Bullwinkle Unbound

George Dishman replied to Jeff Root on November 27 in the
thread "Red shift and homogeneity":

| how about having Rocky and Bullwinkle
| orbiting around each other in space held only by their
| mutual gravitational attraction? Would the total mass
| of the system be greater or less than than the sum of
| their individual masses? I think you need to consider
| both kinetic energy and gravitational potential energy.

George Dishman replied to Jeff Root on December 10 in the
thread "Red shift and homogeneity":

Suppose they are held by a massless rope.

In that case, the mass of the system is greatest when they
are close together, orbiting rapidly.

Right.

For the gravitational case, they remain in orbit only
as long as the sum of the gravitational potential energy
and their kinetic energy is less than zero.

The gravity rope can stretch infinitely, but it has limited
strength. If the kinetic energy is too great for the amount
the rope is stretched, the rope breaks and Rocky is flung away.

Remember the discussion on 'escape velocity'?

Actually I don't recall one which particularly applies here.
Was it with "JosX", or more recent?

It was the one where you (or maybe someone else) said they
couldn't see why 'escape velocity' depended only on the speed
and not on the direction. The standard argument is that the
graviational potential energy is negative. While the sum of
potential and kinetic is less than zero, the particle cannot
escape.

Oh of course! Now this is weird. A thread on escape velocity
started a few days ago in sci.physics, and "Uncle Al" made some
comments which included the clear implication that it depends
on the direction of launch. He said:

| Escape velocity is the minimum initial impulse velocity at and
| normal to the surface (no planetary atmosphere, no rotation)
| necessary to launch a test mass to infinity

This greatly surprised me, and I went looking for the thread
you refer to just a couple of hours before I read your message!
That thread was titled "Esape velocity question" (sic) and took
place last March.

I negligently left off on that thread because the last posts
to me contained too much information for me to be able to deal
with properly. They appear to answer my question/objection,
but I was still uncertain.

I have yet to re-read those posts in detail, but you said
something about centrifugal force on a projectile launched at
an angle. Combining that with something Bill Owen said, I
surmise that the answer is that the horizontal velocity is
turned into vertical velocity. Nobody actually said that to
me, though, and if it is a correct description of what happens,
I'd think you would have, so I'm not sure.

The onus is on me to re-read those posts, and ask further
questions and thank you and Bill for your answers. There was
one statement I made that you and Bill agreed was wrong that
I'm sure was right, so I really should go back to it.

In any case, the idea that the horizontal velocity eventually
becomes vertical velocity seems to make sense and answers my
objection, and John Zinni corrected "Uncle Al",

| It does not need to be normal to the surface. Any old
| direction will do, it will just take a slightly less direct
| route to infinity.

and "Uncle Al" immediately agreed.

So, how does that relate to the discussion we were in here?

So does gravitational potential energy contribute to the
mass of the system? If so the mass should be _less_ than
the sum of R&B if they are in orbit, equal to the sum if
they are passing by on parabolic paths and greater for
hyperbolic trajectories. If it doesn't, the total will
always be greater than the sum of their individual masses.

Since it seems absurd that the total mass could be less than
the sum of the individual masses, you must be trying to get me
to see that potential energy does *not* contribute to the mass
of the system. Is that right?

I hope it *is* right that potential energy doesn't contribute
to the mass, because it feels right, and the contrary feels
wrong. I haven't tried to think through what the consequences
are, though.

Time for a silly diagram to see what a consequence
might be. Assume this is in space so there are no
other influences:

Box 3
+--------------------------+
| |
| Box 1 |
| +---+------+ |
| | o | |
| | |\ | Box 2 |
| | | \ | +-------+ |
| | M \ | | | |
| | G=========R | |
| | M / | | | |
| | | / | +-------+ |
| | |/ | |
| | o | |
| +---+------+ |
| |
+--------------------------+

Two large masses "M" are held apart by ropes over
pulleys "o" fixed to the sides of box 1. They are
allowed to move together under their gravitational
attraction and the ropes turn generator "G". The
power from the generator is fed by the pair of
wires "==" to resistor "R". The energy fed to R
increases its temperature and its mass through
E=mc^2.

Box 3 is a closed system so the total mass must
remain constant.

The conclusion is that the mass of Box 1 must
decrease as the masses get closer together.

It may be a counter-intuitive result but I
think it follows logically.

An absolutely clear illustration. Obviously logic must be
thrown out and discarded as misleading. No more logic.

Okay, the two masses have a greater total mass when they are
separated than when they are close together. The diagram
clearly shows that. It is ridiculous. Absurd. Impossible.

