Science Questions

How much does the Earth weigh?

Sun, 28th Nov 2010

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Question

Ben Shell asked:

One thing I have wondered is how much does earth weigh? How many pounds does earth way, and if we don't yet know, do you think it would ever be possible to guess accurately?

 

KEEP UP THE GREAT WORK!!!

 

Ben

Answer

Chris -   The stated weight – mass, we should more accurately, of the Earth is about 6 x 1024 kilograms.  In other words, if you turn that into tons, it’s 6 with 21 zeros after it, tons.  So pretty heavy, but the big question is and this is where Dave can help me out, how do we actually know that something on the scale of the Earth that we can't physically put on a pair of scales, how do we know how much that weighs because Archimedes famously said, “If you give me a lever long enough and somewhere far enough away to stand, I could lift up the Earth” but how would we have calculated how much the earth actually weighs, Dave?

Dave -   Well, the simple way of doing this is because anything with a mass affects everything around it due to gravity.  If it’s something roughly spherically symmetrical, you can assume that all the mass is in the centre of the planet and it behaves as if it was all the mass is right in the centre, due to some neat bits of maths.  Basically, what you have to do is - if you know how much gravitational force a kilogram of anything will apply to another kilogram of anything, and you know how much force a kilogram of substance is being attracted to the Earth and you know how big the Earth is, you can work out how much mass must be in the Earth.  The second part of that is really easy.  Working out how much force a kilogram produces is really difficult because it’s about 10-11 Newtons at a metre between 2 kilograms.  It’s an incredibly tiny force and it wasn’t done until near the end of 19th century. 

Chris -   Henry Cavendish, wasn’t it?

Dave -   Yup.  Indeed and you can work it out, and then from that, you can work out how heavy the Earth is, and from that, how heavy everything else is in the universe really.

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BenTheH33 asked the Naked Scientists: One thing I have wondered is how much does earth weigh? How many pounds does earth way, and if we don't yet know, do you think it would ever be possible to guess accurately? KEEP UP THE GREAT WORK!!! Ben What do you think? BenTheH33, Tue, 23rd Nov 2010

http://lmgtfy.com/?q=How+much+does+the+Earth+weigh%3F Pikaia, Tue, 23rd Nov 2010

The earth weighs 6.5 x 10^27 in tons. But this is nothing on astronomical terms. Even a spoonful of neutron star would weigh about the same as all the buildings on earth. QuantumClue, Tue, 23rd Nov 2010

david holthaus asked the Naked Scientists: how much does the earth weigh? What do you think? david holthaus, Tue, 23rd Nov 2010

Calculating the mass of the Earth is fairly straightforward, but weight is relevant only in relation to gravity, but perhaps it was mass you were thinking of, in which case it's something like 5.9737 × 10ˆ24kg. Bill S, Tue, 23rd Nov 2010

I found a site that relates to the mass of the earth...
http://hypertextbook.com/facts/2002/SamanthaDong2.shtml
Bill S has it I found this may be an approximate derived calculation since there is mass collecting here from space, constantly.

Let's see if I have it correct
Below: In this definition of Weight I will say earths weight is zero. If this is a true definition...

Weight is the force that results from the acceleration by gravity on the mass of an object.

Sometimes it is defined in operational terms of the weighing process as the force exerted by an object on its support, which also illustrates the condition of weightlessness.

Standing at rest on a weighing scale on Earth, a person's weight equals its mass multiplied by the gravitational acceleration of Earth. However, in a free falling elevator, a scale indicates a zero weight, as no net force is exerted by the body on the support; the person experiences weightlessness. Similarly, in a space craft in orbit around the Earth the weight is also zero, as the orbit represents a free fall. On the surface of the Moon, an object's weight is approximately one sixth of the weight at rest on Earth, as the gravitational force exerted by the Moon is much smaller than that of Earth.

The weight of an object, often denoted W, is defined as the gravitational force exerted on it. It is the product of the mass m of the object and the local gravitational acceleration g: W = m g. In the International System of Units (SI), the unit of measurement for weight is that of force, the newton.

