Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Aaron Thomas on 14/11/2010 13:30:03
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Aaron Thomas asked the Naked Scientists:
Dear Chris,
Thanks for the great show (http://www.thenakedscientists.com/HTML/podcasts/), I am always listening to your pod-casts (http://www.thenakedscientists.com/HTML/podcasts/) which I think are fantastic.
I have a question/s which I hope you can find the answer/s too........
Q. 1.0. Apparently at the earth's core there is an area of zero gravity, how big would this area be? Â
Q. 1.1 If this area exists then would the mass within this area of zero g be affected by the moon or the suns gravity?
Q 1.2 If the moon or the suns gravity causes this area of zero g to move then could this be how our magnetic field is generated? could this be the dynamo?
Sincerely
Aaron Thomas
What do you think?
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Aaron, I don't have the scientific know-how to try to answer your question. No doubt Chris will, but just a thought in the meantime.
1.0 Think of a rotating disc, at the centre is a stationary point. How big is it? My thought is that it must be infinitesimally small. Would the same not apply to the zero-gravity point at the centre of the Earth?
1.1 If this area has measurable mass, it must be accompanied by measurable gravity (?). If it appears to be gravitationally neutral, that must be because of some external influence (?). If it is infinitesimally small, how could we measure any effect it might have?
1.2 Gravity is a two-way process; if this is an area of zero gravity, would it not be un-influenced by external gravitational sources?
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Actually, I believe that there is no absolute center to the earths axes.
There is a wobble that makes the inner core slash around. The core is liquid produced by the gravitational pressure on the surface presses the earths mass.
This wobble also affects the magnetic poles continuously being relocate.
This wobble is caused by the pulling affect of the associated orbit of the moon the earth around the sun and the earth's daily revolution.
This it the same as tethering a bucket of water with a rope and orbiting it in a big circle around your head. You can feel the forces produced at the grip of your hand.
http://alansthoughts.com/resources/2-Why+The+Earth+Wobbles+with+figures.doc
http://en.wikipedia.org/wiki/Polhode
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Q. 1.0. Apparently at the earth's core there is an area of zero gravity, how big would this area be?
When people say that, they usually mean they're assuming the earth is a perfectly uniform sphere. In that case, when you're down below the ground, the bits of earth below you are pulling you down, while the bits above you are pulling you up, and the bits to the side of you are pulling you sideways in various directions. If you're at the point exactly in the center of the earth, the earth is pulling you equally in all directions at the same time, so that the forces all balance each other and you float there, apparently weightless. Of course, there is still gravity from the earth, but it's just perfectly balanced! This would only happen at a tiny point precisely at the earth's center.
Since the earth isn't perfectly spherical and uniform, this point doesn't really exist in practice. Let's assume it does, or at least there's a small region
Q. 1.1 If this area exists then would the mass within this area of zero g be affected by the moon or the suns gravity?
Yes. Only the earth's gravity perfectly cancels at that point. Other sources of gravity outside the earth would effect things at that point.
Q 1.2 If the moon or the suns gravity causes this area of zero g to move then could this be how our magnetic field is generated? could this be the dynamo?
Someone who knows geophysics better can probably answer this question more fully. I believe most of the magnetic field has to do with currents in the liquid outer core of the earth. The outer core is liquid primarily because of the pressure of the earth's outer layers pressing down on it, the residual heat from when the earth formed, and radioactive minerals releasing heat. The moon's gravity squeezes the earth a bit (which you can see quite readily since it's squeezing of the oceans causes tides), so that has some effect too. I don't think it's the main effect, and it acts on the whole earth at once, not just on the very center. The sun's gravity doesn't really squeeze the earth, so it wouldn't have too much of an effect.
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Thanks everyone for the interesting answers to my questions,
So what I think your all saying is;
That there could be an area ,or something?, of neutral, or negative?, gravity in equilibrium with gravity at the centre of the Earth?, having no mass, infinitesimally small, existing in the gravitational central axis?, effected by external gravitational influences, its effects unknown?
