Naked Science Forum
On the Lighter Side => New Theories => Topic started by: gem on 16/04/2010 22:54:13
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Newton did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass.
When showing that a spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center
see link
http://en.wikipedia.org/wiki/Shell_theorem (http://en.wikipedia.org/wiki/Shell_theorem)
Quote from the link
However, since there is partial cancellation due to the vector nature of the force, the leftover component.
So you can not just take the value of gravitational attraction at earths surface and apply inverse square law out in to space.
And for the same reasons you can not derive an accurate value for the weight of the earth from the Michell/Cavendish experiment or a accurate value for the gravitational constant which Einstein included in his calculations
Now i understand that these are absolute fundamentals of physics so it should not be to difficult for any one with relevant training to show the error in my assertions.
for more information see
http://www.thenakedscientists.com/forum/index.php?topic=28473.0 (http://www.thenakedscientists.com/forum/index.php?topic=28473.0)
The above post has gone unchallenged for quite a while, maybe you didn't notice some one was taking a hammer and chisel to the foundations. [;)]
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So you can not just take the value of gravitational attraction at earths surface and apply inverse square law out in to space.
The above seems to be the issue you're trying to raise, but that issue seems to be incomplete to me as it's not clear where you're locating the effective point source origin of the force that should follow the inverse square law.
It would indeed be incorrect to use the inverse square law from an origin on the surface of the Earth, but is that what you're actually getting at? The measured value at the Earth's surface does seem to follow the inverse square law from an imaginary point origin at the Earth's center.
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So you can not just take the value of gravitational attraction at earths surface and apply inverse square law out in to space.
Newton doesn't do this. You may want to learn Newtonian mechanics.
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So you can not just take the value of gravitational attraction at earths surface and apply inverse square law out in to space.
The above seems to be the issue you're trying to raise, but that issue seems to be incomplete to me as it's not clear where you're locating the effective point source origin of the force that should follow the inverse square law.
There is not one point source every particle attracts every other particle from where it actually is in reality.
And the total vector sum of these forces mean that a body's acceleration turns out to be the same as if it had been acted on by a single force towards earths centre.
However as i said previously
Newton did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass.
It would indeed be incorrect to use the inverse square law from an origin on the surface of the Earth, but is that what you're actually getting at? The measured value at the Earth's surface does seem to follow the inverse square law from an imaginary point origin at the Earth's center.
I know that newtons theory takes inverse square law from the centre of the earth and it is the value attraction at earths surface that is used, sorry for not making that clearer.
So to return to newtons error, to give a basic example.
If you had a standard 1 kg mass on a frictionless surface and applied a force of 3 newtons in one direction and also applied a force of 4 newtons at 90 degrees to the other force on the body you would have a resulting acceleration of 5 m/s squared just like a 3-4-5 triangle.
The total force that is applied to the 1 kg mass is however 7 newtons and if that force was applied at a angle of anything less than 90 degrees the amount of acceleration of the mass would be greater as the amount of cancellation of the vector sum would be less.
So if you then consider a body on the earths surface the mass that is attracting it to the centre it is mostly attracting at a vector angle, there is only a very small amount directly in line with a body's direction of acceleration.
The question that then arises is what is the total force that is acting on a body at earths surface as we already know what the resulting acceleration force is, so at what angle is the average cone of attraction and how much cancellation in the vector sum at earths surface.
And when you put space between a body and earths mass lets say one earths radius from earths surface so two radius's from the centre newtons method would bring the force of attraction to one quarter of what it is at the surface with no allowance for the change in the slight lessening of the average angle of attraction.
But when you then bring the physical reality's in that earth is not a homogeneous sphere and you have to allow for the fact half of earths volume is within the last 1315 kilometres of earths radius but the earths core is thought to be a lot denser than the rest of the earth, So lots of variable confusing the issues.
And when you then consider that the total force will only apply close to a its total figure many earth radius's away when the angle becomes much more acute.
As it will have been diluted down hugely by the inverse square of the distance, so may be not surprising it is difficult detect given all the other dynamics of attraction going on in space.
At the core of what i am pointing out are very basic fundamentals of physics although given that gravity is a very weak force i understand how difficult this total force will be to detect.
However D longs paper about experiments to measure gravitational attraction from which the gravitational constant is derived
http://www.springerlink.com/content/u5j4724173336606/ (http://www.springerlink.com/content/u5j4724173336606/)
is about the fact that the majority of results It seems from experimental results show inverse square law violation also.
It appears when looking at the data of experiments made since 1894 they show a dependence on mass separation.
And i say again Einstein included the gravitational constant in his calculations and the fact that G [big G ] is giving results closer to what is observed is due to the fact it came to a higher value for gravitational attraction by accident given the fact these experiments are using high density mass so getting more concentrated distance for the inverse square and more acute angles of attraction.
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Ah right, that's where you're going wrong...
Newton did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass.
...because that's precisely what he did do.
If you had a standard 1 kg mass on a frictionless surface and applied a force of 3 newtons in one direction and also applied a force of 4 newtons at 90 degrees to the other force on the body you would have a resulting acceleration of 5 m/s squared just like a 3-4-5 triangle.
Umm... well you won't actually, because the body is on a surface. You would only get an acceleration corresponding to the sum of the force vectors if the body was in free space.
The total force that is applied to the 1 kg mass is however 7 newtons and if that force was applied at a angle of anything less than 90 degrees the amount of acceleration of the mass would be greater as the amount of cancellation of the vector sum would be less.
...but force is a vector quantity; they have both a value and a direction, and if the directions of the two forces are not the same then they can't simply be added. As you've already shown, the sum of the vector forces is 5N and not 7N.
So if you then consider a body on the earths surface the mass that is attracting it to the centre it is mostly attracting at a vector angle, there is only a very small amount directly in line with a body's direction of acceleration.
Yes, and because Newton realised this, and because he went on to calculate exactly how the forces would sum, that we regard him as being so clever.
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If you had a standard 1 kg mass on a frictionless surface and applied a force of 3 newtons in one direction and also applied a force of 4 newtons at 90 degrees to the other force on the body you would have a resulting acceleration of 5 m/s squared just like a 3-4-5 triangle.
Umm... well you won't actually, because the body is on a surface. You would only get an acceleration corresponding to the sum of the force vectors if the body was in free space.
I think you will find my statement is correct.
So if you then consider a body on the earths surface the mass that is attracting it to the centre it is mostly attracting at a vector angle, there is only a very small amount directly in line with a body's direction of acceleration.
