[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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