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So you can not just take the value of gravitational attraction at earths surface and apply inverse square law out in to space.

Quote from: gem on 16/04/2010 22:54:13So you can not just take the value of gravitational attraction at earths surface and apply inverse square law out in to space.The above seems to be the issue you're trying to raise, but that issue seems to be incomplete to me as it's not clear where you're locating the effective point source origin of the force that should follow the inverse square law.

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.

It would indeed be incorrect to use the inverse square law from an origin on the surface of the Earth, but is that what you're actually getting at? The measured value at the Earth's surface does seem to follow the inverse square law from an imaginary point origin at the Earth's center.

If you had a standard 1 kg mass on a frictionless surface and applied a force of 3 newtons in one direction and also applied a force of 4 newtons at 90 degrees to the other force on the body you would have a resulting acceleration of 5 m/s squared just like a 3-4-5 triangle.

The total force that is applied to the 1 kg mass is however 7 newtons and if that force was applied at a angle of anything less than 90 degrees the amount of acceleration of the mass would be greater as the amount of cancellation of the vector sum would be less.

So if you then consider a body on the earths surface the mass that is attracting it to the centre it is mostly attracting at a vector angle, there is only a very small amount directly in line with a body's direction of acceleration.

Quote from: gem on 18/04/2010 11:17:50If you had a standard 1 kg mass on a frictionless surface and applied a force of 3 newtons in one direction and also applied a force of 4 newtons at 90 degrees to the other force on the body you would have a resulting acceleration of 5 m/s squared just like a 3-4-5 triangle.Umm... well you won't actually, because the body is on a surface. You would only get an acceleration corresponding to the sum of the force vectors if the body was in free space.

Quote from: gem on 18/04/2010 11:17:50So if you then consider a body on the earths surface the mass that is attracting it to the centre it is mostly attracting at a vector angle, there is only a very small amount directly in line with a body's direction of acceleration.Yes, and because Newton realised this, and because he went on to calculate exactly how the forces would sum, that we regard him as being so clever.

I am saying he 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. So therefore giving different sum total results [a variation in the values of acceleration] than would be predicted by applying inverse square law to the value of attraction at earths surface, as at different distances the angle of attraction changes.

Are you, in effect, saying that to determine the net force you really need to sum the individual gravitational forces between all the atoms in the two bodies (taking into account their vectors), and if you did that, you come up with a slightly different law?

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.So you can not just take the value of gravitational attraction at earths surface and apply inverse square law out in to space.

No that would be hugely complex if not impossible,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.http://en.wikipedia.org/wiki/Shell_theoremWhat i am saying is you have to factor in the value of the partial cancellation due to the vector nature of the force, meaning you need to know what the gross value of the attraction is.

because when you get any great distance from a planet you are going to get values of gravitational attraction greater than newtons laws predict.

Can you show the formula(s) you use? The Shell theorem can be derived from vector calculus (for a uniform density sphere) and already takes all the vectors into account. How can your calculation differ if they're doing the same thing?

I don't understand what you mean by "partial cancellation". I presume by "vector nature" you mean that the force is the summation of many forces that are not coincident, but maybe I'm not getting that right either.

Can you show the formula(s) you use?

If you had a standard 1 kg mass on a frictionless surface and applied a force of 3 newtons in one direction and also applied a force of 4 newtons at 90 degrees to the other force on the body you would have a resulting acceleration of 5 m/s squared just like a 3-4-5 triangle.The total force that is applied to the 1 kg mass is however 7 newtons so a partial cancellation of 2 newtonsand if that force was applied at a angle of anything less than 90 degrees the amount of acceleration of the mass would be greater as the amount of partial cancellation of the vector sum would be less.

Also, I don't think you will be able to determine the correct solution without using calculus to integrate the forces, and I rather suspect Newton did that, seeing as he invented calculus to solve problems like this.

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.

He actually did not use calculus in his master work on gravity, because calculus was somewhat contentious at the time. What he did was use really elegant geometrical proofs that made use of vanishing triangles.

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".

If you had 1 newton force acting on a standard 1 kg mass on a frictionless surface in one direction and another 1 newton force acting at 90 degrees to that force you would have a gross force of two newtons and a resulting net acceleration of 1.4 metres per second squared.

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.

All the forces i have described in the example are parallel with the frictionless surface.

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.