On the surface of the Earth, the acceleration due to gravity is approximately constant; this means that the magnitude of an object's weight on the surface of the Earth is proportional to its mass. In situations other than that of a constant position on the Earth, so long as the acceleration does not change, the force it exerts against support in any accelerated frame is proportional to its mass, also. In everyday practical use, therefore, including commercial use, the term weight is commonly used to mean mass.

http://en.wikipedia.org/wiki/Weight maffsolo, Tue, 23rd Nov 2010

The value quoted by QuantumClue is not correct the correct value is 5.9742 × 10^24 kilograms syhprum, Tue, 23rd Nov 2010

We are talking about mass rather that weight, yes?

We seem to have two threads dealing with the same question. Bill S, Tue, 23rd Nov 2010

Good answer, maffsolo, but,David, is that what you were looking for? Bill S, Tue, 23rd Nov 2010



http://www.thenakedscientists.com/forum/index.php?topic=35366.msg331835#msg331835

There seems to be a controversy on terminology with this question We need to call a doctor in to prescribe a solution
Yelling for MEDIC @#$%^& maffsolo, Tue, 23rd Nov 2010

Two identical questions merged - sorry if it's a bit muddled right now! BRValsler, Tue, 23rd Nov 2010



Thank you for the first aid maffsolo, Tue, 23rd Nov 2010



Mine was in tons. Not kilograms. QuantumClue, Tue, 23rd Nov 2010

10^27 tonnes DOES NOT equal 10^24 kilogrammes.  Your answer was 10^6 times too big imatfaal, Tue, 23rd Nov 2010

Oh I see. Well that's not good. QuantumClue, Tue, 23rd Nov 2010

Plug and chug convergence calculator

http://www.onlineconversion.com/weight_all.htm




5.9736e+24 kilogram = 9.4068097813e+23 stone

And I thought Earthy was the third stone from the Sun maffsolo, Tue, 23rd Nov 2010



I apologize, we all make mistakes. That calculator will definately come in useful in the future. QuantumClue, Tue, 23rd Nov 2010



I apologize, we all make mistakes. That calculator will definately come in useful in the future.


No need to appologize I am the master of mistakes.
I can do best with my eyes closed.

I sat here and tried to convert it and I found I was slipping, to double check myself I went to see if my conversion was correct. I was having a bit of a problem with a simple setup and combinding the exponents for a result.

I am here just to keep my gray matter in one mass, I can be trained...:) maffsolo, Tue, 23rd Nov 2010

The weight of the earth is equal to zero because it is in free fall motion around the sun

The earth's MASS however can be determined in a variety of ways

cheers Foolosophy, Tue, 30th Nov 2010

For me, the weight of the earth is equal to mine. I can feel it right now...

The weight is a force : F = m * g CPT ArkAngel, Wed, 1st Dec 2010

(is the earth accelerating around the sun? so from your own Newtonian equation a=0 so F=0,)

The earth is by definition in free fall motion/orbit around the sun

The earth's weight is therefore equal to zero = ie its like being in a weightless environment

The mass of the earth is an intrinsic property of the earth and is constant

(although relativity tells us that mass itself is subject to relativistic effects - for example, the rest mass is different from when the same body is moving - but thats another thread topic)

Foolosophy, Wed, 1st Dec 2010

Foolosophy - yes the earth is accelerating (it's in circular motion; it must be accelerating) a≠0 F≠0.  I agree with the Arkangel - although I think it weighs as much as me rather than as much as him imatfaal, Wed, 1st Dec 2010

do you know why there is weightlessness on an orbiting satellite?

Answer that question and you will understand why the weight of the earth is equal to zero

Weight is a different concept to Mass

(as far as your claim that the earth is accelerating you must remember that acceleration is the change in velocity (ie dv/dt)

how is the earths velocity changing?) Foolosophy, Wed, 1st Dec 2010



So you are claiming that the earth's velocity is changing??

what is driving this dv/dt and in what direction is it occuring?

Foolosophy, Wed, 1st Dec 2010

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.  imatfaal, Wed, 1st Dec 2010


you are still claiming that the earth has weight because its accelerating????

