How does rotational “artificial gravity” differ from normal gravity?
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I am not a physicist, just a curious mind. I was reading a novel by Iain Banks where it was mentioned, that shifting from artificial rotational "gravity" (in space, on a rotating space craft) to real gravity caused some level of discomfort.
And this has me thinking; is there any truth to that? I mean I am aware that reading a science fiction novel does not science make; however it also strikes me as an unlikely story line to inject in there if it was not founded on at least some real theory or actual reality.
So I guess it boils down to this. From the perspective of the individual experiencing it, is there any notable difference from being rotated and thereby experiencing a sensation of gravity, to a person experiencing real gravity (from the attraction of mass)?
newtonian-mechanics newtonian-gravity reference-frames centrifugal-force
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up vote
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I am not a physicist, just a curious mind. I was reading a novel by Iain Banks where it was mentioned, that shifting from artificial rotational "gravity" (in space, on a rotating space craft) to real gravity caused some level of discomfort.
And this has me thinking; is there any truth to that? I mean I am aware that reading a science fiction novel does not science make; however it also strikes me as an unlikely story line to inject in there if it was not founded on at least some real theory or actual reality.
So I guess it boils down to this. From the perspective of the individual experiencing it, is there any notable difference from being rotated and thereby experiencing a sensation of gravity, to a person experiencing real gravity (from the attraction of mass)?
newtonian-mechanics newtonian-gravity reference-frames centrifugal-force
New contributor
I deleted some off-topic comments.
– David Z♦
14 hours ago
add a comment |
up vote
30
down vote
favorite
up vote
30
down vote
favorite
I am not a physicist, just a curious mind. I was reading a novel by Iain Banks where it was mentioned, that shifting from artificial rotational "gravity" (in space, on a rotating space craft) to real gravity caused some level of discomfort.
And this has me thinking; is there any truth to that? I mean I am aware that reading a science fiction novel does not science make; however it also strikes me as an unlikely story line to inject in there if it was not founded on at least some real theory or actual reality.
So I guess it boils down to this. From the perspective of the individual experiencing it, is there any notable difference from being rotated and thereby experiencing a sensation of gravity, to a person experiencing real gravity (from the attraction of mass)?
newtonian-mechanics newtonian-gravity reference-frames centrifugal-force
New contributor
I am not a physicist, just a curious mind. I was reading a novel by Iain Banks where it was mentioned, that shifting from artificial rotational "gravity" (in space, on a rotating space craft) to real gravity caused some level of discomfort.
And this has me thinking; is there any truth to that? I mean I am aware that reading a science fiction novel does not science make; however it also strikes me as an unlikely story line to inject in there if it was not founded on at least some real theory or actual reality.
So I guess it boils down to this. From the perspective of the individual experiencing it, is there any notable difference from being rotated and thereby experiencing a sensation of gravity, to a person experiencing real gravity (from the attraction of mass)?
newtonian-mechanics newtonian-gravity reference-frames centrifugal-force
newtonian-mechanics newtonian-gravity reference-frames centrifugal-force
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New contributor
edited 2 hours ago
Qmechanic♦
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99.2k121781103
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asked 2 days ago
Mark Cassidy
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New contributor
I deleted some off-topic comments.
– David Z♦
14 hours ago
add a comment |
I deleted some off-topic comments.
– David Z♦
14 hours ago
I deleted some off-topic comments.
– David Z♦
14 hours ago
I deleted some off-topic comments.
– David Z♦
14 hours ago
add a comment |
7 Answers
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up vote
52
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I think a rotating frame would have both a centrifugal force, mimicking gravity, and what is called a Coriolis force. So, for example, if you would throw a ball straight up in the air in the rotating space station, you would see it move sideways too, because the outside of a wheel always rotates faster than the inside.
It's possible that the people in the space station could feel this Coriolis force, hence the reason for the discomfort.
New contributor
3
Nice answer. For an instructive homework exercise, try analysing various gravity-testing experiments, such as what happens when balls are dropped from a tall tower set up in a huge rotating space station whose rim rotates with acceleration 1$g$.
– Andrew Steane
2 days ago
1
I like this answer. So, much like changing eyeglasses - the discomfort isn't physical, but you do notice it.
– Dithermaster
2 days ago
9
In an episode of the show "The Expanse", a phrase similar to "in the core where the Coriolis is really bad" is used.
– Davis Yoshida
yesterday
5
Wikipedia has dug out a rule of thumb (belief?) that at 2RPM or below the Coriolis force would be tolerable. 2RPM comes to about $0.2$ radians per second. Meaning that $1g$ or $10 m/s^2$ requires a station with a radius of $250$ meters.
