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Partner cracklover


Oct 9, 2008, 8:25 AM
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jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO


colatownkid


Oct 9, 2008, 8:29 AM
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Re: [curt] tension [In reply to]
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curt wrote:
jdefazio wrote:
JT512 wrote:
A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.
Jay
curt wrote:
Although most of us know what you meant, your statement is incorrect.

JT512 wrote:
..."your rong"...

curt wrote:
..."no yor rong"...

^^^repeat ad-infinitum.

Curt is correct here regarding the semantics of Jay's statement. The rest is like arguing over who is better looking, since the answer depends on who you ask (i.e. what reference frame they are in)...

Fine. Then I'm better looking too. Cool

Curt

i thought that was obvious.


petsfed


Oct 9, 2008, 8:44 AM
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Re: [cracklover] tension [In reply to]
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cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.


jt512


Oct 9, 2008, 9:07 AM
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Re: [petsfed] tension [In reply to]
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petsfed wrote:
cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay


curt


Oct 9, 2008, 9:23 AM
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jt512 wrote:
petsfed wrote:
cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay

Jay,

The problem is that you said the two cases are "indistinguishable" and that simply isn't true. Again, if I plot position versus time for an object moving at constant velocity I will get a straight line. If I plot position versus time for a stationary object, I will get a point. If I now decide to change my frame of reference and "sit" on the object moving at constant velocity, I may indeed opt to say that the "stationary" object is the one in motion, relative to me. It doesn't matter--because if I then create the same two graphs of position versus time they are merely reversed and I can still easily differentiate one graph from the other.

Curt


petsfed


Oct 9, 2008, 9:31 AM
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Both.

It is not the case that the cannonball has both zero and non-zero velocity. Which is what you seem to be implying.

From the same frame of reference, zero and non-zero velocities are clearly different. In fact, so long as you measure both velocities from the same frame of reference (no matter what that reference frame is), the two objects will have different velocities. This should be apparent. To say that v=0 is the same as v=30 is only true if the two measurements are taken from two very specific frames. There is no single reference frame where v=0 is the same as v=30.

Which is (I think) what you're not clarifying in your statement that the two are equivalent.


jdefazio


Oct 9, 2008, 9:53 AM
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jt512 wrote:
Some of you physics guys have lost the big picture... Come on: this is high school stuff...

Agreed.
OMG TEH NUT0NZ L4WZ R TEH S4ME!!! OH NOEZZZ!!!!!!!
---

The recent debate, though, was initiated by the statement:

jt512 wrote:
A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Of course some of us know that you are really talking about Galilean invariance. You do seem like a bright fellow after all! Wink
As a reputed stickler for grammar and semantics, however, you did leave this wide open for a firm-handed bitchslapping.


jt512


Oct 9, 2008, 9:54 AM
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Re: [curt] tension [In reply to]
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curt wrote:
jt512 wrote:
petsfed wrote:
cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay

Jay,

The problem is that you said the two cases are "indistinguishable" and that simply isn't true. Again, if I plot position versus time for an object moving at constant velocity I will get a straight line. If I plot position versus time for a stationary object, I will get a point. If I now decide to change my frame of reference and "sit" on the object moving at constant velocity, I may indeed opt to say that the "stationary" object is the one in motion, relative to me. It doesn't matter--because if I then create the same two graphs of position versus time they are merely reversed and I can still easily differentiate one graph from the other.

Curt

You may be able to differentiate one graph from the other, but it's just begging the question. Those two graphs will completely disagree as to which object is the stationary one. How much more indistinguishable do you want than that? And in any other frame of reference neither object would appear stationary.

I hate to appeal to authority, and hate even more to appeal to Wikipedia as an authority, but the Wikipedia article that colatownkid cited completely agrees with me: "By making rest physically indistinguishable from non-zero constant velocity, Newton's first law directly connects inertia with the concept of relative velocities. ... Simple experiments showed that Galileo's understanding of the equivalence of constant velocity and rest to be correct."

Jay


Partner cracklover


Oct 9, 2008, 10:26 AM
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Jay, you're arguing a point we all agree with. I believe that the trouble is not with what you think, but with what you wrote: "A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable. "

The simplest interpretation of the above is that one cannot distinguish two objects at zero and non-zero velocities (in a given frame of reference) from each other. You merely have to clarify that that is not what you meant, and all is solved (for those even able to follow this).

GO


curt


Oct 9, 2008, 10:29 AM
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jt512 wrote:
curt wrote:
jt512 wrote:
petsfed wrote:
cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay

Jay,

The problem is that you said the two cases are "indistinguishable" and that simply isn't true. Again, if I plot position versus time for an object moving at constant velocity I will get a straight line. If I plot position versus time for a stationary object, I will get a point. If I now decide to change my frame of reference and "sit" on the object moving at constant velocity, I may indeed opt to say that the "stationary" object is the one in motion, relative to me. It doesn't matter--because if I then create the same two graphs of position versus time they are merely reversed and I can still easily differentiate one graph from the other.

Curt

You may be able to differentiate one graph from the other, but it's just begging the question. Those two graphs will completely disagree as to which object is the stationary one. How much more indistinguishable do you want than that?

Indistinguishable would mean the graphs of position versus time for the two objects would look identical--i.e. we can't tell them apart. That is clearly not the case here. In fact (as others have commented) from any given frame of reference the two graphs of position versus time will look different for the two objects. You can go with the wiki quote from colatown kid if you like--but petsfed (who has a degree in physics) jdefazio, robdotcalm and cracklover all disagree with you. I still hope rgold will comment.

Curt


colatownkid


Oct 9, 2008, 10:30 AM
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jt512 wrote:
petsfed wrote:
cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay

i see Jay has read my post about Galilean relativity.


petsfed


Oct 9, 2008, 10:35 AM
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I have a what now?

