Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Pmb on 05/07/2013 00:18:29

Howdy folks,
I might be asking several trivial questions in the near future. I hope you don't mind. I'm going to be thinking a lot about definitions and I'd like to get your thoughts on these subjects.
For example; weight  Most textbooks that I've seen define the weight of a body as the product of a body's mass and the acceleration due to gravity. They define it such that a body in freefall has a weight W = mg. The problem with this is that a body in free fall can be said to be weightless too. So a body can have a nonzero value for its weight and be weightless at the same time. To me that's very weird. I've seen a different definition used by Kenneth Nordtvedt in the Am. J. Phys. He defined weight as the magnitude of the force required to support a body in a gravitational field (some people call this "apparent weight").
Please keep in mind that I'm not asking anybody what the definition of these terms are. I'm asking what you percieved them to be and it you had it your way how would you like them to be defined and why.
Your thoughts?
Thank you everyone in advance.

I have perhaps a smaller problem with the definition of "weight": If we define it as the mass of an object times the acceleration due to the gravitational force, then there is a problem with g  it is obviously variable at different places on the Earth's surface, but should the value of g include the contribution from centrifugal force, or only the one from the differing polar and equatorial radii of the Earth? It seems to me that if we are working in a convenient Earthbound frame of reference, then we will want the former for some purposes but the latter for others.

I have perhaps a smaller problem with the definition of "weight": If we define it as the mass of an object times the acceleration due to the gravitational force, then there is a problem with g  it is obviously variable at different places on the Earth's surface, but should the value of g include the contribution from centrifugal force, or only the one from the differing polar and equatorial radii of the Earth? It seems to me that if we are working in a convenient Earthbound frame of reference, then we will want the former for some purposes but the latter for others.
As I mentioned above it all depends on definition.

If you want to educate the public OR make it easy to educate themselves then you cant beat a clear & concise video
I have learned virtually nothing from books...and almost everything from videos....and ive read plenty of books
Books are by teacher for teachers...that is a general theme running through education.

If you want to educate the public OR make it easy to educate themselves then you cant beat a clear & concise video
That depends on the person. Different people learn differently. I learn much more and more efficiently from reading textbooks on a subject, not from videos. I learn a great deal more by working out problems for myself. Have you ever tried working a problem with pencil, paper and perhaps a calculator? It's wellknown that the only way to learn physics correctly is to work many many problems out by hand. There simply is no substitute for that. In fact that's what I myself am doing right now.
Books are by teacher for teachers...that is a general theme running through education.
That's quite wrong. You can't base what works for everyone else based on your own personal experience. It doesn't work that way.

Weight has the units of Force (ie Newtons in the Metric system, or PoundsForce in the Imperial system).
So bathroom scales measure the downward force of an object (while the seesaw variety of scales has a more direct measure of mass than of weight).
If we place an object on a set of bathroom scales in a lift/elevator:
 It is only if the lift/elevator is stationary that the entire Force=Mass x Gravity is borne by the scales, and the scales indicate what we consider the "normal" weight of the object. In the Imperial system, the weight of the object in PoundsForce equals the mass of the object in Pounds.
 If a lift/elevator accelerating downwards, part of the acceleration due to gravity goes into accelerating the object downwards, and only part of the gravitational acceleration is opposed by the scales, so the object weighs less than "normal".
 Conversely, if a lift/elevator is accelerating upwards, the object weighs more than "normal".
Similarly, if we place an object on bathroom scales at the equator, and increase the rotational speed of the Earth, the object will appear to weigh less. However, we don't normally detect this effect, since the Earth retains a fairly steady rotation, and the oceans and the land of the Earth is already deformed into an oblate spheroid (the geoid (http://en.wikipedia.org/wiki/Geoid)) which takes the rotation of the Earth into account.
For iron weights in air, you can normally ignore this, but air provides an upward buoyancy which reduces the apparent weight of an object below what you expect from the relationship Force = Mass x Gravity. This becomes very significant for light objects, eg a helium balloon can have a negative weight, even though it has a positive mass! Similar comments apply to wood floating in water.
Overall, a more accurate definition of weight would be Weight=Force = (Massm)x(Gravitya), where:
 Mass is the mass of the object
 m is the mass of the displaced medium (which can sometimes be larger than the mass of the object)
 Gravity is the acceleration due to the planet/star's gravity at the relevant altitude
 a is the acceleration of the object due to being placed in a lift, or due to the rotation of the planet (in extreme cases, these can sometimes be larger than the acceleration due to gravity)

From evan au :
Similarly, if we place an object on bathroom scales at the equator, and increase the rotational speed of the Earth, the object will appear to weigh less. However, we don't normally detect this effect, since the Earth retains a fairly steady rotation, and the oceans and the land of the Earth is already deformed into an oblate spheroid (the geoid) which takes the rotation of the Earth into account.
Unfortunately evan these two effects reinforce, rather than cancel  being further from the centre of mass makes the object lighter at the Earth's equator, and so does the centrifugal force. An object is lighter at the equator than at the poles by about 1.3% (quick and dirty calculation that may be in error) which is easily detectable.

