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    I just typed out 30 minutes of algebra to show this, then pressed one wrong key and the browser lost it all. So here is what it showed..

    if you have a uniform sphere of mass m and radius r, and you are a distance R away from it, the gravitational field strength changes depending on whether you model the mass as a single point in the centre with mass m, or two points, each of mass m/2, with centres of mass at the two centres of mass of the semicircles.

    I did a calculation, first with one point and second with the two semicircles and using their centres of mass, the value of g was different, with the second one being dependant on the radius of the sphere.

    So what's wrong? DId I do a calculation wrong?

    If the calculations are all correct then this should mean you can't simplify a sum, for example, which has billions of small particles, and model it as a single point mass, because it will give you a different reading if you add up the value of g resulting from each particle.
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    I imagine it's something to do with the fact that you haven't preserved the symmetry of the spherical shape by placing the point masses there? Surely with something like your second diagram, you'd expect an elliptical sort of gravitational field? You may have more success with four?

    It's also possible that it'll only work when the gaps between the masses tend to zero.

    I don't really know, just a couple of suggestions. Any way, Newton was a clever chap, so I wouldn't thought he got it wrong - maybe Google can help you find a proof?

    I think that it's only possible to simplify "a sum, for example, which has billions of small particles, and model it as a single point mass" if any interactions that the model will undergo are with other models of sufficient distance away from the original that that the innaccuracies that you've found are effectively 0. Otherwise, if you take an extreme situation it's obvious that the mdel won't work. E.g. modelling the Earth as a single point mass is fine if you're doing gravitational attraction between the Earth and the Sun, but if you're doing attraction between the Earth and a single point (e.g. a small heat-proof probe) that's burrowed down to near the centre of the Earth, then a single-point mass will obviously give a different result to doing it the (very) long way: in the single-point model the net attraction will be much stronger than it really is since it doesn't take into account the attraction from all directions.

    That's an extreme case, but I think the general principle still applies in your case. With your diagram, the other mass is close enough that it can still 'see' the difference. Your example takes it to an extreme, however, by splitting the original mass up into 2 perpendicularly to the line between the masses, so because of the inverse square law the difference is probably bigger than the difference between modelling it as a single mass and modelling it as the individual billions of atoms.

    In other words, a model is a model, and is never going to exactly replicate the original. It's point is to provide a balance between replicating it as closely as possible and making the calculations as simple as possible.

    I'm only an A-level student, so the above may be a load of complete *******s, but it seems fairly reasonable.
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Updated: June 12, 2005
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