"Donal" wrote in message
...

"Jeff Morris" wrote in message

Consider an astronaut space walking outside a space station. They both
float
together, feeling no force, although they are both in freefall in their
orbit. If the
astronaut moves to a lower orbit, he will feel a stronger pull and be
drawn in, unless
he speeds up to compensate. If the astronaut moves to a higher orbit, the
force is
reduced. As I said, the force can be calculated without consideration of
momentum.

I don't understant this. In orbit, momentum is the force that balances the
effect of gravity. Without momentum, your astronaut wouldn't "float" - he
would crash straight into the Earth.

Momentum is not a force. You're right that the astronaut has momentum, and that the
force of gravity alters his momentum. In fact, Force is defined by how much it
changes momentum. (Many people learn F=ma; in physics that is normally written as
F=dp/dt, or Force equals the rate of change of momentum with respect to time.)

My point is that you can determine the force on the astronaut without considering his
momentum. To figure out how the force would alter his orbit, you would probably take
momentum into account.

Remember, I'm not trying to calculate the tides, only to show how gravity can cause
two equal size bulges on the Earth.



In fact, I think that your use of the word "float" reveals that you don't
understand the situation at all. Your astronaut wouldn't feel any
difference between a free fall orbit and a headlong race into deepest
space, - would he?

So tell us, what is the difference? Floating in a space station is call "free fall"
because it feels the same as jumping off a cliff.

Furthermore, if he slowed down, then he would still feel
like he was floating -- apart from the temperature, and perhaps the braking
effect of the atmosphere.

Yes, but that's not the point. The point is, if he is in a lower orbit, he
experiences more gravity; in a higher orbit, less gravity. If his speed is not
adjusted to compensate, he will drift further away from the space station. Just like
the tides.


This makes me think that the orbiting "free-fall" astronaut doesn't feel
that he is floating at all.

Haven't you ever seen astronauts floating?

He must feel a constant force as his direction
of travel changes. I wonder if this has been documented on the Internet?

http://science.howstuffworks.com/weightlessness1.htm

What does your physics friend say about this?
He would probably deplore the lack of education in your country.





A purist might say momentum is considered because the mass and velocity of
the every
object in the system is folded together. And, the pure way force is
defined is by how
it changes momentum. But I don't think this is what you're talking about.


I'm not sure. I'm certainly *not* a purist.


Regards

Donal
--

Jeff Morris wrote:
No, you never said it wasn't needed, but you did minimize its importance, claiming
that it can't be used alone to predict the tides. You said:

"Even worse, the site then goes on to use the _differential_ expression
to calculate the ratio of forces between the moon and sun!"

Yes that's right for that site. The differential of the gavity equation
does not give a force it gives the rate of chnage of force with
distance. That's basic calculus.

You even produced a bogus formula to support your claim that centrifugal force varies
across the Earth. This would predict a tidal force 65 times stronger than
differential gravity. If that isn't minimalizing differential gravity, I don't know
what is.


Nothing bogus abot F=m r omaga^2. It's high school physics.


It is quite easy to explain the tides without Centrifugal force - I mentioned the
approach in one of my first posts. There is a gravitational force pulling the Earth
and Moon together. This creates an acceleration such that the Earth is in free fall,
and no net force is felt at the Earth's center. However, there is a larger force on
the Moon side, and a smaller force on the far side. Since the average force has been
accounted for it must be subtracted from these two forces, and the smaller force ends
up being the same magnitude as the larger, but in the opposite direction. Hence, two
tidal bulges, both caused solely by gravity.


Jeff, stop a moment and put what you just said into an equation.
According to you subtraction of a constant from both sides of an
inequality makes an equality?




You, however, have claimed that the variation of the centrifugal force from the near
side to the far creates a force that dominates the tides. Your formula predicts a
force 65 times greater than the differential formula, which would seem to create tides
100 feet or more.