Have I missed something? All I can think to point out is
that the two masses, being pulled together by their mutual
gravitational force, apply force to the ropes and thus to the
ends of the box, thus compressing its sides. When the masses
are falling toward each other, the gravitational force is
increasing, and some of the force is applied to the rotors of
the generators, (two so that the box doesn't rotate) and from
the rotors to the stators, thus generating electric current.
When the two masses are together, the gravitational force
between them is at maximum, and is applied entirely at the
point of contact between them.

I don't see that any of those facts help me.

If the two masses were in orbit around each other at the
starting distance, rather than being held apart by forces in
the ropes and the box, they would have some additional mass
due to their kinetic energy.

If the two masses were in orbit around each other at their
final positions, they would have a *greater* additional mass
because their kinetic energy would be greater.

If the forces in the ropes and the box somehow represent an
amount of mass equal to the mass represented by the kinetic
energy, then I would expect the forces at the point of contact
of the two masses in their final positions to represent a
greater mass, not less.

Am I sufficiently confused yet?

-- Jeff, in Minneapolis

..

"Jeff Root" wrote in message
om...
[much snipped]
In any case, the idea that the horizontal velocity eventually
becomes vertical velocity seems to make sense and answers my
objection, and John Zinni corrected "Uncle Al",

Consider a projectile on a hyperbolic orbit. As it
approaches infinity the angle between its path and
the gravitational force approaches zero so it becomes
a radial motion with the force merely slowing it.

| It does not need to be normal to the surface. Any old
| direction will do, it will just take a slightly less direct
| route to infinity.

and "Uncle Al" immediately agreed.

It is well known but the lingering doubt I think people
have is that some kinetic energy surely remains in the
transverse component because even though the angle is
tending to zero, it is only because the range is tending
to infinity. There is always an angular velocity but it
is angular momentum that is conserved so perhaps you
could use that to see what happens to the energy in the
angular component of the motion. If it goes to zero you
could have your proof.

So, how does that relate to the discussion we were in here?

So does gravitational potential energy contribute to the
mass of the system? If so the mass should be _less_ than
the sum of R&B if they are in orbit, equal to the sum if
they are passing by on parabolic paths and greater for
hyperbolic trajectories. If it doesn't, the total will
always be greater than the sum of their individual masses.

Since it seems absurd that the total mass could be less than
the sum of the individual masses, you must be trying to get me
to see that potential energy does *not* contribute to the mass
of the system. Is that right?

I hope it *is* right that potential energy doesn't contribute
to the mass, because it feels right, and the contrary feels
wrong. I haven't tried to think through what the consequences
are, though.

Time for a silly diagram to see what a consequence
might be. Assume this is in space so there are no
other influences:

Box 3
+--------------------------+
| |
| Box 1 |
| +---+------+ |
| | o | |
| | |\ | Box 2 |
| | | \ | +-------+ |
| | M \ | | | |
| | G=========R | |
| | M / | | | |
| | | / | +-------+ |
| | |/ | |
| | o | |
| +---+------+ |
| |
+--------------------------+

Two large masses "M" are held apart by ropes over
pulleys "o" fixed to the sides of box 1. They are
allowed to move together under their gravitational
attraction and the ropes turn generator "G". The
power from the generator is fed by the pair of
wires "==" to resistor "R". The energy fed to R
increases its temperature and its mass through
E=mc^2.

Box 3 is a closed system so the total mass must
remain constant.

The conclusion is that the mass of Box 1 must
decrease as the masses get closer together.

It may be a counter-intuitive result but I
think it follows logically.

An absolutely clear illustration.

Thank you :-)

Obviously logic must be
thrown out and discarded as misleading. No more logic.

:-o

Okay, the two masses have a greater total mass when they are
separated than when they are close together. The diagram
clearly shows that. It is ridiculous. Absurd. Impossible.

That's the fascinating thing about science, sometimes
it produces a result we feel just has to be wrong, but
if it follows from observation we have to accept it.

Have I missed something?

Not that I am aware of.

All I can think to point out is
that the two masses, being pulled together by their mutual
gravitational force, apply force to the ropes and thus to the
ends of the box, thus compressing its sides. When the masses
are falling toward each other, the gravitational force is
increasing, and some of the force is applied to the rotors of
the generators, (two so that the box doesn't rotate) and from
the rotors to the stators, thus generating electric current.
When the two masses are together, the gravitational force
between them is at maximum, and is applied entirely at the
point of contact between them.

I don't see that any of those facts help me.

