What do we mean by the term weight? Is it an upward force exerted by an object to counteract the downward force of gravity (or) is it the downward force itself? Is the magnitude of the weight always equal to the gravitational force regardless of the frame?
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3 Answers
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the weight of an object is the gravitational force of attraction exerted by the earth. However, the apparent weight is actually equal to the normal contact force.
Why is this, you might ask. well, the thing is you can only feel forces that act on certain parts of your body, like your feet or hands, because they cause your body to exert internal forces in order to maintain its shape. You cannot feel forces that act uniformly on every part of the body, because every part of the body has the same acceleration and no internal forces need to be exerted.
When you are standing on the ground, the normal contact force and the gravitational force are equal. However, you only "feel" the normal contact force as it only acts on your feet; you cannot feel the gravitational force that acts on every single atom of your body. So the normal contact force is the apparent weight.
when you're accelerating upwards on a lift, the normal contact force is greater than the gravitational force. So you're apparent weight is actually greater than you're true weight. When you're in free fall, the normal contact force and thus the apparent weight are zero, hence the misnomer "weightlessness", even though your weight is the unchanged.
note: I mentioned that you cannot feel the earth's gravity because it acts uniformly over ur body. This isn't always the case. As you know, gravity changes with distance. For non massive objects like the earth, the rate of change is very small. However, for massive objects like blackholes, the rate of variation is large and you can actually feel significantly different forces on the upper and lower part of your body, which can kill you. This is known as spaghettification
answered Dec 30, 2020 at 17:43
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$\begingroup$ Forget gravity and suppose in place of earth we place a uniformly charged sphere and we also cover ourselves with a charged surface uniformly and are dropped in gravity free region. Will we feel weightless or as if no force is acting on us ? $\endgroup$– AnkitCommented Dec 30, 2020 at 18:02
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1$\begingroup$ Defining weight as simply a synonym for "force of gravity acting on an object" is very common but it is not the only way to define weight as Professor Walter Lewin explains in this lecture. $\endgroup$– M. EnnsCommented Dec 30, 2020 at 18:05
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2$\begingroup$ @OVERWOOTCH There is no true weight, weight is a variable, you may want to use "mass" instead, as it is invariant in classical physics. $\endgroup$ Commented Dec 30, 2020 at 18:23
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$\begingroup$ im pretty sure weight is invariant too (not in SR ofcourse) $\endgroup$ Commented Dec 30, 2020 at 18:38
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1$\begingroup$ Exact weight of an object can vary in different places on Earth, its weight would be about 1/6 on the Moon, but its mass would be the same in all cases. $\endgroup$ Commented Dec 31, 2020 at 0:36
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Several definitions exists :
ISO 80000-4:2019 definition
Force acting on a body in the gravitational field of Earth,- $\textbf F_g=m\textbf g$, where $m$ is the mass of the body and $\textbf g$ is the local acceleration of free fall.Minuses of such definition :
- Is not applicable to other planets or gravitational bodies other than Earth. For example your weight on Moon,Mars will be different because their mass is different than one of Earth, so different too their free fall accelerations.
- Does not account for your position relative to Earth or gravitational body center. If for example you would be exactly at Earth center - your weight would be zero there, cause all gravitational forces would cancel each other out.
- Local acceleration $\textbf g$ due to gravity is not always an easy task to extract and isolate from other effects. For example if you will measure free fall acceleration of an object,- it will be affected also by :
a) centrifugal force (Earth variations in $r$) b) atmosphere buoyancy c) air drag force d) Earth magnetic field (if an object has some permeability) e) other forces (even Sun light radiation pressure if falling object is very light and can experience it)
Operational definition
Force which object exerts on it's support. (The very act on weighting).Notes :
- Can be zero, even object has rest mass. For example astronauts in an orbital space station does not push station ground or anything due to gravity. Thus they have zero operational weight, but still experience micro-gravity.
- Not exactly clear what force this way measures. It can be that by weighting we are measuring superposition of several forces or something else completely. For example, imagine weighting a ball on a weighing-machine which is attached to an oscillating spring. Then ball weight will be dictated not only by Earth gravity, but by Hook Law also.
Gravitational definition
Force exerted on body by gravitational field, namely $\vec F=m\vec a_g$, where $\vec a_g$ is acceleration due to gravity. Closer to a real thing, but has some small discrepancies too :- Same flaw as in ISO80000 definition. Does not account for a relative distance to gravitational body center. You will not weight the same at North pole and equator, because Earth is not an ideal sphere due to centrifugal force and rotational axis precession.
- Local acceleration $\vec a_g$ due to gravity is not always an easy task to extract and isolate from other effects. For example if you will measure free fall acceleration of an object,- it will be affected also by :
a) centrifugal force (Earth variations in $r$) b) atmosphere buoyancy c) air drag force d) Earth magnetic field (if an object has some permeability) e) other forces (even Sun light radiation pressure if falling object is very light and can experience it)
Better gravitational definition
My preference is that one of Newton. Weight is nothing more than gravitational attraction force between two bodies : $$W = G \frac {Mm}{r^2}$$ Plus is that this definition - universal gravitation law - accounts for object exact distance $r$ from gravitating body center.Still minuses exist:
- Can't be applied for calculating a weight of body which is inside of gravitating body, i.e for example inside of Earth at $-h$ depth below Earth surface. In this case one needs to integrate to find out resulting gravity force.
answered Dec 30, 2020 at 19:06
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$\begingroup$ The universal law of gravitation only holds for point particles or spheres of uniform density. The better 'gravitational' definition is mg(r), where g is the field strength at that point. $\endgroup$ Commented Dec 30, 2020 at 23:23
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$\begingroup$ The universal law of gravitation only holds for point particles or spheres of uniform density Complete nonsense. Calculate free fall acceleration from universal gravitation law by putting Earth radius and mass into equation $g = G M/r^2$, and you'll get an answer which deviates from the standard gravitational acceleration value by only $0.13\%$ relative error. This proves that universal gravity law suits for non-uniform densities objects. $\endgroup$ Commented Dec 31, 2020 at 0:21
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$\begingroup$ As a second contra-argument - You want to say that Newton extrapolated this gravitation law suited only for uniform-density objects, by using Kepler's law extracted from observations of non-uniform densities objects like Earth,Sun ? Don't you see some contradiction here in your argumentation ? $\endgroup$ Commented Dec 31, 2020 at 0:30
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$\begingroup$ I can even explain to you why universal gravity law works determining your weight near Earth surface,- it's because you can approximate non-uniform density Earth as a singularity (point particle) attracting you from a distance $r$. $\endgroup$ Commented Dec 31, 2020 at 8:54
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$\begingroup$ I know why it works. But that doesn't change the fact that it only holds for particles or spheres of uniform density. Moreover, I was talking about a more general gravitational field, not just planets $\endgroup$ Commented Dec 31, 2020 at 11:44
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It is the downward force that any object A does on another object B, or the upward reaction of B on A, that has the same magnitude according to the $3_rd$ Law.
I think it is better definition than the force of earth on the object. Because, in the common use of the word, we say that objects in the ISS are weightless, even being under gravitational attraction.
Sometimes the upward reaction is more apparent, as in the case of a floating baloon, or any other buoyance case. But the downwards force on the fluid is also present.
Even heavier than air vehicles weights on the air, as discussed in Is the weight of the aircraft flying in the sky transferred to the ground?
answered Dec 31, 2020 at 13:14