Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: BenTheH33 on 23/11/2010 10:30:04
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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 (http://www.thenakedscientists.com/HTML/podcasts/)!!!
Ben
What do you think?
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http://lmgtfy.com/?q=How+much+does+the+Earth+weigh%3F
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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.
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david holthaus asked the Naked Scientists:
how much does the earth weigh?
What do you think?
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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.
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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.[1]
Sometimes it is defined in operational terms of the weighing process as the force exerted by an object on its support,[2] 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:[3] 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
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The value quoted by QuantumClue is not correct the correct value is 5.9742 × 10^24 kilograms
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We are talking about mass rather that weight, yes?
We seem to have two threads dealing with the same question. [:P]
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Good answer, maffsolo, but,David, is that what you were looking for?
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We are talking about mass rather that weight, yes?
We seem to have two threads dealing with the same question. [:P]
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 @#$%^&
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Two identical questions merged - sorry if it's a bit muddled right now!
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Two identical questions merged - sorry if it's a bit muddled right now!
Thank you for the first aid
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The value quoted by QuantumClue is not correct the correct value is 5.9742 × 10^24 kilograms
Mine was in tons. Not kilograms.
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10^27 tonnes DOES NOT equal 10^24 kilogrammes. Your answer was 10^6 times too big
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Oh I see. Well that's not good.
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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
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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
I apologize, we all make mistakes. That calculator will definately come in useful in the future.
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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
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...:)
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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
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We discussed this question on our show
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.
Click to visit the show page for the podcast in which this question is answered. (http://www.thenakedscientists.com/HTML/podcasts/show/2010.11.28/) Alternatively, [chapter podcast=2913 track=10.11.28/Naked_Scientists_Show_10.11.28_7576.mp3](https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.thenakedscientists.com%2FHTML%2Ftypo3conf%2Fext%2Fnaksci_podcast%2Fgnome-settings-sound.gif&hash=f2b0d108dc173aeaa367f8db2e2171bd) listen to the answer now[/chapter] or [download as MP3] (http://nakeddiscovery.com/downloads/split_individual/10.11.28/Naked_Scientists_Show_10.11.28_7576.mp3)
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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
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(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)
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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
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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?)
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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
So you are claiming that the earth's velocity is changing??
what is driving this dv/dt and in what direction is it occuring?
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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.
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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.
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?
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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?
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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?
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.
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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.
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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.
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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.
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?
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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.
The F=ma equation is in relation to an accelerating body.
The gravitational force between two bodies can be determined by Newtons law
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwiki.oercommons.org%2Fmediawiki%2Fupload%2F300px-NewtonsLawOfUniversalGravitation.svg.png&hash=e115ecd7d086f805d7c1f7c362479a26)
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.
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"imatfaal" wishes to discuss the centrifugal force of a body in circular motion
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.nennstiel-ruprecht.de%2Fbullfly%2Fcentrif.gif&hash=56a78de0428869d64a762c01dead9c2f)
(And remember the orbit of the earth around the sun isnt strictly circular - its eliptical, but what's a few focii amongst friends hey?)
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the centripetal force is a different thing again
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fastroprofspage.com%2Fwp-content%2Fuploads%2F2006%2F12%2FCF1.jpg&hash=aa4778fea1a7e2dd51290fdbb420a420)
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But most importantly though, Imatfaal needs to acknowledge the fact that the weight of the earth is equal to zero
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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.
The F=ma equation is in relation to an accelerating body.
The gravitational force between two bodies can be determined by Newtons law
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwiki.oercommons.org%2Fmediawiki%2Fupload%2F300px-NewtonsLawOfUniversalGravitation.svg.png&hash=e115ecd7d086f805d7c1f7c362479a26)
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.
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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
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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
...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
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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
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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?
[ Invalid Attachment ]
http://en.wikipedia.org/wiki/Gravitational_constant
Hmmmm no mention of weight there....
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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
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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?
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?
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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
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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?
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
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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.
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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 the it asks what the earth WEIGHS - not its mass.
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?
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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.
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[/quote]
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?
[/quote]
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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
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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.
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.
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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.
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?
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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.
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.)
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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.
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
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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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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
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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 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.
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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...
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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...
[/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.
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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.
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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.
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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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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.
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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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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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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".
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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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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
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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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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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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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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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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? [::)]
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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)
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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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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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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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it is a question of definition...
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it is a question of definition...
No. I think it's a question of lack of definition.
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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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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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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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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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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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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 :)
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And motion can be seen as centrifugal force equalizing gravity in those close orbits, creating a relative mass (considering matter).
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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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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).
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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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).
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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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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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:
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fircamera.as.arizona.edu%2FNatSci102%2FNatSci102%2Fimages%2Fweight.gif&hash=70a4bf7de39f02bac87279aadd2304b7)
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
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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.
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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.
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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
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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.
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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
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.gravitywarpdrive.com%2FNGFT_Equations%2FRelativistic_Mass.gif&hash=fea9138005e06932b8061f97f28f2e7e)
How are you using the term "proper mass"? as in rest mass?
Or as in relation to the Kinetic energy - ie
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fej.iop.org%2Fimages%2F0004-637X%2F536%2F1%2F195%2FFull%2Fdf36.gif&hash=1e602dbc1bf567090da9caab91bf5731)
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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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I use proper mass as the definition of matter, rest mass when we discuss particles. Nice equations :)
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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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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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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.
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.
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.
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.
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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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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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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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. (http://en.wikipedia.org/wiki/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. (http://en.wikipedia.org/wiki/Mass_in_special_relativity)
( 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.
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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.
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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
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.gravitywarpdrive.com%2FNGFT_Equations%2FRelativistic_Mass.gif&hash=fea9138005e06932b8061f97f28f2e7e)
How are you using the term "proper mass"? as in rest mass?
Or as in relation to the Kinetic energy - ie
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fej.iop.org%2Fimages%2F0004-637X%2F536%2F1%2F195%2FFull%2Fdf36.gif&hash=1e602dbc1bf567090da9caab91bf5731)
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I gave a reply to this, but it never processed.
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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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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'.
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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.
But hopefully not your opinion...
Even if his weight is zero, it implies there is no atraction between bodies.
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All that weight is reconfigured into a force. And vice versa.
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All that weight is reconfigured into a force. And vice versa.
Well, it kinda depends. While you are in freefall and rapidly approaching the surface of the Earth, notwithstanding any screaming that might be going on at the time, it's impossible to determine your weight. You can only determine your "Earth weight" when you are static in a direction relative to your center of mass and the Earth's center of mass.
So, the tricky bit is deciding whether you weighed anything during your fall or not. If you can't measure it, arguably, it's anybody's guess.
On the other hand, if you say that you are "weightless" during your fall, you have to wonder why you are falling at all.
The situation is slightly different when it comes to large bodies like the Earth and the Sun, because, when they are falling towards each other, it's actually possible to measure their relative weights based on the center of rotation of the orbiting system (Solar wobble if you like). That situation suggests that you can actually weigh the Earth while it's zipping round the Sun, and, therfore, it aint't weightless.
I think JP had it right when he said it's best to completely avoid the use of the term "weight".
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All that weight is reconfigured into a force. And vice versa.
I think JP had it right when he said it's best to completely avoid the use of the term "weight".
All that weight is reconfigured into a force. And vice versa.
So, the tricky bit is deciding whether you weighed anything during your fall or not. If you can't measure it, arguably, it's anybody's guess.
On the other hand, if you say that you are "weightless" during your fall, you have to wonder why you are falling at all.
why?
photons don't seem to worry too much about their zero mass when moving about - are they weightless?
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Heh, that was a new one
I like that.
"Waiter two ordinary photons, one relative, three Rindler virtual specials and a weightless. And make it snappy, Chop chop"
Well, they are weightless, are they not?
In fact about the only thing there is :)
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Heh, that was a new one
I like that.
"Waiter two ordinary photons, one relative, three Rindler virtual specials and a weightless. And make it snappy, Chop chop"
Well, they are weightless, are they not?
In fact about the only thing there is :)
perhaps
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On the other hand, if you say that you are "weightless" during your fall, you have to wonder why you are falling at all.
why?
photons don't seem to worry too much about their zero mass when moving about - are they weightless?
Urrr.... "weightlessness" is no bar to movement. Once an object is moving, it will continue to move with unchanged velocity until a force acts on it. If no force acts it will stay still/continue to move at constant speed in a straight line for ever (depending on initial conditions). That's Newton, that is.
Anything falling freely under gravity is accelerating all the time. Even if speed is unchanged, as in a circular orbit, the direction changes, which requires an acceleration (one at right angles to the instantaneous direction of travel).
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On the other hand, if you say that you are "weightless" during your fall, you have to wonder why you are falling at all.
why?
photons don't seem to worry too much about their zero mass when moving about - are they weightless?
Urrr.... "weightlessness" is no bar to movement. Once an object is moving, it will continue to move with unchanged velocity until a force acts on it. If no force acts it will stay still/continue to move at constant speed in a straight line for ever (depending on initial conditions). That's Newton, that is.