That's because you've forgetten that the tiode is due to the difference
between forces -how many times do I have to repeat this????


Your explanation that land masses dampen these tides runs counter
to common experience. And, your formula predicts the Sun's contribution is only 1% of
the Moon's, which is clearly not the case.

No it doen't. It's your miscomprehension about the force balance.

As to your claim that the rotation around the barycenter "powers" the tides, well,
that's not true.

Jeff, don't be hysterical. Of course it's true. Furthermore, the
barycenter is not the same distance from all points on earth so the
centrifugal force varies across the earth.

Here is the equation for the force on a mass m of water on the far side
for a non-rotating earth (to keep it clear):

GmM/(R+r)^2 + GmE/r^2 - m(r+s)omega^2


On the near side:

GmM/(R-r)^2 + GmE/r^2 - m(r-s) omega^2

M is the moon mass, E the earth mass, s the distance from the center of
the earth to the barycenter, omega the angular velocity of the
earth-moon pair G the gravitational constant.


Now look very carefully at the three terms in each equation. The first
two are gravity, the third centrifugal. Two terms are different in both
cases. The gravity term is smaller on the far side and the centrifugal
term is bigger. On the near side the gavity is bigger and the
cenbtrifugal is smaller. That is the proof of my argument.

Cheers

Jeff Morris wrote:



Remember, I'm not trying to calculate the tides, only to show how gravity can cause
two equal size bulges on the Earth.



Equal? That'll be hard when it's non-linear. Think about it. Take as
long as you like.

Cheers

"Nav" wrote in message ...


Jeff Morris wrote:
No, you never said it wasn't needed, but you did minimize its importance, claiming
that it can't be used alone to predict the tides. You said:

"Even worse, the site then goes on to use the _differential_ expression
to calculate the ratio of forces between the moon and sun!"

Yes that's right for that site. The differential of the gavity equation
does not give a force it gives the rate of chnage of force with
distance. That's basic calculus.

Simply differentiating the gravitational force is not enough. But the equation
normally given for differential gravity is dimensionally correct.


You even produced a bogus formula to support your claim that centrifugal force
varies
across the Earth. This would predict a tidal force 65 times stronger than
differential gravity. If that isn't minimalizing differential gravity, I don't
know
what is.


Nothing bogus abot F=m r omaga^2. It's high school physics.

That formula is correct. Applying it with a varying "r" is bogus. That's the issue
here, but you haven't noticed it.





It is quite easy to explain the tides without Centrifugal force - I mentioned the
approach in one of my first posts. There is a gravitational force pulling the
Earth
and Moon together. This creates an acceleration such that the Earth is in free
fall,
and no net force is felt at the Earth's center. However, there is a larger force
on
the Moon side, and a smaller force on the far side. Since the average force has
been
accounted for it must be subtracted from these two forces, and the smaller force
ends
up being the same magnitude as the larger, but in the opposite direction. Hence,
two
tidal bulges, both caused solely by gravity.


Jeff, stop a moment and put what you just said into an equation.
According to you subtraction of a constant from both sides of an
inequality makes an equality?

I have no idea what you're saying.

If the force at the center is X, and on the near side is X+D, and on the far side is
X-D, then if you subtract the central force, the near side is D and the far side
is -D.






You, however, have claimed that the variation of the centrifugal force from the
near
side to the far creates a force that dominates the tides. Your formula predicts a
force 65 times greater than the differential formula, which would seem to create
tides
100 feet or more.

That's because you've forgetten that the tiode is due to the difference
between forces -how many times do I have to repeat this????


I just tried to follow all your instructions. You challenged me to work the math, I
did, it produced bogus answers. The difference in the centrifugal force that your
equation predicts is 65 times more that the difference in the gravitational force, so
subtracting still won't help much. Why don't you work out the math? And don't forget
to do it for the Sun also.



Your explanation that land masses dampen these tides runs counter
to common experience. And, your formula predicts the Sun's contribution is only
1% of
the Moon's, which is clearly not the case.