Its a physical fact that the weight of the earth is exaclty equal to zero

reason?? because its in free fall motion around the sun

Its the same reason why astronauts are weightless in orboting space stations

are you disputing these facts? Foolosophy, Wed, 1st Dec 2010



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? Foolosophy, Wed, 1st Dec 2010



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?
I don't think this is technically correct. This is the kind of thing taught in freshmen college, but a deeper understanding of F=Mg suggests otherwise.

F=Mg still applies in space. Technically, astronauts still have a weight because there are gravitational forces acting on the particles of their bodies. QuantumClue, Wed, 1st Dec 2010

For instance, usually in very elementary physics, weight is the force exerted on a mass. The acceleration ''g'' is not zero in space. So setting g=0 for W=Mg is not exactly correct. It is true we experience weightlessness, but there is some debate as to whether this is technically correct. QuantumClue, Wed, 1st Dec 2010

Foolosophy

Firstly - you seem to have drawn back from your claims that the earth isn't accelerating, glad to see you have read up on vectors v scalars and circular motion.  Secondly, you need to understand what free-fall means. 

Free-fall motion is when the only force experienced is that of gravitational attraction.  We feel gravitational attraction on earth - but on earth there is also a normal force (equal and opposite) from the ground; this is why we don't sink through the floor. 

Astronauts in a space station are in orbit - they are in freefall, but it is completely incorrect to say that there is no gravitational attraction towards the earth.  If they were not continually accelerating (due to a force) towards the earth they, and their tin can, would fly off at a tangent.  They weigh something (not as much, but something) in space just as much as they do when they jump in the air when back on planet earth.  Just because there is no normal at an instant in time does not mean that there is no attraction/force - it is that their attraction to the earth is counteracted in a different manner than the normal force that we suffer. imatfaal, Wed, 1st Dec 2010



I am astonished at the polemic here.


The weight of the earth = 0 for the same reason that astronauts are weightless when orbiting the earth in a space station - THEY ARE IN FREE FALL MOTION

Now if you don't like this reality all I can suggest is to take up the matter with Sir Isaac Newton's estate.

ARE you still claiming that the earth's weight ISNT equal to zero??

Perhaps you can use your scalarisation and vectoring techniques to prove your alchemic claim?


Foolosophy, Wed, 1st Dec 2010



The F=ma equation is in relation to an accelerating body.

The gravitational force between two bodies can be determined by Newtons law


where G is the universal gravitational constant = 6.67*10^-11 (Nm^2/kg^2)

(G is just a phyisical constant, its not the same as the "g" in F=mg - you're confusing the two)

For example, when pilots experience g forces in their planes, say 4g, it means that the force they are experiencing is 4 times that of the earths normal gravitational pull. Foolosophy, Wed, 1st Dec 2010

"imatfaal" wishes to discuss the centrifugal force of a body in circular motion


(And remember the orbit of the earth around the sun isnt strictly circular - its eliptical, but what's a few focii amongst friends hey?) Foolosophy, Thu, 2nd Dec 2010

the centripetal force is a different thing again

Foolosophy, Thu, 2nd Dec 2010

But most importantly though, Imatfaal needs to acknowledge the fact that the weight of the earth is equal to zero Foolosophy, Thu, 2nd Dec 2010



The F=ma equation is in relation to an accelerating body.

The gravitational force between two bodies can be determined by Newtons law


where G is the universal gravitational constant = 6.67*10^-11 (Nm^2/kg^2)

(G is just a phyisical constant, its not the same as the "g" in F=mg - you're confusing the two)

For example, when pilots experience g forces in their planes, say 4g, it means that the force they are experiencing is 4 times that of the earths normal gravitational pull.


No offense, but you really do need to work on your units. Yes (a) is acceleration, but g is also an acceleration. This relationship is true. Weight W is also the dimensions of a force exerted on an object. QuantumClue, Thu, 2nd Dec 2010

Learn here:

Equations for a falling body - Wikipedia, the free encyclopediaFor example, Newton's law of universal gravitation simplifies to F = mg, ...
en.wikipedia.org/wiki/Equations_for_a_falling_body - Cached - SimilarShow more results from wikipedia.org QuantumClue, Thu, 2nd Dec 2010



...are you saying that the earth is NOT weightless?

