[time-nuts] Pendulums & Atomic Clocks & Gravity

Ulrich Bangert df6jb at ulrich-bangert.de
Mon May 28 05:30:04 EDT 2007


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: time-nuts-bounces at febo.com 
> [mailto:time-nuts-bounces at febo.com] 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: time-nuts-bounces at febo.com 
> [mailto:time-nuts-bounces at febo.com] 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 
> 
> 
> 
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