No it doen't. It's your miscomprehension about the force balance.

I'm just reporting the numbers your equation predicts.



As to your claim that the rotation around the barycenter "powers" the tides, well,
that's not true.

Jeff, don't be hysterical. Of course it's true. Furthermore, the
barycenter is not the same distance from all points on earth so the
centrifugal force varies across the earth.

This is your basic mistake. Even the web site you pointed to for support explicitly
says this is not true.
Let me repeat it again
http://www.co-ops.nos.noaa.gov/restles3.html

"While space does not permit here, it may be graphically demonstrated that, for such a
case of revolution without rotation as above enumerated, any point on the earth will
describe a circle which will have the same radius as the radius of revolution of the
center-of-mass of the earth around the barycenter. Thus, in Fig. 1, the magnitude of
the centrifugal force produced by the revolution of the earth and moon around their
common center of mass (G) is the same at point A or B or any other point on or beneath
the earth's surface. Any of these values is also equal to the centrifugal force
produced at the center-of-mass (C) by its revolution around the barycenter. This fact
is indicated in Fig. 2 by the equal lengths of the thin arrows (representing the
centrifugal force Fc) at points A, C, and B, respectively."

it continues with:
"While the effect of this centrifugal force is constant for all positions on the
earth, the effect of the external gravitational force produced by another astronomical
body may be different at different positions on the earth because the magnitude of the
gravitational force exerted varies with the distance of the attracting body."

Go on. Read it, think about it. You were quick to cite this page when you thought it
supported you.



Here is the equation for the force on a mass m of water on the far side
for a non-rotating earth (to keep it clear):

GmM/(R+r)^2 + GmE/r^2 - m(r+s)omega^2


On the near side:

GmM/(R-r)^2 + GmE/r^2 - m(r-s) omega^2

M is the moon mass, E the earth mass, s the distance from the center of
the earth to the barycenter, omega the angular velocity of the
earth-moon pair G the gravitational constant.

Nope. This is incorrect. The proper equations a

Farside:
GmM/(R+r)^2 + GmE/r^2 - m s omega^2

On the near side:
GmM/(R-r)^2 + GmE/r^2 - m s omega^2

Further, since GmE/r^2 = m s omega^2, we are left simply with:
Farside: GmM/(R+r)^2
On the near side: GmM/(R-r)^2

This leads to the traditional differential equation 2GmMr/R^3 as shown in
http://mb-soft.com/public/tides.html


Now look very carefully at the three terms in each equation. The first
two are gravity, the third centrifugal. Two terms are different in both
cases.

Nope. Only one term is different.

The gravity term is smaller on the far side and the centrifugal
term is bigger. On the near side the gavity is bigger and the
cenbtrifugal is smaller. That is the proof of my argument.

The proof fails because of a faulty assumption. Sorry Nav, Centrifugal Force is
constant. You can use it if you chose, but it doesn't really change the math and
isn't particularly interesting. The part of the equation that actually produces the
two tides is the differential gravity.

Yes, I should have said "roughly equal." But don't your equations show a huge
difference between the near and far side tides caused by the barycenter being closer
to the near side? Shouldn't that be a clue that something is amiss in your theory?


"Nav" wrote in message ...


Jeff Morris wrote:



Remember, I'm not trying to calculate the tides, only to show how gravity can
cause
two equal size bulges on the Earth.



Equal? That'll be hard when it's non-linear. Think about it. Take as
long as you like.

Cheers

"Jeff Morris" wrote in message
...

"Donal" wrote in message
...

"Jeff Morris" wrote in message

Consider an astronaut space walking outside a space station. They
both
float
together, feeling no force, although they are both in freefall in
their
orbit. If the
astronaut moves to a lower orbit, he will feel a stronger pull and be
drawn in, unless
he speeds up to compensate. If the astronaut moves to a higher orbit,
the
force is
reduced. As I said, the force can be calculated without consideration
of
momentum.