Anything falling freely under gravity is accelerating all the time. Even if speed is unchanged, as in a circular orbit, the direction changes, which requires an acceleration (one at right angles to the instantaneous direction of travel).
You must remember that according to Geezer's Law, "one must wonder why anything is falling at all if it is weightless"
One of the 7 wonders of the modern intellectual realm of challenges
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yor_on.. to go back a few posts (sorry):
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.
Actually, in the case of considering the orbit of the earth and sun, or the freefall of a skydiver, it really doesn't matter at all which definition of "mass" we use since none of the bodies involved are traveling at anything approaching lightspeed and thuse relativistic effects don't have any significant effect.
Fool:
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
Do you, then, believe that there is no force between objects in orbit about one another? That there is no force attracting a satellite toward the earth? I can't see any other interpretation of this post, but am trying one last time to work out what it is you think.
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(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fhome.bway.net%2Frjnoonan%2Fhumans_in_space%2Fgravity.gif&hash=395d49a7e787bbc7ff0689b4eed78c35)
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fabyss.uoregon.edu%2F%7Ejs%2Fimages%2Flec08_2.gif&hash=d045fec35b57481bce73f560aeb26fa0)
...there are people who argue that the centrifugal force is NOT real
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What does the centrifugal force have to do with anything? The centrifugal force is a mathematical convenience in certain reference frames, but let's stick to an earth/satellite system in the reference frame of the earth (or, strictly, the centre of mass of the orbiting system).
Which forces do you then believe are needed to keep the satellite in its orbit? Please explain in your own words without copy-pasting someone else's diagram.
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What does the centrifugal force have to do with anything?
You are asking as to the relevance of the centrifugal force in a classic "earth/satellite" orbit problem??
Interesting......
No wonder you still cannot grasp the simple fact that the earth is weightless as it orbits the sun.
(some equations are very useful in order to make a point clearer - notice how for a given mass and radius a body must attain a certain velocity in order to balance the centrifugal and gravitational forces? Hence free fall orbit conditions = weightlessness
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You are asking as to the relevance of the centrifugal force in a classic "earth/satellite" orbit problem??
Yup. Well, actually, not really. There's no question. In the reference frame of the centre of mass, there is no centrifugal force. It simply does not exist. The centripetal force is what we're interested in. If you don't know what the centripetal force is, look it up. It's key to understanding orbits of all sorts.
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You are asking as to the relevance of the centrifugal force in a classic "earth/satellite" orbit problem??
Yup. Well, actually, not really. There's no question. In the reference frame of the centre of mass, there is no centrifugal force. It simply does not exist. The centripetal force is what we're interested in. If you don't know what the centripetal force is, look it up. It's key to understanding orbits of all sorts.
But you claim that the earth IS NOT weightless as it orbits the sun even though it is free fall motion
You must pay attention Rosy
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fcontent.answcdn.com%2Fmain%2Fcontent%2Fimg%2Foxford%2FOxford_Sports%2F0199210896.centrifugal-force.1.jpg&hash=6870bd3163e215c39a36bd617605a1a3)
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Rosy, as a practical matter forces is a very useful concept, for me though I believe (I'm a believer:) in Einsteins SpaceTime 'geodesics'. According to that concept 'SpaceTime' is bent, wrinkled and bumpy. And while forces is the alternative way to look at it, it's no 'forces' involved for me, except possibly as some 'paths of least resistance' in SpaceTime, to why the planets orbit each other. Well, that's how I see it. To me gravity is no force even though I use the word sometimes, inside apostrophes mostly.
As for the comment on 'mass' that was just a common observation I've made at times, coming up when we discussed 'proper mass' and 'rest mass'. As for the rest of your comment I agree, for what we call a 'free fall' the mass doesn't matter :) all masses live in a 'free fall' depending on how you define your system, sort of?
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Centrifugal force is a fictive force. It only exists in a local frame of reference, such as the guy in the picture swinging the weight. From our frame of reference, a mass in motion will continue in a straight direction unless it is accelerated. In this instance the weight is accelerated toward the guy by a centripetal force applied by the rope. In other words, the guy is applying the only force in the example. There is no centrifugal force unless you confine your frame of reference to the guy. By the way, gravity is also a fictive force that we see from our frame of reference, but from Einstein's frame of reference it is not. In fact, I am pretty sure that Einstein said that the centrifugal frame of reference problem helped him with understanding gravity. I think that this may be what Yor_on is going on about.
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Most things I can think of, except gravity, can be transformed into 'energy', well, with the possible exception of 'anti matter' but if anti matter strikes anti matter there still should be a 'kinetic energy' created inside SpaceTime. If I would define anything as a 'force' then it is light, as it is the purest definition of 'energy' I know of? When you see a orbit 'break down' to the 'force' of gravity, then, from my point of view the geodesic 'pointed' to where the object moves. The only thing opposing geodesics are transforming and expending 'energy', like light is.
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The problem is Foolosophy is in a spacecraft going around the earth and he doesn't feel gravity. Any objects with him appear to have no weight. We are all down to earth and we see him in a free fall around the earth and we say this free fall is due to his weight. Like Geezer said, there is a lack of precise definition of what is the weight. I search the internet but some definitions are more general, speaking only of force and others are more specific, talking about weight relative to the experimenter.
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No CPT, from my view he's following a geodesic, created by gravity as balanced by his 'relative' motion. all orbits in 'free fall' are uniform motions, as soon as you need to 'correct' them by expending energy you are in fact breaking the geodesic path. And 'uniform motion' is a very slippery subject as you in a 'black box scenario', following a geodesic, can't differ between 'speeds', making all 'speeds' equivalent from that perspective. The only truth I think I know in this case to break a geodesic is to 'expend energy'. To define a possible speed you always need coordinates, making those your 'system' and a relative one as such. The only time you can define a speed in this black box is when having a non-uniform acceleration, as I see it. A uniform acceleration is equivalent to gravity, if in a black box scenario.
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Thinking of it, when we say that something accelerates you need to ask if you feel a gravitation. If you don't you're not accelerating, you can't accelerate if following a geodesic. So, as I see it, as the apple falls of the branch, 'accelerating', it's in fact having a 'uniform acceleration' :)
Sorry, not motion.
Or?
I'm not sure about that one, uniform acceleration is equivalent to gravity, that's correct, But from the view point of the apple there is no 'acceleration' separable from a 'uniform motion' as it still follows a geodesic, being 'weight-less'. A tricky one..
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You might possibly expand on the fact that all uniform 'speeds' are equivalent of course, and that way create a proof (?) For this type of acceleration being a 'uniform motion'?
Anyone have a view on this?
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yes, you are right... But he is still in a free fall, but is relative motion prevent him to change of geodesic trajectory. Truly, he feels a small angular acceleration for his change in speed direction around the earth... No?
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On the other hand, if you say that you are "weightless" during your fall, you have to wonder why you are falling at all.
why?
photons don't seem to worry too much about their zero mass when moving about - are they weightless?
Urrr.... "weightlessness" is no bar to movement. Once an object is moving, it will continue to move with unchanged velocity until a force acts on it. If no force acts it will stay still/continue to move at constant speed in a straight line for ever (depending on initial conditions). That's Newton, that is.
Anything falling freely under gravity is accelerating all the time. Even if speed is unchanged, as in a circular orbit, the direction changes, which requires an acceleration (one at right angles to the instantaneous direction of travel).
You must remember that according to Geezer's Law, "one must wonder why anything is falling at all if it is weightless"
One of the 7 wonders of the modern intellectual realm of challenges
Ah yes! Photons, another "bum steer" if ever there was one. However, I'm glad you finally acknowledge my vastly superior intellectual prowess.
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I would still like an answer to these questions, please, Fool:
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?
2.a)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?
2.b)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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It's a good idea CPT, but I don't think so myself. The definition of a true geodesic have to be perfect weightlessness, in a black box scenario. If you introduce a angular gravitational influence, influencing his motion, inside that black box, it's no longer a perfect 'free fall', following a geodesic.
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If you mean that he have two gravitational 'forces' acting on him, but we find that he still floats free in the exact middle of that black box the whole time, then it will be what I call a geodesic though. And now I'm mixing metaphors :)
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Yes, because the acceleration he should feel from the change in speed direction is compensate by gravity. So the sum of both forces is zero...
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Let's put it this way, if you introduce a course change, without acceleration, you should still feel inertia act on you. You should feel that you're not in a 'free fall' any longer, and neither that you are following a geodesic, under that moment inertia knock on your door. And for it to be perfect geodesic, not even the idea of inertia should be there I think.
Looking at it as 'forces' and making a course change, gravity would have to compensate for the inertia created by that course change, and so fluctuate for you to not noticing the course change. In that special case you might think yourself in a geodesic, or 'uniform motion' as it is too. And there is fluctuating gravity associated with gravity waves for example. so maybe?