They don't. The mass of box 2 increases regardless so
whatever distortions you consider for box 1, they only
redistribute the energy between different forms within
the box. The mass of box 1 must reduce regardless.

If the two masses were in orbit around each other at the
starting distance, rather than being held apart by forces in
the ropes and the box, they would have some additional mass
due to their kinetic energy.

Yep.

If the two masses were in orbit around each other at their
final positions, they would have a *greater* additional mass
because their kinetic energy would be greater.

Yep, but wouldn't their potential energy have fallen by
more than the increase in the kinetic increased? That was
the original point.

If the forces in the ropes and the box somehow represent an
amount of mass equal to the mass represented by the kinetic
energy, then I would expect the forces at the point of contact
of the two masses in their final positions to represent a
greater mass, not less.

Am I sufficiently confused yet?

No, not confused, you just find it hard to accept the
answer.

Of course I might be wrong but I don't see where there is
much scope to conserve the total energy without having the
mass of box 1 decrease. It might be appropriate to crosspost
this to sci.physics as more on-topic but perhaps with some
trimming first so I'll leave it just here.

George

(Jeff Root) wrote in message . com...
George Dishman replied to Jeff Root on November 27 in the
thread "Red shift and homogeneity":

| how about having Rocky and Bullwinkle
| orbiting around each other in space held only by their
| mutual gravitational attraction? Would the total mass
| of the system be greater or less than than the sum of
| their individual masses? I think you need to consider
| both kinetic energy and gravitational potential energy.

George Dishman replied to Jeff Root on December 10 in the
thread "Red shift and homogeneity":

Suppose they are held by a massless rope.

In that case, the mass of the system is greatest when they
are close together, orbiting rapidly.

Right.

For the gravitational case, they remain in orbit only
as long as the sum of the gravitational potential energy
and their kinetic energy is less than zero.

The gravity rope can stretch infinitely, but it has limited
strength. If the kinetic energy is too great for the amount
the rope is stretched, the rope breaks and Rocky is flung away.

Remember the discussion on 'escape velocity'?

Actually I don't recall one which particularly applies here.
Was it with "JosX", or more recent?

It was the one where you (or maybe someone else) said they
couldn't see why 'escape velocity' depended only on the speed
and not on the direction. The standard argument is that the
graviational potential energy is negative. While the sum of
potential and kinetic is less than zero, the particle cannot
escape.

Oh of course! Now this is weird. A thread on escape velocity
started a few days ago in sci.physics, and "Uncle Al" made some
comments which included the clear implication that it depends
on the direction of launch. He said:

| Escape velocity is the minimum initial impulse velocity at and
| normal to the surface (no planetary atmosphere, no rotation)
| necessary to launch a test mass to infinity

This greatly surprised me, and I went looking for the thread
you refer to just a couple of hours before I read your message!
That thread was titled "Esape velocity question" (sic) and took
place last March.

I negligently left off on that thread because the last posts
to me contained too much information for me to be able to deal
with properly. They appear to answer my question/objection,
but I was still uncertain.

I have yet to re-read those posts in detail, but you said
something about centrifugal force on a projectile launched at
an angle. Combining that with something Bill Owen said, I
surmise that the answer is that the horizontal velocity is
turned into vertical velocity. Nobody actually said that to
me, though, and if it is a correct description of what happens,
I'd think you would have, so I'm not sure.

The onus is on me to re-read those posts, and ask further
questions and thank you and Bill for your answers. There was
one statement I made that you and Bill agreed was wrong that
I'm sure was right, so I really should go back to it.

In any case, the idea that the horizontal velocity eventually
becomes vertical velocity seems to make sense and answers my
objection, and John Zinni corrected "Uncle Al",

| It does not need to be normal to the surface. Any old
| direction will do, it will just take a slightly less direct
| route to infinity.

and "Uncle Al" immediately agreed.

So, how does that relate to the discussion we were in here?

So does gravitational potential energy contribute to the
mass of the system? If so the mass should be _less_ than
the sum of R&B if they are in orbit, equal to the sum if
they are passing by on parabolic paths and greater for
hyperbolic trajectories. If it doesn't, the total will
always be greater than the sum of their individual masses.

Since it seems absurd that the total mass could be less than
the sum of the individual masses, you must be trying to get me
to see that potential energy does *not* contribute to the mass
of the system. Is that right?

I hope it *is* right that potential energy doesn't contribute
to the mass, because it feels right, and the contrary feels
wrong. I haven't tried to think through what the consequences
are, though.