I don't understant this. In orbit, momentum is the force that balances
the
effect of gravity. Without momentum, your astronaut wouldn't "float" -
he
would crash straight into the Earth.

Momentum is not a force. You're right that the astronaut has momentum,
and that the
force of gravity alters his momentum. In fact, Force is defined by how
much it
changes momentum. (Many people learn F=ma; in physics that is normally
written as
F=dp/dt, or Force equals the rate of change of momentum with respect to
time.)

Ok. What force opposes gravity so that a body may remain in orbit?


My point is that you can determine the force on the astronaut without
considering his
momentum.

In which case, there must be a "force" that is counteracting the effect of
gravity. After all, gravity is trying to pull the orbiting Astronaut
straight towards Earth. There must be another force that is opposing
gravity.



To figure out how the force would alter his orbit, you would probably take
momentum into account.

Remember, I'm not trying to calculate the tides, only to show how gravity
can cause
two equal size bulges on the Earth.

Do you think that centrifugal force plays any part? If so, what do you
think the ratio is between the centrifugal and differential gravity forces?




In fact, I think that your use of the word "float" reveals that you
don't
understand the situation at all. Your astronaut wouldn't feel any
difference between a free fall orbit and a headlong race into deepest
space, - would he?

So tell us, what is the difference?

Acceleration.

Floating in a space station is call "free fall"
because it feels the same as jumping off a cliff.

Furthermore, if he slowed down, then he would still feel
like he was floating -- apart from the temperature, and perhaps the
braking
effect of the atmosphere.

Yes, but that's not the point. The point is, if he is in a lower orbit,
he
experiences more gravity; in a higher orbit, less gravity. If his speed
is not
adjusted to compensate, he will drift further away from the space station.
Just like
the tides.


This makes me think that the orbiting "free-fall" astronaut doesn't feel
that he is floating at all.

Haven't you ever seen astronauts floating?

Yes.... but they are constantly changing direction.... and therefor they
should be aware of the effects of acceleration.



He must feel a constant force as his direction
of travel changes. I wonder if this has been documented on the
Internet?

http://science.howstuffworks.com/weightlessness1.htm

That is a very simplistic explanation. It refers to the fact that the
astronauts will feel the acceleration at take-off, and yet it doesn't seem
to understand that a change of direction is also acceleration.

We humans can detect acceleration. If you sit in an automobile with your
eyes closed, then you can feel an increase or decrease in speed .... or a
change of direction!! As the astronauts are subjected to a constant change
of direction, I suspect that they might not feel that they are completely
free-floating.



What does your physics friend say about this?
He would probably deplore the lack of education in your country.

Ask him anyway!

Perhaps, if you allowed him to read the thread, he might be amazed at your
lack of reading ability. After all, I've already explained that I gave up
Physics at an early stage.


Regards


Donal
--

"Donal" wrote in message
...
I don't understant this. In orbit, momentum is the force that balances
the
effect of gravity. Without momentum, your astronaut wouldn't "float" -
he
would crash straight into the Earth.

Momentum is not a force. You're right that the astronaut has momentum,
and that the
force of gravity alters his momentum. In fact, Force is defined by how
much it
changes momentum. (Many people learn F=ma; in physics that is normally
written as
F=dp/dt, or Force equals the rate of change of momentum with respect to
time.)

Ok. What force opposes gravity so that a body may remain in orbit?

Nothing. The body remains in orbit because it has enough forward velocity (momentum
also, but its the velocity that counts) so that while it "falls" into the Earth, the
forward velocity keeps the body from hitting the Earth. For a low orbit, we normally
thing of circular orbits, but for high orbits they can be quite eccentic. If the
Earth were a "point source" it would only take a small velocity to stay in orbit. Of
course, there are many ways to calculate this - somtimes its done in terms of
"energy," other times as "delta V" so if you want to say the momentum of the body
keeps it from falling into the Earth, that's OK.




