[JP]

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

[gem (OP)]

(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.

[Geezer]

So Newton was right then?

[gem (OP)]

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).

   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

[Geezer]

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.

[gem (OP)]

[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.

[Geezer]

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.

[JP]

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.

[Geezer]

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.

[JP]

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.

[Geezer]

*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? []

[JP]

*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? []

Sssshhhh!  If you give away the secret that we're actually just adding, calculus won't seem so fancy anymore!

[Geezer]

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.

[JP]

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.

[JP]

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.

[Geezer]

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

 [ Invalid Attachment ]

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!)

[gem (OP)]

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.

[Geezer]

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.

[JP]

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
,
they use
,
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.

[Geezer]

Brilliant! That's a really elegant proof. (Why didn't I think of that?  [])