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A quick addition. Although gravity would be neutralized in the center of the earth, the pressure from the whole mass would be enormous.
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A quick addition. Although gravity would be neutralized in the center of the earth, the pressure from the whole mass would be enormous.
The physics of how to overcome this is clearly explained in the Hollywood film 'The Core'! [:D] [:D]
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Aaron - basically, if you could hollow out the centre 10 metres of the earth (insert any number here), and ignoring the fact that you would burn up and that the pressure would be unbearable (and Steve, isn't The Core a terrible film?) , you could float around in there just like you see the astronauts on the international space station. it would have zero gravitational attraction at all points.
I don't understand the second part of your follow up:
having no mass, infinitesimally small, existing in the gravitational central axis?, effected by external gravitational influences, its effects unknown?
Rephrase it and we will try to answer
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A quick addition. Although gravity would be neutralized in the center of the earth, the pressure from the whole mass would be enormous.
The physics of how to overcome this is clearly explained in the Hollywood film 'The Core'! [:D] [:D]
Well, yes. With your unobtanium suit, you'd be just fine!
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A quick addition. Although gravity would be neutralized in the center of the earth, the pressure from the whole mass would be enormous.
The physics of how to overcome this is clearly explained in the Hollywood film 'The Core'! [:D] [:D]
Well, yes. With your unobtanium suit, you'd be just fine!
Damn! It's at the dry cleaners! [;D]
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I think what I am trying to get at is the size of the area of zero "g" and what effect it might have if any. In previous comments someone suggested that this area would be very small. Could it really be big enough to float in? Also once I can find out how big this area is, then I will then be able to possibly visualise it, moving around within the core being effected by the Earths wobble and other gravitational influences, in relation to its position. If the earth is not perfectly spherical and thus causes the center of mass to change then would this point not be moving? I also think that an area which must be as as hot as the surface of the sun at the very center of the earth having no gravity is quite interesting. What also interests me now is how big this area would be in the Sun or even the biggest star VY Canis Majoris? Would this area interact with material in the inner core, meaning what does a substance composed primarily of a nickel-iron alloy at a pressure of 330 to 360 gigapascals and at 5,430 degree centigrade do when in zero "g" would its chemical properties change?
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Aaron - as most of the replies have stated, the shell theorem relates to a hollow sphere. The gravity in a solid sphere is related by 1/r; that is to say that in a mineshaft reaching to the centre of the earth (ignoring impossible heat and pressure) a miner would feel less attraction to the centre of mass as he descended.
you seem to be conflating mathematical ideas with real world problems - we cannot make a hollow room at the centre of the earth, let alone the sun; but we can predict the gravitational attraction if we did.
Theoretically, no matter how big the celestial object and what percentage was hollowed out, as long as it is spherically symmetrical there is no net gravitational attraction within the shell. Practically, on an engineering side there are real experts in these forums and geographically bass, jb and their ilk on the geology board seem to be massively knowledgeable. But you have to ask the right question!
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Sheesh, I don't know what all the fuss is about. Just take your tinfoil suit, your shades, and a couple of beers and it could be great!
Aaron, the actual point of zero gravity would be infinitesimally small, but the region of microgravity (e.g. in close earth orbit) where you couldn't personally tell the difference, would be quite large, perhaps on the order of tens of miles. Steve
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If you hollowed out nearly the whole of the earth and just left the outermost mile (or whatever) as a thin spherical shell, the gravity anywhere inside it would be zero.
It wouldn't just be a point of zero g in the middle; the whole of the region would have zero g.
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Bored chemist, I don't understand what you say at all. Please explain. It seems to me that because of the inverse square law you would feel a much greater attraction of gravity from the side of the hollow sphere that is very close than from the other side that is 7,000 miles away. Steve
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Bored chemist, I don't understand what you say at all. Please explain. It seems to me that because of the inverse square law you would feel a much greater attraction of gravity from the side of the hollow sphere that is very close than from the other side that is 7,000 miles away. Steve
Steve, you need to check out Shell Theorem. I keep getting it wrong myself [;D]
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Bored chemist, I don't understand what you say at all. Please explain. It seems to me that because of the inverse square law you would feel a much greater attraction of gravity from the side of the hollow sphere that is very close than from the other side that is 7,000 miles away. Steve
If, for example, you are very near the bottom of the shell, it's true that the bottom bit is nearer, but most of the shell is above you. More stuff further away balances out less stuff nearer.
