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Author Topic: mass has the tendency to reduce its own gravity as its velocity increases  (Read 10230 times)

Offline olvin

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As per my research i assume that mass reduce its gravity
when its acclerate.
this is the reason why micro gravity effect happens
when aircraft fall in accleration speed ,down side.

person inside the air craft reduces its own gravity force due to speed.
aircraft body also reduce its own gravity force due to speed. 
ther by microgravity is created.   


 

Offline BenV

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Wouldn't that imply that you would also get a microgravity effect when accelerating upwards, away from the Earth?  Or have I misunderstood you?
 

Offline Bored chemist

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"person inside the air craft reduces its own gravity force due to speed."
I don't seem to weigh less when I'm in a plane.
 

Offline Geezer

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I don't seem to weigh less when I'm in a plane.


Well, you might, but it would depend on what the plane happened to be doing. But you don't need to be in a plane to weigh less. All you need to do is stand on your bathroom scales then quickly bend your knees.

I can guarantee this will make you weigh less.

Mind you, I'd like to have a chat with the bright spark that came up with the term "microgravity". Anyone, or any thing, on, or orbiting the Earth, is experiencing pretty much the same gravitational force. There is nothing like the six orders of magnitude difference that the term implies.
 

Offline Bored chemist

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I too would like to read some proper research on this, but I rather doubt that there is any
Proper research would have uncovered things like "I don't weigh more (or less) on a plane" and come to the conclusion that the whole idea doesn't make sense.

(Incidentally, I'm not interested in hearing about the effect of height on gravity or the apparent effect of acceleration)
 

Offline Geezer

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I'm not interested in hearing about the effect of height on gravity or the apparent effect of acceleration.


As weight is always a consequence of acceleration, it's a bit difficult to discuss weight without taking acceleration into account.
 

Offline Bored chemist

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I'm not interested in hearing about the effect of height on gravity or the apparent effect of acceleration.


As weight is always a consequence of acceleration, it's a bit difficult to discuss weight without taking acceleration into account.

With respect to  what am I currently accelerating that explains the dent in the springs in the chair I'm sitting on?
 

Offline Geezer

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With respect to  what am I currently accelerating that explains the dent in the springs in the chair I'm sitting on?


Er, yes.

"An object on the Earth's surface is accelerating relative to the free-fall condition, which is the path an object would follow falling freely toward the Earth's center. It is thus experiencing proper acceleration, even without a change in velocity"

http://en.wikipedia.org/wiki/G-force
 

Offline Bored chemist

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from
http://en.wikipedia.org/wiki/Acceleration
"In physics, acceleration is the rate of change of velocity over time.[1] In one dimension, acceleration is the rate at which something speeds up or slows down."
On the other hand, someone has made up something called
http://en.wikipedia.org/wiki/Proper_acceleration
To make accelerometers look like they give the right answer, even when they are plainly not accelerating.


 

Offline Geezer

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"Weight" is a measure of force. (The springs in your chair deflect according to that force.) You have mass, and a force is acting on you.

Therefore, your acceleration is F/m.
 

Offline Bored chemist

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There is indeed a force acting on me due to the compression of the springs.
There is also an equal but opposite (I'm sure I have heard that phrase before) force acting on me due to gravity.
The two forces cancel out (or, if you prefer, their vector sum is zero)

My acceleration is (F(up)-F(down))/M
i.e. zero.
That's why I'm still here.
 

Offline Geezer

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And how, pray tell, shall we measure this mysterious force of which you speak when you are not restrained by chairs or other means?

If nothing is pressing against you, F=0, which means a=0.
 

Offline Bored chemist

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By measuring my acceleration, because, under those conditions I will actually move.

If you took the chair away one of the two forces would be removed and so there would be a net force and there would be an acceleration.
 

Offline AlmostHuman

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Reading these comments really made my day! Thanks guys! ;)

Have you ever take into account your own gravity pull on those springs... They might be compressing under their own weight caused by your gravity field. J/K

Few days ago, I was passing by group of kids. As I was passing they started picking on the kid with big head... "Hey! Your head is so big that it has its own gravity field!"
 

Offline imatfaal

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Bagsy first in line to pull the chair away as BC and Geezer go to sit down!   Have to agree with BC - in these circumstances acceleration is the change of velocity over time.  The net force is zero, the vectors of gravity and normal are in opposite directions, net force is zero & acceleration is zero.  

