0 Members and 1 Guest are viewing this topic.

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]

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.

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]

The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.

But how does that disprove the shell theorem?

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.

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.

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 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.

Quote from: gem on 29/04/2010 20:40:56This 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.

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 forceAnd this will cause inverse square law violation.

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?

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.

Quote from: gem on 30/04/2010 08:27:08so 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 forceAnd 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.

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.

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,

If Newton et al.

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)?

If Newton et al. got it wrong then it would also seem that the currently accepted mass of the Earth is also wrong.

[/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.

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/s^{2}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/s^{2}).