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You could define it this way too, no matter if you change your 'speed' relative something else, as the ship change course, there will be energy expended. As soon as you introduce 'expending energy' you're breaking the geodesic.
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The problem being that I can think of no way changing course in space without introducing a acceleration, angular or not?
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In the case of the apple falling we have the earth rotating, if it was that you thought of? Or if it was a spinning black hole, but if you're free falling your relative 'motion' should adapt to the frame dragging too I think, keeping you anchored in the middle of that room?
In the case of several spinning VMO:s (very massive objects) SpaceTime would look very weird, if we could color those geodesics so we could see their 'whirls', and as the gravity also should experience 'gravity waves', as a guess, depending on frame dragging and spins? I don't really know if it would be possible to follow a geodesic in such a place?
Awh..
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There is a much simpler way to look at this. Lob a ball into the air so that it travels in a parabolic arc and returns to Earth.
Does the ball have weight while it's in motion, or does it not? Clearly, it has mass, but does it have "weight"?
Other than the fact that the ball returns to Earth in a rather abrupt fashion, there's really no difference between this situation and the situation where a body is in orbit around another body. No doubt my foolish friend will beg to differ and attempt to introduce an entire shoal of "poisson rouge" into the debate [:D]
EDIT: BTW - I don't think there is an answer. Regardless of how you answer the question, I will respond with "OK - devise an experiment that allows us to confirm your theory empirically."
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There is a much simpler way to look at this. Lob a ball into the air so that it travels in a parabolic arc and returns to Earth.
Does the ball have weight while it's in motion, or does it not? Clearly, it has mass, but does it have "weight"?
Other than the fact that the ball returns to Earth in a rather abrupt fashion, there's really no difference between this situation and the situation where a body is in orbit around another body. No doubt my foolish friend will beg to differ and attempt to introduce an entire shoal of "poisson rouge" into the debate [:D]
EDIT: BTW - I don't think there is an answer. Regardless of how you answer the question, I will respond with "OK - devise an experiment that allows us to confirm your theory empirically."
mathematicians dont seem too worried about empirical validations of their proofs
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There is a much simpler way to look at this. Lob a ball into the air so that it travels in a parabolic arc and returns to Earth.
Does the ball have weight while it's in motion, or does it not? Clearly, it has mass, but does it have "weight"?
Other than the fact that the ball returns to Earth in a rather abrupt fashion, there's really no difference between this situation and the situation where a body is in orbit around another body. No doubt my foolish friend will beg to differ and attempt to introduce an entire shoal of "poisson rouge" into the debate [:D]
EDIT: BTW - I don't think there is an answer. Regardless of how you answer the question, I will respond with "OK - devise an experiment that allows us to confirm your theory empirically."
mathematicians dont seem too worried about empirical validations of their proofs
Perhaps, but scientists do.
I'll try again. Is a falling ball weightless or not?
EDIT:
BTW, I notice the dreaded "centrifugal" force has raised its ugly head in this thread. As Rosy points out, it may be used as a mathematical convenience, but don't be fooled into thinking there is any such force. The force that acts on a body to keep it in orbit, or a rock spinning round at the end of a string, is centripetal force.
If you don't believe me, try cutting the string and let me know what happens.
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Geezer
Do they believe the Earth is flat in the USA ?, projectiles only move in a parabolic arc above an infinite flat surface above the spherical Earth the trajectory is elliptical.
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Geezer
Do they believe the Earth is flat in the USA ?
Yes - as a close approximation anyway.
Of course, if you want to be pedantic, the trajectory wouldn't be strictly parabolic or elliptic because of wind resistance. Nor would the ball return in "free fall" for the same reason.
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There is a much simpler way to look at this. Lob a ball into the air so that it travels in a parabolic arc and returns to Earth.
Does the ball have weight while it's in motion, or does it not? Clearly, it has mass, but does it have "weight"?
Other than the fact that the ball returns to Earth in a rather abrupt fashion, there's really no difference between this situation and the situation where a body is in orbit around another body. No doubt my foolish friend will beg to differ and attempt to introduce an entire shoal of "poisson rouge" into the debate [:D]
EDIT: BTW - I don't think there is an answer. Regardless of how you answer the question, I will respond with "OK - devise an experiment that allows us to confirm your theory empirically."
mathematicians dont seem too worried about empirical validations of their proofs
Perhaps, but scientists do.
I'll try again. Is a falling ball weightless or not?
that's the difference between mathematics and science - mathematics is more of an abstract philosophy.
What is the weight of the ball at the point of maximum height in its parabolic trajectory?
When the trajectory of a free falling body that is orbiting another body is projected out into space it goes past the horizon or boundary limits of the central body - if it doesnt and intersects the surface of the body (say the earth) then the 2 bodies will collide.
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When the trajectory of a free falling body that is orbiting another body is projected out into space it goes past the horizon or boundary limits of the central body - if it doesnt and intersects the surface of the body (say the earth) then the 2 bodies will collide.
That is indeed true, and it's why I mentioned the bit about the ball returning to Earth in a rather abrupt fashion.
However, for the purpose of resolving the vexing question regarding the weight of the ball, it doesn't make any difference whether the ball stays in orbit around the Earth for a number of years, or a mere three seconds. Both situations are governed by the same laws of physics.
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When the trajectory of a free falling body that is orbiting another body is projected out into space it goes past the horizon or boundary limits of the central body - if it doesnt and intersects the surface of the body (say the earth) then the 2 bodies will collide.
That is indeed true, and it's why I mentioned the bit about the ball returning to Earth in a rather abrupt fashion.
However, for the purpose of resolving the vexing question regarding the weight of the ball, it doesn't make any difference whether the ball stays in orbit around the Earth for a number of years, or a mere three seconds. Both situations are governed by the same laws of physics.
nobody is challenging the universality of the laws of physics
Its these very laws that define the weight of a free falling body as being equal to zero.
And the earth is in free fall motion around the sun - just like the moon is around the earth.
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Its these very laws that define the weight of a free falling body as being equal to zero.
You keep saying this. It's not clear, though, exactly which law of physics you think tells you that? Would you like to clarify? Because at the moment you're making this assertion over and over again, claiming that it's "pre-university" physics, "accepted", "obvious" when it's nothing of the kind.
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Its these very laws that define the weight of a free falling body as being equal to zero.
You keep saying this. It's not clear, though, exactly which law of physics you think tells you that? Would you like to clarify? Because at the moment you're making this assertion over and over again, claiming that it's "pre-university" physics, "accepted", "obvious" when it's nothing of the kind.
I dont really understand your concern with weightlessness during free fall motion
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I dont really understand your concern with weightlessness during free fall motion
Let me try to lead you through it my objection to your assertion:
When you throw a ball into the air (we'll assume it's going slow enough that air resistance can be ignored), do you consider it to be weightless during its flight?
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I dont really understand your concern with weightlessness during free fall motion
Let me try to lead you through it my objection to your assertion:
When you throw a ball into the air (we'll assume it's going slow enough that air resistance can be ignored), do you consider it to be weightless during its flight?
Are you saying that the weight of this aircraft doesnt change?
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.phys.unsw.edu.au%2F%7Ejw%2Fgraphics%2Ffreefall3.gif&hash=cd4f5656d2368285387afed49755276a)
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Yes. That is precisely what I am saying. The downwards force due to gravity, acting on that aircraft, is unchanged throughout its parabolic flight (its acceleration, downwards, is constant provided we ignore air resistance). I would therefore say that its weight was unchanged.
Would you say differently? Are we back to the question of definitions of weight, or are you really contending that in freefall there is no force due to gravity?
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Yes. That is precisely what I am saying. The downwards force due to gravity, acting on that aircraft, is unchanged throughout its parabolic flight (its acceleration, downwards, is constant provided we ignore air resistance). I would therefore say that its weight was unchanged.
Would you say differently? Are we back to the question of definitions of weight, or are you really contending that in freefall there is no force due to gravity?
well this is a typical flight path for simulating weightlessnes.
People experience near weigthlessness for about 20 seconds at the top of the parabolic flight path.
They experience about 2g
Does their weight remain the same? - interesting
you may have to look up the term APPARENT weight - may be useful
(do you still wish to stand by your statement that planes acceleration downwards is constant? ignoring air resistance?)
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well this is a typical flight path for simulating weightlessnes
I know that. The key word here is "simulating".
Does their weight remain the same... well, it depends how you define weight. I would consider "weight", as I said further up the thread, to be the gravitational force acting on a body. If we are using that definition then yes, it absolutely does.
If you want to consider the experience of a passenger in the plane, of course their experience will equivalent to the experience they would have if they were in a space ship an infinite distance from any other massive object, but that is because their arms and legs are being accelerated under gravity at the same rate as their head and torso, so they are not having to hold themselves upright in the way that one must on earth, and likewise they are accelerating at the same rate as the plane, so the surface they are standing on is not holding them up in the way it would on earth, and a tiny force will push them away from it.
