I have not been following this. However, your formula for the gravitational force should be F = -GMm/r^2, with the acceleration a = -GM/r^2. It appears you wrote the equation for the velocity of an orbit.
LC On Friday, May 6, 2022 at 6:36:38 AM UTC-5 johnk...@gmail.com wrote: > I'm changing the title because I think it's bad form for the title of a > thread to contain the name of a list member, and because I really do feel > like Bill Murray, we've been over this exact same ground over and over > again almost verbatim. I'm (probably foolishly) going to do it one more > time: > > Suppose you wanted to measure the gravitational constant G, how would you > do it? You'd do it the same way Henry Cavendish did it 200 years ago, you'd > use Newton's formula F= √(GM/r) where F is the gravitational force of > attraction between 2 lead balls of equal mass if you chain one of the balls > to the earth's surface and let the other one swing freely. Now you'll need > to determine the mass of the balls, and you can do that by noting how fast > it accelerates when exposed to a known calibration force, for example a > force provided by a precisely made coiled clockspring, but if the energy in > the clockspring is half of what it was in Cavendish's day and the inertia > of the lead ball is also half of what it was in Cavendish's day then the > value of M you will write in your lab notebook will be the same as the > value Cavendish wrote in his lab notebook. And the amount of time it takes > for the freely swinging ball to hit the stationary ball that you write in > your lab notebook will be the same as the time Cavendish found. So the > value of G that you write in your lab notebook will be the same value > Cavendish wrote in his. > > Newton says the orbital velocity of a planet a distance r from the sun is v= > √ (GM/r) , so if G is the same and M is the same (because the inertial > and gravitational mass are always the same) then the orbital speed of a > planet is the same, and the solar system would look the same. And because > the gravitational acceleration on the surface of the earth g= GM/R^2 > where M is the mass of the earth and R is the radius of the earth, g would > still be 9.8 m/s, and force would still equal mass times acceleration. > > The one thing both Newton and Einstein agreed-upon is that gravitational > mass and inertial mass are always exactly the same, and that's why > Aristotle was wrong, something twice as heavy does not fall to the ground > twice as fast; that's also why even physicist who are adamantly opposed to > many worlds and love to badmouth it don't use the argument presented around > here that the solar system would become unstable. Some around here are > arguing in effect that Aristotle was right after all, an object twice as > heavy would fall to the ground twice as fast and we should just ignore 2000 > years worth of progress in physics. Unless somebody says something new that > I haven't already responded to at least twice before (and at this point > that seems unlikely) I'm done with this and Groundhog Day is finally over. > > John K Clark See what's on my new list at Extropolis > <https://groups.google.com/g/extropolis> > dgd > > > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/e9b4adc6-c45f-49f1-89cc-be0c2ed1a95fn%40googlegroups.com.
