Naked Science Forum
On the Lighter Side => New Theories => Topic started by: olvin on 22/08/2011 12:38:00
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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.
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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?
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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.
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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.
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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)
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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.
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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?
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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
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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.
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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.
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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.
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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.
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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.
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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!"
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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
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I'm just amused that the physicists seem to have redefined the word "proper" to mean "bogus".
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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 ;).
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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.
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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 ?
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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.
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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?
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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.
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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.
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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.
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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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I'm not denying the evidence. (and implications that I was doing might be construed as an ad hom)
The evidence is that I'm still sat on this chair.
I have not moved.
My velocity remains zero.
The change in velocity is zero.
The rate of change of velocity is zero
the acceleration is zero.
If you took the chair away then I would experience an acceleration downwards which I would be able to observe, not least when I hit the ground.
"You can assume that your acceleration is caused by a force"
And I do. As you say, if I didn't do that I would be ignoring F=ma.
However I believe in that formula.
And, since I know my mass, I can measure my acceleration (from how fast the world and I get together) and I can calculate the force.
It remains about 700N
That's not an acceleration, its a force; it's called weight.
The fact that it is proportional to mass isn't the issue.
Perhaps you would like to answer the question I posed earlier
"With respect to what am I currently accelerating that explains the dent in the springs in the chair I'm sitting on?"
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If it is merely an acceleration that is completely independent of mass (the Geezer position) and not (what I think BC is contending) the result of a calculation of opposing forces - how does one explain a helium filled balloon. It should accelerate to the floor just as BC's bum does without the chair if all there is a globally applicable acceleration of 9.8ms-2 (and we have moved on from those mad imperial units that only the americans now cling to); of course we know that other forces (not necessarily an acceleration) also play a role; and it is the vector sum of all these forces that determines the motion
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If gravity is a "force", isn't it a bit strange that it is measured in units of acceleration rather than force per mass? Could this possibly be a clue? [::)]
What y'all obviously fail to realize is that "weight" is simply the force necessary to prevent an acceleration. A mass is still being "accelerated" whether it is moving or not, and that is why the springs in your chair are stressed while it is preventing your acceleration.
As I proved earlier, if F=ma, gravity can only be an acceleration. I'm only applying the rules. If you have a problem with this you should complain to your science teacher for giving you a bum steer.
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"If gravity is a "force", isn't it a bit strange that it is measured in units of acceleration rather than "
It isn't
http://en.wikipedia.org/wiki/Gravitational_constant
You seem to be basing your point of view on a false premise.
However it's a moot point. The two sets of units you propose are the same anyway.
Acceleration
m/s/s
Force? Well we both agree that f= ma
so
"force per mass"
has units of force ( which is ma)
kg m/s/s
divided by the units of mass
(that's kg)
so kg m /s /s divided by kg is
m/s/s
which is the same as acceleration.
And, as I have said, I remain a firm believer that F =ma
"What y'all obviously fail to realize is that "weight" is simply the force necessary to prevent an acceleration. "
That's true of any force.
If I push on a rock, the force that stops it accelerating is friction.
Is friction an acceleration?
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Well, if you want to come up with an alternative to General Relativity, knock yourself out.
"From the viewpoint of GR, there is no gravitational force. Rather, in
the absence of electromagnetic and other forces, particles follow the straightest possible
paths (geodesics) through a spacetime curved by mass. Freely falling particles define
locally inertial reference frames. Time and space are not absolute but are combined into
the four-dimensional manifold called spacetime."
http://web.mit.edu/edbert/GR/gr1.pdf
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I'm not trying to come up with an alternative to GR.
I'm trying too get you to explain yourself, but you seem unable or unwilling to do so.
So, for the third time of asking
"With respect to what am I currently accelerating that explains the dent in the springs in the chair I'm sitting on?"
And, while you are at it, why do you think you can add together quantities with different units? And do you consider friction to be an acceleration?
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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?"
You seem to have missed this.
http://www.thenakedscientists.com/forum/index.php?topic=40713.msg365800#msg365800
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I'm not trying to come up with an alternative to GR.
I'm trying too get you to explain yourself, but you seem unable or unwilling to do so.
So, for the third time of asking
"With respect to what am I currently accelerating that explains the dent in the springs in the chair I'm sitting on?"
And, while you are at it, why do you think you can add together quantities with different units? And do you consider friction to be an acceleration?
