How much does the Earth weigh?

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Offline JP

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How much does the Earth weigh?
« Reply #50 on: 02/12/2010 08:05:29 »
Its just a definition thing


It's more fundamental than that. Mass is a property of matter. Weight is a measure of the interaction between matter.

It's still partly a definition thing.  There are two accepted definitions of weight:
1) the force required to keep you stationary against gravity in your particular reference frame, i.e. this is the one that a scale would measure if put under your feet
2) the force exerted on you by gravity (in a Galilean reference frame, I believe).

The big difference is that if I'm in a freely falling elevator, within my reference frame, a scale will read zero, but gravity is still pulling on me with a force of ~700 Newtons.  Both answers are correct, since there are two alternative definitions of weight.  They both agree, however, if I'm standing on the earth's surface.

This is precisely why weight isn't commonly used in physics, and certainly not used unless you make it clear under what circumstances you're using it.

So the answer to the earth's weight could be either zero, or some value obtained by computing the force that the sun exerts on the earth.  What the person who posed the question probably wanted to know was the mass, since I'm not sure what use it would be to know either definition of the weight...

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Actually, the definitions might be even more muddled, since the earth isn't itself a Galilean reference frame due to its orbit and rotation, but I suppose it's close enough...

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Offline Foolosophy

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« Reply #51 on: 02/12/2010 08:13:36 »
Its just a definition thing


It's more fundamental than that. Mass is a property of matter. Weight is a measure of the interaction between matter.

So the answer to the earth's weight could be either zero, or some value obtained by computing the force that the sun exerts on the earth.  What the person who posed the question probably wanted to know was the mass, since I'm not sure what use it would be to know either definition of the weight...

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Actually, the definitions might be even more muddled, since the earth isn't itself a Galilean reference frame due to its orbit and rotation, but I suppose it's close enough...

Is this some sort of quantum defintion of weight? Could be zero could be a positive value?

Would you question that a astronauts weight is equal to zero when in a space station that is in free fall motion around the earth?

Is the astronauts weight either zero or some computed force value?

The definition of weight is unambiguous 
 
« Last Edit: 02/12/2010 08:18:25 by Foolosophy »

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Offline JP

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« Reply #52 on: 02/12/2010 08:20:12 »
Its just a definition thing


It's more fundamental than that. Mass is a property of matter. Weight is a measure of the interaction between matter.

So the answer to the earth's weight could be either zero, or some value obtained by computing the force that the sun exerts on the earth.  What the person who posed the question probably wanted to know was the mass, since I'm not sure what use it would be to know either definition of the weight...

-------------

Actually, the definitions might be even more muddled, since the earth isn't itself a Galilean reference frame due to its orbit and rotation, but I suppose it's close enough...

Is this some sort of quantum defintion of weight? Could be zero could be a positive value?

Would you question that a astronauts weight is equal to zero when in a scape station that is in free fall motion around the earth?

Is the astronauts weight either zero or some computed force value?

The definition of weight is unambiguous 
 

Quantum has nothing to do with any of this.

These are the two textbook definitions of weight.  The definition is indeed ambiguous, as there are two possibilities. 

If you asked me what an astronaut's weight was in the space station, I would ask you which definition you wanted to use, since they would give you two different answers. 

Then I'd probably tell you that asking for the weight is confusing, and that it would be better to ask for force, measured in a particular reference frame, or mass.
« Last Edit: 02/12/2010 08:22:29 by JP »

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Offline Geezer

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« Reply #53 on: 02/12/2010 08:27:34 »
quote author=JP link=topic=35346.msg332910#msg332910 date=1291277129]
Its just a definition thing


It's more fundamental than that. Mass is a property of matter. Weight is a measure of the interaction between matter.

It's still partly a definition thing.  There are two accepted definitions of weight:
1) the force required to keep you stationary against gravity in your particular reference frame, i.e. this is the one that a scale would measure if put under your feet
2) the force exerted on you by gravity (in a Galilean reference frame, I believe).

The big difference is that if I'm in a freely falling elevator, within my reference frame, a scale will read zero, but gravity is still pulling on me with a force of ~700 Newtons.  Both answers are correct, since there are two alternative definitions of weight.  They both agree, however, if I'm standing on the earth's surface.

This is precisely why weight isn't commonly used in physics, and certainly not used unless you make it clear under what circumstances you're using it.

So the answer to the earth's weight could be either zero, or some value obtained by computing the force that the sun exerts on the earth.  What the person who posed the question probably wanted to know was the mass, since I'm not sure what use it would be to know either definition of the weight...

