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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.

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.

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.