Yes, and because Newton realised this, and because he went on to calculate exactly how the forces would sum, that we regard him as being so clever.
I am not disputing newtons genius,
and what he did do was show that a spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center
I am saying he did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass.
So therefore giving different sum total results [a variation in the values of acceleration] than would be predicted by applying inverse square law to the value of attraction at earths surface, as at different distances the angle of attraction changes.
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I am saying he did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass.
So therefore giving different sum total results [a variation in the values of acceleration] than would be predicted by applying inverse square law to the value of attraction at earths surface, as at different distances the angle of attraction changes.
Are you, in effect, saying that to determine the net force you really need to sum the indiviual gravitational forces between all the atoms in the two bodies (taking into account their vectors), and if you did that, you come up with a slightly different law?
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Are you, in effect, saying that to determine the net force you really need to sum the individual gravitational forces between all the atoms in the two bodies (taking into account their vectors), and if you did that, you come up with a slightly different law?
No that would be hugely complex if not impossible,newton already overcame that problem with shell theorem which gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body.
http://en.wikipedia.org/wiki/Shell_theorem (http://en.wikipedia.org/wiki/Shell_theorem)
What i am saying is you have to factor in the value of the partial cancellation due to the vector nature of the force, meaning you need to know what the gross value of the attraction is.
And then allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass, along side using inverse square law.
As i stated earlier
Newton did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass.
So you can not just take the value of gravitational attraction at earths surface and apply inverse square law out in to space.
because when you get any great distance from a planet you are going to get values of gravitational attraction greater than newtons laws predict.
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No that would be hugely complex if not impossible,newton already overcame that problem with shell theorem which gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body.
http://en.wikipedia.org/wiki/Shell_theorem (http://en.wikipedia.org/wiki/Shell_theorem)
What i am saying is you have to factor in the value of the partial cancellation due to the vector nature of the force, meaning you need to know what the gross value of the attraction is.
I don't understand what you mean by "partial cancellelation". I presume by "vector nature" you mean that the force is the summation of many forces that are not coincident, but maybe I'm not getting that right either.
BTW, I'm no hot shot at calculus, but I would think it should be possible to calculate the resultant force from the individual forces. We'd probably have to approximate the densities as uniform though.
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because when you get any great distance from a planet you are going to get values of gravitational attraction greater than newtons laws predict.
Can you show the formula(s) you use? The Shell theorem can be derived from vector calculus (for a uniform density sphere) and already takes all the vectors into account. How can your calculation differ if they're doing the same thing?
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Can you show the formula(s) you use? The Shell theorem can be derived from vector calculus (for a uniform density sphere) and already takes all the vectors into account. How can your calculation differ if they're doing the same thing?
I do not disagree with Shell theorem proving A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center at various distances in space.
I am saying you can not tie those various points in space from the centre of a mass to inverse square law because
Newton did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass.
I don't understand what you mean by "partial cancellation". I presume by "vector nature" you mean that the force is the summation of many forces that are not coincident, but maybe I'm not getting that right either.
Yes that's what i mean,and on the partial cancellation in the example i gave earlier in the 3-4-5 triangle.
If you had a standard 1 kg mass on a frictionless surface and applied a force of 3 newtons in one direction and also applied a force of 4 newtons at 90 degrees to the other force on the body you would have a resulting acceleration of 5 m/s squared just like a 3-4-5 triangle.
The total force that is applied to the 1 kg mass is however 7 newtons so a partial cancellation of 2 newtons
and if that force was applied at a angle of anything less than 90 degrees the amount of acceleration of the mass would be greater as the amount of partial cancellation of the vector sum would be less.
And you will note the further you travel from the mass the more acute gets the angle of interaction.
Can you show the formula(s) you use?
I am only using simple vector analysis and basic geometry and as stated earlier i only differ in the need to factor in the value of the variation of the partial cancellation due to the vector nature of the force, meaning you need to know what the gross value of the attraction is.
And also another aspect of this is as the distance between the mass of the two bodies changes only the particles that are in a direct line with the centre to centre actually travel the distance prescribed by inverse square law relative to each other.
see the diagram i posted on this post on the same subject with a less controversial heading
http://www.thenakedscientists.com/forum/index.php?topic=28473.0 (http://www.thenakedscientists.com/forum/index.php?topic=28473.0)
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If you had a standard 1 kg mass on a frictionless surface and applied a force of 3 newtons in one direction and also applied a force of 4 newtons at 90 degrees to the other force on the body you would have a resulting acceleration of 5 m/s squared just like a 3-4-5 triangle.
The total force that is applied to the 1 kg mass is however 7 newtons so a partial cancellation of 2 newtons
and if that force was applied at a angle of anything less than 90 degrees the amount of acceleration of the mass would be greater as the amount of partial cancellation of the vector sum would be less.
Because the 3 N and 4 N forces are at right angles to each other, there is no component of either force that can be simply added. It would only be legitimate to simply add them together and say the total force is 7 if both forces acted in the same direction.
Consequently, the resultant force of the two forces is 5 N acting in a direction along the line of the hypotenuse of your triangle. There is no "cancellation". You still have two forces, or a resultant force that is the combined effect of both forces.
Also, I don't think you will be able to determine the correct solution without using calculus to integrate the forces, and I rather suspect Newton did that, seeing as he invented calculus to solve problems like this.
As I mentioned earlier, you really are trying to account for all the individual gravitational attractions between all the atoms of the two bodies, because that is the gravitational model. You then have to figure out a mathematical way of doing that. I may be wrong, but I suspect that's exactly what Newton did.
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Also, I don't think you will be able to determine the correct solution without using calculus to integrate the forces, and I rather suspect Newton did that, seeing as he invented calculus to solve problems like this.
He actually did not use calculus in his master work on gravity, because calculus was somewhat contentious at the time. What he did was use really elegant geometrical proofs that made use of vanishing triangles.
As I mentioned earlier, you really are trying to account for all the individual gravitational attractions between all the atoms of the two bodies, because that is the gravitational model. You then have to figure out a mathematical way of doing that. I may be wrong, but I suspect that's exactly what Newton did.
This is exactly what Newton did. His theoretical work is really fantastic and he also made many of his own instruments and carried out excellent experiments. I heartily recommend that those interested in science look at the IB Cohen translation of the Principia that came out about 10 years ago. Great stuff.
