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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?

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.

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.

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}).

Quote from: Geezer on 01/05/2010 20:24:45Gem,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.

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 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.

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.

There is no cancellation of any of the components that produce the gravitational effect.

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.

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 ?.