Naked Science Forum
General Science => General Science => Topic started by: Alan McDougall on 03/07/2008 10:56:43
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Hello,
What is the gravity half way to the core of planet earth.The mass/ density/ gravity ratio would not differ much or would it?
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Surely it would, because their is less matter underneath you pulling you towards it, like the gravity on larger planets is stronger.
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If you tunnelled down to the barycenter, vertical gravitational forces would cancel out when the moon was aligned...
The barycenter (or barycentre; from the Greek βαρύκεντρον) is the point between two objects where they balance each other. In other words, the center of gravity where two or more celestial bodies orbit each other. When a moon orbits a planet, or a planet orbits a star, both bodies are actually orbiting around a point that lies outside the center of the greater body. For example, the moon does not orbit the exact center of the earth, instead orbiting a point outside the earth's center (but well below the surface of the Earth) where their respective masses balance each other.
http://en.wikipedia.org/wiki/Center_of_mass
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If you tunnelled down to the barycenter, vertical gravitational forces would cancel out when the moon was aligned...
Beneath the barycentre say 1 000 kilometres from the surface average, we need an equation.
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Of course we would have to hypothetically solidify the earth to create an equation. Disregard the effect of the sun and moon gravity.
The solution appears a little more complex than I first thought
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The force of gravity inside a perfect sphere (near enough) would be that due to the mass in the sphere below you. So if you say the earth's radius is 6000km and you tunnelled down 1000km the volume of earth beneath you would be (5/6)^3 of the whole earth. Assuming that the earth had constant density (which it does not) then gravity would be reduced by this factor (x0.58). The fact that the core of the earth is more dense means that this would not be true but this is the principle. Of course if you tunneled down to the centre there would be no gravity.
I am puzzled as to the reference to the barycenter for answering this question.
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The force of gravity inside a perfect sphere (near enough) would be that due to the mass in the sphere below you. So if you say the earth's radius is 6000km and you tunnelled down 1000km the volume of earth beneath you would be (5/6)^3 of the whole earth.
What about the mass that is sideways from you? As you tunnel deeper there will be more mass in a horizontal direction, not just above & below.
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The force of gravity inside a perfect sphere (near enough) would be that due to the mass in the sphere below you. So if you say the earth's radius is 6000km and you tunnelled down 1000km the volume of earth beneath you would be (5/6)^3 of the whole earth. Assuming that the earth had constant density (which it does not) then gravity would be reduced by this factor (x0.58). The fact that the core of the earth is more dense means that this would not be true but this is the principle. Of course if you tunneled down to the centre there would be no gravity.
You have forgotten that in this fictional tunnel the gravity from the portion of the Earth above the tunneler would exert an upward force on the tunneler.
I am puzzled as to the reference to the barycenter for answering this question.
I mentioned the barycenter because Alan McDougall's reference to "half way to the core" made me think he was looking for
a gravitational null point similar to Lagrangian point L1...
http://en.wikipedia.org/wiki/Lagrange_point
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The force of gravity inside a perfect sphere (near enough) would be that due to the mass in the sphere below you. So if you say the earth's radius is 6000km and you tunnelled down 1000km the volume of earth beneath you would be (5/6)^3 of the whole earth. Assuming that the earth had constant density (which it does not) then gravity would be reduced by this factor (x0.58). The fact that the core of the earth is more dense means that this would not be true but this is the principle. Of course if you tunneled down to the centre there would be no gravity.
You have forgotten that in this fictional tunnel the gravity from the portion of the Earth above the tunneler would exert an upward force on the tunneler.
No, he's right. Look for Gauss' Theorem.
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Inside a solid sphere of constant density the gravitational force varies linearly with distance from the center,
becoming zero at the center of mass.
http://en.wikipedia.org/wiki/Shell_theorem
So gravity at the bottom of a 1000km vertical tunnel would be 0.83x surface gravity,
(assuming Earth's radius is 6000Km and Earth's density is uniform, which it isn't)
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For a uniform sphere the force of gravity inside the sphere falls of linearly to zero at the centre. So half way down it would be half of what it is at the surface. This is a bit counterintuitive because there is clearly only one eighth of the mass below your feet. (The volume of a sphere is proportional to the cube of the radius) however you are closer to the centre so the effective force increases as the square of the distance so for half the distance the force is four times greater. So four eighths is one half.
