvature of the inside wall of the ball would appear flat as a pancake to the observer.   

If you were floating 4 feet from this wall the gravity of the dense material would be very strong.  The gravity of the opposing wall (100,000 light years away) would be very weak in comparison.

I guess i am having a difficult time wrapping my head around the idea that the gravity of the opposing walls (curved) would cancel out the gravity of the wall 4 feet from you.

I know equations will probably say its true.  But could someone explain in simplified terms as to how this would occur.

[imatfaal]

Raj - it's non-intuitive but here goes. i don't know where you are - but I live close to St Pauls Cathedral in London - google it and check out the dome (and before anyone says I know it's a fake dome).  Well if you go up to the gallery in the dome you can get a much better impression.  Stand with your back to the centre and you can see 50ish metres squared of wall; but stand with your back to the wall and look across the centre and you can see many thousands of square metres.

Ok now to your sphere. You are floating in your sphere, four feet away is the wall, look up look down look left and right - if you haven't cricked your neck you have looked maybe 45degrees down, up, left and right.  Its quite a small area (relative to the whole sphere) that you have looked at.  If you use your spacesuit jets to spin 180 degrees on your long axis the wall is now 4 feet behind you.  Look up, look down, look left and right.  It's big ain't it. Inside a uniform sphere you can think that the bit of the wall i can see is close but small and pulling me forward; but all the wall I cannot see is faraway but big and pulling me backward.  the two forces balance.

In reality it's not just forward and backwards - but up and down and sideways, but the same thing apples   The amazing thing about Newton's Shell theorem or Gauss's law (same thing different maths) is that small and close balances with massive and far away.


It takes a genius like Newton or Gauss to intuitively understand that this works everywhere in the dome and in 3 dimensions - and then to prove it mathematically.  I hope that has helped - as you move towards the edge the amount of mass pulling you back gets bigger.

[CZARCAR]

act kinda like the tides of the oceans?

[RAJ1]

As I move toward the side (4 feet from me) the pull will only become stronger and the pull from the side 100,000 light years away will grow a little weaker.

[Geezer]

As I move toward the side (4 feet from me) the pull will only become stronger and the pull from the side 100,000 light years away will grow a little weaker.

If you think of the sphere as being composed of a series of rings, and consider the gravitational force that each ring exerts on you as you approach the smallest ring, you can see that the closer you get to the smallest ring, (which, in the limit would be a point with no mass), there are fewer rings in front of you and more rings behind you. Not only that, but the rings that are behind you have progessively greater masses until they reach the diameter of the sphere. You can imagine that the masses of the rings behind you increase in a very non-linear fashion as you approach the smallest one.

I think that if you sum the forces exerted by all the rings in front of you you will find that it always equals the sum of the forces exerted by all the rings behind you. If you don't know how to do that using calculus, you might try plugging numbers into a fairly crude spreadsheet model (using rather "chunky" rings). In that case the forces won't balance perfectly, but they should be close enough to demonstrate the principle.

[imatfaal]

As I move toward the side (4 feet from me) the pull will only become stronger and the pull from the side 100,000 light years away will grow a little weaker.

nope.  As Gzr said - less of the wall will be pulling you foward and more pulling you back.

1. Restart your thought experiment from the centre.  at the centre you must agree there is no way you can feel a net force - even if this theorem was wrong everything would still balance at the centre.

2. Move a metre in any direction.  You are a metre closer to the wall directly in front - and the force from that will be a little stronger. You are a metre further away from wall directly to the rear - and the force from that will be a little weaker.  BUT - you need to take in the wall above, below, left and right of you.  There is a metre band that was vertically/below/around you - and previously was neither pulling you forward or backwards BUT NOW it is behind you and pulling you back

3. All these little changes balance out.  If you want to prove it to yourself with Gzr's excel model then let us know and I will provide some ideas (you will have to use components of force and trig).  Personally if I was gonna do a little model I would use lots of points rather than rings - but I am a masochist

[Ken Hughes]

Hi Guys,

Interesting discussion. It's forty years since I did my engineering degree so I'm a little rusty on the math. I will have to take your word for the maths predicting neutral gravitation everywhere inside. I can see that within the hollow sphere, if you move slightly from the centre, forward of you there is an increased attraction due to the closer proximity of the shell, but behind you, although the attraction is weaker due to the greater distances, there is nevertheless more mass doing the attracting. I guess what you're saying is the maths proves these effects are equal and opposite and so always provide equilibrium?