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He actually did not use calculus in his master work on gravity, because calculus was somewhat contentious at the time. What he did was use really elegant geometrical proofs that made use of vanishing triangles.
Offhand, it sounds like vanishing triangles may be a bit like fractals.
Edit: Mind you, I would not have put it past him to invent the vanishing triangles idea as an acceptable metaphor for calculus!
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As I mentioned earlier, you really are trying to account for all the individual gravitational attractions between all the atoms of the two bodies, because that is the gravitational model. You then have to figure out a mathematical way of doing that. I may be wrong, but I suspect that's exactly what Newton did.
I don't disagree,this was my reply on the same point to yourself.
newton already overcame that problem with shell theorem which gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body.
And also to JP
I do not disagree with Shell theorem proving A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center at various distances in space.
and on this point
Because the 3 N and 4 N forces are at right angles to each other, there is no component of either force that can be simply added. It would only be legitimate to simply add them together and say the total force is 7 if both forces acted in the same direction.
correct the net result is 5 metres per second squared.
The gross value of the force is 7 newtons.
There is no "cancellation".
Incorrect, this is a extract from wikipedia on newtons shell theorem.
'However, since there is partial cancellation due to the vector nature of the force,'
http://en.wikipedia.org/wiki/Shell_theorem (http://en.wikipedia.org/wiki/Shell_theorem)
If you had 1 newton force acting on a standard 1 kg mass on a frictionless surface in one direction and another 1 newton force acting at 90 degrees to that force you would have a gross force of two newtons and a resulting net acceleration of 1.4 metres per second squared.
if you changed the angle of the forces to 45 degrees you would have 2 newtons of gross force still, and a resulting net acceleration of 1.8 metres per second squared.
And if you changed the angle to 22.5 degrees you would have 2 newtons of gross force and a resulting net acceleration of 1.95 metres per second squared.
I respectfully suggest people need to take a look at the average vector angle at earths surface where we take the value that we use for the inverse square law and consider how that angle changes the further the distance in to space
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If you had 1 newton force acting on a standard 1 kg mass on a frictionless surface in one direction and another 1 newton force acting at 90 degrees to that force you would have a gross force of two newtons and a resulting net acceleration of 1.4 metres per second squared.
I do not agree. I think you will have a resulting acceleration of 1 m/s^2 in a direction parallel with the frictionless surface.
You would have no acceleration at 90 degrees to the frictionless surface.
Even if the force acting at 90 degrees to the frictionless surface is a billion Newtons, the acceleration of the 1 kg mass will still be 1 m/s^2 parallel with the frictionless surface.
I respectfully suggest you take a look at mechanics and dynamics.
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I do not agree. I think you will have a resulting acceleration of 1 m/s^2 in a direction parallel with the frictionless surface.
You would have no acceleration at 90 degrees to the frictionless surface.
All the forces i have described in the example are parallel with the frictionless surface,the change in angle is relative to the standard 1 kg mass.
To give a example of the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different angles
because as i said before
Newton did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass.
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All the forces i have described in the example are parallel with the frictionless surface.
Well in that case you are applying a force of 2N to the 1kg in a direction parallel with the frictionless surface, and there is no need to discuss vectors at all.
Perhaps a diagram would help?
BTW, you can underline all you want, but you really need to be able to prove your point. Underlining does not constitute proof (unless you believe in Proof by Loud Assertion)
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Well in that case you are applying a force of 2N to the 1kg in a direction parallel with the frictionless surface, and there is no need to discuss vectors at all.
If two or more forces act on the same body,on a frictionless surface and these forces are not parallel with each other then the acceleration of the body is written as the vector sum.
And the gravitational acceleration that we experience on earths surface is the vector sum of the attractive force of earths mass
Since there is a partial cancellation of vector forces it then follows there is a partial cancellation in the forces that give earths gravitational resultant force expressed as the acceleration.
And there will be variation in the amount of cancellation due to changes in angle that earths mass attracts a body at different distances from earths surface
The partial cancellation will be greatest at earths surface diminishing the further out in to space, where as the cancellation becomes total within a shell of homogeneous mass.
Because the net gravitational forces acting on a point mass from the mass elements of the shell totally cancel out.
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But you just told me they were parallel with the frictionless surface. Now you've got me totally confused.
If you sketch a diagram of your model, scan it, and attach it to a post it would help a lot. You can also draw freehand with you mouse under "Create New Diagram", but it's a bit tricky.
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The partial cancellation will be greatest at earths surface diminishing the further out in to space, where as the cancellation becomes total within a shell of homogeneous mass.
Because the net gravitational forces acting on a point mass from the mass elements of the shell totally cancel out.
You're right about vector sums. However, as the posters here have been trying to tell you, this is exactly what Newton did, and this is exactly what the Shell theorem does.
You can show this with brute force calculus, accounting for vector addition of each tiny component of the total force: http://en.wikipedia.org/wiki/Shell_theorem . (Note the sentence beginning "However, since there is partial cancellation due to the vector nature of the force, the leftover component (in the direction pointing toward m) is given by. . .")
You can also do this in a much shorter way with a mathematical trick in vector calculus called Gauss' law: http://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity
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(Note the sentence beginning "However, since there is partial cancellation due to the vector nature of the force, the leftover component (in the direction pointing toward m) is given by. . .")
Yes thanks for that JP i quoted that very extract in my first post on this thread, but have got side tracked defending that there is partial cancellation and showing the amount of that partial cancellation varies at different angles of interaction.
The partial cancellation will be greatest at earths surface diminishing the further out in to space,
You're right about vector sums.
So from that can i read that you agree that the partial cancellation will be greatest at earths surface diminishing the further out in to space,?
However, as the posters here have been trying to tell you, this is exactly what Newton did, and this is exactly what the Shell theorem does.
Yes i believe it does this was my reply on the same point to yourself.
I do not disagree with Shell theorem proving A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center at various distances in space.
but with the caveat that if there is a variation in the partial cancellation due to the vector nature of the force, it therefore follows that there is a variation in the leftover component (in the direction pointing toward m).
And the leftover component is what we measure as gravitational acceleration at earths surface.
So it seems you cannot just take the value of g at earths surface and apply inverse square law out in to space because you are putting a fixed value on to something that actually varies at different angles of interaction.
Making shell theorem results specific to each point that they are calculated only.
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So Newton was right then?
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If what i have postulated here as regards the variation in the partial cancellation due to the vector nature of the force, and that there is also a variation in the leftover component (in the direction pointing toward m),is correct.