Addition:
It belatedly occurred to me that I could probably find polar and equatorial values of g tabulated somewhere.
The polar value is 9.851 m/s/s, while the equatorial value is 9.750 m/s/s, so the different between a polar weight (heavy) and an equatorial weight (light) is actually 1.0%

When I'm president of physics, weight will be defined as the value measured on a scale that's at rest with respect to the weighee. This would mean that "weightless" astronauts in orbit have 0 weight, even though they're being pulled on by gravity.
In terms of teaching the concept of weight, I think it's important to introduce the idea of forces first and emphasize that no one disagrees of the physics of force. Arguments over weight are about defining a shorthand term to describe a particular force, and physicists disagree over which particular force (in which ref. frame) should get this shorthand definition.

Weight has to be the force on a body due to a gravitational field. Thus a lump of iron on a bathroom scale, a helium balloon in air, or a ship floating on the ocean, all have positive weights and the ratio of their weights remains constant regardless of the immersive medium (assuming no variation in g).
The net force on a body in a dense medium depends on Archimedes, and will always be less than or even (in the case of a balloon or aeroplane) in the opposite direction to mg.
This highlights the two important properties of weight  it is a local vector, unlike mass which is a universal scalar.

Weight has to be the force on a body due to a gravitational field.
That would imply that the weight of a body is subjective, i.e. depends on what the gravitational force on the body is. If you had an elevator falling inside an elevator shaft at a rate of 4.95 m/s^{2}. If a person was to step on a weight scale in this elevator then it would read half the weight he'd read if instead he was at home in the bathroom stepping on his bathroom scale and reading it that way. Do you accept this as a consequence of your definition?
Thus a lump of iron on a bathroom scale, a helium balloon in air, or a ship floating on the ocean, all have positive weights and the ratio of their weights remains constant regardless of the immersive medium (assuming no variation in g).

If you define weight to be independent of the immersive medium, then you must measure the weight in a vacuum (or calculate what it would be in a vacuum).
However, some important aspects of weightlessness (eg that whales can breathe when swimming in the ocean in the ocean and that undersea gliders can rise and fall using relatively little power) depend on the immersive medium. Or do you avoid calling this "weightlessness", and instead call it "neutral buoyancy" or similar?

Weight has to be the force on a body due to a gravitational field.
That would imply that the weight of a body is subjective, i.e. depends on what the gravitational force on the body is.
No, then yes! The weight of the body remains the gravitational force on it. Nothing subjective about it because you and I would measure or calculate exactly the same force in a given field. But local, yes: an apple would weigh about 1 newton in the nearearth field, and about 0.4 newton on the surface of Mars.
If you had an elevator falling inside an elevator shaft at a rate of 4.95 m/s^{2}. If a person was to step on a weight scale in this elevator then it would read half the weight he'd read if instead he was at home in the bathroom stepping on his bathroom scale and reading it that way.
Yes, it would show half of the gravitational force. The weight of the person hasn't changed but the reaction force from the floor has halved because the floor is accelerating. In completely free fall, the person of mass M would accelerate at g so his weight remains Mg but the bathroom scale which is also falling at g would read zero. On the other hand if we put the scale on the surface of Mars, it would read about Mg* = 0.38Mg because the guy's weight has changed as g is now g*.
I think a lot of confusion arises by the reference to "space". Objects in a ship in stable earth orbit do not accelerate relative to one another because they are all in free fall under earth's gravity. In deep space, they do not accelerate relative to anything outside the ship because they are not subject to any significant external gravitational field.
When comparing masses in a gravitational field, we do indeed apply buoyancy corrections or preferably make the measurements in a vacuum. In vacuo you can measure the mass or weight of a helium balloon with an ordinary laboratory balance, but it's damn difficult to do it in air!

Having just returned from an ocean cruise to tropical islands, I can confirm that buoyancy is an important measure in its own right!