-------------

Actually, the definitions might be even more muddled, since the earth isn't itself a Galilean reference frame due to its orbit and rotation, but I suppose it's close enough...
[/quote]

Much as I hate to disagree with my learned colleague JP, that's bollocks  [;D]

Weight is what you measure with a calibrated spring, or a comparitive reference on some sort of balance in a gravitational field. It is always a relative measurement. e.g. the Earth equals x Moons

On the other hand, mass can be quantified quite independently of any gravitational field.
« Last Edit: 02/12/2010 08:29:13 by Geezer »
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Offline Geezer

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« Reply #54 on: 02/12/2010 08:35:01 »
I'm not quite sure what happened there, but this is what I tried to post:


Much as I hate to disagree with my learned colleague JP, that's bollocks 

Weight is what you measure with a calibrated spring, or a comparitive reference on some sort of balance in a gravitational field. It is always a relative measurement. e.g. the Earth equals x Moons

On the other hand, mass can be quantified quite independently of any gravitational field.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline Foolosophy

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« Reply #55 on: 02/12/2010 08:54:18 »

On the other hand, mass can be quantified quite independently of any gravitational field.


This is true for "Newtonian non-relativistic mass"

relativistic mass varies with all sorts of variables and conditions - including gravitational fields

As it is now the weight of the earth is zero

You will find that if an examination question asked the student "What is the weight of the earth"? any answer other than zero will be wrong.

« Last Edit: 02/12/2010 08:56:17 by Foolosophy »

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Offline JP

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« Reply #56 on: 02/12/2010 08:57:42 »
It's hard to argue against "that isn't true," so here's some sources:

National Institute of Standards and Technology: Weight is mass in kg times g, which is the gravitational acceleration at the earth's surface~9.8 m/s2.  NIST would disagree with you and say that your weight is the same no matter where you are, with nothing relative about it.  In other words, NIST is saying that weight is equal to mass multiplied by a constant that only makes sense at the earth's surface.
(See: http://physics.nist.gov/Pubs/SP330/sp330.pdf, p. 52)

ISO (International Organization for Standardization) defines it in terms of the apparent gravitational acceleration in some reference frame.  I.e. in their case, you would weigh less on the surface of the moon, and be weightless in a falling elevator.  [I can't find their publication online, but it's in International Organization for Standardization, International Standard ISO 31-3. (1992). “Quantities and units. Part 3, Mechanics.” (Geneva, Switzerland)]

For the problems with weight definitions (and why weight is fairly useless as a technical term) see, for example: http://books.google.com/books?hl=en&lr=&id=CoB5w9Km0mUC&oi=fnd&pg=PA45#v=onepage&q&f=false

also: http://sites.huji.ac.il/science/stc/staff_h/Igal/Research%20Articles/Weight-AJP.pdf

If that doesn't make it clear that the definition ambiguous, I don't know what will.

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Offline Geezer

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« Reply #57 on: 02/12/2010 09:01:55 »
JP beat me to it!


On the other hand, mass can be quantified quite independently of any gravitational field.


This is true for "Newtonian non-relativistic mass"

relativistic mass varies with all sorts of variables and conditions - including gravitational fiels

As it is now the weight of the earth is zero

No it ain't.

Any differences between Newtonian and relativistic masses in these frames are extremely small. Relativity did not cancel out Newton's work, it just refined it in certain situations.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline Foolosophy

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« Reply #58 on: 02/12/2010 09:07:50 »
It's hard to argue against "that isn't true," so here's some sources:

National Institute of Standards and Technology: Weight is mass in kg times g, which is the gravitational acceleration at the earth's surface~9.8 m/s2.  NIST would disagree with you and say that your weight is the same no matter where you are, with nothing relative about it.  In other words, NIST is saying that weight is equal to mass multiplied by a constant that only makes sense at the earth's surface.
(See: http://physics.nist.gov/Pubs/SP330/sp330.pdf, p. 52)

ISO (International Organization for Standardization) defines it in terms of the apparent gravitational acceleration in some reference frame.  I.e. in their case, you would weigh less on the surface of the moon, and be weightless in a falling elevator.  [I can't find their publication online, but it's in International Organization for Standardization, International Standard ISO 31-3. (1992). “Quantities and units. Part 3, Mechanics.” (Geneva, Switzerland)]

For the problems with weight definitions (and why weight is fairly useless as a technical term) see, for example: http://books.google.com/books?hl=en&lr=&id=CoB5w9Km0mUC&oi=fnd&pg=PA45#v=onepage&q&f=false

also: http://sites.huji.ac.il/science/stc/staff_h/Igal/Research%20Articles/Weight-AJP.pdf

If that doesn't make it clear that the definition ambiguous, I don't know what will.