I wonder about the pressure effects also. In an actual planet like the Earth, most of it is fluid (apart from the very thin crust), so there will be a pressure build up, even though the gravitation has been reduced to zero at the centre. Also, the gravitational affects will not be zero everywhere, only at the centre, since there is mass all around, not just at the crust, and if you move off centre a little, the large amount of mass behind you and the small mass in front of you, will pull you back to centre. I believe the condition is termed "meta stable". However, you will be crushed to a pulp by the extreme pressure as well as being fried. Personally, I wouldn't try it.

[Ken Hughes]

Hi again,

I see there is a second part to this thread, namely;- which is the best way to keep your domestic water hot;- leaving the heater on all the time or heating it up on a timer to give you the hot water only when you plan to use it.
If I remember correctly, there is a crossover point, depending on the shape and size of the tank, the conductivity of the insulation and the temperatures inside and outside of the tank, where increasing the insulation thickness, simply increases the area of the outside of the insulation which becomes so large that the reduced temperature of the external surface is overpowered by the greater surface area emitting the heat. So, firstly, let's assume the tank and insulation have been designed to avoid this problem, ie we have not wasted money in over insulating the tank and in losing more heat because of this.

In the case where we leave the heater on, we are simply providing energy over time to replace the heat lost through the insulation due to the temperature difference between the insulation surface and the ambient conditions. This is a constant over a given period, ie heat is lost at a constant rate but if the insulation design is good, this will be at a minimum. We also use energy to replace the hot water we use, presumably via a thermostat.

In the other case, where the heater is on a timer, then when the timer switches off, the rate of heat loss starts to reduce proportionate to the cooling of the tank and we save energy since less heat is lost as the tank cools. When the timer cuts in, we heat the tank up again from a lower temperature. This time we replace the reduced heat lost due to radiation, plus the heat in whatever hot water we use.

The answer to the question therefore is;- "IT DEPENDS"    []

It depends on the design of the system and how much and how often you want to use the water. The decision is one of balancing the convenience of having hot water on demand at any time of day or otherwise frequently, against the small savings to be made from the temperature differences between a constantly hot tank and a gradually cooling tank.

Keeping the heater on is very slightly less efficient, but not noticably so. Personally, I always keep the heater on and have never experienced major heating bills compared to people who don't. I always get hot showers, any time I want.

[imatfaal]

Hi Ken - Oh Yeah youre right - the hollow earth bit of the thread is a little misleading - the shell has to be uniform and spherically symmetrical too!  Whilst the idea of an enormous spherical sphere is obviously ludicrous - the isotropic and homogeneous layout of space at a large scale means that the Shell Theorem (also called Newton's or Gauss's)  is one of the most basic tools in cosmology - to simplify massively if you are in an enormous cloud of stars and galaxies and it approximates homogeneity (each bit is the same) and isotropy (each direction looks the same) then the shell theorem can apply.

[Ken Hughes]

Hi,

Surely the application of the shell theory to the whole universe has to be an approximation, perhaps practical for very small densities inside the hypothetical sphere?

[Phractality]

When dealing with the whole universe, you have to make some assumptions, and it is always best to know what your assumptions are and not forget them. A has already been stated, we are assuming the distribution of mass of the universe (at the largest scales) is isotropic and homogeneous. Now, we must decide whether we are looking at a finite quantity of contained within a finite volume of expanding space (which started as a point singularity; i.e. big bang), or whether there is an infinite amount of mass contained in an infinite volume of space (which started at infinite density, but has always been infinitely large), or some other scenario.

In the big bang scenario, the whole space-time continuum is warped by gravity, because there is an outermost hyper-spherical shell, and gravitational potential is a function of 4D distance from the center (i.e. from the big bang). In the second scenario, there is no outermost shell (in any of the four dimensions), gravitational potential is the same everywhere, and so, there is no large-scale warp of the space-time continuum. (It is often said that the calculation of critical density considers only density and not size, but the truth is, the formula is valid only if you assume finite size.)

"Other scenarios" might include: A universe that has always had finite density but infinite size and no beginning or end. A lumpy universe which just happens to be homogeneous and isotropic within our Hubble sphere. Multiverses, God done, it, etc.

[yor_on]

You know Raj, the universe is weird. The idea of 'infinity' applied topologically may just become a dough nut, or a sphere, on where you can walk but never arrive to any endpoint. It's geometries, and as such restricted only by those 'laws' we can think up, allowing topologies to change shape like if twisting/bending them, or turning them inside out, etc.

Newtons shells as I got it.

==


You better read this one too A better explanation as mine might seem a little confused. It's not that 'self sufficient'(?) to understand how he thought that masses took themselves out. But he had one direction left in his shell theorem as I understands it? and applied on a universe such a direction must exist too, which it doesn't, as far as I know that is