– Jyrki Lahtonen
yesterday
6
@Dithermaster "So, much like changing eyeglasses - the discomfort isn't physical" it is absolutely physical, for both cases. Changing glasses results in the lens muscles having to work in different ways, which tires them. Having uneven forces on your body is also physical.
– UKMonkey
yesterday
|
show 5 more comments
up vote
23
down vote
I'm speculating, but the speculation is based on actual physics :).
Your physical experience of gravity on a planet and artificial gravity at the outside of a rotating wheel might be different based on the following.
The force you feel from a planet is $G*m_{you}*M_{planet}/r^2$ (Gravitational constant times your mass times the mass of the planet, divided by the distance $r$ from you to the center of the planet, squared.
The force you feel from the rotating wheel is $m_{you}*omega^2r$ (your mass times the angular velocity (squared) times $r$, the distance from you to the center of the wheel).
So, suppose you are on a planet (which would normally have a very large value of $r$--meaning, you are a long way from its center), and you are seated, then you stand up. Your head has moved from $r$ meters to $r+1$ meters (your head is now 1 meter farther from the center of the planet). So, on earth, you've moved from about 6.4 million meters away to about 6.4 million meters...plus one! That's going to make a change in the force on your head that's probably way too small for you to notice.
On a man-made rotating wheel, you're going to have a much smaller value of $r$ (assuming the wheel is way less than the size of a planet). So $r-1$ meters (keep in mind, when you stand up inside the rotating wheel, your head is closer to the hub of the wheel, so it's a change to $r-1$ instead of $r+1$ as it would be on the planet) might be different enough from $r$ meters to be something you feel, and, if you spent a lot of time there, or were born there, or whatever, you would get used to things (like your head) being "lighter" when you stand up. If that was your "normal", then it might feel really strange to you when that didn't happen in Earth's gravity.
New contributor
1
Isn't this why such craft have to be pretty large?
– RonJohn
yesterday
3
The term of art for the effects you're talking about is tidal forces.
– Michael Seifert
yesterday
2
@RonJohn Yes, but there's an economy to consider. E.g. it would be nice if trips to space didn't require such high acceleration as in modern rockets, but it's more economical to train a few specialists to handle those accelerations than to fly rockets at lower accelerations. The same way, the rotating ships would be built as small as possible for a given tolerable level of discomfort for most of their users. Maybe at a radius of 200 meters, noöne would notice the rotation - but 200 meters is a pretty bulky ship (and it would only work on the outer edge anyway!).
– Luaan
yesterday
You should also account for "running widdershins" making your heavier. The effect seems to scale down with the square root of the radius, so it might persist longer than the linear height change impact.
– Yakk
20 hours ago
add a comment |
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7
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For a non-technical answer, remember when you were a kid on the playground? (Yes, I know I'm making what's perhaps a parochial assumption.) If you sat on the merry-go-round (this: https://en.wikipedia.org/wiki/Roundabout_(play) ) and got the other kids to push it around really fast, you could feel the "gravity" pulling you outwards. But because you were also going around in a tight circle, the fluid in your ears sloshed around, and so you got dizzy.
Now scale this up to a moderately-sized space station. You might still have some effect on the ears from rotation (how much depends on the size), but because you've been there a long time, your body has adapted to this as being normal. When you shift to "real" gravity, the rotation effect goes away, but to your body this is now NOT normal.
(Whether this would actually happen I can't say: AFAIK no one has tried it, but it's certainly plausible enough for SF :-))
The distance scale could be such that the rotation rate is very small, say once per 24 hours. Ear-related effects would then be too small to matter.
– Andrew Steane
yesterday
2
@AndrewSteane It depends on two things: 1) how big your habitat is and 2) how fast its spinning. The smaller it is, the faster it has to spin in order to generate 1 G of gravity on the outer surface as well as causing a steeper gradient (i.e. if your habitat is 12 feet in diameter, then your head experiences 0 G and your feet 1 G; an extreme scenario).
– Draco18s
yesterday
7
@AndrewSteane one rotation per 24 hours would require a radius of ~2 million km for 1G
– trapper
yesterday
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4
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You would be unlikely to notice any difference unless the spacecraft is fairly small.
For example with 50m radius there is only a 2% difference between 50m and 49m. The station in this case would be spinning at 4.25 rpm to generate 1G.
New contributor
2
2% per metre is quite a lot. A 2m tall, 80kg person upon standing up, would be thrown forwards with a 3kg force and vertical as sensed by their inner ear would vary by up to 18 degrees as you did so, depending upon which way you were facing relative to the direction of travel. That should be enough to stumble or fall if you expected it to go one way and it went the other.
– Robert Frost
12 hours ago
I can't even imagine what kind of math lead you to those conclusions.
– trapper
11 hours ago
1
Sine X approximates X for small X so simply multiply mass by percentage difference for an instant approximation. Simples.