But you just said that weight is the product of mass and "g" and your weight will be the same wherever you are???

g is a variable isnt it? what is the value of g at the centre of the earth? What is your weight at the centre of the earth?


When you see an astronaut floating in a space station are they experiencing weightlessness or masslessness?


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Offline JP

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« Reply #59 on: 02/12/2010 09:14:13 »
Here's the problem, which is expressed well by the NIST and ISO disagreement on standards.  Both say the equation for weight is

W=gm, where W is weight, m is mass and g is a number. 

NIST says that g = 9.8 m/s2 no matter where you are and what you're doing.

ISO says that you need to tell me where you're defining weight in order to define g.  If you're in a freely falling elevator, g is zero.  If you're standing on the earth's surface, g=9.8, and if you're standing on the moon, g=9.8/6.

That's why there is ambiguity. 

As for the astronaut, they have mass.  Whether they are weightless or not depends how you define weight.  NIST says they are never weightless, while ISO says they are weightless, in their own reference frame.  They might not be weightless in some other reference frame.

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Offline Geezer

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« Reply #60 on: 02/12/2010 09:27:52 »
They don't get it.

You "weigh" something by comparing the force it exerts in a gravitational field (which is virtually inescapable) with the force exerted by another "thing".
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline Geezer

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« Reply #61 on: 02/12/2010 10:12:56 »
BTW, the term "weightless", as frequently applied to objects that are orbiting the Earth at a particular speed, may be slightly suspect.

I am reasonably confident that if those objects were to stop in orbit they would immediately attain significant weight.
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Offline Foolosophy

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« Reply #62 on: 02/12/2010 12:12:24 »
BTW, the term "weightless", as frequently applied to objects that are orbiting the Earth at a particular speed, may be slightly suspect.

I am reasonably confident that if those objects were to stop in orbit they would immediately attain significant weight.

wow - you change the conditions and you get a different result.

Why not stop the earth and take the earth to a scale situated on Jupiter and weigh it?

The question is "How much does the earth weigh"?

In its current free falling orbit around the sun the earth's weight is equal to zero.

Look at the physical theory - not some ISO standards used to make weighing carrots or salmon more easier to understand. Their terms of reference dont include commerce on the planet Venus
« Last Edit: 02/12/2010 12:16:16 by Foolosophy »

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Offline JP

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« Reply #63 on: 02/12/2010 13:44:59 »
They don't get it.

You "weigh" something by comparing the force it exerts in a gravitational field (which is virtually inescapable) with the force exerted by another "thing".

If I had to pick a definition or be shot, I'd go with something like that.  My personal preference is to skip all the sillyness about defining weight and stick to forces and masses, which are far less ambiguous.  If the OP had asked what the mass of earth was, the question would have been much easier!

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Offline Foolosophy

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« Reply #64 on: 02/12/2010 14:01:45 »
They don't get it.

You "weigh" something by comparing the force it exerts in a gravitational field (which is virtually inescapable) with the force exerted by another "thing".

If the OP had asked what the mass of earth was, the question would have been much easier!

nothing wrong with a little bit of intellectual wrestling

the worst thing that can result is that something is learnt


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Offline peppercorn

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« Reply #65 on: 02/12/2010 15:55:35 »
nothing wrong with a little bit of intellectual wrestling

the worst thing that can result is that something is learnt

Yeah, sometimes you learn how obstinate some posters are! [::)]
...More like an intellectual 100-years-war with some...

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Offline Bored chemist

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« Reply #66 on: 02/12/2010 18:20:36 »
Actually, NIST says
"The kilogram is the unit of mass; it is equal to the mass of the international prototype
of the kilogram;
2. The word “weight” denotes a quantity of the same nature as a “force”: the weight of a
body is the product of its mass and the acceleration due to gravity; in particular, the
standard weight of a body is the product of its mass and the standard acceleration due
to gravity;
3. The value adopted in the International Service of Weights and Measures for the
standard acceleration due to gravity is 980.665 cm/s2, value already stated in the laws
of some countries."

So they talk about a "standard weight" that is 9.80665 times the mass (in KG) but the weight might be anything.

I note with amusement that a work titled "The international system of units" uses a non-SI unit for the standard value of g.
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Offline Geezer

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« Reply #67 on: 02/12/2010 19:45:11 »

In its current free falling orbit around the sun the earth's weight is equal to zero.