I am not saying that the earth is massless (its mass is an instrinsic property)

I am just saying that the earth is in free fall motion aroudnd the sun - so its weight is equal to zero

Similar to being weightless in a space station that is orbiting the earth in free fall motion

Weight and Mass are not equivalent Foolosophy, Thu, 2nd Dec 2010

What is the difference between mass and weight?

Mass is a measure of how much matter an object has. Weight is a measure of how strongly gravity pulls on that matter. Thus if you were to travel to the moon your weight would change because the pull of gravity is weaker there than on Earth but, your mass would stay the same because you are still made up of the same amount of matter.

Answered by: A. Godbehere, High School Student, Port Perry

Imagine yourself out is space away from any gravitational field, with a bowling ball in your hands. Let it go and it just floats in front of you. Without gravity, it has no weight. Now grab it again and shake it back and forth. That resistance to being moved is inertia, and mass measures how much inertia an object has. Inertia does NOT depend on gravity.

Mass is determined only by the amount of matter contained in an object.

Any two masses exert a mutual attractive force on each other. The amount of that force is weight. A one kilogram mass on the Earth's surface results in 2.2 pounds of force between the mass and the Earth, so we say the mass weighs 2.2 pounds. That same one kilogram mass on the Moon, because of the Moon's lower mass, results in only about 1/3 pounds of mutual force.

Just remember that the weight of an object depends on where it is, while its mass stays the same.

Answered by: Paul Walorski, B.A., Part-time Physics Instructor

http://www.physlink.com/education/askexperts/ae321.cfm
=============================

Ironic that weight is always referenced to a surface ground plane ...

So what ground plane is the Earth's weight   referenced to ????

I have heard the that a free fall in space still has Micro Gravity, what is meant by that?

Is it earth's gravity affect on the freefall?





http://en.wikipedia.org/wiki/Gravitational_constant

Hmmmm no mention of weight there....



maffsolo, Thu, 2nd Dec 2010



So you agree that the earth is in free fall motion around the sun and therefore its weight is equal to ZERO??

Why are astronauts weightless in space stations that are orbiting the earth? Foolosophy, Thu, 2nd Dec 2010



So you agree that the earth is in free fall motion around the sun and therefore its weight is equal to ZERO??

Why are astronauts weightless in space stations that are orbiting the earth?


Yea how can you weigh the earth against the earth.

I kind of like learning to staying away from the association of the words "weightless" with "freefall" in the same breath...
Reason being, it is easy to misinterpret the slang...

"There is no gravity in space."  FALSE  If there were no gravity in space, the space shuttle would not be able to orbit the Earth, the moon would not orbit the Earth, and the Earth would not orbit the Sun.  The reason we tend to think of there being no gravity in space is that we have seen movies of the astronauts being "weightless".  They aren't actually weightless, they are still being pulled down by gravity but they and the space shuttle are in a constant state of freefall around the Earth.  So they seem to be weightless as a result of the falling - just as you would seem weightless if you were in an elevator when the cable broke.

http://www.regentsprep.org/regents/physics/phys01/unigrav/default.htm
maffsolo, Thu, 2nd Dec 2010

maffsolo

what is the weight of the astronauts in the spcae stations when they are orbiting the earth in free fall motion?

They are weightless - their mass doesnt change (although strictly speaking their relativistic masss does change when compard to their rest mass - but that is neglible for the speed that they are travelling at and really a topic for another thread)

I doubt whether you can find a physicist in the world today that would dispute the fact that the earth is weightless as it orbits the sun.

I am just responding to the actual question posed in this thread - and it asks what the earth WEIGHS - not its mass. Foolosophy, Thu, 2nd Dec 2010


I see your point!

I believe you can mathmatically hypothisize a value that can represent a weight.
In saying that, I also contest that, that value has no scientific value except for saying wow thats heavy.