You're accelerating in time.
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Geezer,
that purports to be an explanation based on redefining the word "acceleration" to include standing still or alternatively failing to explain that what you meant was this odd thing called "proper acceleration"
Is it worth asking you to reply to the other questions about dimensional analysis?
Mike,
Thanks for actually answering the question.
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BC
You're welcome.
You might think that saying "You're accelerating in time" is obvious but it goes much deeper.
If gravity and acceleration are equivalent then gravity is no more than an acceleration but an acceleration in what? It has to be non-directional in a spatial sense and that only leaves an acceleration in time. Acceleration contains a time factor, therefore any change in the passage of time will affect acceleration. The passage of time is slower on the surface of the Earth becoming progressively faster the further away you get. If the passage of time is slowest on the Earth and the Earth is traveling through time then the Earth must be accelerating in time.
Therefore, mass accelerating through time causes the effect of gravity.
See http://www.thenakedscientists.com/forum/index.php?topic=40746.msg365767#msg365767
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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?"
And, while you are at it, why do you think you can add together quantities with different units? And do you consider friction to be an acceleration?
Ok, I'll steal Geezer's thunder a bit. The acceleration is with respect to an inertial reference frame. If you were free-falling, you wouldn't dent the springs. Since you're accelerating with respect to a free-falling reference frame, you dent the springs.
For the second bit, of course force isn't acceleration. But calculations within accelerating reference frames require the use of fictitious forces if you want F=ma to hold (centrifugal force is the classic example). Fictitious forces aren't real forces and are due entirely to acceleration: if you properly write out Newton's laws, accounting for the accelerating coordinate system, you can eliminate the fictitious forces and explain their effects in terms of accelerating reference frames. Friction wouldn't be an acceleration since it can appear in an inertial reference frame.
If you assume that gravity is a fictitious force, you notice right away that the force of gravity disappears when you're in free fall. If you then figure out the proper way to change variables to non-free-falling coordinate systems, you can eliminate the force of gravity and describe it in terms of accelerations of coordinate systems with respect to free-fall. This is what GR does.
All that Geezer seems to be pointing out here is that if you use GR, you can explain the force of gravity as a fictitious force that comes from not properly accounting for accelerating coordinate systems. Of course, you can also describe gravity as a force and use Newtonian mechanics quite well in most cases that appear in daily life.
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Now we have that sorted out, perhaps we can get back to the OP's point.
They contend that "person inside the air craft reduces its own gravity force due to speed."
And I disagree, because I don't seem to weigh less when I'm in a plane.
In a plane I am travelling fast so, WRT me, the earth is travelling fast. But it still seems to hold the same attraction for me.
As I said, we are not talking about the effect of altitude on gravity here, nor the effect of the plane rising or falling.
Just to keep it simple we are on a plane in level flight (i.e. at a constant height WRT the Earth's C of G).
Given that starting point, it seems to me that the OP is mistaken.
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Now we have that sorted out, perhaps we can get back to the OP's point.
They contend that "person inside the air craft reduces its own gravity force due to speed."
I agree. Speed shouldn't effect gravitational force.
And I disagree, because I don't seem to weigh less when I'm in a plane.
In a plane I am travelling fast so, WRT me, the earth is travelling fast. But it still seems to hold the same attraction for me.
As I said, we are not talking about the effect of altitude on gravity here, nor the effect of the plane rising or falling.
Just to keep it simple we are on a plane in level flight (i.e. at a constant height WRT the Earth's C of G).
Given that starting point, it seems to me that the OP is mistaken.
I think what the poster might be describing (though the question was worded in a confusing manner) is that you appear weightless when in orbit. As your plane goes faster and faster maintaining a constant distance from the earth's center of mass, your weight as measured on a scale on board the plane will decrease. We can argue about whether that's a proper definition of weight, but so far as you and the scale are concerned, it's reading a lower number. If your plane is actually a spaceship, and you reach a stable orbit, the scale reads zero. But it's not your speed causing this. It's the fact that you have to be falling to maintain a constant distance from the earth's center of mass, and falling makes you appear to weigh less on that scale.
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I think what the poster might be describing (though the question was worded in a confusing manner) is that you appear weightless when in orbit.
But you don't need to be in orbit. To become weightless, all you need to do is jump so that your feet are no longer in contact with anything.
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I think what the poster might be describing (though the question was worded in a confusing manner) is that you appear weightless when in orbit.
