Huh, You're right. Picturing it more clearly now, The vectors do always add up to equal the same as the pull at the center of the small sphere.
So the g field is a constant inside a spherical hollow in a sphere with constant density. That being said, next time this comes up, you might want to summarize a little bit differently, because what threw me off was it appeared you were taking R1 and R2 to be constants, instead of taking the vector sum to be constant. --- "Stephen A. Lawrence" <[EMAIL PROTECTED]> wrote: > > > Merlyn wrote: > > Stephen A. Lawrence wrote: > > > > > > > > The field at any point inside a uniform sphere of > > density > > rho is > > > > F = -(4/3)*pi*G*rho*R > > > > where "R" is the _radius vector_ from the center > of > > the > > sphere to the point where we're finding the field. > > > > For the big sphere, let the radius vector be R1. > For > > the > > small (cut-out) sphere let the radius vector be > R2. > > (Note that they point from different origins, but > > that's > > OK, all we care about are the direction and > length.) > > Then the net field anywhere inside the small > (cut-out) > > sphere will be > > > > F(total) = -(4/3)*pi*G*rho*(R1 - R2) > > > > But (R1 - R2) is a _constant_, and is just the > vector > > from the center of the big sphere to the center of > the > > small sphere. > > > > So the force is also a constant, proportional to > the > > distance between the spheres' centers, pointing > along > > the > > line which connects the small sphere's center to > the > > big > > sphere's center. > > ----------------------------------- > > > > Not so. > > R1 and R2 are NOT the actual radii of the spheres, > but > > the radii to the point of measurement. > > Exactly. Outside the sphere, the field goes as > 1/R^2. > > Inside the sphere it goes as R. > > On the surface of a uniform sphere of radius R the > field strength is > (4/3)G*pi*rho*R. Here's why: > > The mass is (4/3)*pi*rho*R^3 (just the volume > times the density) > > The field behaves as though all the mass is > concentrated at the center, > at on point, and the field strength goes as > > G*M/R^2 > > Plugging in the value for the mass, this is > > (4/3)*G*pi*(R^3/R^2) = (4/3)*G*pi*R > > > This is > > because any spherical shell of constant density > has no > > net gravitational effect on an object within it, > > Right, which is why the field strength at any point > _inside_ a uniformly > dense sphere at a distance "Rc" from the center is > the same as the field > strength on the surface of an equally dense sphere > of radius "Rc". > > Or, in other words, _inside_ the sphere, the field > goes as > > (4/3)*G*pi*rho*R > > where R is the distance from the center of the > sphere to the point where > we're measuring the field (_not_ the distance to the > surface of the sphere). > > Since the field points directly toward the center of > the sphere, if we > replace the distance to the center with the vector > "[R]" which points > from the center to the point where we're measuring > the field, then the > actual field at each point will be > > -(4/3)*G*pi*rho*[R] > > and I seriously wish I had overbars to make this > look more readable. > > > so > > you only need the mass of the spherical volume > with > > radius equal to your distance from the center of > mass. > > Thus I can measure the gravitational field > strength > > along a constant radius from the center of the > large > > sphere (R1 constant) but at different locations > within > > the volume of the small sphere (R2 variable) and > > achieve different results. > > It seems that way, doesn't it? But keep in mind > that R1 and R2 are > vectors, so neither one is constant in this case -- > R1 has constant > _magnitude_ but its direction is varying. > > You need to draw a fairly careful picture here to > get an idea of what's > going on. > > Draw a circle around the big sphere's center, such > that the circle > passes through the small sphere's center. > > At the center of the small sphere, the gravity is > "normal" for the big > sphere -- the small sphere contributes nothing. > > As we move off along the circle, the small sphere > starts to contribute > something. But at the same time, the big sphere's > _angle_ of pull > changes. Initially, the contribution of the small > sphere is (nearly) > perpendicular to the big sphere's pull, and it just > cancels the angle > change in the big sphere's pull. > > As we get farther and farther from the little > sphere's center, the small > sphere's "push" is no longer perpendicular to the > big sphere's "pull". > It's actually contributing to the force along the > line between their > centers, at the same time that it's cancelling the > pull perpendicular to > that line. > > After thinking about it for a while I decided it at > least _seemed_ > possible, and I stopped looking for a flaw in the > math... :-) > > > > > Merlyn > > Magickal Engineer and Technical Metaphysicist > > Merlyn Magickal Engineer and Technical Metaphysicist __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