And as the leftover component is what we measure as gravitational acceleration at earths surface.
It means that when newton took the value of g at earths surface where the partial cancellation will be greatest and said that you could then apply inverse square law to that value from earths centre, is the error.
Because the physical reality's will not allow it to be so.IE for that value to diminish in accordance with inverse square law.
And i believe this is what they have been detecting when studying the results of the experiments done since Cavendish did his first experiment to present day.
Below is an extract from a paper from this area of science
(see http://physics.uci.edu/gravity). (http://(see http://physics.uci.edu/gravity).)
1 Introduction
In 1974 Daniel Long published a paper Why do we believe Newtonian gravitation at laboratory
dimensions? (Long 1974), comparing measurements of G made since 1894 with
various mass separations r. His plot of G values as a function of r strongly suggested a
dependence on mass separation. Two years later, Long reported an experiment of his own
(Long 1976) which used a torsion balance to compare the forces produced by source masses
at distances of 4.5 cm and 29.9 cm. Long’s experiment used ring-shaped source masses,
exploiting the fact that the force on a test mass at a certain point on the axis of a ring source
mass is at an extremum and thus is quite insensitive to error in its position relative to the
ring. Daniel Long reported that the ratio of the torque produced by the more distant ring to
that produced by the nearer ring exceeded the Newtonian prediction by (0.37 ± 0.07)%, a
result consistent with the distance dependence found in his analysis of G measurements.
So Newton was right then?
That Geezer is THE question
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Gem,
If you can provide a simple vector diagram that shows how the neglected forces work, it would be enormously helpful. If I understand what you have been saying at all, it should not be very complicated.
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[diagram=585_0][diagram=586_0]
OK Geezer in diagram 1 we have a overhead view of two one newton forces acting on a one KG weight on a frictionless surface at 90 degrees to each other giving a net resulting acceleration of 1.4 metres per second squared so 70 percent of the gross input potential.
And in diagram 2 we have the same 1 newton forces acting on the 1 KG weight at a angle of 45 degrees to each other giving a net resulting acceleration of 1.8 metres per second squared so 90 percent of the gross input potential.
[diagram=587_0]
And in diagram 3 we have a example of the angle of interaction getting more acute and any mass that is not in line with the centre line C/L Note, As the distance between the mass of the two bodies changes only the particles that are in a direct line with the centre to centre actually travel the distance prescribed by inverse square law relative to each others centres.
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Ah! Nice pix. Thanks.
From your previous description I thought only one of the forces acted along a frictionless surface. I see what you were getting at now.
I'll get back to you.
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And in diagram 3 we have a example of the angle of interaction getting more acute and any mass that is not in line with the centre line C/L Note, As the distance between the mass of the two bodies changes only the particles that are in a direct line with the centre to centre actually travel the distance prescribed by inverse square law relative to each others centres.
This is all gravitational force? The forces between points of mass all still follow a r-2 law, which I think we're agreeing on. If all your masses are tiny, then the r you use is the distance between the point masses. If the masses are large, then you have to sum over all the tiny elements making each of them up, making sure to add the vectors appropriately as you indicated.
Do you disagree with any of this?
I think what people (and Newton) were saying is that the addition of all these vectors greatly simplifies if one mass is really tiny and the other is a uniform sphere, since all this vector addition reduces to treating all the large-sphere mass as if its concentrated at the center of the sphere. The vector addition isn't wrong, but you can do the vector addition in a much simpler way.
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I think what Gem is getting at is that the angles between all the individual gravitational forces that sum to the total force vector vary with the distance between the two objects. At a great distance, the angles would hardly matter at all. As the distance diminishes, the angles have a greater effect.
I'm pretty sure Newton's model takes that into account, but I'm not that familiar with how he actually did it.
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Newton's model tells you to sum* gravitational forces as vectors. It's only in very particular cases that the symmetry of things lets you eliminate the vector sums. (You remove them from the sums by making symmetry arguments to account for the vector nature of the forces rather than actually having to sum over the vector components.)
Although I haven't read Newton's original work, so this is all based on how they teach Newtonian gravity these days. I think Newton did something similar.
*In most problems you're dealing with continuous objects, so instead of summing, you integrate.
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*In most problems you're dealing with continuous objects, so instead of summing, you integrate.
Well, yes. But isn't integration really just a way of adding all the little bits together without having to do the actual adding? [:D]
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*In most problems you're dealing with continuous objects, so instead of summing, you integrate.
Well, yes. But isn't integration really just a way of adding all the little bits together without having to do the actual adding? [:D]
Sssshhhh! If you give away the secret that we're actually just adding, calculus won't seem so fancy anymore!
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Wait a minute! The mists are finally beginning to clear.
If you want to calculate the gravitational force at the surface of the Earth, you have to integrate all the "individual" forces acting on the body using the inverse square law, taking into account the force vectors. If you do that, you'll get a number that is the same as the observed gravitational force. If you assume the forces all act towards the center of the Earth, you'll calculate a value that is substantially greater than the observed value.
Newton never said you can neglect the angles did he? Only when the two objects are sufficiently distant can you use an approximation and assume that all the forces act between the centers of mass of the objects.
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That's what I was getting at. Summing (or integrating) all the individual masses using the r-2 law along with the proper vectors will always get you the right answer. Newton didn't say you could neglect the angles/vectors. You can simplify the vector sums by utilizing symmetry and/or vector calculus tricks.
To determine forces between two extended objects, you have to sum over the forces between all pairs of points on both objects, making it a very complicated equation. It's much easier if we treat one object as a point (which is a good approximation if it's much smaller than the other object). Then we only need to vector sum over the forces generated by all the points of the large object on the small object.
There are a variety of cases in which this simplifies. If the objects are very far apart, then all the vectors point more or less in the same direction and you can just assume they all point towards the center of the large object without much error. If the large object is a uniform sphere, then you know from symmetry that the total force on the small object has to point directly to its center, and you can take care of the vector sum using some mathematical tricks. The result in that case is that mathematically you can treat all the mass of the large object as being concentrated at its center and use a 1/r2 law from there. The vector addition still holds, but because of the symmetry you can do the addition ahead of time and get a simple looking answer. In the uniform sphere case, your objects don't have to be far apart to make the simplification. It's the symmetry of the problem that allows you to do it.
You should get similar simplifications in other symmetric cases, but since planets/stars tend to form into roughly spherical objects, that's the case that often gets cited.
