[gem (OP)]

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

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

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. []

[LeeE]

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.

[PhysBang]

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.

[gem (OP)]

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/
 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.

[LeeE]

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.

[gem (OP)]

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.
 

[Geezer]

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?

[gem (OP)]

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

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. 

[Geezer]

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

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.

[JP]

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?

[gem (OP)]

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

[Geezer]

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.

[PhysBang]

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.

[Geezer]

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!

[gem (OP)]

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

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

[Geezer]

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.

[gem (OP)]

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.

[Geezer]

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)

[gem (OP)]

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.

[Geezer]

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.