Time for a silly diagram to see what a consequence
might be. Assume this is in space so there are no
other influences:

Box 3
+--------------------------+
| |
| Box 1 |
| +---+------+ |
| | o | |
| | |\ | Box 2 |
| | | \ | +-------+ |
| | M \ | | | |
| | G=========R | |
| | M / | | | |
| | | / | +-------+ |
| | |/ | |
| | o | |
| +---+------+ |
| |
+--------------------------+

Two large masses "M" are held apart by ropes over
pulleys "o" fixed to the sides of box 1. They are
allowed to move together under their gravitational
attraction and the ropes turn generator "G". The
power from the generator is fed by the pair of
wires "==" to resistor "R". The energy fed to R
increases its temperature and its mass through
E=mc^2.

Box 3 is a closed system so the total mass must
remain constant.

The conclusion is that the mass of Box 1 must
decrease as the masses get closer together.

It may be a counter-intuitive result but I
think it follows logically.

An absolutely clear illustration. Obviously logic must be
thrown out and discarded as misleading. No more logic.

Okay, the two masses have a greater total mass when they are
separated than when they are close together. The diagram
clearly shows that. It is ridiculous. Absurd. Impossible.

Have I missed something? All I can think to point out is
that the two masses, being pulled together by their mutual
gravitational force, apply force to the ropes and thus to the
ends of the box, thus compressing its sides. When the masses
are falling toward each other, the gravitational force is
increasing, and some of the force is applied to the rotors of
the generators, (two so that the box doesn't rotate) and from
the rotors to the stators, thus generating electric current.
When the two masses are together, the gravitational force
between them is at maximum, and is applied entirely at the
point of contact between them.

I don't see that any of those facts help me.

If the two masses were in orbit around each other at the
starting distance, rather than being held apart by forces in
the ropes and the box, they would have some additional mass
due to their kinetic energy.

If the two masses were in orbit around each other at their
final positions, they would have a *greater* additional mass
because their kinetic energy would be greater.

If the forces in the ropes and the box somehow represent an
amount of mass equal to the mass represented by the kinetic
energy, then I would expect the forces at the point of contact
of the two masses in their final positions to represent a
greater mass, not less.

Am I sufficiently confused yet?

-- Jeff, in Minneapolis

I have posted on thread 'Mass of a coiled Spring' question as to does
gravitational potential energy contribute to the mass. Unc had
previously claimed that a coiled spring could be shown by calorific
experiment to contain more mass than an unstressed one. I then
suggested that a piece of rock subject to this type of experiment in a
valley should give a different result from an identical piece at the
mountain top-- silence........ Then sugestion (from others) that
potential gravitational energy does not contribute mass, and I await
comment on what then if gravity is used to compress the spring!

As for Rocky and Bullwinkle: maybe you are mistakingly considering
them as individuals, when they are always just one system, with the
center of the system not doing anything (as the center of gravity
between)

I have thought for some time that this effect may be overlooked in
binary star systems, when considering the causes of the EMR behaviour

Jim G

.

"George Dishman" wrote in message ...


That's the fascinating thing about science, sometimes
it produces a result we feel just has to be wrong, but
if it follows from observation we have to accept it.


Funny,funny,funny.

It may be fascinating because of the utter stupidity of determining
that if the motion of the local stars takes 23 hours 56 min 04 sec
then this must be the rotation of the Earth through 360 deg .

You may be the most backward people to set foot on the planet,truly !.

I have crossposted this to sci.physics as well as sci.astro.

(Jim Greenfield) wrote in message . com...
(Jeff Root) wrote in message . com...
George Dishman replied to Jeff Root on November 27 in the
thread "Red shift and homogeneity":

| how about having Rocky and Bullwinkle
| orbiting around each other in space held only by their
| mutual gravitational attraction? Would the total mass
| of the system be greater or less than than the sum of
| their individual masses? I think you need to consider
| both kinetic energy and gravitational potential energy.
....
Time for a silly diagram to see what a consequence
might be. Assume this is in space so there are no
other influences:

Box 3
+--------------------------+
| |
| Box 1 |
| +---+------+ |
| | o | |
| | |\ | Box 2 |
| | | \ | +-------+ |
| | M \ | | | |
| | G=========R | |
| | M / | | | |
| | | / | +-------+ |
| | |/ | |
| | o | |
| +---+------+ |
| |
+--------------------------+

Two large masses "M" are held apart by ropes over
pulleys "o" fixed to the sides of box 1. They are
allowed to move together under their gravitational
attraction and the ropes turn generator "G". The
power from the generator is fed by the pair of
wires "==" to resistor "R". The energy fed to R
increases its temperature and its mass through
E=mc^2.

