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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_theorem [Links inactive - To make links active and clickable, login or click here to register]What 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.

The partial cancellation will be greatest at earths surface diminishing the further out in to space, where as the cancellation becomes total within a shell of homogeneous mass. Because the net gravitational forces acting on a point mass from the mass elements of the shell totally cancel out.

(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. . .")

Quote from: gem on 22/04/2010 22:40:47The 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,?Quote from: JP on 23/04/2010 05:21:50However, 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.Quote from: gem on 20/04/2010 20:00:13I 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.

The partial cancellation will be greatest at earths surface diminishing the further out in to space, You're right about vector sums.

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.

So Newton was right then?

And in diagram 3 we have a example of the angle of interaction getting more acute and any mass that is not in line with the centre line C/L Note, As the distance between the mass of the two bodies changes only the particles that are in a direct line with the centre to centre actually travel the distance prescribed by inverse square law relative to each others centres.

*In most problems you're dealing with continuous objects, so instead of summing, you integrate.

Quote from: JP on 27/04/2010 05:23:09*In most problems you're dealing with continuous objects, so instead of summing, you integrate.Well, yes. But isn't integration really just a way of adding all the little bits together without having to do the actual adding? []

If you want to calculate the gravitational force at the surface of the Earth, you have to integrate all the "individual" forces acting on the body using the inverse square law, taking into account the force vectors. If you do that, you'll get a number that is the same as the observed gravitational force. If you assume the forces all act towards the center of the Earth, you'll calculate a value that is substantially greater than the observed value.

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.

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.