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You need to look into the area of a chord, and then extend that to three dimensions to give you a volume. Assuming constant density throughout the volume (which isn't the case) will get you started.
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SoulSurfer, you are right. I forgot to include the inverse square law and shorter distance to the centre.
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The barycentre of the Earth/Moon system is deep beneath the surface of the earth and so might effect gravity. This is the reason I said we should not consider this in the thread.
Alan
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RD
So gravity at the bottom of a 1000km vertical tunnel would be 0.83x surface gravity,
(assuming Earth's radius is 6000Km and Earth's density is uniform, which it isn't)
Your answer is the best. Gravity does not decease like a lot of people surmise. Many think that half way down the gravity would be half.
If we condensed the mass of the earth into half of its present volume what would the gravity be then? If we had the means to compress the earth into a tiny volume of about 10 centermeters would result in a mini black hole.
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Alan, your answer seems somewhat bizarre given the original question. If you compressed the earth to half its radius then the gravity at the surface would be higher by a factor of 4 because the mass is the same but you are half the distance to the centre. If (and this was stated) the earth was constant density and you tunnel down to half the radius, the mass beneath you is 1/8th but you are half the distance to the centre. This gives half the gravity. I can do the maths (though rather boring to most) that shows that the mass in the shell above you exactly cancels. This is not obvious but is an easy shortcut when dealing with inverse square fields. It is true that the earths core is considerably more dense than the average density so this would not be the right answer (we do not know the density distribution) but I think the spirit of the question was to explore the principles and the physics.
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Grayham,
Yes this is me going of on a walkabout tangent instead of keeping to the topic. Compressing the Earths mass until it became a black hole just popped into my unbridled mind from nowhere.
"Sorry" I will try to keep myself in check
Regards
Alan
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Graham,
Alan, your answer seems somewhat bizarre given the original question. If you compressed the earth to half its radius then the gravity at the surface would be higher by a factor of 4 because the mass is the same but you are half the distance to the centre. If (and this was stated) the earth was constant density and you tunnel down to half the radius, the mass beneath you is 1/8th but you are half the distance to the centre. This gives half the gravity. I can do the maths (though rather boring to most
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I can also I do the maths, but Grayham, what about the huge half radius above you, are you not effected by it? Read SoulSurfers post!!
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The gravitational force anywhere inside a uniform thick spherical shell, or a spherical cavity inside a volume of indefinite size is zero.
This is one of the most counterintuitive facts about gravity.
The reason is that the gravitational forces from all the other elements of the structure balance out and sum to zero.
This means that for a uniform universe of indefinite extent the net gravitational force everywhere is zero but the moment any sort of irregularity appers these irregularities tend to grow whether they are areas of higher or lower density. A gravitating universe is always unstable and dynamic.
Also the only universe in which there can be orbits with a long term metastability is one with an inverse square law of gravity.
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SS,
The gravitational force anywhere inside a uniform thick spherical shell, or a spherical cavity inside a volume of indefinite size is zero.
This is one of the most counterintuitive facts about gravity.
The reason is that the gravitational forces from all the other elements of the structure balance out and sum to zero.
This means that for a uniform universe of indefinite extent the net gravitational force everywhere is zero but the moment any sort of irregularity appers these irregularities tend to grow whether they are areas of higher or lower density. A gravitating universe is always unstable and dynamic.
Also the only universe in which there can be orbits with a long term metastability is one with an inverse square law of gravity
My question was not about a sphere on indefinate size or volume, it is about our 10 000 kilometer Earth sphere of 10 000 kilometers.
I also do not get your point balancing out as zero, this would only happen at the middle of the earths core.
Perhaps, I am just getting old and gray and a little slow on the uptake.
I think Grayham has still come the closest
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Alan, the key point to note is that the field INSIDE a spherical shell of uniform density is everywhere equal to zero. This is not very intuitive but results from the field being an inverse square law. This is true for an electric field from a spherical shell of uniform charge too. It reqires a page or so of calculus to show this. If you accept this then you can see that it would also apply to a concentric set of spherical shells which can represent all the mass above you as you tunnel down. The gravitational effect of all these concentric shells can be ignored leaving just the mass of the sphere beneath you having any effect.
