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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...
 

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 »
 

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..
 

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 »
 

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
 

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 »
 

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.
 

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.
 

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. 

 

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 »
 

Offline Foolosophy

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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.

 

Offline rosy

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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.
 

Offline Foolosophy

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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

 

Offline rosy

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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?

 

Offline Foolosophy

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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?
 

Offline rosy

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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?
 

Offline Foolosophy

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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 »
 

Offline rosy

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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".

Quote
(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
 

Offline imatfaal

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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. 
 

Offline yor_on

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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 »
 

Offline Geezer

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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?"
 

Offline rosy

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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.*

Quote
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).
 

Offline yor_on

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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 :)
 

Offline Geezer

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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.

 
 

Offline yor_on

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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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