Er, if it's weightless, why is it falling?  [::)]
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline CPT ArkAngel

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« Reply #68 on: 03/12/2010 03:47:17 »
g = G * M / R^2

where

G is he gravitational constant
M is the mass of the earth
R is the radius of the earth

Between an object on the surface of the earth and the earth,

m * g =  - (M * a)

where

m is the mass of the object
g is the acceleration constant due to gravity of the earth (not really a constant)
M is the mass of the earth
a is the acceleration due to the gravity of the object (constant only if the object is spherical and uniformly dense)
« Last Edit: 03/12/2010 04:57:15 by CPT ArkAngel »

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Offline Foolosophy

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« Reply #69 on: 03/12/2010 05:31:43 »

In its current free falling orbit around the sun the earth's weight is equal to zero.


Er, if it's weightless, why is it falling?  [::)]

Do you know how satellites orbit another body?

(hint: look at their trajectory and project it through space - does it go past the earths horizon or intersect with the earths surface

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Offline Foolosophy

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« Reply #70 on: 03/12/2010 05:38:12 »
g = G * M / R^2

where

G is he gravitational constant
M is the mass of the earth
R is the radius of the earth

Between an object on the surface of the earth and the earth,

m * g =  - (M * a)

where

m is the mass of the object
g is the acceleration constant due to gravity of the earth (not really a constant)
M is the mass of the earth
a is the acceleration due to the gravity of the object (constant only if the object is spherical and uniformly dense)


so do you agree that the earth is in free fall motion around the sun and so its weight is equal to zero?

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Offline Geezer

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« Reply #71 on: 03/12/2010 05:55:40 »

In its current free falling orbit around the sun the earth's weight is equal to zero.


Er, if it's weightless, why is it falling?  [::)]

Do you know how satellites orbit another body?


Yes I do, and I also know that you are ducking my question.
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Offline CPT ArkAngel

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« Reply #72 on: 03/12/2010 06:37:22 »
it is a question of definition...

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Offline Geezer

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« Reply #73 on: 03/12/2010 06:50:23 »
it is a question of definition...

No. I think it's a question of lack of definition.
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Offline Foolosophy

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« Reply #74 on: 03/12/2010 11:30:02 »

In its current free falling orbit around the sun the earth's weight is equal to zero.


Er, if it's weightless, why is it falling?  [::)]

Do you know how satellites orbit another body?


Yes I do, and I also know that you are ducking my question.

You mean your question about why something is falling when its weightless??

The question itself reveals something about you

I am not sure whether you know what that is Geezer

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Offline Foolosophy

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« Reply #75 on: 03/12/2010 12:04:55 »
I wonder what Geezer's weight would be if he was on one of the Voyager probes heading towards the Ort cloud?

In fact how much does the Voyager porbe weigh at the moment?

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Offline QuantumClue

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« Reply #76 on: 03/12/2010 13:05:05 »
I wonder what Geezer's weight would be if he was on one of the Voyager probes heading towards the Ort cloud?

In fact how much does the Voyager porbe weigh at the moment?

You don't understand what he means, nor have you understood what I addressed either. All matter will possess a relative weight when measured against the acceleration of another body. If the acceleration is cancelled, then we experience what appears to be weightlessness. But as I have explained, this does not reduce W=Mg to zero, because g is never truely zero. Not only that, but your mass would contradict g since g would reduce M to zero too. I take it you have a zero mass too then?

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Offline Foolosophy

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« Reply #77 on: 03/12/2010 13:29:19 »
I wonder what Geezer's weight would be if he was on one of the Voyager probes heading towards the Ort cloud?

In fact how much does the Voyager porbe weigh at the moment?

You don't understand what he means, nor have you understood what I addressed either. All matter will possess a relative weight when measured against the acceleration of another body. If the acceleration is cancelled, then we experience what appears to be weightlessness. But as I have explained, this does not reduce W=Mg to zero, because g is never truely zero. Not only that, but your mass would contradict g since g would reduce M to zero too. I take it you have a zero mass too then?

You still cannot grasp the fundamental difference between mass and weight - they are not indentical and interchangeble quantities.

Just because you experience weigthlessness does not mean your MASS which is an intrinsic property vanishes into thin air.

Can't you accept that you made an error in challenging the simple "high school level Newtonian Physics" fact that the earth's weight is equal to zero as it orbits the sun?

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Offline QuantumClue

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« Reply #78 on: 03/12/2010 13:38:50 »
I wonder what Geezer's weight would be if he was on one of the Voyager probes heading towards the Ort cloud?

In fact how much does the Voyager porbe weigh at the moment?

You don't understand what he means, nor have you understood what I addressed either. All matter will possess a relative weight when measured against the acceleration of another body. If the acceleration is cancelled, then we experience what appears to be weightlessness. But as I have explained, this does not reduce W=Mg to zero, because g is never truely zero. Not only that, but your mass would contradict g since g would reduce M to zero too. I take it you have a zero mass too then?