If nothing is pressing up against me (firstly I have spent another night on the pull in vain) then the net force is almost entirely the force of gravity and I will fall towards the centre of mass of the earth - the force is calculated with F=GMearthMmeR^-2 , to get the acceleration divide the force by my mass a=GMearthR^-2; at the earth's surface this is damn near constant at 9.8ms^-2.  This will agree with timing of Geezer tumbling onto the floor after the chair is pulled away

 

Offline Bored chemist

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I'm just amused that the physicists seem to have redefined the word "proper" to mean "bogus".
 

Offline AlmostHuman

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I'm just amused that the physicists seem to have redefined the word "proper" to mean "bogus".


And it's not even modern physics... This (proper acceleration) even makes sense. I was working on project that involves (water) pressure control. And I was bit confused about constant air pressure (where did it go?) and water pressure. Also, my instruments were hitting negative side of scale pretty often. I guess that was "proper" pressure ;).
 

Offline JP

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I'm just amused that the physicists seem to have redefined the word "proper" to mean "bogus".

Well, if you consider where the proper acceleration is really useful, it's not bogus at all.  It's a useful concept in general relativity, where it's natural to describe observers moving in inertial reference frames (free fall).  Proper acceleration is acceleration from the point of view of these free-falling frames. 

But in the case of someone sitting on the chair, I agree that it's a bit silly to  analyze it in terms of proper acceleration.
 

Offline Geezer

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By measuring my acceleration, because, under those conditions I will actually move.

If you took the chair away one of the two forces would be removed and so there would be a net force and there would be an acceleration.

But you can't measure the force that way.

If there is a force accelerating you, not only would you feel it acting on your body, but it would have to be the product of your mass times your acceleration. However, in this case, your acceleration has nothing to do with your mass, so it can't be a force that's causing the acceleration.

Or are you saying F≠ma ?
« Last Edit: 26/08/2011 19:45:07 by Geezer »
 

Offline Bored chemist

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"But you can't measure the force that way."
Says who?
I could, were I so minded measure the distance between my hindquarters and the ground as a function of time, calculate the acceleration (it will be close to 8.9 m/s/s) and since I know that M is 70 Kg, I can calculate F which will be close to 700N).
Why do you think that's a problem?
I can also, if I choose, measure the deflection of the springs in the chair and calculate the force from that.

While I'm at it,
"If there is a force accelerating you, not only would you feel it acting on your body"
OK, with practice I could use the strength of that feeling to measure the force.

"However, in this case, your acceleration has nothing to do with your mass, "
Oh yes it has, it's just that F is proportional to M so A is constant- about 9.8 m/s/s

"so it can't be a force that's causing the acceleration."
Yes it is; that force concerned is generally called weight. It's what would cause me to accelerate downwards if the chair stopped supplying a force that holds me up.

"Or are you saying F≠ma ?"
No, of course not.
I'm saying that there are several forces involved and A is proportional to the sum of those forces.
If the two forces cancel out, I don't accelerate.


 

Offline Geezer

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"But you can't measure the force that way."
Says who?
I could, were I so minded measure the distance between my hindquarters and the ground as a function of time, calculate the acceleration (it will be close to 8.9 m/s/s) and since I know that M is 70 Kg, I can calculate F which will be close to 700N).

Says me.

You are in "free fall" accelerating towards the Earth. Your measurements always produce the same result, even if you vary your mass. What do you suppose you are measuring?
 

Offline Geezer

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the net force is almost entirely the force of gravity and I will fall towards the centre of mass of the earth - the force is calculated with F=GMearthMmeR^-2 , to get the acceleration divide the force by my mass a=GMearthR^-2; at the earth's surface this is damn near constant at 9.8ms^-2.  This will agree with timing of Geezer tumbling onto the floor after the chair is pulled away



But, there aint'a no sanity clause, and there aint'a no gravitational force either. (Come to that, there aint'a a no centrifugal force, but that's another story.)

F=ma, right?

A body (anybody in fact) has the same acceleration in free fall. From this we are forced to conclude that the acceleration due to gravity is a constant. (32 ft/sec/sec in your superior English units)

Therefore, if it is a force that is causing the acceleration, it is always proportional to mass.

But when you plug that into F=ma, it takes mass out of the picture completely -

something x mass = acceleration x mass
something = acceleration

and you end up concluding that either F does not equal ma, or gravity simply produces acceleration without any force. That's why I said "weight is always a consequence of acceleration".