/still hasn't finished college


(This post was edited by petsfed on Oct 9, 2008, 10:35 AM)


jdefazio


Oct 9, 2008, 10:42 AM
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petsfed wrote:
I have a what now?

/still hasn't finished college

Pffft. Then, I guess...STFU physics N00B! WinkWinkWink


jt512


Oct 9, 2008, 10:43 AM
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petsfed wrote:
From the same frame of reference, zero and non-zero velocities are clearly different. In fact, so long as you measure both velocities from the same frame of reference (no matter what that reference frame is), the two objects will have different velocities. This should be apparent. To say that v=0 is the same as v=30 is only true if the two measurements are taken from two very specific frames. There is no single reference frame where v=0 is the same as v=30.

True, but there is another frame of reference in which the velocities of the two objects are switched, and both frames of reference are equally valid.

Jay


sungam


Oct 9, 2008, 10:44 AM
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BTW, just to stir the pot, both your sailor and your landlubber are rong.
The cannon ball is staying still (duh) and the other things are moving around it.


curt


Oct 9, 2008, 10:47 AM
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petsfed wrote:
I have a what now?

/still hasn't finished college

Well, you're close, right? And it is a degree in physics--isn't it?

Curt


colatownkid


Oct 9, 2008, 10:49 AM
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sungam wrote:
BTW, just to stir the pot, both your sailor and your landlubber are rong.
The cannon ball is staying still (duh) and the other things are moving around it.

oh snap.


curt


Oct 9, 2008, 10:49 AM
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jt512 wrote:
petsfed wrote:
From the same frame of reference, zero and non-zero velocities are clearly different. In fact, so long as you measure both velocities from the same frame of reference (no matter what that reference frame is), the two objects will have different velocities. This should be apparent. To say that v=0 is the same as v=30 is only true if the two measurements are taken from two very specific frames. There is no single reference frame where v=0 is the same as v=30.

True, but there is another frame of reference in which the velocities of the two objects are switched, and both frames of reference are equally valid.

Jay

But, they will always be different--and hence not "indistinguishable."

Curt


petsfed


Oct 9, 2008, 11:00 AM
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curt wrote:
petsfed wrote:
I have a what now?

/still hasn't finished college

Well, you're close, right? And it is a degree in physics--isn't it?

Curt

Yeah, I have to take circuits in the spring, and then I'll have a BS in Astrophysics. As opposed to a strong desire to bs about physics.


jt512


Oct 9, 2008, 11:24 AM
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curt wrote:
jt512 wrote:
petsfed wrote:
From the same frame of reference, zero and non-zero velocities are clearly different. In fact, so long as you measure both velocities from the same frame of reference (no matter what that reference frame is), the two objects will have different velocities. This should be apparent. To say that v=0 is the same as v=30 is only true if the two measurements are taken from two very specific frames. There is no single reference frame where v=0 is the same as v=30.

True, but there is another frame of reference in which the velocities of the two objects are switched, and both frames of reference are equally valid.

Jay

But, they will always be different--and hence not "indistinguishable."

Curt

I think you may be forgetting what I said was indistinguishable. What I said was that rest and constant velocity are indistinguishable. That the two objects have different velocities is clear. But if I were to hand you both graphs you would have no way to distinguish between which one is at rest and which one isn't. That's because the distinction is meaningless in Newtonian physics. The question implies that Newtonian velocity is an absolute quantity, which it isn't; it is a relative quantity.

Jay


Valarc


Oct 9, 2008, 11:40 AM
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jt512 wrote:
petsfed wrote:
From the same frame of reference, zero and non-zero velocities are clearly different. In fact, so long as you measure both velocities from the same frame of reference (no matter what that reference frame is), the two objects will have different velocities. This should be apparent. To say that v=0 is the same as v=30 is only true if the two measurements are taken from two very specific frames. There is no single reference frame where v=0 is the same as v=30.

True, but there is another frame of reference in which the velocities of the two objects are switched, and both frames of reference are equally valid.

Jay

I'm no Rgold, but this is korrekt.

Although, if you listen to the cosmologists, we should measure everything relative to the cosmic microwave background, which would give a preferred, or (egads!) "absolute" frame of reference to which we should compare things.

I've never liked cosmologists, personally.


jdefazio


Oct 9, 2008, 11:44 AM
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petsfed


Oct 9, 2008, 11:45 AM
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Re: [jt512] tension [In reply to]
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jt512 wrote:
The question implies that Newtonian velocity is an absolute quantity, which it isn't; it is a relative quantity.

Jay

Which should be readily apparent from the definition of an inertial frame, the kind of reference frame that Newtonian mechanics is derived for. Any inertial reference frame is basically equivalent to any other in terms of how the laws of physics work.

There are non-inertial frames of reference though.



Thank you xkcd.org


petsfed


Oct 9, 2008, 11:46 AM
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Re: [Valarc] tension [In reply to]
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Valarc wrote:
I've never liked cosmologists, personally.

That's ok. They think you smell funny.


Partner cracklover


Oct 9, 2008, 11:49 AM
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Re: [jt512] tension [In reply to]
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jt512 wrote:
robdotcalm wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

On the other hand, both cars were moving at approximately 66,600 mi/hr around the sun, as were you. Still think you can distinguish constant velocity from zero velocity?

Jay

Here is the issue, in a nutshell: The phrase "at rest" or "in motion" only have meaning within a given frame of reference. So when robdot says he can tell the difference, he is correct, from his frame of reference.

While it is true that for newtonian physics to work, a frame of reference is not required, that doesn't mean that one cannot have a frame of reference and speak intelligently about the motions of objects within that frame. This is the point which JT is unwilling to acknowledge.

GO

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