This experience is not, in fact, the absence of weight (in the sense of a downward force due to gravity, which is the sense in which I have, as I explained at the outset, been using it all along) it is merely subjective "weightlessness".
(do you still wish to stand by your statement that planes acceleration downwards is constant? ignoring air resistance?)
In the absence of air-resistance*, then yes, the plane's acceleration (not either its speed or its velocity, but simply its acceleration) is constant.
What do you think are (would be) the changes in the plane's acceleration (in the absence of air resistance).
*and assuming that the plane's trajectory is short enough that the difference in the distance from the earth's centre between the top and bottom of the trajectory is small enough not to make a significant difference to the distances between the centres of mass
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I think it is time to stop feeding the troll - he clearly isn't interested in a debate or exchange of ideas.
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No trolls here I think :)
The definition of weightlessness may vary with your standpoint. Newton considered a free fall weightless. "They are accelerated by gravity toward the Earth, but their inertia in the direction tangential with their path results in a curved path around the planet. In essence, they are always missing the planet in their fall toward it." Einstein agrees that a free fall can be seen as an absence of gravity.
But it's possible to see it as Rosy does too I think, her idea seems one of 'geodesics' with one difference, where I see each geodesic as its own 'path' one might imagine it as a layered onion of geodesics growing from 'proper mass' like the Earth. With each layer representing a certain magnitude of gravitational 'force' depending on its distance from Earth. Well, this is my interpretation of it based on geodesics.
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But it's a strange onion in that all layers end at the same surface :) (Other planets excepted) But that it does in my 'geodesics' too. The only way to avoid that is to add a motion to the object orbiting which then will give a new geodesic becoming adapted to the circumference of Earth. And that this becomes a geodesic too is proven just by the 'weightlessness' perceived, as I see it.
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Newton considered a free fall weightless.
Ah yes! So, you don't need to have any sort of lateral motion to be in free fall. Soooo, if you are plunging straight towards the surface of the earth you would also be weighless, which you might consider slightly paradoxical under the circumstances.
Thinks: "Hmmm?? This is interesting. How come I'm accelerating towards the surface of the Earth if I'm supposed to be weightless?"
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No trolls here I think :)
Do you? I'm still, just, reserving judgement.. my test case is the Fool's interpretation of the acceleration of an object subject only to gravity.. so subject to negligible frictional forces and travelling a distance short enough that the gravitational field is effectively constant, so as in the case of a ball thrown in the air. If he thinks the acceleration of the ball is anything other than constant, then I'm with imatfaal.*
Newton considered a free fall weightless. "They are accelerated by gravity toward the Earth, but their inertia in the direction tangential with their path results in a curved path around the planet. In essence, they are always missing the planet in their fall toward it."
You've lost me. I assume that quote is attributed to Newton.. and it certainly agrees with my understanding of Newton's conclusions... but where does your conclusion that Newton considered freefall "weightless" come from? That in freefall an individual will experience "weightlessness" is implicit in Newton's (and indeed Galileo's) conclusions. But that is not the same as not being subject to a net force, which so far as I can make out (and it's not being made very clear) is the Fool's contention.
But it's possible to see it as Rosy does too I think, her idea seems one of 'geodesics' with one difference, where I see each geodesic as its own 'path' one might imagine it as a layered onion of geodesics growing from 'proper mass' like the Earth. With each layer representing a certain magnitude of gravitational 'force' depending on its distance from Earth. Well, this is my interpretation of it based on geodesics.
I don't claim my understanding of mechanics goes much beyond Newton, but I do know enough to know that to interpret and make highly accurate, quantitative predictions regarding this system of orbiting satellites doesn't require that we invoke relativistic masses or other mathematically demanding abstractions, and certainly I think this harping on geodesics distracts from the matter at hand.
* By messing about with which reference frame you consider, you could probably conclude that the answer is 2.476 or, indeed, a fish. But the Fool's insisted repeatedly that we're talking about Newtonian/high school level physics, and if so then we really have to stick to high school physics' conventional reference frame (that of the center of mass of the system).
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Geezer, in the quote I cited, your lateral force is equivalent by "their inertia in the direction tangential with their path" as I see it? I like that English btw :) Sounds so nice, like 'heavenly bodies'
As for being weightless under a direct fall towards the Earth, consider yourself inside that (in)famous black box. then tell me how you would differ being weightless in a orbit, constantly 'missing' Earth as Newton expressed it, (due to your added lateral motion), from being inside that black box free falling towards Earth? I don't see how to be able to do it?
What I might argue(?) is that as the frame dragging created by Earths rotation might express itself slightly different, but considering the time frame I doubt one would have time to test that.
A net force Rosy?
As a force you can consider gravity to have a magnitude and then a net force can only be zero when the gravity acting on your proper mass is balanced by an equally strong 'force' acting in the oposite direction. And that is when you are standing on the floor, at which time your net force is null. But I have to admit that I've read it as you were debating weightlessness? Seen as a pure geodesics there is no 'force' of course. But hey, we live in world where we use that word a lot :)
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As for being weightless under a direct fall towards the Earth, consider yourself inside that (in)famous black box. then tell me how you would differ being weightless in a orbit, constantly 'missing' Earth as Newton expressed it, (due to your added lateral motion), from being inside that black box free falling towards Earth? I don't see how to be able to do it?
LOL! You wouldn't be able to tell the difference. That's my point. [:D]
The discussion about orbits and aeroplanes is interesting, but not particularly helpful. We only need to resolve the most simple case to resolve the issue.
The simplest case is - What happens when you are falling towards the Earth? Are you weightless, or not?
The answer is going to depend on your definition of weight. I think weight is a comparison of relative mass. Now, when you are falling towards the Earth, a weighing device is not going to be much use obviously, so you could say that because it's impossible to weigh an object while it's in free fall, it is therefore "weightless".
But hold on just a minute. That's not quite true. It is possible to weigh an object in free fall (or in orbit for that matter) because the gravitational attraction also accelerates the other body, so measuring the accelerations of the two bodies allows us to compare their masses. According to my definition, this is also known as weighing them!
It would be a bit difficult to do this where the masses of the two objects are vastly different, but it's not so hard to do it with the Earth and the Sun. Therefore, it is possible to weigh the Earth while it is in free fall around the Sun, so maybe Newton was wrong after all.
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It seems we are mixing the concept of weightlessness with net forces and gravity then? Neither Newton nor Einstein said that there was no gravity in Space, it's just that where Newton saw gravity as a 'force' Einstein found it to be a geodesic instead. But gravity is everywhere in SpaceTime, as long as the 'expansion' won't be able to grow 'locally faster' than gravity can compensate for. That as gravity, according to Einstein, moves with the speed of light. I'm not sure if that's possible though?
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Inside my black box?
hmmm :)
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Yor_on:
Q- How to tell free fall straight down from free fall in orbit while in a black box? A- Coriolis force (also a fictive force).
Steve
EDIT-- I withdraw this pending some hard thought, or help.
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It seems we are mixing the concept of weightlessness with net forces and gravity then?
I don't think so. Anyway, the question was, "How much does the Earth weigh?"
I believe I have proposed a satisfactory method of actually weighing it even while it's in orbit around the Sun. If I can weigh it, I think it's reasonable to assume that it's not weightless. [;D]
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Let me see if I got it right. You mean that by measuring the orbit of the moon knowing the mass of Earth, you can make a educated guess about the mass of the moon, That seems correct to me. But assuming that I didn't know the mass of the Earth too, then measured the distance of the moons orbit around Earth, I would have a he** of a problem 'splitting' the number I got into two, also you could get the same distance/orbit, if so, with different masses it seems to me? But you're thinking of having one known value, and then the distance I presume?
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No Steve, you're correct, I forgot to specify there.
But I think it's included in the 'frame dragging' as such
Heh ::))
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Rereading you, missed the way you split the Q.
Yep, I think you're spot on.
If we imagine someone in a geodesic in deep space he won't have a Coriolis force acting on him.
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Rosie I think you are right, I looked and I didn't find any proofs for that he considered it 'weightless'. What he did consider it as was that the object in orbit was 'pulled over the horizon' due to its speed with the net forces taking themselves out, as I understands it?
In a way that is semantics though as when they 'equal out' they are equivalent to being weightless as I see it? There is a difference though in that Einstein went one step further, exchanging those net forces for his geodesics, and stated that when you were weightless it was actually possible to consider it equivalent to there being no gravity at all.
Gotta love physics :)
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Let me see if I got it right. You mean that by measuring the orbit of the moon knowing the mass of Earth, you can make a educated guess about the mass of the moon, That seems correct to me. But assuming that I didn't know the mass of the Earth too, then measured the distance of the moons orbit around Earth, I would have a he** of a problem 'splitting' the number I got into two, also you could get the same distance/orbit, if so, with different masses it seems to me? But you're thinking of having one known value, and then the distance I presume?
It's a simple beam balance. If you can determine the "fulcrum" (the point about which bodies both rotate) you can determine their relative weights. The unit of weight would be relative e.g. the Earth weight = x Moons, but weight is always a relative measurement, so that makes no difference. The point is, if you can determine the weight of a body, how can it be considered weightless?
