Why Is an Object's Weight Equal to the Force of Gravity It Experiences?

[PeroK]

Personally, I don't use the concept of weight in physics. You need the concepts of force and mass, but weight is is a superfluous and often ill-defined concept. It's possible of course to define the weight of a body of mass $m$ as simply $W = mg$. Often, however, astronauts in the space station are described as weightless - which clearly contradicts this definition.

Note that Earth's gravitational pull extends beyond its atmosphere. In fact, theoretically, to infinity.

 

[adjurovich]

Personally, I don't use the concept of weight in physics. You need the concepts of force and mass, but weight is is a superfluous and often ill-defined concept. It's possible of course to define the weight of a body of mass $m$ as simply $W = mg$. Often, however, astronauts in the space station are described as weightless - which clearly contradicts this definition.

Note that Earth's gravitational pull extends beyond its atmosphere. In fact, theoretically, to infinity.

The idea used here is that “Earth’s force of gravity” works as a concept only very close to the surface of Earth. It happens to work this way:​

$F_g = mg$

where $g$ is the strength of Earth’s gravitational field:

$g = G \dfrac{M}{R^2}$
​

As long as there is no additional height $h$ added to the Earth’s radius, g is approximately 9.8 or 9.81 depending on where you measure it. But unique value can be chosen. I find it a bit odd to use two different forces to explain literally the same thing

When it comes to astronauts, my interpretation is that they in fact do not exert the force on other objects since they are floating so they are weightless,
however they do not truly experience 0 gravity state unless they are far enough from other planets / stars. However theoretically speaking it would be impossible, but in practice you can the neglect small quantities​

 

[jbriggs444]

My interpretation is that "g" is whatever free fall acceleration you accept by adopting a non-inertial local reference frame.

Here on earth is is convenient to adopt a local frame where the soil is unmoving. In a space craft it is convenient to adopt one where the lab bench is unmoving.

An object's weight is its mass times the free fall acceleration that you have accepted.

Note that the free fall acceleration relative to the Earth's surface is not in general given by $\frac{GM_e}{r^2}$.

 

[Ibix]

When it comes to astronomers, my interpretation is that they in fact do not exert the force on other objects since they are floating so they are weightless,

This would be wrong. A spacecraft in orbit around a planet is clearly experiencing the force of gravity - that's why it's in orbit. But it, and its crew, are "weightless" as commonly described.

The point is that all of them are (approximately) affected the same way by gravity so they have zero acceleration with respect to each other so do not see effects we commonly think of as due to gravity. That's why they are called "weightless". But the "approximate" part is neglecting tidal forces, and sufficiently precise experiment can detect them. So even "weightless" you can, if sufficiently careful, detect the presence of gravity.

 

[adjurovich]

This would be wrong. A spacecraft in orbit around a planet is clearly experiencing the force of gravity - that's why it's in orbit. But it, and it's crew, are "weightless" as commonly described.

The point is that all of them are (approximately) affected the same way by gravity so they have zero acceleration with respect to each other so do not see effects we commonly think of as due to gravity. That's why they are called "weightless". But the "approximate" part is neglecting tidal forces, and sufficiently precise experiment can detect them. So even "weightless" you can, if sufficiently careful, detect the presence of gravity.

I will point something out. Weight is defined, here where I live, as the force that object exerts on other objects when it’s standing (for example on soil or a box) / attached to rope. I don’t think astronomers can stand? In that case they do not exert “weight” on other objects.

However I don’t think I said they don’t experience gravity? They surely do. But what I meant to say is that if a space ship was placed in some spot $B$ that is very far from any planet / star it would almost experience zero gravity.

 

[Ibix]

I don’t think astronomers can stand?

Do you mean astronauts? Astronomers use telescopes, astronauts fly spaceships.

 

[adjurovich]

Do you mean astronauts? Astronomers use telescopes, astronauts fly spaceships.

Thanks for pointing out, it made me laugh really hard

 

[jbriggs444]

I will point something out. Weight is defined, here where I live, as the force that object exerts on other objects when it’s standing

As I've indicated, I take "weight" as the downward inertial force on an object arising from an accelerating reference frame, from gravity or from both taken together.

Normally, this is a distinction without a difference. For a stationary object subject to no forces other than the support force of interest, the interaction force of the object against its support will be equal to the net inertial force on the object.