– Robert Frost
11 hours ago
1
You can’t just multiply numbers randomly though. 3kg is not a ‘force’, and your centre of mass while standing is at hip level, not 2m off the ground.
– trapper
11 hours ago
1
You're obviously right re kg not being a force but if rotation is generating 1g at the circumference then mass at the circumference is isometric with weight on earth, so I was talking in terms of the weight of 3kg on Earth.
– Robert Frost
8 hours ago
add a comment |
up vote
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Experiencing rotational forces and fixed direction gravity at the same time would be weird.
A person under the influence of gravity experiences a constant acceleration. A person in a rotating reference frame experiences a constant magnitude acceleration, but the direction is changing constantly.
This means that if you are experiencing both at once, and the axis of rotation is not parallel to the direction of gravity, the total acceleration that you feel will be constantly fluctuating. It's more or less equivalent to the fact that if you swing a bucket on a rope in a vertical circle, the tension in the rope is higher when the bucket is near the ground than when it is at the top of the swing.
Depending on how fast the rotation of your station is, this could make the transition period feel like a rollercoaster.
Of course, the logical way to transition reference frames would be to leave one, enter zero-g, then enter the second. That would avoid the roller coaster effect. But if they skipped that process then I could easily see people emptying their stomachs during the process.
New contributor
1
Sorry, but this is incorrect. Imagine swinging a bucket on a rope in a horizontal circle.
– Beta
yesterday
@Beta, well, it depends on which way the station is rotating. You could organize the transition in a logical, non-rollercoaster manner. But you don't have to.
– Arcanist Lupus
yesterday
2
I hope you aren't referring to orientation relative to a planet -- the only linear acceleration on the station would be due to its translational rocket engine.
– amI
yesterday
The tension on a rope on a bucket increases and decreases because you are standing on a planet experiencing its gravitational field. That does not apply in this situation.
– msouth
3 hours ago
add a comment |
up vote
0
down vote
In my opinion there is mechanical difference in which the rotation affects you in those two cases (you rotate on planet while not on poles). On planets surface the mass pulls you inward and the planetary rotation lessens the force applied to you. On the station the rotation works the other way, basically creating gravity from nothing.
Have a nice day.
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Please have a look at "Why don't spaceships have artificial gravity" by SciShow Space on YouTube. It explains the topic way better than I could ever do.
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Hi and welcome to Physics Stack Exchange. However, please note that 'link only' answers such as this are generally frowned upon. A good answer should be able to stand 'on its own two feet'. This would probably be more appropriate for a comment.
– Time4Tea
1 hour ago
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
– Time4Tea
1 hour ago
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7 Answers
7
active
oldest
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7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
52
down vote
I think a rotating frame would have both a centrifugal force, mimicking gravity, and what is called a Coriolis force. So, for example, if you would throw a ball straight up in the air in the rotating space station, you would see it move sideways too, because the outside of a wheel always rotates faster than the inside.
It's possible that the people in the space station could feel this Coriolis force, hence the reason for the discomfort.
New contributor
3
Nice answer. For an instructive homework exercise, try analysing various gravity-testing experiments, such as what happens when balls are dropped from a tall tower set up in a huge rotating space station whose rim rotates with acceleration 1$g$.
– Andrew Steane
2 days ago
1
I like this answer. So, much like changing eyeglasses - the discomfort isn't physical, but you do notice it.
– Dithermaster
2 days ago
9
In an episode of the show "The Expanse", a phrase similar to "in the core where the Coriolis is really bad" is used.
– Davis Yoshida
yesterday
5
Wikipedia has dug out a rule of thumb (belief?) that at 2RPM or below the Coriolis force would be tolerable. 2RPM comes to about $0.2$ radians per second. Meaning that $1g$ or $10 m/s^2$ requires a station with a radius of $250$ meters.
– Jyrki Lahtonen
yesterday
6
@Dithermaster "So, much like changing eyeglasses - the discomfort isn't physical" it is absolutely physical, for both cases. Changing glasses results in the lens muscles having to work in different ways, which tires them. Having uneven forces on your body is also physical.
– UKMonkey
yesterday
|
show 5 more comments
up vote
52
down vote
I think a rotating frame would have both a centrifugal force, mimicking gravity, and what is called a Coriolis force. So, for example, if you would throw a ball straight up in the air in the rotating space station, you would see it move sideways too, because the outside of a wheel always rotates faster than the inside.
It's possible that the people in the space station could feel this Coriolis force, hence the reason for the discomfort.
New contributor
3
Nice answer. For an instructive homework exercise, try analysing various gravity-testing experiments, such as what happens when balls are dropped from a tall tower set up in a huge rotating space station whose rim rotates with acceleration 1$g$.