My point is that you can determine the force on the astronaut without
considering his
momentum.

In which case, there must be a "force" that is counteracting the effect of
gravity. After all, gravity is trying to pull the orbiting Astronaut
straight towards Earth. There must be another force that is opposing
gravity.


See above. If the body were motionless (WRT Earth), it would fall directly in. But
if it has any velocity, its has a chance of missing it. The reason why I say velocity
is important, and not momentum, is that two bodies of different mass (and hence,
different momentum) will float together in orbit. Of course, if you want to calculate
the amount of fuel to burn, momentum becomes important.

On Earth, we always have air resistance, and other forms of friction, so momentum is
more significant.




To figure out how the force would alter his orbit, you would probably take
momentum into account.

Remember, I'm not trying to calculate the tides, only to show how gravity
can cause
two equal size bulges on the Earth.

Do you think that centrifugal force plays any part? If so, what do you
think the ratio is between the centrifugal and differential gravity forces?


It is possible to look at the problem without considering Centrifugal Force. However,
even if you use CF, it is a constant, and equal to the net gravitation force. So if
it helps to expain how there can be a force away from the moon, that's OK, but you
must remember that it is the gravitation force that varies, so that's where the
interesting math comes from.





In fact, I think that your use of the word "float" reveals that you
don't
understand the situation at all. Your astronaut wouldn't feel any
difference between a free fall orbit and a headlong race into deepest
space, - would he?

So tell us, what is the difference?

Acceleration.

No, the acceleration is the same, more or less. (Not counting the difference in
distance from the Earth, or air resistance, etc.) The only real difference is that
the astronaut has enough velocity (hopefully) to miss the Earth as he falls.



Floating in a space station is call "free fall"
because it feels the same as jumping off a cliff.

Furthermore, if he slowed down, then he would still feel
like he was floating -- apart from the temperature, and perhaps the
braking
effect of the atmosphere.

Yes, but that's not the point. The point is, if he is in a lower orbit,
he
experiences more gravity; in a higher orbit, less gravity. If his speed
is not
adjusted to compensate, he will drift further away from the space station.
Just like
the tides.


This makes me think that the orbiting "free-fall" astronaut doesn't feel
that he is floating at all.

Haven't you ever seen astronauts floating?

Yes.... but they are constantly changing direction.... and therefor they
should be aware of the effects of acceleration.


I must admit this subtlety has perplexed me - clearly the don't feel the G force,
since it the same as a car in a tight turn. But I keep thinking it should be
detectable, if only because the path is curving.



He must feel a constant force as his direction
of travel changes. I wonder if this has been documented on the
Internet?

http://science.howstuffworks.com/weightlessness1.htm

That is a very simplistic explanation. It refers to the fact that the
astronauts will feel the acceleration at take-off, and yet it doesn't seem
to understand that a change of direction is also acceleration.

We humans can detect acceleration. If you sit in an automobile with your
eyes closed, then you can feel an increase or decrease in speed .... or a
change of direction!! As the astronauts are subjected to a constant change
of direction, I suspect that they might not feel that they are completely
free-floating.


Of course, from a General Relativity, Gravity Well point of view, the obital path is a
straight line in curved space. I should know the answer here - let me cogitate ...



What does your physics friend say about this?
He would probably deplore the lack of education in your country.

Ask him anyway!

Actually, it was Scout's friend. However, you should remember I majored in physics
and worked for NASA doing spacecraft navigation. I may be rusty now, but 25 years ago
I really knew this stuf!



Perhaps, if you allowed him to read the thread, he might be amazed at your
lack of reading ability. After all, I've already explained that I gave up
Physics at an early stage.

I haven't forgotton that. It was just a little dig since usually you Brits complain
about our sorry education.

Jeff Morris wrote:



I must admit this subtlety has perplexed me - clearly the don't feel the G force,
since it the same as a car in a tight turn. But I keep thinking it should be
detectable, if only because the path is curving.

