Bill Beam wrote: >>> Not true. >>> Very simple experiments will show occupants of the satellite that they >>> are in a non-inertial reference frame. (Release a few test masses >>> about the cabin and you will observe that they move/accelerate for no >>> apparent reason, unless the satellite is in free fall which you'll know soon >>> enough,) The experimenter must conclude that the satellite is undergoing >>> acceleration due to the influence of an attractive (gravitational) field. >>> >>> Just because NASA calls it 'microgravity' doesn't make it true. It means >>> NASA is wrong. Weightlessness is not the same as zero-g. >>> >>> >>> >> Only, if you insist on sticking to Newtonian physics with all its >> attendant problems. >> > > This discussion began as a classical problem. The relativistic effects > are many orders of magnitude smaller than Newtonian (v/c=2.6e-5). > For example: A test mass released on the Earth side of the satellite > cabin will advance in its own orbit a few mm/sec faster than one released > on the far side due to purely classical differences in orbits. Easily > observable > without even using a timepiece. > > Once your feet leave the ground, not even Newtonian mechanics is > intuitive. Who would have thought that 'putting on the brakes' to > leave orbit would cause a satellite to speed up.... > > The existence of "privileged" frames of reference in Newtonian mechanics and special relativity has always seemed problematic, especially when it may be observationally difficult to identify such a frame of reference. Fortunately the inertial reference frames are not imbued with the same privileges by general relativity. >>>> Pendulum clocks fail to work, given an initial push they will just >>>> rotate around the pivot, provided the pivot suitably constrains the >>>> motion of the pendulum (ie a shaft running in a set of ball or roller >>>> bearings or similar and not a knife edge pivot). >>>> > > Run the numbers - depends on how hard the push. > Consider sheeparding of material in Saturn rings by small moons. > > Surely the forces involved are impractically low for a real pendulum. >> A finite gradient doesn't imply that the field itself is nonzero, except >> of course towards the extremeities of the satellite. >> > > Of course it does. > > If g=0 everywhere in the neighborhood of a point then the gradient is zero. > Else, what is the meaning of gradient? > > Grad not zero implies field not uniform implies not(field zero everywhere). > > I should have been more explicit I meant a finite gradient at a point does not imply the field is no zero at that point.Bill Beam
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