[time-nuts] Pendulums & Atomic Clocks & Gravity

Bill Hawkins bill at iaxs.net
Sun May 27 17:17:32 EDT 2007


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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