But you don't need to be in orbit. To become weightless, all you need to do is jump so that your feet are no longer in contact with anything.
Whilst you are accelerating upwards you weigh more. At the top of the jump you weigh the same and only on the way down do you start to weigh less. It's got nothing to do with your feet touching the floor. It's all to do with acceleration in space-time.
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I think what the poster might be describing (though the question was worded in a confusing manner) is that you appear weightless when in orbit.
But you don't need to be in orbit. To become weightless, all you need to do is jump so that your feet are no longer in contact with anything.
Whilst you are accelerating upwards you weigh more. At the top of the jump you weigh the same and only on the way down do you start to weigh less. It's got nothing to do with your feet touching the floor. It's all to do with acceleration in space-time.
Not really. You are literally "weightless". It's only possible to "weigh" something when it is in contact with something else. You do weigh more while your legs are accelerating your mass and your feet are still in contact with the ground, but as soon as they leave the ground, you are weightless.
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I think what the poster might be describing (though the question was worded in a confusing manner) is that you appear weightless when in orbit.
But you don't need to be in orbit. To become weightless, all you need to do is jump so that your feet are no longer in contact with anything.
True. I was trying to figure out what the OP meant. He mentioned flying in a plane and microgravity, so I thought about orbits. If you free fall and weigh yourself on a scale, you get zero weight.
The force of gravity isn't reduced in any case (according to Newton).
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I think what the poster might be describing (though the question was worded in a confusing manner) is that you appear weightless when in orbit.
But you don't need to be in orbit. To become weightless, all you need to do is jump so that your feet are no longer in contact with anything.
Whilst you are accelerating upwards you weigh more. At the top of the jump you weigh the same and only on the way down do you start to weigh less. It's got nothing to do with your feet touching the floor. It's all to do with acceleration in space-time.
Not really. You are literally "weightless". It's only possible to "weigh" something when it is in contact with something else. You do weigh more while your legs are accelerating your mass and your feet are still in contact with the ground, but as soon as they leave the ground, you are weightless.
Geezer
You are quite right if using the purely gravitation explanation "defining the weight of an object as the force measured by the operation of weighing it (using a force-sensitive scale, such as a spring scale), in vacuum."
but there is another definition defining it as "When used to mean force, its magnitude (a scalar quantity), often denoted by an italic letter W, is the product of the mass m of the object and the magnitude of the local gravitational acceleration g;[3] thus: W = mg."
http://en.wikipedia.org/wiki/Weight
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Mike,
Wikipedia isn't always correct. The g in W=mg is constant which means that W must be proportional to mass, but it clearly is not. If you stand on a scale and bend your knees, your weight varies substantially.
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Gm1m2/r^2 = m1a
a = Gm2/r^2, not really a constant, only for a specific geodesic orbit it is constant.
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Mike,
Wikipedia isn't always correct. The g in W=mg is constant which means that W must be proportional to mass, but it clearly is not. If you stand on a scale and bend your knees, your weight varies substantially.
Surely W=mg is just a way of knowing the weight of an object without actually 'weighing' it. When you bend your knees your rest mass remains the same but you have added an inertial mass component. Likewise, when you jump off the ground the action of jumping gives you inertial mass which is 'heavier' than your rest mass. At the top of the jump you have lost all of the inertial mass and your weight returns to normal.
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Surely W=mg is just a way of knowing the weight of an object without actually 'weighing' it.
W=mg is an approximation for weight assuming that is only valid in particular cases. It's an unreliable method of determining weight in general. It's not easy to alter your mass, but it's very easy to alter your weight.
It's a definition thing. Weight is a measure of net force on an object. If the forces change, the weight has to change.
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Weight is a measure of the force that local gravity exerts on something.
So, while W=mg, g isn't a constant.
Currently I weigh about 700 N, but if I moved to the moon, I would weigh about a sixth of that.
If I moved upstairs I would weigh slightly less than I do here.
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Weight is a measure of the force that local gravity exerts on something.
So, while W=mg, g isn't a constant.
Currently I weigh about 700 N, but if I moved to the moon, I would weigh about a sixth of that.
If I moved upstairs I would weigh slightly less than I do here.
True. Not only that, but g does not take into account the force produced by the atmosphere that's trying to lift you off the scales.
A mass of 1 kg can have a negative weight on Earth if it happens to be hydrogen gas.