That is part of how bathroom scales work. They require the user to stand still and ignore the buoyancy from bathroom air.

Questions (about language usage, not about physics):

For you, does your "weight" increase as you jump into the air and as you land from a jump? Are you "weightless" in between?

For you, does your "weight" decrease when you are wading in a swimming pool?

 

[adjurovich]

Questions (about language usage, not about physics):

For you, does your "weight" increase as you jump into the air and as you land from a jump? Are you "weightless" in between?

For you, does your "weight" decrease when you are wading in a swimming pool?

1. Yes, you are weightless during the free fall
2. Yes, it does decrease when you are wading in swimming pool

 

[cianfa72]

The point is that all of them are (approximately) affected the same way by gravity so they have zero acceleration with respect to each other so do not see effects we commonly think of as due to gravity. That's why they are called "weightless". But the "approximate" part is neglecting tidal forces, and sufficiently precise experiment can detect them. So even "weightless" you can, if sufficiently careful, detect the presence of gravity.

They are affected the same way by gravity whether the gravitational field is either uniform or null (flat spacetime). In that case there is no spacetime curvature, i.e. no tidal forces or geodesic deviation.

 

[adjurovich]

They are affected the same way by gravity whether the gravitational field is either uniform or null (flat spacetime). In that case there is no spacetime curvature, i.e. no tidal forces or geodesic deviation.

I don’t think Sir Isaac Newton had an idea of what spacetime curvature is?

 

[Herman Trivilino]

I am not from the USA / UK and here we have kinda different terminology. Weight is usually considered to be the pulling force exerted on rope / force that body exerts on surface (the one of which reaction pair is the normal force $\vec{N}$).

Weight is the force that would make a body accelerate at a rate equal to the local freefall acceleration.

However we have the other thing that I wouldn’t be sure how to directly translate, probably something like Earth’s force of gravity? The force that Earth exerts on all objects in its atmosphere. That’s why I started this thread quite some time ago

The force of gravity.

The two are very nearly the same and are often erroneously equated. The weight is actually a bit less than the gravitational force due to the effect of Earth's spin.

 

[Herman Trivilino]

1. Yes, you are weightless during the free fall

Right, because the local freefall acceleration is zero.

2. Yes, it does decrease when you are wading in swimming pool

No. Weight doesn't include the effects of buoyancy.

Freefall is, by definition, in a vacuum.

 

[Herman Trivilino]

It's possible of course to define the weight of a body of mass m as simply W=mg. Often, however, astronauts in the space station are described as weightless - which clearly contradicts this definition.

Not if you go with the officially-sanctioned definition of weight. It's the force that would make a body accelerate at a rate equal to the local freefall acceleration. Therefore it's zero in the orbiting ISS because $g$ is zero.

 

[PeroK]

Not if you go with the officially-sanctioned definition of weight. It's the force that would make a body accelerate at a rate equal to the local freefall acceleration. Therefore it's zero in the orbiting ISS because $g$ is zero.

That's interesting. What precisely is "officially sanctioned"?

 

[jbriggs444]

That's interesting. What precisely is "officially sanctioned"?

Following a reference from NIST, I found ISO 80000-4 where:

Item: 4-9.2
Name: weight
Symbol: $F_g$
Quantity: Force (Item 4-9.1) acting on a body in the gravitational field of Earth.$$F_g=mg$$where m (item 4-1) is the mass of the body and $g$ is the local acceleration of free fall (ISO 80000-3)
Unit: $$N$$ $$\text{kg} \text{ m} \text{ s}^{-2}$$
Remark: In colloquial language the term "weight" continues to be used where mass is meant. This practice should be avoided.
Weight is an example of a gravitational force. Weight comprises not only the local gravitational force but also the local centrifugal force due to the rotation of the Earth

This definition seems pretty restrictive. Using this definition, one could not meaningfully ask "what would your weight be on the moon?".

However, this is not the only "officially sanctioned" meaning.

In commerce (as opposed to in science, engineering or physics pedagogy) in the U.S. the term "weight" or, in particular, "net weight" has a government mandated operational definition as a mass, not a force. This is just as well since we would not want a bottle of ketchup or a box of Cheerios to have the correctness of its label depend on the city in which it is vended.

 

[Herman Trivilino]

That's interesting. What precisely is "officially sanctioned"?

International Organization for Standardization, International Standard ISO 31-3, Quantities and Units, Part 3: Mechanics (ISO, Geneva, Switzerland, 1992).