You seem to be having a slight problem with this orbiting thing. "Free fall" applies whenever a body is not constrained and falls towards another body. It happens every time both your feet leave the ground. You don't have to be orbiting the Earth to experience it. Whether you are actually orbiting the Earth, or running for a bus, it makes no difference. You are either weightless in both cases, or not weightless in both cases. You can't be weightless in one case but not in the other case.
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Gotta love physics :)
Indeed!
To semanticise (my word) this weight/mass polemic trivialises and misrepresents what objects in free fall motion "feel". The fact that astronauts are weightless in space stations is not a matter of semantics - actually its a very accurate and apt description of their physical state (and supported by the laws of physics)
If you wish to stop the earth and place it on a set of scales under some known gravitaional field then I am sure that you will get a value for the earth's weight.
But that is another question
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Gotta love physics :)
But that is another question
No it isn't. It was the question.
Did you actually read any of this? The weighless condition an astronaut "feels" is exactly the same condition that you feel when you are falling to Earth from any height, and that, BTW, is entirely supported by the laws of physics.
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To semanticise (my word) this weight/mass polemic trivialises and misrepresents what objects in free fall motion "feel".
To semanticise this topic is over-complicating what is actually an extremely simple question. You all agree on the laws of physics, but are arguing over definitions of the word "weight." Since there is no universally accepted definition, as pointed out earlier, what's the point in arguing that your particular definition is the right one?
By the way, an interesting thing about the astronaut in orbit around the earth is that he's not in an inertial reference frame. If he drops an object it will appear to hover next to him, but if he instead throws it along the direction of his orbit, he'll see it curve up away from him rather than in a straight line. An astronaut in deep space would see the ball travel in a straight line if he threw it in any direction. I was needlessly confused for a while while learning physics, since both were described as "weightless." It was only after learning Newtonian mechanics that I actually realized that the concept of weight isn't really all that descriptive...
You can make things even more complicated if you consider one of those artificial-gravity space station designs that rotates so that the centripetal force is equivalent to the force of gravity on earth. The same problem applies that although you would "feel" right, the motion of objects isn't identical to being on the earth, since you're no longer in an inertial reference frame.
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To semanticise (my word) this weight/mass polemic trivialises and misrepresents what objects in free fall motion "feel".
To semanticise this topic is over-complicating what is actually an extremely simple question. You all agree on the laws of physics, but are arguing over definitions of the word "weight." Since there is no universally accepted definition,
Untrue - there isnt any controversy over what weightlessness means.
The earth must by definition be wieghtless as it orbits the sun in free fall motion.
This fact is irrelevant to the earth's mass which is an intrinsic property of matter
(although one can argue over relativistic mass which is a thread topic in itself)
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To semanticise this topic is over-complicating what is actually an extremely simple question. You all agree on the laws of physics, but are arguing over definitions of the word "weight." Since there is no universally accepted definition,
Untrue - there isnt any controversy over what weightlessness means.
[/quote]
I said the definition of weight was controversial, not weightlessness. You can argue with me on that point, but you'd be wrong.
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To semanticise this topic is over-complicating what is actually an extremely simple question. You all agree on the laws of physics, but are arguing over definitions of the word "weight." Since there is no universally accepted definition,
Untrue - there isnt any controversy over what weightlessness means.
I said the definition of weight was controversial, not weightlessness. You can argue with me on that point, but you'd be wrong.
[/quote]
????
By your very own logic if the definition of weight is controversial then weightlessness must also be controversial.
But in the case of a free falling orbiting body, its weight must equal to zero.
I am sure that its mass is not offended
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Well Geezer, I thought I knew what it was until stumbling on this thread :)
Never mind, I'll stick to my geodesics then :)
Well, in a way I see your point there. Let's agree on that planets have a 'weight' even if relative, as I see it. Proper mass isn't though. Cause if it is we will have to redefine the whole darn thing with matter I think
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the proof for that is simple.
Place Earth on Jupiter, then ask it if it feel a weight :) If it complains do the same with it on the moon, and ask it if it feels better now :)
doctors order..
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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.
the issue of scalar and vector quantities isnt disputed by anybody in here.
its about weightlessness of bodies in free fall.
are you still disputing the basic scientific fact that the earth is in free fall motion around the sun and therefore weightless by definition?
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Well, in a way I see your point there. Let's agree on that planets have a 'weight' even if relative, as I see it. Proper mass isn't though. Cause if it is we will have to redefine the whole darn thing with matter I think
Are you certain about that?
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.quantonics.com%2FQuantonics%2520Site%2520GIFs%2FRelativistic_Mass_Equation.gif&hash=e271c2967b6a80306edd0e615afe27b3)
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the issue of scalar and vector quantities isnt disputed by anybody in here.
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?)
This quote and others from the first page of this thread make it fairly clear you do not understand the vector nature of velocity and acceleration
its about weightlessness of bodies in free fall.
I think you should give your definition of free fall - because it it quite clearly not the same as the more accepted version of the motion of a body which only experiences force from gravitational attraction
are you still disputing the basic scientific fact that the earth is in free fall motion around the sun and therefore weightless by definition?
I dispute your assertion that one follows from the other. you have provided no proof or argument other than saying it is a basic fact. Please provide the definitions that confirm your assertion.
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To semanticise this topic is over-complicating what is actually an extremely simple question. You all agree on the laws of physics, but are arguing over definitions of the word "weight." Since there is no universally accepted definition,
Untrue - there isnt any controversy over what weightlessness means.
I said the definition of weight was controversial, not weightlessness. You can argue with me on that point, but you'd be wrong.
By your very own logic if the definition of weight is controversial then weightlessness must also be controversial.
No. That's by your logic, which is still wrong no matter how emphatically you state otherwise.
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:) Foolosophy ::))
You gonna mix relativistic mass into this *** thread?
This thread will explode soon ::))
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I liked the way you explained your equation by the way.
It's a good habit.
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and yes, I'm as sure as one can be, only depending on what the boffins might cook up next :) Proper mass is a very defined property, hopefully correct too.
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Eh, thinking of it, anyone disagreeing?
(counting down 3 .. 2.99 ... 2.98 ... \B00M/)
And some day I will grow up.
But not tonight :)
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I'm afraid I have to weigh in on this one more time.
I'm less concerned about the definition thing than I am about consistency. The conditions effecting a body in orbit are no different from the conditions effecting a ball that's moving in a parabolic (or elliptical) arc because somebody threw it in the air.
If a body in orbit is "weightless", a ball thrown in the air must also be "weightless". Everyone keeps wanting to discuss the complicated case of orbiting without understanding the uncomplicated case of chucking a !@#$%^& ball in the air. And, if it follows that a thrown ball is weightless, it also follows that a human is weightless every time they lose contact with Terra Firma, or not! Either way, I'm pretty sure the various cases have to be consistent (unless somebody invented a new branch of Dynamics without telling me about it.)
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"Weight is the force that results from the acceleration by gravity on the mass of an object.[1]
Sometimes it is defined in operational terms of the weighing process as the force exerted by an object on its support,[2] 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."
Take it up with the wiki instead.
They have a 'talk page', that's what I would do.
From Weight. (http://en.wikipedia.org/wiki/Weight)
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Answers.com I like, a lot. Lots of info on most subjects. This is some of what they have to say about 'Zero Gravity'.
"Newton proposed the law of universal gravitation, which states that two bodies of matter in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
According to this law, even a small increase in the distance between bodies will produce a large decrease in the gravitational force, since the force decreases with the square of the distance. As a body moves from the Earth's surface to a location an infinite distance from the Earth, the gravitational force approaches zero and the body approaches weightlessness. In the true sense, a body can be weightless only when it is an infinite distance from all other objects.
Weightlessness is also defined as a condition in which no acceleration, whether of gravity or any other force, can be detected by an object or organism within the system in question.
According to Albert Einstein's principle of equivalence, there is no way to distinguish between the forces of gravitational fields and the forces due to inertial motion. When a gravitational force on a body is opposed by an equal and opposite inertial force, a weightless state is produced.
This is based on the fact that the mass that determines the gravitational force of a body is the same as the mass related to the acceleration produced by an inertial force of any kind. These inertial forces have no external physical origin, but are the consequences of an accelerated state of motion.
Because of inertia, a moving object always tends to follow a straight line. When a person swings a bucket by the handle in a large circle, he or she feels a pull on his or her hand, because inertial force (also called centrifugal force in this case) tends to keep the bucket moving in a straight line, while the bucket holder exerts a counterforce constraining the bucket to move along the circle.
A similar situation exists in a spaceship orbiting the Earth 200 mi (320 km) above the Earth's surface, where the gravitational field is only slightly weaker than at sea level. The ship, in free fall with negligible atmospheric drag, is pulled toward the Earth by the Earth's gravitational attraction force, while the inertial or centrifugal force of the moving ship is directed radially outward from the Earth; consequently, the force of gravity on the orbiting ship is opposed and nullified by the centrifugal force, and apparent weightlessness results."
