[gem (OP)]

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.

[Geezer]

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?

[gem (OP)]

 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.

[PhysBang]

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.

[Geezer]

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.

[gem (OP)]

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.

[gem (OP)]

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?

[PhysBang]

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.

[JP]

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?

[Geezer]

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.

[gem (OP)]

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?

[gem (OP)]

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.

[JP]

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=F_surface/r²,

where F_surface 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?

[gem (OP)]

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.

[Geezer]

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)?

[LeeE]

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.

[Geezer]

If Newton et al.

Don't you mean Al as in Al Einstein?

[LeeE]

Oops! nope - I meant Ethal.

[gem (OP)]

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. 

[JP]

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.