Why do people say that the person would be weightless?

[rakeru]

Homework Statement

If someone is in an elevator that is in free fall, why do people say that the person would be weightless? Wouldn't the person weigh mass times the gravitational acceleration?

 

[QED Andrew]

Yes, the weight of the person is equal in magnitude to the product of the mass of the person and the magnitude of the acceleration due to gravity. Describing the person as weightless is inaccurate. It is more accurate to say the person is experiencing "apparent weightlessness", which means the normal force acting on the person is zero. The person is not truly weightless because a gravitational force still acts. But the person's sensations are exactly the same as though the person were in outer space without a gravitational force at all. The person and the elevator fall together with the same acceleration, so nothing pushes the person against the floor or walls of the elevator.

 

[rakeru]

Ooh, okay. So it's not really weightless.. it just feels like it is. So a scale would read zero but only because there's no normal force? Wait but in space there is gravitational force.

 

[Chestermiller]

To expand on QED Andrew's response, there is no test that the person within the elevator can do to convince himself that he is in anything but a gravity-free environment.

 

[rakeru]

But there really is gravity, right? Or is there no gravity because he has no way of telling that there's no gravity? Is that what 'reference frames' refers to?

 

[Chestermiller]

But there really is gravity, right? Or is there no gravity because he has no way of telling that there's no gravity? Is that what 'reference frames' refers to?

These questions take you to the very edge of the envelope between Newtonian gravitational physics and General Relativity. In Newtonian physics, gravity is considered a real force on an object. In General Relativity, there is no such force as gravity, and the effects we observe in free fall are a manifestation of the curvature of the space-time continuum (induced by the presence of a massive body). If you want to learn more about this, read up on Einstein's Equivalence Principle (to begin with).

Chet

 

[Andrew Mason]

Yes, the weight of the person is equal in magnitude to the product of the mass of the person and the magnitude of the acceleration due to gravity. Describing the person as weightless is inaccurate. It is more accurate to say the person is experiencing "apparent weightlessness", which means the normal force acting on the person is zero. The person is not truly weightless because a gravitational force still acts. But the person's sensations are exactly the same as though the person were in outer space without a gravitational force at all. The person and the elevator fall together with the same acceleration, so nothing pushes the person against the floor or walls of the elevator.

It depends on how one defines weight. Not everyone defines it the same way.

If you define weight of a body as its mass x local acceleration of the body in free fall toward the earth, (which is the ISO definition) so that a body weighs less at the equator than at higher latitudes due to the rotation of the earth, then an orbiting astronaut is actually weightless.

If you define weight as Weight as the force of Earth's' gravity on a body (ie F_g = m GM/r² where m is the body's mass and M is the mass of the Earth and r is the length of the radial vector from the centre of the Earth to the body), then an orbiting astronaut is not weightless.

So to answer this question the OP has to tell us how he is defining weight.

AM

 

[rakeru]

These questions take you to the very edge of the envelope between Newtonian gravitational physics and General Relativity. In Newtonian physics, gravity is considered a real force on an object. In General Relativity, there is no such force as gravity, and the effects we observe in free fall are a manifestation of the curvature of the space-time continuum (induced by the presence of a massive body). If you want to learn more about this, read up on Einstein's Equivalence Principle (to begin with).

Chet

Wow.. Just.. Yep. Thanks

 

[rakeru]

It depends on how one defines weight. Not everyone defines it the same way.

If you define weight of a body as its mass x local acceleration of the body in free fall toward the earth, (which is the ISO definition) so that a body weighs less at the equator than at higher latitudes due to the rotation of the earth, then an orbiting astronaut is actually weightless.

If you define weight as Weight as the force of Earth's' gravity on a body (ie F_g = m GM/r² where m is the body's mass and M is the mass of the Earth and r is the length of the radial vector from the centre of the Earth to the body), then an orbiting astronaut is not weightless.

So to answer this question the OP has to tell us how he is defining weight.

AM

So will the weight using mass times the local acceleration of a person on the surface of the Earth be the same as the weight found using the universal gravitation? How would I know which to use?

 

[rakeru]

Okay I just tested it with a 60 kg person and both were the same!

 

[Andrew Mason]

So will the weight using mass times the local acceleration of a person on the surface of the Earth be the same as the weight found using the universal gravitation? How would I know which to use?

Not quite the same. At the equator, an object in free-fall is rotating at an angular speed of ω = 2π/3600*24 rad/sec. So the centripetal force is Fc = mω²r. The gravitational force is mGM/r². Since in the ISO standard, the weight is determined by the second time derivative of its radial distance to the centre of the earth, you have to subtract the centripetal force.

$weight = m\ddot{r}= m(GM/r^2 - \omega^2r) = m(g - \omega^2r)$

Since ω²r = .034 and g = 9.81 (m sec^-2) at the equator, you would need to be able to measure the apparent weight very accurately to see the difference. At the equator, the apparent g (which is the ISO standard) is 9.78 m sec^-2.

The ISO standard is the easiest to use since it is what a spring scale would measure.

AM

 

[rakeru]

Not quite the same. At the equator, an object in free-fall is rotating at an angular speed of ω = 2π/3600*24 rad/sec. So the centripetal force is Fc = mω²r. The gravitational force is mGM/r². Since in the ISO standard, the weight is determined by the second time derivative of its radial distance to the centre of the earth, you have to subtract the centripetal force.

$weight = m\ddot{r}= m(GM/r^2 - \omega^2r) = m(g - \omega^2r)$

Since ω²r = .034 and g = 9.81 (m sec^-2) at the equator, you would need to be able to measure the apparent weight very accurately to see the difference. At the equator, the apparent g (which is the ISO standard) is 9.78 m sec^-2.

The ISO standard is the easiest to use since it is what a spring scale would measure.

AM

I don't understand that, but hopefully in the future I will.. I guess if the object is on the surface I would just use mg, but if the object is far from the surface I'd use GmM/r^2 ?

 

[voko]

Ooh, okay. So it's not really weightless.. it just feels like it is. So a scale would read zero but only because there's no normal force? Wait but in space there is gravitational force.

The reason why one is "weightless" in space is exactly the same one is in an elevator - it is the free fall. The only difference between the "space" and "non-space" is that in space one is usually very far from anything else, so can be falling for extended periods of time without other effects, or is falling with a great velocity, which makes it possible to orbit around other things, again avoiding the "non-space" consequences of falling.

 

[Andrew Mason]

I don't understand that, but hopefully in the future I will.. I guess if the object is on the surface I would just use mg, but if the object is far from the surface I'd use GmM/r^2 ?

By definition, g = GM/r². Whether you would use weight = m(g-Fc) or weight = mg depends on the definition of weight that you wish to use.

AM