From weightlessness. (http://www.answers.com/topic/weightlessness)
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I have been following and enjoying this argument, and I now believe that JP is correct. There is a relatively simple confusion regarding definitions of what the word “weight” really means. Definitions can be important. The Wiki article is pretty good and it discusses some of the ambiguities. It says “Weight is the force that results from the acceleration by gravity on the mass of an object.” Force is mass X acceleration (of gravity in this case). So here is my take on this:
Weight in everyday life is relatively simple. You bring your trout to a hanging scale (having cleverly forced a few stones down its mouth) and watch how much the calibrated spring stretches to proclaim your trophy. So where is the spring scale in the earth sun system? As the earth orbits the sun it is being constantly accelerated at a right angle to its direction of movement by the centripetal force resulting from the mutual attraction between the earth and sun, and this force on the sun is evidenced by a wobble in the earth’s orbital plane. We on the earth feel weightless, relative to the sun, because we are being accelerated by the same amount as our environment, but we and the earth are very certainly under the effect of a great force. It isn’t a great stretch (intended) to see the orbit and wobble as the stretch indicator of the spring in this scale.
Again for effect- If we are falling to earth (or in orbit around the earth) we may feel weightless because we and all our internal organs and all our baggage are falling at the same rate, but this doesn’t mean that there isn’t a considerable force acting on us, and there is considerable evidence of the force in this situation. F= m x a.
If one wishes to define weight as meaning the force exerted by a mass in a specific gravity field, but only if the mass is stationary relative to the gravity field, then OK. Yor_on, you have to be very careful when using the concept of centrifugal force because it can lead you astray, as it has the author of the piece you linked. Steve
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"When a person swings a bucket by the handle in a large circle, he or she feels a pull on his or her hand, because inertial force (also called centrifugal force in this case) tends to keep the bucket moving in a straight line, while the bucket holder exerts a counterforce constraining the bucket to move along the circle. " This Steve?
And where?
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'Pseudo force' or not, that's also a question where people seem to have different definitions. And that I know :) Been there done that as they say, long time ago :)
All centrifugal 'forces' are 'straight lines' constrained, as far as I understands it. With the constraint being the 'centripetal force' as i see it?
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Like me on the inside of a hollow tube in space, the tube spinning. Then I have a 'gravity' but the 'gravity' can also be seen as a centrifugal force constricted by a centripetal force that in this case is the wall of the tube. Am I right?
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"When a person swings a bucket by the handle in a large circle, he or she feels a pull on his or her hand, because inertial force (also called centrifugal force in this case)
No they don't [:D]. What they feel is a reaction to the centripetal force that's necessary to maintain the bucket in a circular path. Just as "there ain'ta no sanity clause", "there ain'ta no centrifugal force" either.
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well, a matter of opinion it is :)
But the quote comes from Answers.com.
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For those of you that enjoy Newton. Why not take a look at Newton's Philosophiae Naturalis Principia Mathematica, from Stanford university encyclopedia of philosophy. (http://plato.stanford.edu/entries/newton-principia/) I love that site, so many cool thoughts about 'everything'.
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Newton surprises me :) We talk about it as singular 'forces' but he seemed to have another view.
"It is important to recognize that, in calling the referents of the defined terms “quantities,” Newton is assigning them to the ontological category of quantity in Aristotle's sense. Thus force and motion are quantities that have direction as well as magnitude, and it makes no sense to talk of forces as individuated entities or substances.
Newton's laws of motion and the propositions derived from them involve relations among quantities, not among objects. In place of “no entity without identity,” we have “no quantity without definite proportions;”[19] and the demand on measurement is to supply values that unequivocally yield an adequate approximation to these definite proportions."
Didn't know that.
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well, a matter of opinion it is :)
But the quote comes from Answers.com.
Well, if somebody discovered the existence of centrifugal force, they must have done so in the last 45 years or so, 'cos it sure didn't exist when I was being taught this stuff. Either that, or it's a reflection on the deplorable state of education these days that causes educators to adopt a "politically correct" stance so they don't offend the "true believers".
If we run around inventing special names for all the reactions to actual forces, we're going to get into a real pickle.
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Yor_on:
There is no controversy among physicists regarding centrifugal force. Centrifugal force is actually just a calculation shortcut when designing devices that are spinning at high speed. When you look at these situations as a physicist does, not from any specific viewpoint (e.g. that of the guy holding on to the spinning weight), it is obvious that when gravity acts on a mass moving perpendicular to the direction of the acceleration of gravity, there is only a centripetal force (gravity). As I already said, a moving object will continue unless acted upon, that is, accelerated by a force (centripetal in this case).
The Answers.com article goes sideways when it says that the ship “is pulled toward the Earth by the Earth's gravitational attraction force, while the inertial or centrifugal force of the moving ship is directed radially outward from the Earth; consequently, the force of gravity on the orbiting ship is opposed and nullified by the centrifugal force, and apparent weightlessness results."
A ship in this situation if pulled by gravity that is momentary, or continuous, or either but much weaker or stronger, will result in some specific orbital change, such as flying away, or crashing, or a different higher or lower orbit, but in all these situations the ship and its inhabitants will continue to be weightless. They would not even be able to detect a change without navigational instruments. The Answers author seems to think that there is some specific centrifugal force value that will result in a delicate balance with centripetal force that will in turn result in weightlessness, but this is not true and one doesn’t make this mistake if they keep in mind that there is actually no centrifugal force.
Steve
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Yeaahh Steve! (I was beginning to wonder if my memory was even more dodgy than I already know it is.) What I do remember is that our Physics teacher (aka "Bilko", due to an uncanny resemblance to the Sargeant of the same name) would beat the tar out of anyone who proposed the existence of "centrifugal" force.
This is another good example of why it's a good idea not to believe everything you read on the Internet (unless, of course, the author happens to be yours truly).
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Yeaahh Steve! (I was beginning to wonder if my memory was even more dodgy than I already know it is.) What I do remember is that our Physics teacher (aka "Bilko", due to an uncanny resemblance to the Sargeant of the same name) would beat the tar out of anyone who proposed the existence of "centrifugal" force.
This is another good example of why it's a good idea not to believe everything you read on the Internet (unless, of course, the author happens to be yours truly).
So were you frequently beaten up in class?
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This is another good example of why it's a good idea not to believe everything you read on the Internet
including Wikileaks
(what hope is there for this Physics chat forum if the most posts for a thread topic is "How much does the earth weigh?)
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Well that's why I avoid those expressions normally. To me if you spin something, you are accelerating it, breaking its geodesic path, and expending energy, but also confining its motion from 'any' of the geodesics it otherwise would have. And the 'engine' is the guy rotating. But I remember having this discussion before, and there seems to be different views on what consists of 'fictitious forces'.
So if you looked at my example with that tube spinning, would you say that I interpreted the 'forces' wrongly? I don't like 'forces' at all :)
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Rereading :) To me it always have an importance who expends the energy. Ignoring that may give you an 'objective idea' of what 'forces' there exist, but to me it's always something 'inducing' something when it comes to spending energy.
Maybe there exist examples where you can't say who's doing that though? It's still to early in the morning for me :)
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Yep, I'm gonna get me ass out of here for a while. Need to raid a supermarket..
C000FFFFEEEE
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Back, three cups after, keyed up and ready. Throw it at me, I can take it :)
(But be quick, or I will need more coffee.. :))
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To late, off to the kitchen..
(Quick metabolism)
I think?
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:) Foolosophy ::))
You gonna mix relativistic mass into this *** thread?
why not - we are all friends here arent we?
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Yep..
And those that ain't I'll ...
:)
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Yor_on:
Either you have had too much coffee or I haven't had enough because I don't understand what you are talking about. It almost sounds like you are saying that an object can't produce a force due to gravity because no work is done.
Steve
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It almost sounds like you are saying that an object can't produce a force due to gravity because no work is done.
Steve
You could say that
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.physics.ohio-state.edu%2F%7Ephysedu%2Fmapletutorial%2Ftutorials%2Fwork_eng%2Fimages%2Fwork_eng30.gif&hash=d81cf808a80a4e79db43a874dfa1b689)
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Steve, I guess I'm saying that :)
Think of it this way
A apple falls to the ground. From the viewpoint of Earth the apple is accelerating but from the viewpoint of the apple it's just following a geodesic. Which one is correct? To me it's the geodesic, and it's easy to see what constitutes one, it's, ahem, weightless :) So the acceleration you observe is not a 'uniform acceleration' but just a geodesic that relative a proper mass will bend 'SpaceTime' into a steeper slope, making the apple behave as it is accelerating to you looking at it, having the same 'frame of reference' as Earth.
We had a discussion before here on TNS discussing if a accelerating charge would radiate, where I think we discussed this type of acceleration? But maybe you have a counter-example proving ah, otherwise :) It's the way I look at it now :) But I'm not cast in iron :) We're all here to learn I think.