You still cannot grasp the fundamental difference between mass and weight - they are not indentical and interchangeble quantities.

Just because you experience weigthlessness does not mean your MASS which is an intrinsic property vanishes into thin air.

Can't you accept that you made an error in challenging the simple "high school level Newtonian Physics" fact that the earth's weight is equal to zero as it orbits the sun?


What definition of weight are you working from? In my texbook definition, weight is proportional to mass. I never said that they were identical quantities, if they were, the equation W=Mg would have been inconsistent dimensionally as W=M.

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Offline yor_on

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« Reply #79 on: 03/12/2010 13:42:30 »
Foolosophy - perhaps if you spent a moment with a basics physics text or even on wikipedia; the recommended reading topic is vector quantities with reference to velocity and acceleration  (magnitude and direction).  You will soon learn that scalars such as speed are not same as vectors such as velocity.  Both forms of reference will also have a section on circular motion - that will fill the most obvious gaps. 

Are you still claiming that the earth has a value for WEIGHT?

The fact is that the earth is in free fall motion around the sun and so its weigth is equal to exactly zero - its weightless.

Why do astronauts experience weightlessness in orbiting space stations?

Are you disputing this simple high school physics assigment?

It's quite simple. Proper mass is matter, weight is 'gravity'. What makes 'gravity' is proper mass, relative mass and momentum. Foolosophy has it right. But if you're moving in a close orbit around the earth it's your speed making your 'weight less', not that you're without 'gravity', and so a 'weight', if meeting a surface. Without that speed, the closer your orbit is to the Earth, the sooner you would start to spiral down. As for being 'weightless' in the form of there being no 'gravitational influences' where you are? Don't know if that one exist in SpaceTime, I thing gravity is 'everywhere' myself.

you need to see that being weightless is not the absence of 'gravity'. It's rather where 'gravity' acts on it itself, our astronaut in the center of that action, equaling itself out. To make it even clearer look up Lagrange points, where you will become 'still' relative the solar-system. Outside such you will 'drift' towards the highest gravitational potential, or 'slide' if you like :)

=
And motion can be seen as centrifugal force equalizing gravity in those close orbits, creating a relative mass (considering matter).
« Last Edit: 03/12/2010 13:49:56 by yor_on »
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Offline Foolosophy

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« Reply #80 on: 03/12/2010 13:52:22 »
Foolosophy - perhaps if you spent a moment with a basics physics text or even on wikipedia; the recommended reading topic is vector quantities with reference to velocity and acceleration  (magnitude and direction).  You will soon learn that scalars such as speed are not same as vectors such as velocity.  Both forms of reference will also have a section on circular motion - that will fill the most obvious gaps. 

Are you still claiming that the earth has a value for WEIGHT?

The fact is that the earth is in free fall motion around the sun and so its weigth is equal to exactly zero - its weightless.

Why do astronauts experience weightlessness in orbiting space stations?

Are you disputing this simple high school physics assigment?

It's quite simple. Proper mass is matter, weight is 'gravity'. What makes 'gravity' is proper mass, relative mass and momentum. Foolosophy has it right. But if you're moving in a close orbit around the earth it's your speed making your 'weight less', not that you're without 'gravity' and so a weight. Without that speed, the faster the closer your orbit is to the Earth the sooner you would start to spiral down. As for being 'weightless' in the form of there being no 'gravitational influences' where you are? Don't know if that one exist in SpaceTime, I thing gravity is 'everywhere' myself.

you need to see that being weightless is not the absence of 'gravity'. It's rather where 'gravity' acts on it itself, our astronaut in the center of that action, equaling itself out. To make it even clearer look up Lagrange points, where you will become 'still' relative the solar-system. Outside such you will 'drift' towards the highest gravitational potential, or 'slide' if you like :)

nicely put

(your orbital speed is the key - for a satellite to be weigthless its speed must be high enough that if you project its trajectory as it orbits the earth, this trajectory must not intersect the earth's surface. That is, the trajectory must be past the earths horizon and into space. Galileo infered that with his parabolic projections of objects - some centuries ago. One way to look at it is the satellite is moving fast enough to be in effect "forever falling" past the earths horizon)

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Offline rosy

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« Reply #81 on: 03/12/2010 14:00:00 »
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Can't you accept that you made an error in challenging the simple "high school level Newtonian Physics" fact that the earth's weight is equal to zero as it orbits the sun?

Foolosophy... you are completely wrong on this point and making a donkey of yourself.

Weight is a force, and the only sensible definition of an object's weight, unless it is an object on the earth's surface and subject to the earth's gravity, is the net force on it due to gravity (thus your weight on the moon is less than your weight on the earth).