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I can even tell you basically how to do it. Think about a ring of mass and an object lying along the ring's axis, like I show in the figure. You know that two points on opposite sides of the ring (shown in red) each pull towards each other with an inverse square law. When you add the vectors, because of symmetry, the resulting vector points directly towards the center of the ring. Now you can add up all points around the ring. Since each point has a corresponding point on the opposite side, the entire force is directed exactly towards the ring's center. The magnitude of it is pretty easy to calculate as well.
[diagram=588_0]
Next, you can consider a disk of mass by shading in the ring with mass. Since a disk can be sliced into a bunch of rings, and each ring contributes a force pointing along its axis, the entire disk contributes to a force pointing along its axis.
Finally, you consider the whole earth to be made up of a bunch of disks of different sizes, stacked along the axis.
Anyway, if you didn't follow the whole argument, I hope at least the bit about the ring made sense to you. You can do the entire ring calculation and get a force without having to actually do vector addition of each point on the ring. You need to know how two points on opposite sides of the ring sum to give you a force along the ring's axis, and then you simply have to sum around the ring, knowing that your final force has to point along the axis, from symmetry.
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If you want to calculate the gravitational force at the surface of the Earth, you have to integrate all the "individual" forces acting on the body using the inverse square law, taking into account the force vectors. If you do that, you'll get a number that is the same as the observed gravitational force. If you assume the forces all act towards the center of the Earth, you'll calculate a value that is substantially greater than the observed value.
I take it back! According to http://en.wikipedia.org/wiki/Earth%27s_gravity
"Note that this formula only works because of the mathematical fact that the gravity of a uniform spherical body, as measured on or above its surface, is the same as if all its mass were concentrated at a point at its centre."
I suspect that it's because of this
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If the object is sitting at the "North Pole", it will be subjected to equal attractive forces from either side of the "equator". The B force is attenuated by distance whereas the A force is attenuated by angle. I think that the two forces acting along the vertical axis will be equal.
I have left the proof as an exercise for the reader (because the algebra defeated me!)
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This is all gravitational force? The forces between points of mass all still follow a r-2 law, which I think we're agreeing on.
Yes every particle follows inverse square law
Newton never said you can neglect the angles did he? Only when the two objects are sufficiently distant can you use an approximation and assume that all the forces act between the centers of mass of the objects.
There is no allowance for the change in angle of interaction from earths surface to a hundred earths radius's away in the way the strength of earths gravity field is calculated at present.
JP in your diagram,
For the inverse square law to work at different distances away, the force arrows in black have to stay at the same ratio to the vector sum 'arrow in red' for every pair of particles at what ever distance away.
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Well, I figured out why I could not produce a general proof that the two forces in my diagram are equal. It's because they are not! It's more complicated than said.
However, I still think the statement in Wiki is correct, but I'd like to see the proof.
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http://en.wikipedia.org/wiki/Shell_theorem
Derivation is there. They eliminate explicit vector sums by using the symmetry of the sphere to do them. Rather than each force being
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fmath%2F2%2F7%2F1%2F271c7a94be4496a99e2534e2ceae7751.png&hash=c2658a3d6899a5aed6c1be0988232407),
they use
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fmath%2F6%2F3%2F8%2F63898081c163a2fc2c132f68be1bf017.png&hash=647e36df940e13c559a7ee34fafc6b6e),
where the sine accounts for the vector nature of the force and uses the fact that that is the only part of the force (pointing towards the earth's center) that isn't canceled by another force.
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Brilliant! That's a really elegant proof. (Why didn't I think of that? [;D])
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I can even tell you basically how to do it. Think about a ring of mass and an object lying along the ring's axis, like I show in the figure. You know that two points on opposite sides of the ring (shown in red) each pull towards each other with an inverse square law. When you add the vectors, because of symmetry, the resulting vector points directly towards the center of the ring. Now you can add up all points around the ring. Since each point has a corresponding point on the opposite side, the entire force is directed exactly towards the ring's center. The magnitude of it is pretty easy to calculate as well.
[diagram=588_0]
'pretty Easy'
JP in your diagram,
For the inverse square law to work at different distances away, the force arrows in black have to stay at the same ratio to the vector sum 'arrow in red' for every pair of particles at what ever distance away.
The above statement is correct otherwise you would have a variation in the value of where the sine accounts for the vector nature of the force and uses the fact that that is the only part of the force (pointing towards the earth's center) that isn't canceled by another force.
But i will make it even easier could anyone point out any pair of point masses on any of the rings that make up a sphere where the force arrows in black stay at the same ratio to the vector sum 'arrow in red' at what ever distance away.
[other than on the rings axis]
I think that you will find the vector sum arrow in red will increase in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all your calculations.
Making a mockery of applying inverse square law to the value of attraction at the surface of the sphere.
And also can anyone highlight any pairs of point masses on any of the rings that make up a sphere that travel the same distance from the object mass as the centre of the ring. [other than on the rings axis]
Also Making a mockery of the symmetry of the sphere argument as well as applying inverse square law to the value of attraction at the surface of the sphere.
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But i will make it even easier could anyone point out any pair of point masses on any of the rings that make up a sphere where the force arrows in black stay at the same ratio to the vector sum 'arrow in red' at what ever distance away.
[other than on the rings axis]
Of course they don't. The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance. The "ratio" that you speak of will only be the same for points that have the same angle.
But how does that disprove the shell theorem?
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The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.
Absolutely hold that thought
But how does that disprove the shell theorem?
Because of this statement which is physically impossible
JP in your diagram,
For the inverse square law to work at different distances away, the force arrows in black have to stay at the same ratio to the vector sum 'arrow in red' for every pair of particles at what ever distance away.
This is because when newton applied inverse square law to the strength of earths gravitational attraction at earths surface he fixed that value as the one to be used at all distances from the surface.
And if you take a close look at all the pairs of point masses even the ones on the far side of the sphere [which we call earth] they are all contributing to the value of the gravitational force that is measured on earths surface at a angle to the axis,
so the attraction is along the axis of the centre of the rings.
Not from the centre of the sphere, IE each point mass attracts from where it is in reality.
Which means the value of gravitational attraction at earths surface is a vector sum only, [ie a net force] of all the pairs of point masses that make up the rings and disks all the way through the earth.
So if we then bring your statement back in to consideration.
The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.
Remembering that what we are measuring on earths surface is only vector sum so only a proportion of the gross force.
So the error is Newton did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass,when he set the value at earths surface to be the one to be used in all calculations at all distances for the inverse square law.
