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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.
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)?
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.
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.
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 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,
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 ?.