Box 3 is a closed system so the total mass must
remain constant.

The conclusion is that the mass of Box 1 must
decrease as the masses get closer together.
....
I have posted on thread 'Mass of a coiled Spring' question as to does
gravitational potential energy contribute to the mass. Unc had
previously claimed that a coiled spring could be shown by calorific
experiment to contain more mass than an unstressed one.

It would have greater mass because you have stored energy in
it by compressing the spring. That energy can be recovered
by letting the spring expand against a force but it is not
gravitational potential energy when in the spring.

I then
suggested that a piece of rock subject to this type of experiment in a
valley should give a different result from an identical piece at the
mountain top-- silence........

Erm, well, yes, I can undertsand that. How do you propose
to store the energy? Is it a springy piece of rock? Why
would it matter where you do the experiment?

Then sugestion (from others) that
potential gravitational energy does not contribute mass,

That is certainly germaine to this discussion. It does not
affect the mass of the individuals but I believe it affects
the mass of the system as you say later.

and I await
comment on what then if gravity is used to compress the spring!

The method you use to compress the spring doesn't matter, it
is the fact that energy is stored in the EM bonds between the
atoms and m=e/c^2 that changes the mass of the spring.

As for Rocky and Bullwinkle: maybe you are mistakingly considering
them as individuals, when they are always just one system, with the
center of the system not doing anything (as the center of gravity
between)

Yes, that is the key, but perhaps you missed the earlier posts.
The question is how does the mass of the system compare to the
sum of the masses of the individuals in two different scenarios:

1) R&B are relatively far apart so gravitational effects are
negligible. They are rotating about a common point at high
speed tethered by a (massless) rope. Jeff and I agree the
mass of the system must be _greater_ than the sum of the
masses of R&B because the kinetic energy of their motion
must be included.

2) R&B are relatively close together and are orbiting about a
common point held only by mutual gravitational attraction
(there is no rope). I think the mass of the system must be
_less_ than the sum of the masses of R&B because the fact
that they are in a stable orbit means the sum of the kinetic
energy and the gravitational potential energy must be negative.

If gravitational potential energy does not contribute to the mass
of the system then the total in (2) would be greater as in (1).
My diagram above attempts to make the case that it does contribute
based on conservation of total energy in a closed system, box 3.

Note that in none of this has anyone suggested that the mass of
either Rocky or Bullwinkle would be affected, only that of the
system.

I have thought for some time that this effect may be overlooked in
binary star systems, when considering the causes of the EMR behaviour

I was going to mention that in my last post and snipped it at the
last minute. For example in the binary system measured by Hulse
and Taylor, energy is being lost in the form of gravitational
radiation. That means that in theory the total mass of the system
must be falling, but I guess other effects, such as the loss of
energy through the light that allows us to see those stars, are
much greater so I doubt this effect could be measured.

However, it is significant in cosmology since certain models have
the (negative) gravitational potential energy exactly equal in
magnitude to the (positive) energy in the form of radiation and
matter. The zero total means there is no question of where the
energy came from to create the universe.

George

(George G. Dishman) wrote in message om...
I have crossposted this to sci.physics as well as sci.astro.

(Jim Greenfield) wrote in message . com...
(Jeff Root) wrote in message . com...
George Dishman replied to Jeff Root on November 27 in the
thread "Red shift and homogeneity":

| how about having Rocky and Bullwinkle
| orbiting around each other in space held only by their
| mutual gravitational attraction? Would the total mass
| of the system be greater or less than than the sum of
| their individual masses? I think you need to consider
| both kinetic energy and gravitational potential energy.
It would have greater mass because you have stored energy in
it by compressing the spring. That energy can be recovered
by letting the spring expand against a force but it is not
gravitational potential energy when in the spring. ...
I have posted on thread 'Mass of a coiled Spring' question as to does
gravitational potential energy contribute to the mass. Unc had
previously claimed that a coiled spring could be shown by calorific
experiment to contain more mass than an unstressed one.



I then
suggested that a piece of rock subject to this type of experiment in a
valley should give a different result from an identical piece at the
mountain top-- silence........

Erm, well, yes, I can undertsand that. How do you propose
to store the energy? Is it a springy piece of rock? Why
would it matter where you do the experiment?

The rock is held up by the mountain. I just wonder does the higher
rock have more mass due to its altitude (and increased gravitational
potential energy)

Then sugestion (from others) that
potential gravitational energy does not contribute mass,

That is certainly germaine to this discussion. It does not
affect the mass of the individuals but I believe it affects
the mass of the system as you say later.

and I await
comment on what then if gravity is used to compress the spring!

