How much does the Earth weigh?

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Offline yor_on

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« Reply #100 on: 03/12/2010 18:17:52 »
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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Offline QuantumClue

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« Reply #101 on: 03/12/2010 18:29:03 »
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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Offline QuantumClue

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« Reply #102 on: 03/12/2010 18:30:01 »
All that weight is reconfigured into a force. And vice versa.

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Offline Geezer

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« Reply #103 on: 03/12/2010 20:22:20 »
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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Offline Foolosophy

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« Reply #104 on: 04/12/2010 07:27:15 »
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?

« Last Edit: 04/12/2010 07:32:22 by Foolosophy »

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Offline yor_on

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« Reply #105 on: 04/12/2010 09:51:55 »
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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Offline Foolosophy

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« Reply #106 on: 04/12/2010 10:46:42 »
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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Offline rosy

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« Reply #107 on: 04/12/2010 11:34:30 »
Quote
On the other hand, if you say that you are "weightless" during your fall, you have to wonder why you are falling at all.

Quote
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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Offline Foolosophy

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« Reply #108 on: 04/12/2010 12:11:38 »
Quote
On the other hand, if you say that you are "weightless" during your fall, you have to wonder why you are falling at all.

Quote
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"

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Offline rosy

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« Reply #109 on: 04/12/2010 13:00:32 »
yor_on.. to go back a few posts (sorry):
Quote
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:
Quote
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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Offline Foolosophy

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« Reply #110 on: 04/12/2010 13:20:56 »



...there are people who argue that the centrifugal force is NOT real

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Offline rosy

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« Reply #111 on: 04/12/2010 13:31:38 »
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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Offline Foolosophy

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« Reply #112 on: 04/12/2010 13:42:45 »
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
« Last Edit: 04/12/2010 13:54:56 by Foolosophy »

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Offline rosy

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« Reply #113 on: 04/12/2010 14:26:11 »
Quote
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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Offline Foolosophy

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« Reply #114 on: 04/12/2010 14:36:50 »
Quote
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


« Last Edit: 04/12/2010 14:44:22 by Foolosophy »

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Offline yor_on

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« Reply #115 on: 04/12/2010 15:59:43 »
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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« Reply #116 on: 04/12/2010 16:34:12 »
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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Offline yor_on

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« Reply #117 on: 04/12/2010 17:28:18 »
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.
« Last Edit: 04/12/2010 17:35:40 by yor_on »
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Offline CPT ArkAngel

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« Reply #118 on: 04/12/2010 17:33:50 »
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.
« Last Edit: 04/12/2010 17:39:43 by CPT ArkAngel »

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Offline yor_on

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« Reply #119 on: 04/12/2010 17:45:56 »
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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Offline yor_on

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« Reply #120 on: 04/12/2010 18:04:05 »
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..
==

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?
« Last Edit: 04/12/2010 18:18:51 by yor_on »
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Offline CPT ArkAngel

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« Reply #121 on: 04/12/2010 18:14:26 »
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?
« Last Edit: 04/12/2010 18:26:28 by CPT ArkAngel »

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Offline Geezer

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« Reply #122 on: 04/12/2010 18:21:55 »
Quote
On the other hand, if you say that you are "weightless" during your fall, you have to wonder why you are falling at all.

Quote
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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Offline rosy

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« Reply #123 on: 04/12/2010 18:26:47 »
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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Offline yor_on

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« Reply #124 on: 04/12/2010 18:37:13 »
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.

==

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 :)
« Last Edit: 04/12/2010 18:42:28 by yor_on »
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Offline CPT ArkAngel

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« Reply #125 on: 04/12/2010 19:07:29 »
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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Offline yor_on

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« Reply #126 on: 04/12/2010 19:19:36 »
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?
==

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.
==

The problem being that I can think of no way changing course in space without introducing a acceleration, angular or not?
« Last Edit: 04/12/2010 19:29:13 by yor_on »
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Offline yor_on

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« Reply #127 on: 04/12/2010 19:52:28 »
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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Offline Geezer

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« Reply #128 on: 04/12/2010 20:38:13 »
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."
« Last Edit: 04/12/2010 21:15:44 by Geezer »
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Offline Foolosophy

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« Reply #129 on: 05/12/2010 07:10:55 »
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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Offline Geezer

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« Reply #130 on: 05/12/2010 18:20:12 »
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.
« Last Edit: 05/12/2010 18:45:01 by Geezer »
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Offline syhprum

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« Reply #131 on: 05/12/2010 19:12:30 »
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.
syhprum

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Offline Geezer

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« Reply #132 on: 05/12/2010 19:21:26 »
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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Offline Foolosophy

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« Reply #133 on: 06/12/2010 00:22:02 »
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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Offline Geezer

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« Reply #134 on: 06/12/2010 05:10:44 »
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.
« Last Edit: 06/12/2010 05:13:28 by Geezer »
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« Reply #135 on: 06/12/2010 10:52:01 »
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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« Reply #136 on: 06/12/2010 11:51:24 »
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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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« Reply #137 on: 06/12/2010 12:56:48 »
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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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« Reply #138 on: 06/12/2010 13:18:40 »
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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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« Reply #139 on: 06/12/2010 13:59:11 »
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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?

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« Reply #140 on: 06/12/2010 14:34:19 »
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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« Reply #141 on: 06/12/2010 14:43:44 »
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?)
« Last Edit: 06/12/2010 14:45:44 by Foolosophy »

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« Reply #142 on: 06/12/2010 15:38:30 »
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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".

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(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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« Reply #143 on: 06/12/2010 17:01:14 »
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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« Reply #144 on: 06/12/2010 19:51:46 »
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.
==

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.


« Last Edit: 06/12/2010 20:09:08 by yor_on »
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« Reply #145 on: 06/12/2010 20:14:55 »

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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« Reply #146 on: 06/12/2010 20:25:47 »
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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.*

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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.  

Quote
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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« Reply #147 on: 06/12/2010 21:14:17 »
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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« Reply #148 on: 06/12/2010 21:35:41 »
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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« Reply #149 on: 06/12/2010 21:38:00 »
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?
« Last Edit: 06/12/2010 21:40:55 by yor_on »
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