Bill, > You have a cannon aimed at 45 degrees into the sky, for > maximum altitude. You fire the cannon, but the projectile > falls back to Earth. So you increase the amount of gunpowder > (propellant) which causes the projectile to fall to the Earth > farther away. You keep increasing the propellant, and finally > the projectile falls around the Earth in a circular path.
This picture is perfectly ok and I used it myself in an article in the German cq-DL ham magazine. Most important: The picture does NOT use centrigugal forces at all to explain the motion. If you were to explain: Are centrifugal forces something that you would use to explain the motion of an cannonball? When do they start to exhibit? Best regards Ulrich Bangert > -----Ursprüngliche Nachricht----- > Von: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] Im Auftrag von Bill Hawkins > Gesendet: Sonntag, 27. Mai 2007 23:18 > An: 'Discussion of precise time and frequency measurement' > Betreff: Re: [time-nuts] Pendulums & Atomic Clocks & Gravity > > > Ulrich, > > Just returned from Mannheim, where the gravitational force on > my body increased by ten pounds, due to the fine food. > > A NASA film explains the motion of satellites in the following way: > > You have a cannon aimed at 45 degrees into the sky, for > maximum altitude. You fire the cannon, but the projectile > falls back to Earth. So you increase the amount of gunpowder > (propellant) which causes the projectile to fall to the Earth > farther away. You keep increasing the propellant, and finally > the projectile falls around the Earth in a circular path. > > The interesting thing is that the mass of the projectile does > not affect the height of the orbit. Only the circular > velocity determines the orbital altitude. When an astronaut > "drops" a bolt from an assembly, it does not change orbital > altitude because its mass is much smaller than the space > station's mass. Instead, it maintains the same altitude > unless it was given some small change in velocity. > > Obviously, the mass and gravity are still present and F=Ma, > for all practical purposes. What happens is that the > acceleration 'a' goes to zero because it is countered by a > rotational acceleration, defined by the square of the > rotational velocity times the distance between the centers of > gravity of the Earth and the satellite. > > Set the rotational acceleration equal to gravitational > acceleration, and you have an equation that determines the > distance as a function of rotational velocity and the > (constant) gravitational force. > > The fact that some people call the rotational acceleration > 'fictitious' does not alter the results of the equation. > > Sorry, I don't accept the idea that the Earth and the Moon > repel each other with only gravitational forces. > > Regards, > Bill Hawkins > > > -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf > Of Ulrich > Bangert > Sent: Sunday, May 27, 2007 12:59 PM > To: 'Discussion of precise time and frequency measurement' > Subject: Re: [time-nuts] Pendulums & Atomic Clocks & Gravity > > Didier, > > > gravitational forces, so do objects in Lagrange points. These points > > represent areas where the centrifugal forces compensate for > > gravity.... > > I am almost sure that this will again produce me a lot of > trouble in answering a lot of people but the idea that there > are centrifugal forces which compensate for gravity are one > of the BIGGEST misconcepts that one may have in physics at > all although it is quite common and you may find statements > like that eben in (bad) physics textbooks. > > Centrifugal forces are so called fictitious forces which are > only observed from within accelerated systems. Normal physics > is done in inertial systems. In an inertial system consisting > of earth and an satellite there are only TWO forces > available: The gravity force by which earth attracts the > satellite and the gravitational force by which the satellite > attracts earth. They are of the same magnitude but of > opposite direction. That is the reason why the "sum of > forces" is zero for the closed system consisting of earth and > satellite. There is no place for any other force like > centrifugal or so because there is no counterforce available > that would make the sum of forces zero i case a centrifugal > force would exist. In case you like to discuss it a bit > please go on but be prepared that I will to blow your > arguments into little bits. A good idea to start with is to > look after what Newton's first law is saying about the > behaviour of a body for which all forces compensate each > other. Is that what a satellite does??? > > 73 Ulrich, DF6JB > > > > _______________________________________________ > time-nuts mailing list > [email protected] > https://www.febo.com/cgi-> bin/mailman/listinfo/time-nuts > _______________________________________________ time-nuts mailing list [email protected] https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