We can hypthetically evaluate...

We know the earths mass we know the earths gravitational acceleration.
Let's just say we like to know, if an object was the same mass of the earth and it were sitting on the surface of the earth, hypothetically speaking, how much will the object weigh? maffsolo, Thu, 2nd Dec 2010

It would certainly be quite hard to find bathroom scales that were capable of weighing it. Boots maybe?

You'd also need a really big bathroom. Geezer, Thu, 2nd Dec 2010

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

We can hypthetically evaluate...

We know the earths mass we know the earths gravitational acceleration.
Let's just say we like to know, if an object was the same mass of the earth and it were sitting on the surface of the earth, hypothetically speaking, how much will the object weigh?

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

what do you mean by the "earth's gravitational acceleration"??? g?

You can hypothecise all you wish but the question in htis thread is "how much does the earth weigh"

the answer is zero because the earth is in free fall motion around the sun

What is the weight and mass of an astronaut on a free falling space station?

What is the weight of a pilot under 4g flight conditions?

When you jump from an aeroplane with a parachute (hopefully) and you reach a terminal velocity (ie stop accelerating) your weight is equal to zero - you are experiencing weightlessness.

If you want to put another earth on top of our earth and weigh it you can - but both earths will be weightless as they hurl around the sun

I dont understand why people are disputing this in here?

Its just a definition thing

Foolosophy, Thu, 2nd Dec 2010



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

I suppose it's possible to weigh the Earth on a sort of beam balance if you compare it with the mass of the Sun, or the Moon. For example, we might say that the Earth has a weight of x Moons. We should be able to determine the null point of the fulcrum from the Earth's Lunar wobble. Geezer, Thu, 2nd Dec 2010



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

I suppose it's possible to weigh the Earth on a sort of beam balance if you compare it with the mass of the Sun, or the Moon. For example, we might say that the Earth has a weight of x Moons. We should be able to determine the null point of the fulcrum from the Earth's Lunar wobble.


Its not about weighing the earth whilst its static

THe question is "what is the weight of the earth"?

Are you disputing the fact that the earth's weight is equal to zero?
Foolosophy, Thu, 2nd Dec 2010



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

I suppose it's possible to weigh the Earth on a sort of beam balance if you compare it with the mass of the Sun, or the Moon. For example, we might say that the Earth has a weight of x Moons. We should be able to determine the null point of the fulcrum from the Earth's Lunar wobble.


Its not about weighing the earth whilst its static

THe question is "what is the weight of the earth"?

Are you disputing the fact that the earth's weight is equal to zero?



I really don't know if I'm disputing any facts or not.

Weight is a measure of gravitational attraction due to mass. The Moon and the Earth can be weighed against each other, therefore, they are not weightless. (If you can find an alternative definition for weight, you should be able to prove me wrong.) Geezer, Thu, 2nd Dec 2010



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

I suppose it's possible to weigh the Earth on a sort of beam balance if you compare it with the mass of the Sun, or the Moon. For example, we might say that the Earth has a weight of x Moons. We should be able to determine the null point of the fulcrum from the Earth's Lunar wobble.


Its not about weighing the earth whilst its static

THe question is "what is the weight of the earth"?

Are you disputing the fact that the earth's weight is equal to zero?



I really don't know if I'm disputing any facts or not.

Weight is a measure of gravitational attraction due to mass. The Moon and the Earth can be weighed against each other, therefore, they are not weightless. (If you can find an alternative definition for weight, you should be able to prove me wrong.)


This is incorrect reasoning (if you dont mind me being abrupt)

The moon is also in free fall motion aroudn the earth so its weight is zero also.

If you weighed yourself on earth and then on the moons surface the weight values will differ (by a factor of 6). But you are not in free fall motion when youre being weighed.

Its to do with the fact that the earth is in free fall orbit around the sun - just like a space station orbiting the earth Foolosophy, Thu, 2nd Dec 2010



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... jpetruccelli, Thu, 2nd Dec 2010



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 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 
Foolosophy, Thu, 2nd Dec 2010



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. jpetruccelli, Thu, 2nd Dec 2010

quote author=JP link=topic=35346.msg332910#msg332910 date=1291277129]


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...