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This is because when newton applied inverse square law to the strength of earths gravitational attraction at earths surface he fixed that value as the one to be used at all distances from the surface.
Well, no he didn't. One has only to look at the famous "moon test" of Book III of the Principia to see that the force of gravity on the moon from the Earth is less than that at the surface.
And if you take a close look at all the pairs of point masses even the ones on the far side of the sphere [which we call earth] they are all contributing to the value of the gravitational force that is measured on earths surface at a angle to the axis,
so the attraction is along the axis of the centre of the rings.
As basic geometry tells us it should be.
Not from the centre of the sphere, IE each point mass attracts from where it is in reality.
Except that the centre of the sphere is on the axis of the centre of the rings and, geometrically, it doesn't matter whether or not all the mass is along the axis or concentrated at the centre.
Remembering that what we are measuring on earths surface is only vector sum so only a proportion of the gross force.
So the error is Newton did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass,when he set the value at earths surface to be the one to be used in all calculations at all distances for the inverse square law.
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Gem,
Because of symmetry, the gravitational force acting on a body is going to act along a line from that body towards the center of the Earth. I think (at least I hope) we all agree about that.
So, the only question is, how far along that line does the mass of the Earth appear to act when all the forces are resolved?
Shell theory proves it's half way across the diameter of the Earth along that line.
If our point mass is at the "north pole" that means the sum of all the forces along the axis produced by the upper hemisphere is equal to the sum of all the forces produced by the lower hemisphere.
It's not so difficult to understand why that might be. The forces that the matter in the Earth exert near the body are large because the distance is small, however, as many of them act at a large angle relative to the axis, the resultant force that they produce along the axis is greatly reduced.
On the other hand, the forces coming from the lower hemisphere are reduced because the distance is greater, but the angles are also much smaller, so a much greater component of the force acts along the axis.
The theorem proves that without a doubt. If you want to convince anyone that Newton was wrong, you'll have to explain what is wrong with the math.
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Except that the centre of the sphere is on the axis of the centre of the rings and, geometrically, it doesn't matter whether or not all the mass is along the axis or concentrated at the centre.
Shell theory proves it's half way across the diameter of the Earth along that line.
If you pretend that all of earths mass is at the centre in a point mass you eliminate this effect from being taken in to account
The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.
for every point mass contained within the earth as you leave earths surface.
If our point mass is at the "north pole" that means the sum of all the forces along the axis produced by the upper hemisphere is equal to the sum of all the forces produced by the lower hemisphere.
The theorem proves that without a doubt. If you want to convince anyone that Newton was wrong, you'll have to explain what is wrong with the math.
As you point out it is only the vector sums that equal to the centre ,and as you put space between a point mass and the earth this effect
The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.
Actually continues to happen.
[because all of earths mass is not concentrated at the centre]
so to show what is wrong with the maths.
I think that you will find the vector sum arrow in red in JP's diagram , will increase in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all calculations done for any pair of point Masses contained within the sphere acting on the said point mass object. IE a greater percentage of the potential gross force
And this will cause inverse square law violation.
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This is because when newton applied inverse square law to the strength of earths gravitational attraction at earths surface he fixed that value as the one to be used at all distances from the surface.
Well, no he didn't. One has only to look at the famous "moon test" of Book III of the Principia to see that the force of gravity on the moon from the Earth is less than that at the surface.
Its my understanding Newton had worked backwards from a known and relatively accurate value for the rate of fall of objects on Earth and applied inverse square law from earths centre using that value. are you saying he did not?
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Newton took the measured rate of fall of the moon and applied an inverse square law to it, determining what it's rate of fall would be at the surface of the Earth. Thus he is comparing the strength of terrestrial gravity at the surface of the Earth to the strength of an inverse square force holding the moon in orbit. He is explicitly assuming that the force holding the moon in orbit obeys the inverse square law.
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Gem,
I'm still thoroughly confused as to what you're saying Newton claimed. So let's assume the earth is a uniform sphere.
1) Are you saying that Newton claimed that the earth's gravity measured above the earth's surface is proportional to the inverse square of the distance between the center of the earth and the object?
or
2) Are you saying that what Newton did was to compute the force of gravity of something at the earth's surface and then say that the force of gravity on an object raised off that surface is proportional to that force multiplied by the inverse square distance between the earth's surface and the object?
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so to show what is wrong with the maths.
I think that you will find the vector sum arrow in red in JP's diagram , will increase in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all calculations done for any pair of point Masses contained within the sphere acting on the said point mass object. IE a greater percentage of the potential gross force
And this will cause inverse square law violation.
I think what you are saying here is that as the object gets further from the earth, the angles will reduce, so the component of the force acting between the centers (the attractive force times the cosine of the angle) will increase with distance.
Assuming I got that right, then yes, I agree it will. However, the gravitational attraction will also diminish while the distance increases as the inverse of the square of the distance, so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.
The theorem takes that into account, and, it turns out that, regardless of the distance between the Earth and the object, when the forces are all integrated, they produce the same force that would be produced if all the mass of the Earth was concentrated in one point at its center.
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Gem,
I'm still thoroughly confused as to what you're saying Newton claimed. So let's assume the earth is a uniform sphere.
1) Are you saying that Newton claimed that the earth's gravity measured above the earth's surface is proportional to the inverse square of the distance between the center of the earth and the object?
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I know that newtons theory takes inverse square law from the centre of the earth and it is the value attraction at earths surface that is used, sorry for not making that clearer.
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It's still a bit unclear. The value of attraction at the earth's surface is used for what? It sounds like you're saying that Newton's claim is that the force of gravity, F, on an object is given by
F=Fsurface/r2,
where Fsurface is the force of gravity on that object at the earth's surface and r is the distance from the center of the earth to the object. Is this the equation that you're saying Newton used?
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Geezer your last post is right on the crux of the matter.
so to show what is wrong with the maths.
I think that you will find the vector sum arrow in red in JP's diagram , will increase in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all calculations done for any pair of point Masses contained within the sphere acting on the said point mass object. IE a greater percentage of the potential gross force
And this will cause inverse square law violation.
I think what you are saying here is that as the object gets further from the earth, the angles will reduce, so the component of the force acting between the centers (the attractive force times the cosine of the angle) will increase with distance.
Assuming I got that right, then yes, I agree it will. However, the gravitational attraction will also diminish while the distance increases as the inverse of the square of the distance, so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.
absolutely right.
and, it turns out that, regardless of the distance between the Earth and the object, when the forces are all integrated, they produce the same force that would be produced if all the mass of the Earth was concentrated in one point at its center.
absolutely right again i posted this statement much earlier.