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Yor_on, when the apple hits the ground it certainly exerts a force, doesn't it? Further, for it to fall at all a force had to be applied to accelerate it.
As an aside, my understanding of the use of the word "geodesic" apparently doesn't match your usage. Please explain. Steve
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In what way do you mean Steve?
A geodesic, to me, that is :) is the path any uniformly moving object, not expending energy, will take in SpaceTime, including when 'accelerating' due to gravity. The last one ('accelerating' due to gravity) is how I see it though, but that I said too? And in mathematics a geodesic is the idea of a shortest path between two points, depending on the 'surface' described. Like Earth, or on a paper it will express itself differently to us observing both, but they are both the shortest 'paths' and they both 'save energy'.
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Although 'shortest' is a matter of definition I agree. I like to describe it in form of expending energy myself. And yes, you can see it as a 'force' if you like but to me gravity is no force, and that also include the apple falling. As for it's 'potential energy' relative Earth it exist.. In the impact, not before as I see it. To me it's as relevant as discussing that apples 'potential energy' relative the moon when it comes to describing it from its own frame of reference, while being 'weight-less' in a 'free fall', well, to me that is :)
As in a black box scenario you would have no way of differing whether you were falling towards the moon, Earth, or just coasting the space-lanes. Coriolis force and frame-dragging excluded that is. They might tell you that space was twisted :) Maybe gravity waves can do the same though? Which would make the equivalence absolute, all as I see it?
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A way to see my definition of 'acceleration' is to look at it from the point of 'energy spent'. the apple doesn't expend any energy and so I think of it as following a geodesic. Does that mean that its 'room time geometry' doesn't change?
Assume that it is free falling towards a Black Hole, or a neutron star. I would expect its 'room time geometry' to change if so, but it still doesn't expend any energy, it's the room time itself that are 'twisted' where it is as I see it.
Now, does that make sense?
It depends on how you look at motion, and matter I admit.
To me it make a twisted sense though :)
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And if you want to argue that by spending a finite amount of energy you can 'force' the whole of SpaceTime to 'age' into entropic equilibrium you will make me very curious, I don't see how that is possible? And that is what you will need to do if treating it as a 'force' I think? (Ah, this is when discussing 'normal acceleration, expending energy)
If we now use that idea on our apple, not expending any energy what so ever?
Then if I look at as a 'force' it will without expending any of its energy, as measured by you in a 'exact same' free fall beside it, 'twist' our SpaceTime? And that's one step further into a very interesting subject, filled with traps. I'm sure we can create the thread of threads debating this one :)
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Yor_on:
It appears to me that you are using some terms that you have redefined to mean something different than common usage. This makes communication difficult. My version of Webster’s dictionary says that geodesic is “the shortest line between two points that lies in a given surface.” This is a pretty tight definition and you should use it accurately. It is possible that I am just ignorant, and if so I am quite willing to be educated, but I need something more than your opinion regarding how you like to use the term.
Gravity, from a Newtonian perspective, is a force between two objects related to their mass and distance. You can measure this force directly, yourself, with an object that has a known mass. This force of gravity is often measured in newtons, a unit of force. You can measure the force of gravity by observing how an object is accelerated, whether free falling straight down or in orbit. If you, or I, were in a closed room, moving or not, under the effect of gravity or an inertial frame, or many thousands of light years from any massive object, we could detect what our situation was with some relatively simple but highly accurate instruments that measure force. Just because you, or I, in our relatively imperfect bodies would not feel any force doesn’t mean anything.
Now if all you are doing is switching to an Einsteinian frame of reference in which gravity is a fictive force, like centrifugal and coriolis forces, then OK, but I still don’t see what your point is in the context of this thread. This is because- If you wish to say that something in free fall among some massive bodies is not acted upon by any force, because gravity is a fictive force, then you also have to say that when you stand on a scale you are weightless because you aren’t really exerting a force on the scale. If somebody asks me how much something weighs (e.g. the earth) I would either try to pin down what their frame of reference was, or try to use the most obvious one to answer their question. In addition to word meanings, it is also important to define your frame of reference in order to communicate accurately and, then, to be consistent.
Steve
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'Weight' is what you feel when 'proper mass' is acting on (propping up) 'proper mass' :)
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Also there is its equivalence to a uniform acceleration at one G. If you was in that spaceship and weighted yourself the scale would give you the same weight as on Earth. Now you might want to say that as the spaceship moves, energy is expended and a 'force' created. But it is in a 'black box scenario' only that this equivalence exist. Any other way you will know that you are 'traveling', and so destroy its equivalence as I understands it.
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Thinking of it again :) I better add that all acceleration, in where you expend energy, will give you a 'weight'. but no 'acceleration' will if you are following a geodesic, as falling of the ladder, or that apple falling. And that's how I differ it.
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Anyway, my 'proper mass' bending the 'room time geometry' for me as I stand on the scale. The scale answering as if my 'proper mass' was a 'force' applied on it. You can also see the scale as a sort of intermediary between me and the planet it rests on, reacting to me, as it is 'nulled' versus that planets gravity well. That is, being at Zero when nothing is placed on it. As I see it, gotta admit it sounds kind of zen that one :)
Yep it does. I think I need to stop eating vegetarians, vegetarian?.
Anyway, think of a laser shooting that beam at the moon. It's a ruler in a sense and the path it takes will be the straightest possible, although, its path 'bends'? Why? Because of 'SpaceTime Geodesics', they bend 'Space'. In fact, anything moving will follow those weird ahem, 'ley-lines' (Druid is my middle name)
And that's the way I see it, except I call them Geodesics, not ley-lines ( Too many fantasy novels there :)And I agree, they are very different from the concept of 'gravity's force', but they suit my way of looking at it.
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Proper mass will create a 'gravity' when 'unmoving'. Not that we can prove anything to be so, except as compared relative something else. That's a unique 'property' belonging only to 'proper mass'. So I bend 'SpaceTime', just as you do, creating 'gravity wells'. And when two objects of 'proper mass' are placed together they 'attract' each other. What the scale measures is that 'attraction', graded in some common nominator to us, be it pounds or kilos. But it's not a 'force' even though it sounds like one. Our preconceptions is based on a Euclidean geometry in where that lasers beam 'shouldn't bend', but we know it will. We live in a place where matter 'distort SpaceTime', just as momentum, relative mass and 'pure energy' does. As it is I'm not entirely sure that it is the momentum 'bending space'? After all, don't we expect the 'momentum' to 'push'? Can it do both? Isn't 'energy' the better description for whats bends/distorts 'space' when two beams 'travel' parallel to each other so that they ahem, 'gravitate' towards each other?
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Thinking of it, that's a nice concept illustrating the idea. Photons are 'mass less' according to mainstream physics, so why do they bend? And should I then expect them to weight something if they do so? If you think they don't weigh, and accept that they still gravitates towards each other then you will see the concept I'm thinking of. It's a concept relating to everything we know, even anti-matter as I understands it.
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Photons are 'mass less' according to mainstream physics, so why do they bend?
Photons do however possess a value for momentum, p and do impart pressure.
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fhyperphysics.phy-astr.gsu.edu%2Fhbase%2Frelativ%2Fimgrel%2Fpphot.gif&hash=21625f10f57702297f867ff3e94a7881)
Photons bend because according to the theory of general relativity, space is bent or warped in the presence of a body's mass and so the photons follow the warped space. That's what gravity is, the warping/distortion effect of space (and time) by mass etc.
.........interesting
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Yes, but the momentum is also what is expected to 'push' on 'stuff', isn't it?
That seems to make them unique in some motto. when proper matter 'push' we call it 'kinetic energy' and it can be described in form of 'springs' compressing and recoiling as 'proper mass' meets 'proper mass', But if we talk about light it seems as if the momentum is expected to both be able to 'push' and 'attract', aka distorting SpaceTime?
Or what am I missing?
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I should stop using 'attract' huh :)
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Yor_on:
OK, now I think I understand what you are about. In the future I will, as much as I can bear, try not to disturb you in your secret room.
Steve
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Heh. I think I see what Yor_on was getting at. Geodesics are shortest-paths (actually they're extremal paths, so they could be longest paths too) within general relativity's curved space-time. Freely falling objects follow geodesics and gravity isn't treated as a force within that framework.
I'm not sure that defining weight is really useful within that framework, but maybe you could define it as the force required to hold an object at a constant position? I can't imagine a use for defining a concept of weight in GR, though...
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Yeah, I found myself in deep trouble trying to find a decent definition of weight JP. And it's 'energy' not 'momentum' that distorts 'SpaceTime' isn't it? I have used momentum so long that suddenly it felt as if a photon only was its momentum there :) Senility my middle-name (I'm a collector of those too:)
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Got any link to that?
extremal paths I mean?
