Why are astronauts weightless?

  • Thread starter Forge
  • Start date
79
6
Read about Microgravity
 

A.T.

Science Advisor
9,733
1,565
What's the difference between
1) a fish floating
5) an astronaut in the space station
Which of the above could be described as 'weightless'?
Floating in water and in space is not quite the same, because of the non-uniform density of the body.
 
5,598
39
These are ridiculous statements, completely contrary to conventional teaching and text book explanations.
Why not just rewrite the Wikipedia article as well??
 

cjl

Science Advisor
1,760
347
What's the difference between
1) a fish floating
2) A skydiver with parachute not open
3) A skydiver with parachute open
4) A passenger in a free falling lift
5) an astronaut in the space station

Which of the above could be described as 'weightless'?
I would make the case that only #4 and #5 could be considered "weightless", since those are the only cases of free fall listed above (since a skydiver, even with no parachute, has substantial air resistance). Floating in water is subjectively similar, but it is not the same, since the buoyant force acts on the surface of the object, while gravity acts on the whole volume.

I would also say that based on the equivalence principle, free fall and weightlessness are the same thing - in the absence of tidal forces, there is no difference of any significance.
 
1,506
17
Why not just rewrite the Wikipedia article as well??
I have text books that have stood the test of time.
Don't need wiki
It is the worst thing to have happened to education....full of irrelevant information and incorrect facts....check their explanation of faradays laws. Who checks wiki....how much does it cost??
You get what you pay for in this world
Goodbye
 
79
6
Obviously he is trolling
 
79
6
The term "net gravity" is not used there. What is your definition of it?
The total gravitational force is near zero in orbit. Weight is dependent on gravity. Gravity is acceleration. If you are going at 17k relative to someone on Earth in LEO you are not accelerating. No acceleration and your weight is zero. It doesn't matter if your speed relative to an observer on the Earth is really fast. The force it takes to move an object is its mass times it's acceleration. If it is in a microgravity environment wear the acceleration is near zero than the mass is almost meaningless and therefore it weighs just about nothing. The reason it seems conter-intuitive is that we have just become used to constantly accelerating through spacetime at 9.8mps2 or whatever is here in this planet. That's why we have weight on the surface even when standing still.
 
19,465
3,894
The astronaut is undergoing centripetal acceleration, so there obviously is a net force on him. Bringing GR into it will probably just confuse the OP.
I don't really think you should have dismissed what Naty1 said like that. We know from GR that there really is no net force acting on him when he is in free fall (which I'm sure you are aware of). For a question like this, the OP certainly should at least be made aware that there is a pre-relativistic model of what is happening, and a (now believed by most to be correct) GR version of what is happening, and that the two descriptions are very different.
 

Andrew Mason

Science Advisor
Homework Helper
7,536
317
The total gravitational force is near zero in orbit.
Can you explain that in terms of F = GMm/r^2 ?

Weight is dependent on gravity.
Ok. But it is also dependent on there being a force applied to the body to keep it from accelerating due to gravity.

Gravity is acceleration. If you are going at 17k relative to someone on Earth in LEO you are not accelerating.
So there is no centripetal acceleration in this circular motion?

AM
 
Last edited:

A.T.

Science Advisor
9,733
1,565
For a question like this, the OP certainly should at least be made aware that there is a pre-relativistic model of what is happening, and a (now believed by most to be correct) GR version of what is happening, and that the two descriptions are very different.
I have no problem with explaining it in different contexts. But the explanation should clearly state which statements apply to which model, and which reference frame. The OP makes it quite clear that he wants an explanation for the frame, where the orbiting astronaut is in uniform circular motion. In that frame the net force and thus the acceleration are obviously not zero or near zero.
 

A.T.

Science Advisor
9,733
1,565
The total gravitational force is near zero in orbit.
Can you provide a formula to calculate that "total gravitational force" which is near zero for an object in orbit?
 
79
6
I have no problem with explaining it in different contexts. But the explanation should clearly state which statements apply to which model, and which reference frame. The OP makes it quite clear that he wants an explanation for the frame, where the orbiting astronaut is in uniform circular motion. In that frame the net force and thus the acceleration are obviously not zero or near zero.
GR is more accurate but often Newtonian physics are enough to explain it to someone. And yes the acceleration (g-force) IS practically nil. The only way F = 0 when the mass hasn't changed is if you multiply by (almost) zero.






Can you provide a formula to calculate that "total gravitational force" which is near zero for an object in orbit?
I'm not ignoring this, but my orbital mechanics book is at home and I am at work so let me get back to this later.


BTW you can attack my definitions and ask more and more questions until you can find something to trip me up on but I know that an object in orbit is not only in unaccelerated flight, but are weightless because of the microgravity environment, and that g-force/acceleration are two sides of the same coin. Whether or not it has to do with lack of force or opposing forces it doesn't matter they are weightless because they are in a zero-g environment. If you have a better explanation I would like to hear it.
 

Bandersnatch

Science Advisor
2,840
1,695
GR is more accurate but often Newtonian physics are enough to explain it to someone. And yes the acceleration (g-force) IS practically nil. The only way F = 0 when the mass hasn't changed is if you multiply by (almost) zero.

I'm not ignoring this, but my orbital mechanics book is at home and I am at work so let me get back to this later.

BTW you can attack my definitions and ask more and more questions until you can find something to trip me up on but I know that an object in orbit is not only in unaccelerated flight, but are weightless because of the microgravity environment, and that g-force/acceleration are two sides of the same coin. Whether or not it has to do with lack of force or opposing forces it doesn't matter they are weightless because they are in a zero-g environment. If you have a better explanation I would like to hear it.
Seriously, this is secondary school-level.