'I do not disagree with Shell theorem proving A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center at various distances in space.'
so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.
The theorem takes that into account,
this is the crux of the matter i don't believe that at various distance the increased effectiveness of the force associated with the reduction in angle has been taken into account.
I believe what has been done is show , regardless of the distance between the Earth and the object, when the forces are all integrated, they produce the same force that would be produced if all the mass of the Earth was concentrated in one point at its center.
And concluded that therefore it is reasonable to apply inverse square law to a measured value of gravitational acceleration.[so this is the error]
Because as i have pointed out earlier the gravitational acceleration we measure is a vector sum only IE a net figure, at the background of that net force there is a gross force that will manifest a greater percentage of its potential force the more acute gets the angle of interaction.
This is will be demonstrated [as in JPs diagram] by the vector sum arrow in red increasing in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all calculations for all point mass pairs calculated from where they are in reality.
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Gem,
Are you saying that the inverse square law only works for point masses but it is in error when the object has real dimensions (like the Earth for example)?
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Gem:
If Newton et al. got it wrong then it would also seem that the currently accepted mass of the Earth is also wrong.
If we treat the Earth as a point mass and work out the acceleration at the radius of the Earth's surface we get the same value of acceleration that is actually measured at the Earth's surface...
Using a= -(G*M)/r2
where...
G = the gravitational constant = 6.67428e-11
M = the mass of the Earth (kg) = 5.9736e24
r = Earth's equatorial radius (m) = 6.3781e6
a = the resulting acceleration (towards the center of the Earth)
we get an acceleration of -9.800718 m/s2
which, when the centripetal reduction due the the equatorial rotation is taken into account, is just about what is actually measured (the average acceleration at the equator is surface 9.780327 m/s2).
However, if the accepted mass of the Earth is incorrect then all the stuff we've launched up into orbit wouldn't be where it's supposed to be: geostationary satellites would drift and GPS would be wildly inaccurate.
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If Newton et al.
Don't you mean Al as in Al Einstein?
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Oops! nope - I meant Ethal.
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Gem,
Are you saying that the inverse square law only works for point masses but it is in error when the object has real dimensions (like the Earth for example)?
Yes i believe a unified field theory requires it.
Which then calls into question the validity of big G.
As its name of gravitational constant becomes a bit of a misnomer because with different density's of of mass but the same value of mass within a homogeneous sphere the angles of attraction will vary, [due to variation in the lengths of radius ] causing tests of gravitational constant to vary.
If Newton et al. got it wrong then it would also seem that the currently accepted mass of the Earth is also wrong.
Yes it would seem so, if you look at the average angles of attraction between the large mass spheres in Cavendish's original experiment and the small mass spheres there respective sizes being approximately 12 inch spheres and 2 inch spheres.
Then their angle of attraction between there respective point masses would be more acute than the comparison that was made between the small lead sphere and the earth.
Meaning that there will have been a greater pro rata value of attraction in the experiment causing a slight under value of the density of the earth.
[/ftp]
However, if the accepted mass of the Earth is incorrect then all the stuff we've launched up into orbit wouldn't be where it's supposed to be: geostationary satellites would drift and GPS would be wildly inaccurate.
Geostationary satellites are not used for GPS they are used for communication.
However if your point is one of the altitude that they are supposedly stationary at and what the value of gravitational acceleration is at that distance in space from the earth then your point is a valid one .
But it is not as easy as just applying the equations as there are large variations in the values of attraction that already exceed the effect i am suggesting.
Because while a geostationary orbit should hold a satellite in fixed position above the equator, orbital perturbations, such as by the Moon, IE the earths orbit around its barycentre, and from the fact that the Earth is not an exact sphere cause slow but steady drift away from the geostationary location. Satellites correct for these effects with station-keeping maneuvers
where...
G = the gravitational constant = 6.67428e-11
M = the mass of the Earth (kg) = 5.9736e24
r = Earth's equatorial radius (m) = 6.3781e6
a = the resulting acceleration (towards the center of the Earth)
we get an acceleration of -9.800718 m/s2
which, when the centripetal reduction due the the equatorial rotation is taken into account, is just about what is actually measured (the average acceleration at the equator is surface 9.780327 m/s2).
Yes the variation of measured g with latitude at sea level ranges from about 9.78 m/s at the equator to over 9.83 at the poles, so allowing for the fact that the radius is 21k/m less at the poles you should get a gravitational acceleration force that is approximately 3 millimetres per/second squared less.
However the recorded measured values don't confirm this and this is thought to be down to the fact that the earth is not a homogeneous sphere and gets denser the closer to the center.
Although big G seems to be close to the values at earths surface it is not exact,So not the last word on the matter.
And as i have said earlier in this post it is as the distance from the mass increases that this gross force will manifest itself but it will obviously be diluted by the inverse square of the distance but as a percentage of the force it will not be inconsiderable.
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Gem, could you clearly write out the equation you think Newton erroneously used, indicating what all the constants and variables mean? Then could you point out what needs to change in his equation in order to make it right? It's easy to understand the general idea of what you're saying but extremely hard to nail down mathematical details that would lead to disagreements with Newton's theory based on your explanations with no math to back them up.
As it is, when I try to apply the vector sum techniques you mention, I get something that agrees with Newton.
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Gem,
Are you saying that the inverse square law only works for point masses but it is in error when the object has real dimensions (like the Earth for example)?
Yes i believe a unified field theory requires it.
Gem,
The Shell Thorem proves that a spherical object of uniform density produces exactly the same gravitational force as a point object of the same mass. If the gravitational forces they produce are different as you say, there must be something wrong with the math in Shell Theorem, but you also agree that Shell Theorem is correct.
Apparently the foundations laid down by Newton are still quite secure.
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There is a difference in the following two statements.
The Shell Theorem proves that a spherical object of uniform density produces exactly the same gravitational force as a point object of the same mass.
I think you will find that shell theorem does not actually prove that, what it proves is.
a spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre ,after allowing for the partial cancellation due to the vector nature of the force.
IE the value of the cancellation is not included.
In your statement above there is no cancellation so no variation in cancellation at different distances.
So what you posted here does not get considered.
so to show what is wrong with the maths.