The method you use to compress the spring doesn't matter, it
is the fact that energy is stored in the EM bonds between the
atoms and m=e/c^2 that changes the mass of the spring.

As for Rocky and Bullwinkle: maybe you are mistakingly considering
them as individuals, when they are always just one system, with the
center of the system not doing anything (as the center of gravity
between)

Yes, that is the key, but perhaps you missed the earlier posts.
The question is how does the mass of the system compare to the
sum of the masses of the individuals in two different scenarios:

1) R&B are relatively far apart so gravitational effects are
negligible. They are rotating about a common point at high
speed tethered by a (massless) rope. Jeff and I agree the
mass of the system must be _greater_ than the sum of the
masses of R&B because the kinetic energy of their motion
must be included.

The "massless" rope is tensioned, as is a compressed spring. Otherwise
R&B would fly apart.

2) R&B are relatively close together and are orbiting about a
common point held only by mutual gravitational attraction
(there is no rope). I think the mass of the system must be
_less_ than the sum of the masses of R&B because the fact
that they are in a stable orbit means the sum of the kinetic
energy and the gravitational potential energy must be negative.

If gravitational potential energy does not contribute to the mass
of the system then the total in (2) would be greater as in (1).
My diagram above attempts to make the case that it does contribute
based on conservation of total energy in a closed system, box 3.

Note that in none of this has anyone suggested that the mass of
either Rocky or Bullwinkle would be affected, only that of the
system.

I have thought for some time that this effect may be overlooked in
binary star systems, when considering the causes of the EMR behaviour

I was going to mention that in my last post and snipped it at the
last minute. For example in the binary system measured by Hulse
and Taylor, energy is being lost in the form of gravitational
radiation. That means that in theory the total mass of the system
must be falling, but I guess other effects, such as the loss of
energy through the light that allows us to see those stars, are
much greater so I doubt this effect could be measured.

However, it is significant in cosmology since certain models have
the (negative) gravitational potential energy exactly equal in
magnitude to the (positive) energy in the form of radiation and
matter. The zero total means there is no question of where the
energy came from to create the universe.

George

Thanks George

Jim G

"Jim Greenfield" wrote in message
om...
(George G. Dishman) wrote in message
om...
(Jim Greenfield) wrote in message
. com...
I then
suggested that a piece of rock subject to this type of experiment in a
valley should give a different result from an identical piece at the
mountain top-- silence........

Erm, well, yes, I can undertsand that. How do you propose
to store the energy? Is it a springy piece of rock? Why
would it matter where you do the experiment?

The rock is held up by the mountain. I just wonder does the higher
rock have more mass due to its altitude (and increased gravitational
potential energy)

No, AFAIK the rock has the same mass as if it was lower
and so does the planet. The mass of the system is probably
the same because raising the rock from the bottom needed
energy. Put the other way round, if you throw the rock
of the mountain into a pool. there is a loss of potential
energy that is converted first into kinetic as the rock
gains speed while falling, then into heat as it is slowed
in the water. Total energy remains constant so total mass
also remains constant.

Yes, that is the key, but perhaps you missed the earlier posts.
The question is how does the mass of the system compare to the
sum of the masses of the individuals in two different scenarios:

1) R&B are relatively far apart so gravitational effects are
negligible. They are rotating about a common point at high
speed tethered by a (massless) rope. Jeff and I agree the
mass of the system must be _greater_ than the sum of the
masses of R&B because the kinetic energy of their motion
must be included.

The "massless" rope is tensioned, as is a compressed spring. Otherwise
R&B would fly apart.

I was assuming the rope did not stretch. Energy is force
times distance (the integral if the force varies) so if
the rope doesn't stretch there is no energy stored.

Thanks George

Pleasure.

George

"George Dishman" wrote in message ...
"Jim Greenfield" wrote in message
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(George G. Dishman) wrote in message
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I then
suggested that a piece of rock subject to this type of experiment in a
valley should give a different result from an identical piece at the
mountain top-- silence........

Erm, well, yes, I can undertsand that. How do you propose
to store the energy? Is it a springy piece of rock? Why
would it matter where you do the experiment?

The rock is held up by the mountain. I just wonder does the higher
rock have more mass due to its altitude (and increased gravitational
potential energy)

No, AFAIK the rock has the same mass as if it was lower
and so does the planet. The mass of the system is probably
the same because raising the rock from the bottom needed
energy. Put the other way round, if you throw the rock
of the mountain into a pool. there is a loss of potential
energy that is converted first into kinetic as the rock
gains speed while falling, then into heat as it is slowed
in the water. Total energy remains constant so total mass
also remains constant.