Or, as Geezer says, check out the theorem.
http://en.wikipedia.org/wiki/Shell_theorem#Inside_a_shell
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Thanks guys, you made my day.
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If, for example, you are very near the bottom of the shell
Which bit is the bottom??? [:P]
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If, for example, you are very near the bottom of the shell
Which bit is the bottom??? [:P]
The other end,from the top.
Of course, it doesn't really matter but it makes it easier to refer to the stuff above you and below you rather than trying to keep it abstract.
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If the fact that we are in so called free-fall orbit around the sun would indicate that this area of zero gravity is not effected by the gravity of the moon or sun then;
Why do we have tides on the surface of the planet where gravity is at full strength? Is this not effected by the pull of the moons gravity? If the moon can exert such an influence on the surface of the Earth then would it also do something to the huge area within the Earth that is less effected by gravity?
Professor Brian Cox seems to think and has stated that gravity is responsible for the commencement of the reaction at the centre of stars. Also this extract: "In the nebular hypothesis, the majority of the mass of the dust cloud collects at the center. The intense gravitational forces present ultimately lead to nuclear fusion taking place. As most of the matter initially present in the nebula is hydrogen, the process of hydrogen burning takes place." (University of Winnipeg)
How can the above be correct when at the centre of a large mass such as the Sun, nebulae, there is essentially zero gravity in total contradiction to this theory! So I would hypothesise that nuclear fusion must take place at the point where gravity is the strongest, on the surface?
Finally, maybe if you were able to find your way to the centre of a large interstellar object you would not float at all in zero gravity but your atoms would all get pulled apart by the forces pulling on you in every direction. I imagine this would not be a very pleasant experience, but does it explain nuclear fusion?
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How can the above be correct when at the centre of a large mass such as the Sun, nebulae, there is essentially zero gravity in total contradiction to this theory! So I would hypothesise that nuclear fusion must take place at the point where gravity is the strongest, on the surface?
Ah! You may be missing an important point here. Gravity at the center of a star is zero, but the pressure is enormous. The pressure is caused by gravity pulling all the material towards the center. The pulling force (gravity) reduces near the center, but the enormous mass of material above the center creates the pressure.
The pressure results in fusion, but gravity was still responsible for creating the pressure.
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Thanks geezer, so it is the pressure combined with heat that causes fusion and this is a result of the forces of gravity bringing the mass together in the first place. Ahhhh now it makes sense.
What about the tide thing and the Moon with relation to the zero gravity area? Any suggestions......
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Aaron - Newtons Shell theorem (the one that tells us, amongst other things, that gravity within the above mentioned shell is zero) does specify that the material of the shell is symmetrical and spherical for zero gravitational attraction to apply. The presence of the moon and sun (and to a much much lesser extent) the other planets do disrupt this symmetry. I believe, but I am not absolutely sure, that the effect of the moon could be worked out using simple F = G.m1.m2/r2 - and I dont think it would be very much at all.
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I cannot help notice that everyone is using empty shells to describe the gravity on earth. Why is that please?
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Well, life's really just a bag o'shells in the end, ain't it?
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Think of a 'point of gravity'. Then decide how you think it 'radiates'. I know it doesn't relatively seen, but the shell theorem is a Newtonian approach, and quite useful.
When I think of it like that I get all those arrows pointing equally 'out in space' from every point of our original point :)
And their 'magnitude/strength' of 'gravity' should (in a empty space) be equal from any position on that 'point'. A little like our common Erinaceus europaeus (hedgehog) turning itself into a bristling ball of spikes. Then you just need to mark up each of those arrows in equal measures, and finally, at every point, 'roof them over'.