I think that you will find the vector sum arrow in red in JP's diagram , will increase in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all calculations done for any pair of point Masses contained within the sphere acting on the said point mass object. IE a greater percentage of the potential gross force
And this will cause inverse square law violation.
I think what you are saying here is that as the object gets further from the earth, the angles will reduce, so the component of the force acting between the centers (the attractive force times the cosine of the angle) will increase with distance.
Assuming I got that right, then yes, I agree it will. However, the gravitational attraction will also diminish while the distance increases as the inverse of the square of the distance, so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.
So do you still agree it will?.
JP the error i am highlighting is i don't believe it is possible for the gravitational force to diminish according to the inverse square law. so i am not saying there is a error in his equation.
The error is to think it proved that you could apply inverse square law because of it.
Are you aware of any aspect of accounting for a increased value of the force?
so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.
The theorem takes that into account,
Does it ?
And if we all agree that the value of gravitational attraction we measure at the earths surface is a vector sum what percentage is the amount of cancellation prior to that net figure at that point?.
What figure does shell theorem give to it, IE what does shell theorem state the gross force is.
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a spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre ,after allowing for the partial cancellation due to the vector nature of the force.
Gem,
All you are saying is that a body in the Earth's gravitational field is subject to multiple forces. Each force can be resolved into two components. One component can be considered to act between the center of the Earth and the body. The other component can be considered to act at right angles to the first force.
The "cancellation" that you refer to only applies to the components that act at right angles to the direction of the gravitational attraction. Because they act at right angles to the gravitational force, they can have no influence on the gravitational force at all. There is no cancellation of any of the components that produce the gravitational effect.
I suspect the problem here is not so much with any error that Newton (or Einstein for that matter) made, but more to do with your lack of understanding of how to properly add vectors.
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There is no cancellation of any of the components that produce the gravitational effect.
I thought we already covered partial cancellation due to the vector nature of the force,you are contradicting 'JP' 'wikipedia'and it would seem yourself.
I think what you are saying here is that as the object gets further from the earth, the angles will reduce, so the component of the force acting between the centers (the attractive force times the cosine of the angle) will increase with distance.
Assuming I got that right, then yes, I agree it will.
And this
so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.
The theorem takes that into account,
Because forces are vectors, the affect of an individual force upon an object is often canceled or partially canceled by the affect of another force.
The theorem takes that into account,
Could you show where increased effectiveness of the force associated with the reduction in angle is actually accounted for ?.
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Could you show where increased effectiveness of the force associated with the reduction in angle is actually accounted for ?.
Sure. The force is proportional to the cosine of the angle. As the angle decreases, the cosine of the angle increases. You'll see the theorem takes that into account.
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Gem,
Everyone here agrees with you that vector addition is how to properly account for gravitational forces from objects that aren't points. What we disagree with you on is that Newton didn't account for this vector addition. He did. There's a mistake somewhere in your calculations in between your first (correct) statement that the forces are vectors and must be added as such and your (incorrect) conclusion that this shows that Newton made a mistake. Since you're trying to make a mathematical point, we can't do much more than tell you that you've made a mistake somewhere unless you're willing to actually show us your calculations.
By the way, Geezer is pointing out the correct way to do vector addition that agrees with what Newton said.
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Gem,
Perhaps you are thinking that it will require more mass in the case of sphere to achieve the same gravitational force that a point mass would produce because, in the case of a sphere, some of the force produced by its mass is not contributing anything to the gravitational force?
It is quite true that some of the mass of the sphere is producing forces that tend to stretch an object within its gravitational field "sideways" (tending to make the object "fatter" ). That being the case you might conclude that the gravitational force has to be somewhat reduced.
However, realize also that, in the case of the sphere, a great deal of its mass is closer to the object than would be the case for a point mass, and because the force exerted is proportional to the inverse of the distance squared, this conveniently compensates for all the "lost" components of the vectors. Consequently a point mass ends up exerting exactly the same gravitational force that a sphere (of the same mass) would exert.
It's not intuitively obvious (to me anyway) that this would be the case, but the math seems to prove that it is quite true.
As JP says, if there is a flaw in this, the only way to get at it is to find a flaw in the math. I suspect it's not feasible to solve this problem by inspection, although there may well be alternative mathematical methods that could be applied. I think Physbang mentioned that Newton did use an alternative method.
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OK all i know i must be testing your patience, because it seems like i am agreeing with newtons shell theorem on one hand but saying there is an error on the other.
But hang in there and i will try to isolate what i perceive to be the error.
So, shell theorem says that the magnitude of the gravitational force is the same as that of a point mass in the centre of the shell with the same mass taking into account all the vector sums and all the cancellations.
And yes i believe it does, so whats the problem?.
The problem is the gravity force that we measure at the surface of the earth has two aspects magnitude, and a direction.
And because newton showed with shell theorem that the magnitude of the force was equal to a point mass coming from the centre that has been the assumed to be the direction also.
But shell theorem does not prove direction it proves magnitude of the force.
In your diagram the points A and B if they had opposite points in the other half IE a mirror image you would have two cones under your body of mass on earths surface one cone of average interaction for the northern hemisphere and one cone of average interaction for the southern hemisphere.
So the average direction cone of the force will be some where between the two.
And it is this aspect of the direction of the force that will cause variation.
In that at various distance the increased effectiveness of the force associated with the reduction in angle has to be taken into account.
Because at earths surface there is a force field that is equal to the magnitude of all of earths mass equivalent to that of a point mass in the centre.
Because of its direction we are unable to measure all of its force as weight or gravitational acceleration
[ Invalid Attachment ]
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Gem,
I give up.
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Gem,
The shell theorem works by specifically taking the vector nature of the forces into account when adding them. The reason you keep claiming it's wrong is precisely the reason why it's right. We've made many attempts to explain this to you, but you continue to post the same argument over and over again without actually looking at how your argument is precisely what makes the derivation work. I highly recommend that you take some time to try to read and understand a derivation of the shell theorem from the basic principles of adding vectors of the gravitational field. The wikipedia link (http://en.wikipedia.org/wiki/Shell_theorem) is one of the best online sources I've seen for it, although basic physics textbooks generally have a similar derivation. The full derivations aren't necessarily easy since they generally involve calculus, but you can at least see how the vector addition comes in without needing calculus.
I'm going to lock this thread at this point, since it is demonstrably false that the shell theorem (and Newton) are wrong based on the reasoning you're providing, and because the thread is going in circles now. If you have questions about the derivation of the shell theorem, feel free to start another thread to ask them.
Cheers,
JP (moderator)