"The weight of a body in a specified reference system is that force which, when applied to the body, would give it an acceleration equal to the local acceleration of free fall in that reference system."

Note that when we write the expression $w=mg$ we can use the above standard in which case $g$ is the local free fall acceleration and $m$ is the inertial mass.

An alternative definition would be to refer to $w$ as the gravitational force, which would make $m$ the gravitational mass, and people using this definition would refer to $g$ as the acceleration due to gravity. They would then refer to the above as the "apparent weight".

The difference is of course due to Earth's spin. My observation is that the free fall acceleration varies from about 9.747 9.781 N/kg at the equator to 9.832 N/kg at the poles. This is a difference of about 0.051 N/kg. About 2/3 of this difference is due to Earth's spin, the other 1/3 being due to the equatorial bulge.

People who round $g$ to 9.81 N/kg or 9.80 N/kg are using values that include the effects of Earth's spin. Yet some of them will erroneously refer to it as the acceleration due to gravity. If they were really using the acceleration due to gravity they would not be including the effects of Earth's spin and the value of $g$ would vary from about 9.781 9.815 N/kg at the equator to 9.798 9.832 N/kg at the poles, due to the equatorial bulge. No way they'd get the rounded-off values they use! unless they insisted that it rounds to 9.80 N/kg at the poles, and to heck with what it is right here where we're using it!

Note: Edited to correct mistakes made in the calculations.

 

[Herman Trivilino]

In commerce (as opposed to in science, engineering or physics pedagogy) in the U.S. the term "weight" or, in particular, "net weight" has a government mandated operational definition as a mass, not a force. This is just as well since we would not want a bottle of ketchup or a box of Cheerios to have the correctness of its label depend on the city in which it is vended.

And that would be especially true if it were a bar of gold instead of a bottle of ketchup or a box of cereal.

 

[Herman Trivilino]

And by the way, while I'm on my little rant, I'll point out that the pound Avoirdupois, which is the pound used throughout the USA and many other places, is by international treaty, exactly 0.453 592 37 kg.

In other words, it's officially a unit of mass, not force.

Throughout the ages the pound with its various definitions has always been a unit of mass. It is called a unit of weight only in the sense described in the last paragraph of Post #46. It is not, and never has been, a unit of force, despite what many textbooks would have you believe.

Why would merchants buy and sell commodities based on a unit of measure that varies with location?!

 

[jtbell]

Throughout the ages the pound with its various definitions has always been a unit of mass. It is called a unit of weight only in the sense described in the last paragraph of Post #46. It is not, and never has been, a unit of force, despite what many textbooks would have you believe.

Nevertheless, the informal use of "pound" as a unit of force, instead of the more precise term "pound-force", is rather common. For example:

ACDelco digital torque wrench (amazon.com)

Easy to read LCD display with measurements (kg-cm, N-m, in-lb. ft-lb)

which also has "kg" instead of "kgf".

 

[Herman Trivilino]

Nevertheless, the informal use of "pound" as a unit of force, instead of the more precise term "pound-force", is rather common.

It's nearly ubiquitous in introductory physics textbooks. Even NASA uses it. Nevertheless, there is no officially-sanctioned definition of the pound-force.

For example:

ACDelco digital torque wrench (amazon.com)



which also has "kg" instead of "kgf".

The kilogram-force was officially dropped by the SI as of, IIRC, 1960. Prior to that a so-called standard value of $g$ equal to 9.806 65 N/kg was used to define the kilogram-force. It was simply an invention designed to allow the measurement of mass using units of force. There is no place on Earth where $g$ has a value of 9.806 65 N/kg for any significant length of time.

Unfortunately, no such standard value was ever adopted for the definition of the pound-force, although the above value appears to be used by NASA for their definition. So they end up defining the pound-force as the product of 0.453 592 37 kg and 9.806 65 N/kg, resulting in an unjustifiably large number of significant digits.

 

[jbriggs444]

There is no place on Earth where $g$ has a value of 9.806 65 N/kg for any significant length of time.

According to the Wiki article the figure was the result of a measurement (made in 1888 in Paris) and a correction (specified in the previous year by the CIPM) to obtain a corresponding estimate applicable to 45 degrees latitude).

Wisely, no one bothered to correct the figure to account for more accurate gravitational measurements obtained in the years since.