Why do people say that the person would be weightless?

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In summary, the conversation discusses the concept of weightlessness in an elevator in free fall. It is determined that the person in the elevator is not truly weightless, but rather experiencing apparent weightlessness due to the zero normal force acting on them. The conversation also delves into the differences between the Newtonian and General Relativity views on gravity, and how different definitions of weight can affect the measurement. It is mentioned that the weight using mass times the local acceleration of a person on the surface of the Earth may not be the same as the weight found using the universal gravitation, as the rotation of the Earth can affect the centripetal force acting on the person.
  • #1
rakeru
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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?
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
rakeru said:
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
 
  • #7
QED Andrew said:
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 Fg = 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
 
  • #8
Chestermiller said:
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
 
  • #9
Andrew Mason said:
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 Fg = 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?
 
  • #10
Okay I just tested it with a 60 kg person and both were the same!
 
  • #11
rakeru said:
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.

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

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
 
  • #12
Andrew Mason said:
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.

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

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 ?
 
  • #13
rakeru said:
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.
 
  • #14
rakeru said:
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
 

1. Why do people say that the person would be weightless?

People say that a person would be weightless in certain situations, such as in outer space or during freefall, because they are experiencing a lack of gravitational force. In these situations, the person's weight is essentially canceled out, causing them to feel weightless.

2. Is it possible for a person to be truly weightless?

Technically, no. Weightlessness is a sensation caused by the absence of gravity, but gravity still exists even in outer space. However, in certain situations, such as orbiting the Earth or being in freefall, the effects of gravity are negligible, making it feel like a person is truly weightless.

3. How do astronauts experience weightlessness in space?

Astronauts experience weightlessness in space because they are constantly falling towards the Earth at the same rate as their spacecraft. This creates a state of perpetual freefall, where the effects of gravity are canceled out and the astronauts feel weightless.

4. What does it feel like to be weightless?

Being weightless can feel like floating or flying, as there is no sensation of weight or pressure on the body. Some people also describe it as feeling like they are constantly falling or being in a state of weightless relaxation.

5. How does weightlessness affect the human body?

The lack of gravitational force in weightlessness can cause changes in the human body, such as a decrease in muscle and bone mass, changes in blood flow, and fluid shifts. However, these effects are temporary and can be counteracted with proper exercise and nutrition.

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