Seems quite serious, the ones I found :)
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And it's 'energy' not 'momentum' that distorts 'SpaceTime' isn't it? I have used momentum so long that suddenly it felt as if a photon only was its momentum there :) Senility my middle-name (I'm a collector of those too:)
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Got any link to that?
extremal paths I mean?
Seems quite serious, the ones I found :)
One of the points of GR is that it works no matter what reference frame you're in. In other words, the basic equations of GR have to work no matter what frame you're in, and in different frames, energy and momentum can change. The correct formulation of GR requires using a stress-energy tensor, which includes energy and momentum as well as the flow of energy and momentum in all four space-time coordinates and shear stresses. In lay terms, this means that energy and momentum might be enough in one reference frame, but they aren't necessarily enough in all reference frames. I can't go into a lot more detail than that, since I don't know GR in a lot of detail, but I do know that energy and momentum alone aren't enough except in special cases.
As for extremal paths and geodesics, I get it from something like this:
http://en.wikipedia.org/wiki/Geodesics_as_Hamiltonian_flows#Geodesics_as_an_application_of_the_principle_of_least_action
I see geodesics described as "extremal paths," "shortest paths," and "the closest thing to a straight line in a curved space." I've been thinking about them as extremal paths because that's a technique used to solve a lot of problems in classical mechanics as well as to make approximations in quantum mechanics and electromagnetism. In those cases, you look for paths that are basically the shortest or longest, so I've been assuming that's how GR works. I haven't found a derivation proving that a geodesic can be the longest path, but neither have I seen one proving that it's the shortest...
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Well, I guess it depend on how you would define a 'longest path' wouldn't it? You could easily see a geodesic as a 'longest path', if you define it from a 'flat plane' I presume? But then again, that's not what you're talking about, is it?:) It's more of a rigorous mathematical definition, or approximative technique I guess in where one have to take in consideration the type of 'space' described in that path, whatever I may mean by that :)
I've always like the idea of energy myself, it seems to fit somehow. But there is also the idea of least force or action, maybe they all are equally correct. But for any action taken in SpaceTime I would expect you to have to expend energy. So I'll keep it :)
As for the tensions and stress inside some specific location of SpaceTime? That should mean other gravitational 'influences' acting on the photons path, except their own I presume? And possibly also covering how to define it depending on where you are watching the 'events'? Different 'frames of reference' in SpaceTime observing different causality chains?
Awh, any which way, time for me to get some sleep here.
My pleasure JP :)
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Well, I guess it depend on how you would define a 'longest path' wouldn't it?
I don't think so. The "extremal path" description comes from the mathematical description that the derivative of the path length with respect to slight variations in the path equals zero. If a derivative of a path equals zero, all you know is that it's either a minimum or a maximum length path, but not which until you get further information.
I don't know if this same treatment holds up in GR. I'm confused by the fact that shortest path, extremal path and closest path to a straight line are all used...
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Ok :)
A mathematical definition in where certain attributes of your equation need to equal zero for it to fit the description of a extremal path, But how?
You have two points that you want to connect 'A' and 'B', then what?
If you look at the path between those two, practically you will need to consider what kind of 'space' they exist in right? Like drawing it on a paper the shortest way is a '2-dimensional flat plane' And in SpaceTime you have three (and time).
What more do you need to weight in for getting a description?
And how do you get to a longest path in SpaceTime?
Assuming that it is a 'sphere' the longest path have to be go around?
Or am I just confusing the concept?
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Thinking of the concept of 'force'. That is F=ma right?
And what it refers to is the concept of 'inertia' which is a objects tendency to stay in whatever motion, or position, it has unless acted upon by some external 'influence'. 'Influence' as we want to include gravity in the concept.
(Thinking about it, I should avoid to say external 'influence' and just call it 'change' instead. That as you can turn on your space engine which is entirely internal acting at the walls of your engine, 'pushing' at them.)
Still, what it all comes down to has to be the idea of 'energy expended'.
Correct me if you think I'm wrong here.
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Here you can find another explanation of that apple, explained in form of geodesics.
Straight lines in a curved SpaceTime. (http://www.relativitet.se/spacetime1.html)
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I don't know, maybe you could look at gravity as an expression of time too?
Light that have no proper mass and is intrinsically time-less will have a 'shorter' SpaceTime path than Matter as 'proper mass' have a time-dilation, as seen from a outside 'frame of reference', and so have a longer Space-Time path? As even though with a 'time dilation' matter would still have a arrow 'ticking' intrinsically whilst light would have none?
Or if you like to see light as 'mass' too, relative or momentum, there still would be a 'mass difference'. But then I would say that it was light having to travel the 'longer path' according to the 'outside observer' as with a bigger 'time dilation' matter should have to be the 'faster'? As the idea of a intrinsic timelessness would be gone?
And the equivalence should hold for Lorentz contraction too, as observed from the 'inside' of their respective frames, assuming that light also meet that phenomena of course? Which it should if you assume a mass to it. But in the first case I'm not sure what light 'sees'?
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One way around it would be to give only proper mass a time dilation, and ignore relative mass and momentum as having any importance for a time dilation? Then you can let light have a 'mass' and still not be affected by any intrinsic time? Light is so strange :)
I like it best not propagating at all :)
But if I do so I would have to assume that 'motion' in itself have some time dilating property, unrelated to relative mass and momentum? Nahh, that one was too weird..
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Or maybe not?
Momentum shouldn't have a time dilation?
Because if it did it should invalidate the photons timelessness, shouldn't it?
And then we have relative mass left?
which in a way is a equivalence to 'potential energy'.
And I don't like 'potential energy' as a 'force' :)
Heh.
Which leaves 'motion'
And if i assume that light don't 'move'.
But matter does?
With should make 'motion' time dilation in itself.
Stop throwing things at me :)
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Maybe 'energy' need to be looked at though?
We can't say that 'matter' gains 'energy' as such by motion. That is, the atoms doesn't start to 'jiggle' in my spaceship just because it 'moves' faster, do they? But there have to be a relation anyway. But I do 'bend' SpaceTime, contracting it, and also getting a time dilation as seen from 'outside' my frame of reference.
Could I assume that I gain 'energy' by a Lorentz contraction?
That is that if I assumed that the energy in the 'field' around my ship due to motion have increased?
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What more do you need to weight in for getting a description?
And how do you get to a longest path in SpaceTime?
Assuming that it is a 'sphere' the longest path have to be go around?
You look at the length of the path from point A to point B as the path taken varies. Then you look for places where slight variations in the path don't change its length. This happens at a local maxima or minima of the path length. I don't know if it's possible in GR for an object to take a maximal path, but in other cases which use the same mathematics these paths are important.
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Sweet, now I see what you mean :)
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I'm not sure I follow your question, yor_on.
In SR, velocity alone determines time dilation. Mass isn't useful in doing so. Velocity is also determined by the ratio of energy to momentum, so you could determine time dilation if you know energy and momentum. For photons velocity is always c and the energy-momentum ratio is always constant. Of course, there is no mathematical theory for the point-of-view of photons, so no physical theory is going to answer the question of how they experience the universe. [:)]
In GR, time dilation is also caused by curvature of space-time, so mass does get involved, but as I mentioned elsewhere it gets involved through the stress-energy tensor, so the fundamental quantities are still energy and momentum. I don't know the details as well in GR, though.
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Nah it's just me having assumed that they all was involved in it, well, not momentum maybe :) I haven't thought of it like this before. Strange..
But it makes sense, if it is this way then I don't need the concept of 'potential energy' I think? And neither do I need 'relative mass'?
That is because the 'potential energy' is a direct effect of motion and not 'potential' at all but very real. It's not so much as invalidating the concept, more like you have a 'sea' of energy that when contracted will increase. But it fits well in with how I think 'potential energy' is described with the difference of it not being 'potential' but 'real' :) If you get my drift..
And to then 'fit it' against other 'frames of reference' you just need to remember that all of them have their own 'SpaceTime', moving relative you or not. In them they will observe the same effects as you do. And as all of them are 'uniquely' their own, with a 'own SpaceTime' aka 'frame of reference' the 'potential energy' you expect to be released with a impact will be there as a relation.
It's moving me into a really weird universe if it's correct. Every 'frame of reference' a direct expression of a unique 'room time geometry' possibly swimming in a sea of 'energy' that we only get access to through 'motion' but still to us representing a 'whole 'smooth' SpaceTime' as far as we can observe having no 'borders' between what we know to be those 'frames of reference'. And then we have matter? A black hole create a very strange 'room time geometry', doesn't it? Or a Neutron star? And the Bekenstein bound as a description of the amount of 'states' possible, and possibly also 'energy' available(?), in your 'frame of reference'
But hey :)
It's late..
And momentum is then a very specialized description of photons 'pushing'.
And suddenly those Rindler observers starts to make a new sense..
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I need to get some sleep huh :)
Anyway, for me its opening new ground as I reasoned it out all by myself ::))
However wrong I may be :)
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And thanks JP, you have a flair for making your explaining succinct.
It's a very nice sign of knowing what one speaks of..
Same as me in fact... N 0t :)