In Response To: Heh we asked for it, but.... (DeusExMachina =PN=)
> I would venture to say you included an extra factor in your
> calculation: correct me if I'm wrong, but didn't you mention
> calculating the inertial (whatever you call non-rotationally
> induced) gravity of the ring? I don't think this is necessary.
> Can't you apply Gauss's law (I think it is) to mass as well as
> charge? When, in highschool physics, you deal with charge
> distributions, say to calculate electric field from a solid
> & charged sphere, you integrate radially, but you only deal
> with the charge closer to the center than your drho... this is
> the feature that allows you to assume the electric field is zero
> within bubbles, and that the field is zero within the sphere.
> Wouldn't that indicate (if it generalizes to all r^-2
> accelerations) that if we are closer to the center of the ring
> than its 'ground' we'd basically only be fealing a centripetal
> acceleratio from mountains that are 'higher' than us?
> If so, then the same thinking that lets that shell make a whole
> loop (that there are no hills) would leave us with the ability
> to disregard the mass of the ring in our net centripetal force
> calculations... Doesn't that make everyone feel much better? ;)
Yes. The same thing can be done using gravitational force and potential energy. Only mass closer to the gravitational center exerts gravitational force. If you are 'inside' the earth, for example, only the mass beneath your feet would exert gravity, and you can completely disregard everything 'above' you. The calculations can be easily adapted to a ring.
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