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. Geezer, Thu, 2nd Dec 2010

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.
Geezer, Thu, 2nd Dec 2010



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.

Foolosophy, Thu, 2nd Dec 2010

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. 

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. jpetruccelli, Thu, 2nd Dec 2010

JP beat me to it!



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. Geezer, Thu, 2nd Dec 2010



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?

Foolosophy, Thu, 2nd Dec 2010

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. jpetruccelli, Thu, 2nd Dec 2010

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". Geezer, Thu, 2nd Dec 2010

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. Geezer, Thu, 2nd Dec 2010



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 Foolosophy, Thu, 2nd Dec 2010



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! jpetruccelli, Thu, 2nd Dec 2010



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

Foolosophy, Thu, 2nd Dec 2010



Yeah, sometimes you learn how obstinate some posters are!
...More like an intellectual 100-years-war with some... peppercorn, Thu, 2nd Dec 2010

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. Bored chemist, Thu, 2nd Dec 2010



Er, if it's weightless, why is it falling?  Geezer, Thu, 2nd Dec 2010

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)
CPT ArkAngel, Fri, 3rd Dec 2010



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
Foolosophy, Fri, 3rd Dec 2010



so do you agree that the earth is in free fall motion around the sun and so its weight is equal to zero? Foolosophy, Fri, 3rd Dec 2010



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. Geezer, Fri, 3rd Dec 2010

it is a question of definition... CPT ArkAngel, Fri, 3rd Dec 2010



No. I think it's a question of lack of definition. Geezer, Fri, 3rd Dec 2010



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 Foolosophy, Fri, 3rd Dec 2010

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? Foolosophy, Fri, 3rd Dec 2010



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? QuantumClue, Fri, 3rd Dec 2010



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?
Foolosophy, Fri, 3rd Dec 2010



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. QuantumClue, Fri, 3rd Dec 2010



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). yor_on, Fri, 3rd Dec 2010



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) Foolosophy, Fri, 3rd Dec 2010



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).



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 rosy, Fri, 3rd Dec 2010



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).



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?

Foolosophy, Fri, 3rd Dec 2010



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 Foolosophy, Fri, 3rd Dec 2010



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
Foolosophy, Fri, 3rd Dec 2010

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. yor_on, Fri, 3rd Dec 2010



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 Foolosophy, Fri, 3rd Dec 2010

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. yor_on, Fri, 3rd Dec 2010



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



Foolosophy, Fri, 3rd Dec 2010

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 :) yor_on, Fri, 3rd Dec 2010

I use proper mass as the definition of matter, rest mass when we discuss particles. Nice equations :) yor_on, Fri, 3rd Dec 2010



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

interesting because one can argue that even "rest mass" is relative (lol)   Foolosophy, Fri, 3rd Dec 2010



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) Foolosophy, Fri, 3rd Dec 2010



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.



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.


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.




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.



Indeed. And since that is, to all intents and purposes, the weight of the earth, I have thereby answered the OP's question. rosy, Fri, 3rd Dec 2010



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? yor_on, Fri, 3rd Dec 2010

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?
rosy, Fri, 3rd Dec 2010

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. yor_on, Fri, 3rd Dec 2010



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




QuantumClue, Fri, 3rd Dec 2010

I gave a reply to this, but it never processed. QuantumClue, Fri, 3rd Dec 2010

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. Geezer, Fri, 3rd Dec 2010

I think I can present a argument speaking for my interpretation of 'proper mass'. Imagine yourself inside that 'box' described above, being 'at rest' with one of the 'particles'. When being so you will subtract the 'added' 'mass' as all motion can be seen, or transformed, into heat, and also as 'energy'. Being 'at rest', unmoving relative the particle will allow you to see it in its 'original state' and that state is also what I would call its 'rest mass' or if you like 'invariant mass' and those definitions are the equivalence to a piece of matter being 'proper mass', invariant in all 'frames of reference'. yor_on, Fri, 3rd Dec 2010

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