The earth's weight, as it orbits the sun, can be calculated as the force exerted on it due to gravity and the mass of the sun (which keeps it in its circular orbit rather than shooting off in a straight line as it would in the absence of any force).

F = -G*m1*m2/r^2

G is 6.67 x 10^-11
The sun's mass is 1.98 x 10^30 kg
The earth's mass is 5.97 x 10^24 kg
The sun-earth distance is about 1.50 x 10^11 m

The earth's weight, the force exerted on it by the sun, is thus 4 x 10^22 N.

Whether it is in orbit, or freefall, or even if it were held perfectly still in some difficult-to-imagine way involving sky-hooks this "weight" would be unaltered (provided the distance from the sun were held constant).

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Do you know how satellites orbit another body?

(hint: look at their trajectory and project it through space - does it go past the earths horizon or intersect with the earths surface

You very clearly fail utterly to understand how satellites orbit. They orbit because they are falling all the time. Have a play with these simultations and come back when you are better informed, or less arrogant, or (vain hope) both.

http://www.daveansell.co.uk/?q=node/26

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Offline Foolosophy

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« Reply #82 on: 03/12/2010 14:52:49 »
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Can't you accept that you made an error in challenging the simple "high school level Newtonian Physics" fact that the earth's weight is equal to zero as it orbits the sun?

Foolosophy... you are completely wrong on this point and making a donkey of yourself.

Weight is a force, and the only sensible definition of an object's weight, unless it is an object on the earth's surface and subject to the earth's gravity, is the net force on it due to gravity (thus your weight on the moon is less than your weight on the earth).

The earth's weight, as it orbits the sun, can be calculated as the force exerted on it due to gravity and the mass of the sun (which keeps it in its circular orbit rather than shooting off in a straight line as it would in the absence of any force).

F = -G*m1*m2/r^2

G is 6.67 x 10^-11
The sun's mass is 1.98 x 10^30 kg
The earth's mass is 5.97 x 10^24 kg
The sun-earth distance is about 1.50 x 10^11 m

The earth's weight, the force exerted on it by the sun, is thus 4 x 10^22 N.

Whether it is in orbit, or freefall, or even if it were held perfectly still in some difficult-to-imagine way involving sky-hooks this "weight" would be unaltered (provided the distance from the sun were held constant).

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Do you know how satellites orbit another body?

(hint: look at their trajectory and project it through space - does it go past the earths horizon or intersect with the earths surface

You very clearly fail utterly to understand how satellites orbit. They orbit because they are falling all the time. Have a play with these simultations and come back when you are better informed, or less arrogant, or (vain hope) both.

http://www.daveansell.co.uk/?q=node/26

So you dispute the fact that the earth is in free fall orbit around the sun and that its weight by definition is equal to zero?

Its elementary pre-University physics - actually Galileo understood this simple fact and he had very little instrumentation and theory to rely on

What's your excuse?


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Offline Foolosophy

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« Reply #83 on: 03/12/2010 15:16:12 »
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Can't you accept that you made an error in challenging the simple "high school level Newtonian Physics" fact that the earth's weight is equal to zero as it orbits the sun?

The earth's weight, the force exerted on it by the sun, is thus 4 x 10^22 N.


Congratulations, you calculated the gravitational force of attraction between two bobies - in this case between the sun and the earth.

lol

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Offline Foolosophy

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« Reply #84 on: 03/12/2010 15:36:12 »
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Can't you accept that you made an error in challenging the simple "high school level Newtonian Physics" fact that the earth's weight is equal to zero as it orbits the sun?

Foolosophy... you are completely wrong on this point and making a donkey of yourself.



You dont even understand what you have calculated.
Look at this diagram:


Notice how the object with mass "m" on the surface of the earth is assumed to be stationary?

If that same body was free falling towards the surface of the earth, what would be its weight then? Zero right?

Well the earth is in free fall motion around the sun therefore its weight = zero
« Last Edit: 03/12/2010 15:38:55 by Foolosophy »

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Offline yor_on

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« Reply #85 on: 03/12/2010 15:57:45 »
Hmm, maybe I should have read it all before writing :)

To me weight is a relation between two objects of 'proper mass', 'relative mass' or 'momentum'. Also it can be a result of centrifugal motion, and rotation aka spinning. So yes, one can look at it your way too Rosy. But where weight is a 'relation' between objects, 'proper mass' is described as being a intrinsic property, unchanging in times arrow, as long as I'm not on a diet, as I understands it?