(1) How tight is the turn and (2) how big is the imbalance between
centrifugal and garvitational forces for such a small body?



Of course, from a General Relativity, Gravity Well point of view, the obital path is a
straight line in curved space. I should know the answer here - let me cogitate ...


Eat some bran? :-P

Cheers

"Jeff Morris" wrote in message
...

"Donal" wrote in message
...

In fact, I think that your use of the word "float" reveals that you
don't
understand the situation at all. Your astronaut wouldn't feel any
difference between a free fall orbit and a headlong race into
deepest
space, - would he?

So tell us, what is the difference?

Acceleration.

No, the acceleration is the same, more or less. (Not counting the
difference in
distance from the Earth, or air resistance, etc.) The only real
difference is that
the astronaut has enough velocity (hopefully) to miss the Earth as he
falls.


That was a very revealing answer.

Earlier in the thread you were confident enough about your position to
question my lack of education.

Now you seem to think that an object travelling at "x" miles an hour in a
straight line ("headlong into space") has the same acceleration as a body
travelling at "x" miles an hour in orbit.

It's probably time that you consulted your physics partner. Before you let
him read what you have written, you should make sure that there is a
cloakroom near the PC. Otherwise, have a potty close at hand - because he
is really going to **** himself when he reads your words.



Haven't you ever seen astronauts floating?

Yes.... but they are constantly changing direction.... and therefor they
should be aware of the effects of acceleration.


I must admit this subtlety has perplexed me - clearly the don't feel the G
force,
since it the same as a car in a tight turn. But I keep thinking it should
be
detectable, if only because the path is curving.

Yeah. It's called acceleration - a bit like G-force.

Please note that "a bit" = "exactly" in European understatement.




He must feel a constant force as his direction
of travel changes. I wonder if this has been documented on the
Internet?

http://science.howstuffworks.com/weightlessness1.htm

That is a very simplistic explanation. It refers to the fact that the
astronauts will feel the acceleration at take-off, and yet it doesn't
seem
to understand that a change of direction is also acceleration.

We humans can detect acceleration. If you sit in an automobile with
your
eyes closed, then you can feel an increase or decrease in speed .... or
a
change of direction!! As the astronauts are subjected to a constant
change
of direction, I suspect that they might not feel that they are
completely
free-floating.


Of course, from a General Relativity, Gravity Well point of view, the
obital path is a
straight line in curved space. I should know the answer here - let me
cogitate ...

No, you don't know the answer. Trust me. Despite my dreadful education, I
am confident that someone who doesn't understand the basic principles of
acceleration is incapable of getting their head around the TGR. Cogitation
would be a complete waste of your time.





What does your physics friend say about this?
He would probably deplore the lack of education in your country.

Ask him anyway!

Actually, it was Scout's friend. However, you should remember I majored
in physics
and worked for NASA doing spacecraft navigation. I may be rusty now, but
25 years ago
I really knew this stuf!

GULP!!!

So why do you not seem to understand the difference between "velocity" and
"speed"?





Perhaps, if you allowed him to read the thread, he might be amazed at
your
lack of reading ability. After all, I've already explained that I gave
up
Physics at an early stage.

I haven't forgotton that. It was just a little dig since usually you
Brits complain
about our sorry education.

Little digs are very welcome. I appreciate a good insult - and calling me
"you Brits" is definitelay a reasonable insult. You're not lagging too far
behind Joe! He called me a Brit about six months ago.


Regards


Donal
--

"Donal" wrote in message
...

"Jeff Morris" wrote in message
...

"Donal" wrote in message
...

In fact, I think that your use of the word "float" reveals that you
don't
understand the situation at all. Your astronaut wouldn't feel any
difference between a free fall orbit and a headlong race into
deepest
space, - would he?

So tell us, what is the difference?

Acceleration.

