Why Do Astronauts Feel Weightless in Orbit?

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Objects in orbit around Earth are in uniform circular motion, experiencing centripetal force and acceleration towards the planet's center, which leads to a sensation of weightlessness. Astronauts do not feel weight because both they and their spacecraft are in free fall, with gravity acting equally on both, resulting in no net forces acting on them. Weightlessness can also be experienced in other scenarios, such as jumping from a tall building, where gravity acts but there is no upward force counteracting it. The perception of weight is linked to the forces acting on the body, and in free fall, there is no deformation felt, making it seem as though gravity is absent. Discussions around weightlessness often involve clarifying the difference between mass and weight, as well as the implications of gravitational forces in various contexts.
  • #31
technician said:
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.
 
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  • #32
Naty1 said:
Why not just rewrite the Wikipedia article as well??
I have textbooks 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
 
  • #33
Obviously he is trolling
 
  • #35
A.T. said:
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.
 
  • #36
A.T. said:
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.
 
  • #37
p1l0t said:
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:
  • #38
Chestermiller said:
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.
 
  • #39
p1l0t said:
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?
 
  • #40
A.T. said:
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.
A.T. said:
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.
 
  • #41
p1l0t said:
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:
F_g=F_c=G \frac{Mm}{R^2}=mRω^2≠0
 
  • #42
Bandersnatch said:
Seriously, this is secondary school-level.

For uniform circular motion over body M at radius R:
F_g=F_c=G \frac{Mm}{R^2}=mRω^2≠0

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
 
  • #43
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 textbooks 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.
 
  • #44
technician said:
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.
 
  • #45
The original post claimed they are accelerating. They are not. They are at a constant speed of probably around 17, 000ish mph
 
  • #46
(Relative to an observer on the surface)
 
  • #47
p1l0t said:
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.

p1l0t said:
(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).
 
  • #48
Bandersnatch said:
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...
 
  • #49
p1l0t said:
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.

p1l0t said:
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.
A.T. said:
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.
 
  • #50
Bandersnatch said:
Towards the centre of the Earth, of course. Vectors, remember?
It's pure Newton, too.

Lol then how do they maintain altitude?
 
  • #51
p1l0t said:
Lol then how do they maintain altitude?
Well, they have such a high tangential velocity that they "fall" towards the Earth at the same rate as they "fly away".

I don't know what else to tell you, these are not high-level concepts. Should I direct you to some physics book maybe?
Resnick&Halliday Physics part I, chapter 4-4 ("uniform circular motion").
 
  • #52
Bandersnatch said:
Well, they have such a high tangential velocity that they "fall" towards the Earth at the same rate as they "fly away".

I don't know what else to tell you, these are not high-level concepts. Should I direct you to some physics book maybe?
Resnick&Halliday Physics part I, chapter 4-4 ("uniform circular motion").

Yeah so the net force is what? Wait for it...

ZERO!
 
  • #53
p1l0t said:
Yeah so the net force is what? Wait for it...




ZERO!
And your answer is... Wait for it...

Incorrect!
 
  • #54
Doc Al said:
And your answer is... Wait for it...

Incorrect!

If you are falling away just as fast you are falling in then are you not weightless?
 
  • #55
p1l0t said:
If you are falling away just as fast you are falling in then are you not weightless?
The term "weightless" is something of a misnomer. Something is "weightless" because there is no supporting force, not because there is no weight. An astronaut in the space shuttle is still being pulled by Earth's gravity. It is Earth's gravity which keeps them in orbit. They feel "weightless" because both astronaut and shuttle are in free fall.

Since they are accelerating (moving in a circle) about the earth, there must be a net force on them. There is--gravity!

You can experience "weightlessness", albeit for a short period, by jumping off a cliff. Your weight--the pull of gravity--doesn't disappear.
 
  • #56
Doc Al said:
The term "weightless" is something of a misnomer. Something is "weightless" because there is no supporting force, not because there is no weight. An astronaut in the space shuttle is still being pulled by Earth's gravity. It is Earth's gravity which keeps them in orbit. They feel "weightless" because both astronaut and shuttle are in free fall.

Since they are accelerating (moving in a circle) about the earth, there must be a net force on them. There is--gravity!

You can experience "weightlessness", albeit for a short period, by jumping off a cliff. Your weight--the pull of gravity--doesn't disappear.

You still have mass, yes. Weight, no. They are not accelerating. They are at a constant speed. If they are accelerating they would have weight. How much force does it take to move an astronaut? Practically none because he is weightless. F = M * 0
 
  • #57
p1l0t said:
You still have mass, yes. Weight, no. They are not accelerating. They are at a constant speed. If they are accelerating they would have weight. How much force does it take to move an astronaut? Practically none because he is weightless. F = M * 0
There is more to acceleration than changing speed. As has already been mentioned, changing direction of motion is also acceleration. You need to learn a little physics before being so adamant with your opinions.

Look up centripetal acceleration. Something moving in a circle is accelerating and that requires a force. (Try driving your car in a circle on a patch of ice. No friction to provide the centripetal acceleration, so you won't be able to turn.)
 
  • #58
Doc Al said:
There is more to acceleration than changing speed. As has already been mentioned, changing direction of motion is also acceleration. You need to learn a little physics before being so adamant with your opinions.

Look up centripetal acceleration. Something moving in a circle is accelerating and that requires a force. (Try driving your car in a circle on a patch of ice. No friction to provide the centripetal acceleration, so you won't be able to turn.)

My physics level maybe needs some learnin' but I still contend that astronauts are weightless in microgravity because of lack of acceleration compared to that which we have at the surface. It maybe because of many opposing accelerating forces, but the astronauts are weightless because with no acceleration (or canceling opposing forces whatever) the net force it takes to move them is going to be near zero.
 
  • #59
I highly suggest you google Einsteins equivelence principle. Bandersnatch an Doc have been giving you the correct answers. So maybe after reading up on the equivelence principle will aid your understanding
 
  • #60
Mordred said:
I highly suggest you google Einsteins equivelence principle. Bandersnatch an Doc have been giving you the correct answers. So maybe after reading up on the equivelence principle will aid your understanding

Doesn't that support my argument? Isn't the whole idea of the principal that falling bodies are bound by non-gravitational forces only?
 

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