Your weight is a 'relative thing', but your 'proper mass' is your own in all 'frames of reference'. I would say you are describing the same thing from different 'systems', and that you both have a point.
==

There is one simple way of defining 'weightlessness' though. If we all agree that we don't notice any weight under that period of time we fall from the Eiffel-tower, and then take a look at what that trajectory means from the perspective of relativity I think it can be described as following a geodesic? So, is the Earth following a geodesic or does it expend 'energy'?

That doesn't invalidate the relation you describe though, but it might be a definition we can agree on?

Ps: don't ask me to do it again, it hurts.
« Last Edit: 03/12/2010 16:13:20 by yor_on »
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Offline Foolosophy

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« Reply #86 on: 03/12/2010 16:28:51 »
Your weight is a 'relative thing', but your 'proper mass' is your own in all 'frames of reference'.

interesting point - although "weight" is not relative in the way you describe.

Even mass is relativistic - faster you go the greater your mass is

But as we define weightlessness, the earths free fall orbit around the sun is a classic example of a moving body experiencing weightlessness

I really dont know what everybody is getting all worked up about
« Last Edit: 03/12/2010 16:31:14 by Foolosophy »

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Offline yor_on

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« Reply #87 on: 03/12/2010 16:33:24 »
Hmm :) To me that's mixing relative mass, or momentum with 'proper mass'. In fact I believe that what I wrote is the correct definition of 'proper mass, well, as far as I know. 'Proper mass' is assumed to always be the same, in all 'frames of reference', be it at the EV of a black hole, or on Earth.
« Last Edit: 03/12/2010 16:36:31 by yor_on »
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Offline Foolosophy

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« Reply #88 on: 03/12/2010 16:45:26 »
Hmm :) To me that's mixing relative mass, or momentum with 'proper mass'. In fact I believe that what I wrote is the correct definition of 'proper mass, well, as far as I know. 'Proper mass' is assumed to always be the same, in all 'frames of reference', be it at the EV of a black hole, or on Earth.

by relativistic mass I mean the Einsteinian meaning



How are you using the term "proper mass"? as in rest mass?

Or as in relation to the Kinetic energy - ie



« Last Edit: 03/12/2010 16:47:56 by Foolosophy »

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Offline yor_on

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« Reply #89 on: 03/12/2010 16:48:19 »
Sorry about the choice of words, I'm kind of tired. I should have used 'proper mass is a invariant intrinsic property in all frames of reference'. I'm getting sloppy here, dangerous with you Guys and Gals :)
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Offline yor_on

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« Reply #90 on: 03/12/2010 16:50:54 »
I use proper mass as the definition of matter, rest mass when we discuss particles. Nice equations :)
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Offline Foolosophy

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« Reply #91 on: 03/12/2010 16:51:45 »
Sorry about the choice of words, I'm kind of tired. I should have used 'proper mass is a invariant intrinsic property in all frames of reference'. I'm getting sloppy here, dangerous with you Guys and Gals :)

So in the same sense as "rest mass"???

interesting because one can argue that even "rest mass" is relative (lol)  

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Offline Foolosophy

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« Reply #92 on: 03/12/2010 16:53:11 »
I use proper mass as the definition of matter, rest mass when we discuss particles. Nice equations :)

Not my equations - I can only lay claim to one set of so called "novel" equations (and they are trivially pathetic to say the least)

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Offline rosy

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« Reply #93 on: 03/12/2010 16:54:11 »
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So you dispute the fact that the earth is in free fall orbit around the sun and that its weight by definition is equal to zero?

Of course I don't, and I would hope that anyone who read what I wrote without seeking creatively to mis-understand it would have appreciated that.

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If that same body was free falling towards the surface of the earth, what would be its weight then? Zero right?

No. Wrong, wrong, and wrong again. Weight is a force. It had units of force. Nothing in freefall can be weightless, else it wouldn't accelerate (and it does!!)

The earth is in free fall orbit around the sun. That does not equate to "having no weight", any more than astronauts in freefall orbit around the earth on the ISS "have no weight", any more than a parachutist jumping out of a plane "has no weight". They experience no net force relative to their immediate surroundings (the ISS is also in freefall), so they experience "weightlessness", it is (to them) indistinguishable from truly not experiencing a weight due to gravity, but this is an illusion, just as the "weightlessness" experienced when falling from a great height is an illusion. The weight, the force, still acts.

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You will find that if an examination question asked the student "What is the weight of the earth"? any answer other than zero will be wrong.

Which examination board would this be? If you give me their name and contact details, and the relevant details of the examination syllabus on which this might occur (actually the contact details are optional, I can no doubt google for them) I would be more than happy to take it up with them.


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Its elementary pre-University physics - actually Galileo understood this simple fact and he had very little instrumentation and theory to rely on

Nonsense. Galileo established that the acceleration due to freefall is independent of mass, but that is because the mass term in the weight cancels with the mass term in F = m*a which gives the force (the weight) required to produce a given acceleration.