Looking on the web came up with a nunber of sites that show this calculation. Here is one that may explain:
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm
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I see so weight decreases at a constant rate down to zero at the core.
Many thanks guys!!
What if your drilled a hole right through the earth and let a peace of lead fall, would it oscillate betwen the surface points at oppsite sides of the earths circumference (assuming there was a vacuum)
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Alan McDougall
I also do not get your point balancing out as zero, this would only happen at the middle of the earths core.
Inside a hollow spherical shell, wherever you are, the gravitational potential is zero; well known and easy to show the sums. Inside a solid sphere, the bits outside you can be looked at as a set of shells so still zero.
The gravitational force is proportional to the distance from the centre - also very easy to show. This means that a body would oscillate with simple harmonic motion if it were dropped down a hole through the centre. (Only a thought experiment) The time period would be about 90minutes. This also happens to be the time for a low Earth orbit.
As far as the barycentre is concerned, measuring your weight there would give a constant value; at any other point there would be 'tidal' forces due to the Moon and Earth's mutual orbit.
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The gravity at the surface will be 1g and it will be 0g at the centre.
All you need now is somebody that can remember their calculus and you'll have your answer.
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No calculus needed, really.
The force is proportional to the mass of the sphere beneath your feet (which is 4πρr3/3)
ρ is density - we can assume it's fairly uniform.
It's also proportional to 1/r2 because of the inverse square law.
Between the two that makes it proportional to r. QED.
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If the gravitational force was 0 gravity at the core and 1 g at the surface, there would be no reason for the denser materials to be attracted towards the core.
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Sophie, the reason for the calculus is only that it is not obvious that the field is zero everywhere inside a spherical shell. Knowing this means you don't need the calculus, but if you need to prove this, then I think you do need calculus.
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sophiec,
No calculus needed, really.
The force is proportional to the mass of the sphere beneath your feet (which is 4πρr3/3)
ρ is density - we can assume it's fairly uniform.
It's also proportional to 1/r2 because of the inverse square law.
Between the two that makes it proportional to r. QED
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You answer is the the best and easy to comprehend , but the others also valid
Alan
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Sophie, the reason for the calculus is only that it is not obvious that the field is zero everywhere inside a spherical shell. Knowing this means you don't need the calculus, but if you need to prove this, then I think you do need calculus.
I agree.
There is a fairly good 'arm waving' argument. Rather than calculating the force on you (sitting in the middle, somewhere), you think of the force on the surface of the shell (equal and opposite - so it's valid) and equate the gravitational, inverse square law, to the effect of ISL on illumination from a light source. Two cones with the same solid angle, pointing in opposite directions from you, will ' illuminate' the shell with an equal amount of energy. The light is spread out on the distant bits and less spread out in the nearer bits. The total light flux hitting each end of the 'double cone' will be the same, though, and, by analogy, the total gravitational forces on the two areas formed by the two cones of light will also be the equal and opposite. Forces on the shell and, therefore, also on you, will be balanced in any direction. It's easiest to visualise if you are on a diameter but it works for any position - you don't need to consider the actual shape which the light beams form on the inside of the shell.
(I made this up on my own and I'm quite pleased with it!)
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Oh yes, Sophie. That's very good and why it works is because a spreading flux and the inverse square law are effectively the same in Euclidean space; hmm, may be the same in other spaces but I am not even going to think about that one. I wonder how it would do as a physics undergrad answer?
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If, as has been shown, the force is proportional to the radius then an object dropped into this "idealised" hole (no moon, full of vacuum- the earth being solid with equal density all the way down) would exhibit simple harmonic motion.
The maths for that is well known so I will leave it as an exercise for the reader.
And, Sophie, That's a seriously good proof.
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Lovely when you don't have to do anything other than wave your arms about.
I gave it some further thought and , first of all, it seemed that the argument would apply for any shape of shell. Then I realised that the shell has to be spherical if the density is uniform - for
F = Gm1m2/r2
to apply.
I have written this elsewhere (and there was a TNS Podcast, too) but the period would be the same (90 minutes) for small, spherical asteroid and a pea sized object moving through a hole through the centre as long as the average density was the same as the Earth's. What an experiment to try, one day.
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Go to my other thread on gravity.It relates to a hollow shell like earth etc