Now you got the 'shells', well, as I see it :)
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And yes, the arrows should in reality point 'inwards' towards that 'point', not 'outwards'. But when you're inside, having the shells surrounding you, then the arrows becomes (at least) twofold as I see it.
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yor_on, thanks for the explanation, I was thinking that as gravity is variable within a solid sphere, like the earth, that a shell would fail to describe this. I suppose if it is uniformly variable then this shell theory would be adequate, I'm not sure so please can you help me?
When I imagine all these arrows originating from a single point I take that point to be the centre of the earth. So as far as I can envisage there would be no arrows to start with and they would slowly get bigger and bigger as they make there way towards to surface of the earth reaching maximum strength on the surface and then slowly getting smaller and smaller as they head off into space. If you could average out the size of the arrows then you would eventually get a shell I suppose although how thick this shell is would be relative to the mass of the subject?
Geezer, Is the universe just a bag of shells too? [:P]
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Hmm, I think we are seeing it differently :) I was just using it as an idea for how to construct different shells, building up a mass. And there I used the arrows for getting my poles to construct my shells on. You shouldn't take it as the theorem itself. Newton first decided to treat Earth as an evenly distributed mass, and that decision he made looking at the direction of falling objects. They all seemed to point to the middle of the Earth, which then gave him the direction to work from.
Then he thought of earth as one thin shell only, and decided that, if you looked at a 'moon' close to this thin Earth-shell, the Earth's gravitational attraction on the moon had to be the sum of all vectors that could bear on it. Right under the moon the Earth-moon attraction would be strongest, and the further out to the 'rim' of our shell-Earth you went, as seen from that moon (in 2-D, sort of), the weaker that attraction should be.
Thinking like that he created an equation stating that the sum of all those vectors had to be equal to if this shell-earth had all its mass placed at/in the center of itself. And if that moon had been inside the shell all those vectors would cancel themselves out leaving no measurable 'gravity'. And that's because the shells 'gravity vectors' would now work on that moon in the middle from all 'sides/points' of that Earth-shell, if we think of how he defined those first vectors.
I was wrong thinking of it as a twofold arrow. To him all mass of the Earth could be seen as laying in that center, meaning that if you dug a shaft into the earth you could ignore the mass above you as you went down. After all, according to his idea the mass would behave as if it was concentrated at that center of the earth.
Now we take my idea of all those shells again, and define each one as if all their 'mass' all was placed in the center. Phieew, I should have looked at it before telling you how I thought of it.. I actually though I knew it :) But its a long time since I looked at that one.
But you can look at it this way too, when you dig your way down the 'forces' of the gravitation coming from the sides, acting at you, take themselves out as they meet at your position, so the the force left is the one pointing down, possibly?
I'm not even sure if that's correct actually :) But it seems like it would fit this idea.
He was one very smart guy.
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But I think you can use the arrows and a twofold direction too? You just need to see where they meet the moon, there you will have the 'gravity arrows' working both directions 'attracting'. That is, if you imagine that first 'point' from where they go out to be the center of the earth. I'm not sure but I think it would be the same, as if you think of the arrows hitting the moon and then move the moon closer and closer to that point until they merge. In the merge there can't be any arrows pointing anywhere.
Does that make any sense? You would need to have a decided equilibrium between the moon and the Earth though, and then move that as a lever, reaching a maximum on the surface to then thin off, getting into a 'negative phase' as it continues past? Nah, it doesn't make sense. How would one get to that idea from my arrows?
Hah, Maybe you could, that is if you defined gravity as needing something to act on.. Then you could suspect that as the objects merged there wouldn't be anything to ... Naaah :)
But your concept of the arrows getting thicker as they reach the surface to then get thinner as gravity weaken again makes sense. The only problem being that we know it does so :)
Newton reasoned his way without knowing.
I'm getting jealous here :)
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Geezer, Is the universe just a bag of shells too? [:P]
Sigh!
Probably.
It's a play on the words, "Life is but a bagatel". In other words, "it's a crap shoot".
So, probability has a lot to do with things on Earth, and also in the Universe.