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Congratulations, you calculated the gravitational force of attraction between two bobies - in this case between the sun and the earth.

Indeed. And since that is, to all intents and purposes, the weight of the earth, I have thereby answered the OP's question.

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Offline yor_on

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« Reply #94 on: 03/12/2010 17:04:17 »
Sorry about the choice of words, I'm kind of tired. I should have used 'proper mass is a invariant intrinsic property in all frames of reference'. I'm getting sloppy here, dangerous with you Guys and Gals :)

So in the same sense as "rest mass"???

interesting because one can argue that even "rest mass" is relative (lol)  


How would you argue then?
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Offline rosy

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« Reply #95 on: 03/12/2010 17:07:45 »
A couple of questions, Fool...

1. If you were in a rocket, accelerating upwards from the earth's surface, would your weight have increased relative to when you were stationary?

If you think your weight would not have changed, how is this different to the situation of being in a rocket in freefall orbit about the earth?

If you think your weight would have changed, if the rocket were accelerating not upwards, but sideways at a tangent to the earth's surface, what would your weight be then? Would it still pull you toward the earth's centre or would it suddenly have a "backwards" component?

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Offline yor_on

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« Reply #96 on: 03/12/2010 17:25:57 »
I know one thing. when discussing 'mass' in physics it's important to agree on what the he* we are discussing :) People use the same synonyms for totally different properties at times it seems.

This is what I mean with 'proper mass'

"The invariant mass, intrinsic mass, proper mass or just mass is a characteristic of the total energy and momentum of an object or a system of objects that is the same in all frames of reference."

From Mass in special relativity.
 
And somewhere a long time ago I learnt that it was only when discussing particles one should use 'rest mass'? But as you point out, and a simple web search can show one, there seems to exist different interpretations of what 'rest mass' mean.

But I think you can find support for my interpretation in Invariant mass.

( Or maybe not :)

"The invariant mass is another name for the rest mass of single particles."

Anyway :) I would like to see your arguments for rest mass being 'relative'. It's always nice with new ideas, and to me that's a new one.

And.

"If a stationary box contains many particles, it weighs more in its rest frame, the faster the particles are moving. Any energy in the box (including the kinetic energy of the particles) adds to the mass, so that the relative motion of the particles contributes to the mass of the box. But if the box itself is moving (its center of mass  is moving), there remains the question of whether the kinetic energy of the overall motion should be included in the mass of the system.

The invariant mass is calculated excluding the kinetic energy of the system as a whole (calculated using the single velocity of the box, which is to say the velocity of the box's center of mass), while the relativistic mass is calculated including invariant mass PLUS the kinetic energy of the system which is calculated from the velocity of the center of mass."

Which is how I see it too.

==

Although rereading my quote I'm slightly in disagreement with calling an added 'speed' of particles inside your 'system', as that 'box' of particles becomes here, an added 'proper mass'. It's a matter of correct definitions to me. To me all relativistic mass is 'relativistic mass'.

But as you define a 'system' you create imaginary borders for your needs. So the box overall 'mass' might increase with heat, but if the 'proper mass' would increase then it seems to me that it would invalidate the definition of 'proper mass' being invariant in all 'frames of reference' like if putting our box in a oven, or a sun.

So with that exception I agree to it as a definition.
« Last Edit: 03/12/2010 17:57:53 by yor_on »
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Offline QuantumClue

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« Reply #97 on: 03/12/2010 18:00:55 »
Hmm :) To me that's mixing relative mass, or momentum with 'proper mass'. In fact I believe that what I wrote is the correct definition of 'proper mass, well, as far as I know. 'Proper mass' is assumed to always be the same, in all 'frames of reference', be it at the EV of a black hole, or on Earth.

Pleasel, from now on, just leave your arguement, or I will respectfully ask the mods to keep this kind of thinking to ATM.

by relativistic mass I mean the Einsteinian meaning



How are you using the term "proper mass"? as in rest mass?

Or as in relation to the Kinetic energy - ie





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Offline QuantumClue

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« Reply #98 on: 03/12/2010 18:02:36 »
I gave a reply to this, but it never processed.

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Offline Geezer

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« Reply #99 on: 03/12/2010 18:05:04 »
Fool's definition of weight may be correct, but he'll also have to be consistent. That means for example, that every time he accelerates his mass and jumps a few millimeters in the air, he is weightless (strictly speaking he'd need to be in a vacuum of course, which, come to think of it, might not be such a bad idea.)

I will refrain from expressing an opinion on what kind of person he is.
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