Interesting question. I think there must be a flaw here but I sure can't see what it is. The problem I have with it is that it violates the equivalence principle, which, as far as I know, is, for simple mechanics problems, compatible with the Newtonian gravity model, which is what you're dealing with here.
If the conclusion here is correct, then someone in a falling elevator could tell that they were falling by throwing a gyroscope (in a sealed box) across the room. Depending on its orientation, it would not go straight -- it would curve up or down. I don't think that's supposed to happen. (Of course, the person in the elevator could also tell from tidal effects that they were falling -- water drops will be distorted.) On 12/10/2010 09:25 PM, David Jonsson wrote: > Hi > > I was calculating and found a strange thing. It seems like a rotating > molecule is less effected by gravity compared to a non rotating molecule. > > Imagine a diatomic molecule at the equator of a rotating planet. The > molecules has its axis of rotation parallel to the planets axis. It > seems to me that the centrifugal force on the molecule is different > when the molecule rotates and when it is not rotating. > > Speed of surface of the rotating planet = v > Radius of planet = r > Speed of the atoms in the rotating molecule = u > Mass of molecule = m ,(m/2 for each atom) > > In the non rotating case the centrifugal force on the gas molecule > becomes: > f = m v^2 / r > > If the molecule rotates the centrifugal force is different on the two > atoms it consists of. Lets take the case when the molecule is vertical > like this > > O <---- rotation of upper atom > I > O ----> rotation of lower atom > > <---- rotation of planet > ----------Planet surface------- > > The centrifugal force on the upper atom becomes > fu = m / 2 * (v+u)^2 / r > and on the lower > fl = m / 2 * (v-u)^2 / r > > adding the forces together to find the net effect gives > f = fu + fl = m / 2 * (v+u)^2 / r + m / 2 * (v-u)^2 / r = > = m / 2r * ((v+u)^2+(v-u)^2) = > = m / 2r * (v^2+2uv+u^2+v^2-2uv+u^2) = > = m / r * (v^2+u^2) > > The effect is of course smaller at other positions of the molecule and > the mean value over an entire revolution would be somewhat lower but > still higher than the case when the molecule is not rotating. > > It seems that the centrifugal force on the molecule is higher when it > rotates. Since the centrifugal force is opposed to gravity it means > that the rotating molecule would be less affected by gravity than the > rotating one. Can this really be the case? It has to apply to all > rotating matter and not only molecules as for example a rotating > planet around a star. > > Have I done something wrong? > > David > > David Jonsson, Sweden, phone callto:+46703000370 >