It's convenient to think of "the force of gravity" because that works very well in lots of situations, but it has limitations that can get us into a bit of a pickle if we are not careful. I don't think this should be a great surprise because gravity still remains a very mysterious thing, and whether we like it or not, our chairs are continuously producing a force on our bums that counteracts gravitational acceleration.
« Last Edit: 27/08/2011 04:07:19 by Geezer »
 

Offline Bored chemist

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"But you can't measure the force that way."
Says who?
I could, were I so minded measure the distance between my hindquarters and the ground as a function of time, calculate the acceleration (it will be close to 8.9 m/s/s) and since I know that M is 70 Kg, I can calculate F which will be close to 700N).

Says me.

You are in "free fall" accelerating towards the Earth. Your measurements always produce the same result, even if you vary your mass. What do you suppose you are measuring?

Nonsense.
I can see the ground coming towards me if I fall towards it.
As I already said "I could, were I so minded measure the distance between my hindquarters and the ground "
so why do you ask "What do you suppose you are measuring?"?

If the chair is there I stay at the same height and I can feel the force on my butt.
If the chair goes away those facts change.
What you are saying is that I cannot tell if the chair is there or not.
How absurd are you prepared to be?

Incidentally, the fact that gravitational mass and inertial mass are the same (as far as we know ) is just an oddity of the universe.

Have a look at electrostatics.
I can get two conductors of different masses, but the same charge. I can put them near the earth (presumed to be much heavier and also conducting).
They are attracted to the earth with a force proportional to the charge squared and the reciprocal of the distance above the (approximately) flat surface of the earth.
Because they have different masses they have different accelerations in spite of the fact that they have the same charge.
If we were to look at electrons instead then, because they all have the same charge and the same mass they would all have the same acceleration (from a given height).
Now imagine that I (magically) come up with some other material that has the same charge^2 to mass ratio as an electron.
It will fall at the same speed as the electrons.

The important factor for rate of fall is the ratio of the charge (squared) to the mass.

It's not that the mass has been written out of the equation; it's still there. It's just that when you look at the ratio of the rates of fall, it cancels out.


The thing is that, unlike electrostatic charges; with gravity, the ratio of "propensity to move" to "difficulty of getting it to move" is the same for any object.
It's called mass.
Just because it cancels out in some  pair of equations doesn't mean it isn't real.


Look at de Broglie's equation.
He took E=MC^2 and E=h nu
and said that the E in both cases is the energy so
MC^2 =h nu
and he got the Nobel prize for it (after rearranging it in terms of momentum).
Just because the E term has been eliminated doesn't mean that energy has ceased to exist.

" our chairs are continuously producing a force on our bums that counteracts gravitational acceleration."
Wrong, by dimensional analysis.
A force (like that from the chair) can counteract another force (like weight) but it cannot counter an acceleration because you can't do the vector sum on quantities with different units.

The fact that we don't fully understand one of those forces doesn't matter.



« Last Edit: 27/08/2011 10:49:08 by Bored chemist »
 

Offline Geezer

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"But you can't measure the force that way."
Says who?
I could, were I so minded measure the distance between my hindquarters and the ground as a function of time, calculate the acceleration (it will be close to 8.9 m/s/s) and since I know that M is 70 Kg, I can calculate F which will be close to 700N).

Says me.

You are in "free fall" accelerating towards the Earth. Your measurements always produce the same result, even if you vary your mass. What do you suppose you are measuring?

Nonsense.
I can see the ground coming towards me if I fall towards it.
As I already said "I could, were I so minded measure the distance between my hindquarters and the ground "
so why do you ask "What do you suppose you are measuring?"?


You can deny the evidence if you like, but all your experiment will tell you is that you are subject to a constant acceleration that is independent of mass. You are not measuring a force, you are measuring an acceleration.

You can assume that your acceleration is caused by a force, but you can't escape from the fact that the "force" is proportional to mass. When you apply that in F=ma, you end up with units of acceleration only.

BTW BC, please knock off the ad hominem editorials.
« Last Edit: 27/08/2011 19:16:23 by Geezer »
 

Offline Geezer

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We seem to have drifted a long way from the original topic. A more specific question here;

http://www.thenakedscientists.com/forum/index.php?topic=40810.msg366118#msg366118
 

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