No, the acceleration is the same, more or less. (Not counting the
difference in
distance from the Earth, or air resistance, etc.) The only real
difference is that
the astronaut has enough velocity (hopefully) to miss the Earth as he
falls.


That was a very revealing answer.

Earlier in the thread you were confident enough about your position to
question my lack of education.

Now you seem to think that an object travelling at "x" miles an hour in a
straight line ("headlong into space") has the same acceleration as a body
travelling at "x" miles an hour in orbit.

Actually, I was saying that any body under the influence of Earth's gravity will
"feel" the same force, more or less. It doesn't matter whether its falling off a
cliff, in orbit, or leaving orbit into space. The pull of the Earth is the same,
adjusting of course, for the distance.

It turns out that most non-technically inclined people think that these are three
completely different situations (as they are, in some respects) but in all three the
body is being accelerated by gravity exactly the same.

Sorry Donal, that's physics. If you had taken a physics course, you might understand
that. Your continued rant is making you look rather silly.




It's probably time that you consulted your physics partner. Before you let
him read what you have written, you should make sure that there is a
cloakroom near the PC. Otherwise, have a potty close at hand - because he
is really going to **** himself when he reads your words.


Actually, my partner was given a PhD in Astrophysics only if he promised to continue
as a programmer.

How about my former boss, he won the Nobel Prize in Physics a few years ago.





Haven't you ever seen astronauts floating?

Yes.... but they are constantly changing direction.... and therefor they
should be aware of the effects of acceleration.


I must admit this subtlety has perplexed me - clearly the don't feel the G
force,
since it the same as a car in a tight turn. But I keep thinking it should
be
detectable, if only because the path is curving.

Yeah. It's called acceleration - a bit like G-force.

Please note that "a bit" = "exactly" in European understatement.




He must feel a constant force as his direction
of travel changes. I wonder if this has been documented on the
Internet?

http://science.howstuffworks.com/weightlessness1.htm

That is a very simplistic explanation. It refers to the fact that the
astronauts will feel the acceleration at take-off, and yet it doesn't
seem
to understand that a change of direction is also acceleration.

We humans can detect acceleration. If you sit in an automobile with
your
eyes closed, then you can feel an increase or decrease in speed .... or
a
change of direction!! As the astronauts are subjected to a constant
change
of direction, I suspect that they might not feel that they are
completely
free-floating.


Of course, from a General Relativity, Gravity Well point of view, the
obital path is a
straight line in curved space. I should know the answer here - let me
cogitate ...

No, you don't know the answer. Trust me. Despite my dreadful education, I
am confident that someone who doesn't understand the basic principles of
acceleration is incapable of getting their head around the TGR. Cogitation
would be a complete waste of your time.

OK, why don't you explain to us? How does someone in orbit determine they are
travelling a curved path. No fair using outside references. No fair looking for
effects over time. What sort of device will instantaneously tell you the local
gravitational field? I'm not saying its impossible, but its not obvious.








What does your physics friend say about this?
He would probably deplore the lack of education in your country.

Ask him anyway!

Actually, it was Scout's friend. However, you should remember I majored
in physics
and worked for NASA doing spacecraft navigation. I may be rusty now, but
25 years ago
I really knew this stuf!

GULP!!!

So why do you not seem to understand the difference between "velocity" and
"speed"?

Where did I talk about that??? You're losing it Donal.

So, please explain to all of us: what is the difference between velocity and speed,
and how was that relevent to anything I said?








Perhaps, if you allowed him to read the thread, he might be amazed at
your
lack of reading ability. After all, I've already explained that I gave
up
Physics at an early stage.

I haven't forgotton that. It was just a little dig since usually you
Brits complain
about our sorry education.

Little digs are very welcome. I appreciate a good insult - and calling me
"you Brits" is definitelay a reasonable insult. You're not lagging too far
behind Joe! He called me a Brit about six months ago.


Oh, sorry, I forgot. That explains your lack of education.