Compessing the spring required (work) energy input as well. If gravity
is used to perform the compression, a larger piece of rock is needed
at the top of the mountain than in the valley, as gravity is less at
the higher level. But the same amount of energy (mass) is stored in
the spring (if the calorific thing is correct). I think this shows
that potential (gravitational) energy SHOULD contribute mass to the
higher rock, and R&B's individual masses would be greater at distance
due to the increased potential (for gravity to accellerate them)

Yes, that is the key, but perhaps you missed the earlier posts.
The question is how does the mass of the system compare to the
sum of the masses of the individuals in two different scenarios:

1) R&B are relatively far apart so gravitational effects are
negligible. They are rotating about a common point at high
speed tethered by a (massless) rope. Jeff and I agree the
mass of the system must be _greater_ than the sum of the
masses of R&B because the kinetic energy of their motion
must be included.

The "massless" rope is tensioned, as is a compressed spring. Otherwise
R&B would fly apart.

I was assuming the rope did not stretch. Energy is force
times distance (the integral if the force varies) so if
the rope doesn't stretch there is no energy stored.

Don't you mean "work is force times distance"- tension in the rope is
equivalent to tension in the spring, is to tension in the mountain
supporting the rock

,......and think about bringing opposing magnetic poles close- work is
done, but if I lock them in proximity, and then do the "calorific mass
equivalence test", will I see an increase in the mass of the magnets?)

Jim G

"Jim Greenfield" wrote in message
om...
"George Dishman" wrote in message
...
"Jim Greenfield" wrote in message
om...
(George G. Dishman) wrote in message
om...
(Jim Greenfield) wrote in message
. com...
I then
suggested that a piece of rock subject to this type of experiment
in a
valley should give a different result from an identical piece at
the
mountain top-- silence........

Erm, well, yes, I can undertsand that. How do you propose
to store the energy? Is it a springy piece of rock? Why
would it matter where you do the experiment?

The rock is held up by the mountain. I just wonder does the higher
rock have more mass due to its altitude (and increased gravitational
potential energy)

No, AFAIK the rock has the same mass as if it was lower
and so does the planet. The mass of the system is probably
the same because raising the rock from the bottom needed
energy. Put the other way round, if you throw the rock
of the mountain into a pool. there is a loss of potential
energy that is converted first into kinetic as the rock
gains speed while falling, then into heat as it is slowed
in the water. Total energy remains constant so total mass
also remains constant.

Compessing the spring required (work) energy input as well.

That was the point, the spring was compressed but the rock
wasn't. You can nitpick that everything is compressible to
some degree but I took it that you changed from a spring to
a rock to indicate it was non-compressible.

If gravity
is used to perform the compression, a larger piece of rock is needed
at the top of the mountain than in the valley, as gravity is less at
the higher level. But the same amount of energy (mass) is stored in
the spring (if the calorific thing is correct).

The force compressing the spring is due to gravity so the
source of the energy is the gravitational field, not the rock.

I think this shows
that potential (gravitational) energy SHOULD contribute mass to the
higher rock, and R&B's individual masses would be greater at distance
due to the increased potential (for gravity to accellerate them)

The "potential for gravity to accellerate them" also suggests
the energy is in the gravitational field, not in the rock.

Yes, that is the key, but perhaps you missed the earlier posts.
The question is how does the mass of the system compare to the
sum of the masses of the individuals in two different scenarios:

1) R&B are relatively far apart so gravitational effects are
negligible. They are rotating about a common point at high
speed tethered by a (massless) rope. Jeff and I agree the
mass of the system must be _greater_ than the sum of the
masses of R&B because the kinetic energy of their motion
must be included.

The "massless" rope is tensioned, as is a compressed spring. Otherwise
R&B would fly apart.

I was assuming the rope did not stretch. Energy is force
times distance (the integral if the force varies) so if
the rope doesn't stretch there is no energy stored.

Don't you mean "work is force times distance"-

'work done' is another phrase meaning energy. If you press on
a spring it gets shorter. The end moves against a force so work
is done and energy transferred from whatever creates the force
into the spring. When you press on a rock in comparison, there
is no significant energy stored.

tension in the rope is
equivalent to tension in the spring, is to tension in the mountain
supporting the rock

If the tension is "in the mountain supporting the rock", then
it would be the mass of the mountain that would increase, not
that of the rock.

,......and think about bringing opposing magnetic poles close- work is
done, but if I lock them in proximity, and then do the "calorific mass
equivalence test", will I see an increase in the mass of the magnets?)