For uniform circular motion over body M at radius R:
[tex]F_g=F_c=G \frac{Mm}{R^2}=mRω^2≠0[/tex]
 
79
6
Seriously, this is secondary school-level.

For uniform circular motion over body M at radius R:
[tex]F_g=F_c=G \frac{Mm}{R^2}=mRω^2≠0[/tex]
Thanks for that, I don't remember those off the top of my head. Just the simple ones like Newtons Second Law where the net force equals mass times acceleration. Fnet = M * A
 
1,506
17
but I know that an object in orbit is not only in unaccelerated flight
This statement is completely wrong. In orbit is the result of a centripetal force/acceleration.

but are weightless because of the microgravity environment,

The only way this statement can make any sense to me is if it relates to astronauts in the orbiting station. If you are born in an orbiting station and spend your life in an orbiting station there is no way that you can experience 'weight' in the conventional way that we mean....something that can be measured on bathroom scales. If astronauts read their text books they will find the explanation has something to do with 'free fall' which has something to do with centripetal force.

In basic physics lessons it is required that students can explain the readings on bathroom scales in an accelerating and decellerating lift.
We should be pleased that the physics principles are the same for all.
 

A.T.

Science Advisor
9,733
1,565
but are weightless because of the microgravity environment,

The only way this statement can make any sense to me is if it relates to astronauts in the orbiting station.
His statement probably relates to the rest frame of the astronauts center of mass, which in Newtonian context is an non-inertial frame. Therefore there are inertial forces which cancel gravity.

Why is it a bad answer to the OPs question?
1) The OP asked about the frame where the astronaut is in circular motion, if you chose to explain it in a different frame you should explicitly state and justify this.
2) Non-inertial frames and inertial forces up which are not needed to answer the question about a frame independent effect.
3) Being at rest in some frame says nothing about feeling acceleration. In the non-inertial rest frame of an accelerating car the passengers are at rest too, but they do fell the acceleration. The reason is that the force from the seats is applied to their backs only, while gravity is a applied uniformly to the astronaut, as stated on page 1 several times.
 
79
6
The original post claimed they are accelerating. They are not. They are at a constant speed of probably around 17, 000ish mph
 
79
6
(Relative to an observer on the surface)
 

Bandersnatch

Science Advisor
2,840
1,695
The original post claimed they are accelerating. They are not. They are at a constant speed of probably around 17, 000ish mph
Force, acceleration and velocity are vectors. As such, they can change direction without changing the magnitude. This is exactly the case with uniform circular motion as per OP's question. Constant acceleration at right angle to the velocity vector causes it to change direction but not the magnitude.

This is acceleration.

Take the equation I've provided earlier, and divide everything by m. Now you've got gravitational acceleration equal to centripetal acceleration which is not equal 0.

Additionally, the speed(i.e., the magnitude of the velocity vector), is not necessarily 17k mph. This value is a function of the radius of the orbit, which has not been specified in the OP.

(Relative to an observer on the surface)
Which is an odd choice of reference frame, for reasons explained by A.T. just above your posts.
It is also not true, as the relative speed changes due to the varying angles between the orbiting body's velocity vector and the one of the observer "riding" on the rotating surface.
The only case when it's constant w/r to the surface observer, would be the geostationary orbit(and it'd be equal to 0 then).
 
79
6
Force, acceleration and velocity are vectors. As such, they can change direction without changing the magnitude. This is exactly the case with uniform circular motion as per OP's question. Constant acceleration at right angle to the velocity vector causes it to change direction but not the magnitude.

This is acceleration.

Take the equation I've provided earlier, and divide everything by m. Now you've got gravitational acceleration equal to centripetal acceleration which is not equal 0.

Additionally, the speed(i.e., the magnitude of the velocity vector), is not necessarily 17k mph. This value is a function of the radius of the orbit, which has not been specified in the OP.


Which is an odd choice of reference frame, for reasons explained by A.T. just above your posts.
It is also not true, as the relative speed changes due to the varying angles between the orbiting body's velocity vector and the one of the observer "riding" on the rotating surface.
The only case when it's constant w/r to the surface observer, would be the geostationary orbit(and it'd be equal to 0 then).
It's not EXACTLY zero but the net force IS almost negligible. That is why they are weightless. If they were accelerating they would feel g-force. If they are at a constant speed and not changing altitude which way do you propose they are accelerating? This is why I chose to explain this with Newtonian physics rather than GR because I don't think he cares about time dilation...
 

Bandersnatch

Science Advisor
2,840
1,695
If they are at a constant speed and not changing altitude which way do you propose they are accelerating? This is why I chose to explain this with Newtonian physics rather than GR because I don't think he cares about time dilation...
Towards the centre of the Earth, of course. Vectors, remember?
It's pure Newton, too.

It's not EXACTLY zero but the net force IS almost negligible. That is why they are weightless. If they were accelerating they would feel g-force.
3) Being at rest in some frame says nothing about feeling acceleration. In the non-inertial rest frame of an accelerating car the passengers are at rest too, but they do fell the acceleration. The reason is that the force from the seats is applied to their backs only, while gravity is a applied uniformly to the astronaut, as stated on page 1 several times.
 
79
6

Related Threads for: Why are astronauts weightless?

Replies
6
Views
5K
  • Last Post
Replies
5
Views
708
Replies
3
Views
907
Replies
10
Views
7K
Replies
65
Views
10K
Replies
18
Views
8K
Replies
2
Views
3K
Top