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New Theories / Is it publishing ?
« on: 06/10/2014 18:08:26 »
HI all, on this part of the forum it states : Got a new theory on something? Post your hypotheses here... is this equivalent to publishing ?
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I understand that gravity not only pulls you towards the Earth, but it also exerts a force that tends to pull "outwards"Yes it causes partial cancellation of gravitational force.
Why would they get an outward force? I might be missing it, but if every point on your body is being pulled (as a result of the net force) towards the earth's center,
I think it might be different if you don't treat the Earth as a point mass.
Yes last time i looked it didn't appear to be one.
I think it might be different if you don't treat the Earth as a point mass.
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.
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,
The theorem takes that into account,
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 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.
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,
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.
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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/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).
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.
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,
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.
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?
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.
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.
The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.
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.
The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.
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?
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 force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.
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.
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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 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.
So Newton was right then?
(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. . .")
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.
Well in that case you are applying a force of 2N to the 1kg in a direction parallel with the frictionless surface, and there is no need to discuss vectors at all.
I do not agree. I think you will have a resulting acceleration of 1 m/s^2 in a direction parallel with the frictionless surface.
You would have no acceleration at 90 degrees to the frictionless surface.
As I mentioned earlier, you really are trying to account for all the individual gravitational attractions between all the atoms of the two bodies, because that is the gravitational model. You then have to figure out a mathematical way of doing that. I may be wrong, but I suspect that's exactly what Newton did.
newton already overcame that problem with shell theorem which gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body.
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.
Because the 3 N and 4 N forces are at right angles to each other, there is no component of either force that can be simply added. It would only be legitimate to simply add them together and say the total force is 7 if both forces acted in the same direction.
There is no "cancellation".