No, the energy is stored in the magnetic field, not in the
magnets. I believe the mass of the system would change but
not that of the magnets.

Incidentally, a small technical note: gravitational effects
are a result of what is called the "stress-energy tensor" so
in fact pressure creates gravitational effects as well as
mass. I don't know GR well enough to comment further but I
thought I would mention it in case just to be complete. I
don't think it affects our discussion.

George

"George Dishman" wrote in message ...
"Jim Greenfield" wrote in message
om...
"George Dishman" wrote in message
...
"Jim Greenfield" wrote in message
om...
(George G. Dishman) wrote in message
om...
(Jim Greenfield) wrote in message
. com...

The rock is held up by the mountain. I just wonder does the higher
rock have more mass due to its altitude (and increased gravitational
potential energy)

No, AFAIK the rock has the same mass as if it was lower
and so does the planet. The mass of the system is probably
the same because raising the rock from the bottom needed
energy. Put the other way round, if you throw the rock
of the mountain into a pool. there is a loss of potential
energy that is converted first into kinetic as the rock
gains speed while falling, then into heat as it is slowed
in the water. Total energy remains constant so total mass
also remains constant.

Compessing the spring required (work) energy input as well.

That was the point, the spring was compressed but the rock
wasn't. You can nitpick that everything is compressible to
some degree but I took it that you changed from a spring to
a rock to indicate it was non-compressible.

Compression of the rock is not the issue: it is comparison of the
different energies- that stored in the spring due to compression, and
that (equal amount) of energy expended by lifting the rock against
gravity.

If gravity
is used to perform the compression, a larger piece of rock is needed
at the top of the mountain than in the valley, as gravity is less at
the higher level. But the same amount of energy (mass) is stored in
the spring (if the calorific thing is correct).

The force compressing the spring is due to gravity so the
source of the energy is the gravitational field, not the rock.

Maybe- but gravity has contributed indirectly mass to the spring (if
the calorific experiment re increased mass due to the stress is
correct). Gravity has increased the mass of the spring, but did the
increase come from the rock or the gravitational field?

I think this shows
that potential (gravitational) energy SHOULD contribute mass to the
higher rock, and R&B's individual masses would be greater at distance
due to the increased potential (for gravity to accellerate them)

The "potential for gravity to accellerate them" also suggests
the energy is in the gravitational field, not in the rock.

Yes, that is the key, but perhaps you missed the earlier posts.
The question is how does the mass of the system compare to the
sum of the masses of the individuals in two different scenarios:

1) R&B are relatively far apart so gravitational effects are
negligible. They are rotating about a common point at high
speed tethered by a (massless) rope. Jeff and I agree the
mass of the system must be _greater_ than the sum of the
masses of R&B because the kinetic energy of their motion
must be included.

The "massless" rope is tensioned, as is a compressed spring. Otherwise
R&B would fly apart.

I was assuming the rope did not stretch. Energy is force
times distance (the integral if the force varies) so if
the rope doesn't stretch there is no energy stored.

Don't you mean "work is force times distance"-

'work done' is another phrase meaning energy. If you press on
a spring it gets shorter. The end moves against a force so work
is done and energy transferred from whatever creates the force
into the spring. When you press on a rock in comparison, there
is no significant energy stored.

As above, I was lifting the rock, not compressing it.

tension in the rope is
equivalent to tension in the spring, is to tension in the mountain
supporting the rock

If the tension is "in the mountain supporting the rock", then
it would be the mass of the mountain that would increase, not
that of the rock.

Hasn't the mountain just become the equivalent of the mechanism
restraining the stressed spring?

,......and think about bringing opposing magnetic poles close- work is
done, but if I lock them in proximity, and then do the "calorific mass
equivalence test", will I see an increase in the mass of the magnets?)

No, the energy is stored in the magnetic field, not in the
magnets. I believe the mass of the system would change but
not that of the magnets.

If released, these magnets would behave very much like the spring. The
increased mass of the spring then must be due to the forces between
the iron atoms (obviously) in the spring, but outside the iron in the
case of the magnets?

Incidentally, a small technical note: gravitational effects
are a result of what is called the "stress-energy tensor" so
in fact pressure creates gravitational effects as well as
mass. I don't know GR well enough to comment further but I
thought I would mention it in case just to be complete. I
don't think it affects our discussion.

It might.If pressure contributes to increased gravity, and gravity
produces pressure, we have the makings of a perpetual motion machine,
or a black hole!?

Merry Christmas, George
(and I get to open my presents a day before you!!!)