– Andrew Steane
2 days ago
1
I like this answer. So, much like changing eyeglasses - the discomfort isn't physical, but you do notice it.
– Dithermaster
2 days ago
9
In an episode of the show "The Expanse", a phrase similar to "in the core where the Coriolis is really bad" is used.
– Davis Yoshida
yesterday
5
Wikipedia has dug out a rule of thumb (belief?) that at 2RPM or below the Coriolis force would be tolerable. 2RPM comes to about $0.2$ radians per second. Meaning that $1g$ or $10 m/s^2$ requires a station with a radius of $250$ meters.
– Jyrki Lahtonen
yesterday
6
@Dithermaster "So, much like changing eyeglasses - the discomfort isn't physical" it is absolutely physical, for both cases. Changing glasses results in the lens muscles having to work in different ways, which tires them. Having uneven forces on your body is also physical.
– UKMonkey
yesterday
|
show 5 more comments
up vote
52
down vote
up vote
52
down vote
I think a rotating frame would have both a centrifugal force, mimicking gravity, and what is called a Coriolis force. So, for example, if you would throw a ball straight up in the air in the rotating space station, you would see it move sideways too, because the outside of a wheel always rotates faster than the inside.
It's possible that the people in the space station could feel this Coriolis force, hence the reason for the discomfort.
New contributor
I think a rotating frame would have both a centrifugal force, mimicking gravity, and what is called a Coriolis force. So, for example, if you would throw a ball straight up in the air in the rotating space station, you would see it move sideways too, because the outside of a wheel always rotates faster than the inside.
It's possible that the people in the space station could feel this Coriolis force, hence the reason for the discomfort.
New contributor
New contributor
answered 2 days ago
Eric David Kramer
52113
52113
New contributor
New contributor
3
Nice answer. For an instructive homework exercise, try analysing various gravity-testing experiments, such as what happens when balls are dropped from a tall tower set up in a huge rotating space station whose rim rotates with acceleration 1$g$.
– Andrew Steane
2 days ago
1
I like this answer. So, much like changing eyeglasses - the discomfort isn't physical, but you do notice it.
– Dithermaster
2 days ago
9
In an episode of the show "The Expanse", a phrase similar to "in the core where the Coriolis is really bad" is used.
– Davis Yoshida
yesterday
5
Wikipedia has dug out a rule of thumb (belief?) that at 2RPM or below the Coriolis force would be tolerable. 2RPM comes to about $0.2$ radians per second. Meaning that $1g$ or $10 m/s^2$ requires a station with a radius of $250$ meters.
– Jyrki Lahtonen
yesterday
6
@Dithermaster "So, much like changing eyeglasses - the discomfort isn't physical" it is absolutely physical, for both cases. Changing glasses results in the lens muscles having to work in different ways, which tires them. Having uneven forces on your body is also physical.
– UKMonkey
yesterday
|
show 5 more comments
3
Nice answer. For an instructive homework exercise, try analysing various gravity-testing experiments, such as what happens when balls are dropped from a tall tower set up in a huge rotating space station whose rim rotates with acceleration 1$g$.
– Andrew Steane
2 days ago
1
I like this answer. So, much like changing eyeglasses - the discomfort isn't physical, but you do notice it.
– Dithermaster
2 days ago
9
In an episode of the show "The Expanse", a phrase similar to "in the core where the Coriolis is really bad" is used.
– Davis Yoshida
yesterday
5
Wikipedia has dug out a rule of thumb (belief?) that at 2RPM or below the Coriolis force would be tolerable. 2RPM comes to about $0.2$ radians per second. Meaning that $1g$ or $10 m/s^2$ requires a station with a radius of $250$ meters.
– Jyrki Lahtonen
yesterday
6
@Dithermaster "So, much like changing eyeglasses - the discomfort isn't physical" it is absolutely physical, for both cases. Changing glasses results in the lens muscles having to work in different ways, which tires them. Having uneven forces on your body is also physical.
– UKMonkey
yesterday
3
3
Nice answer. For an instructive homework exercise, try analysing various gravity-testing experiments, such as what happens when balls are dropped from a tall tower set up in a huge rotating space station whose rim rotates with acceleration 1$g$.
– Andrew Steane
2 days ago
Nice answer. For an instructive homework exercise, try analysing various gravity-testing experiments, such as what happens when balls are dropped from a tall tower set up in a huge rotating space station whose rim rotates with acceleration 1$g$.
– Andrew Steane
2 days ago
1
1
I like this answer. So, much like changing eyeglasses - the discomfort isn't physical, but you do notice it.
– Dithermaster
2 days ago
I like this answer. So, much like changing eyeglasses - the discomfort isn't physical, but you do notice it.
– Dithermaster
2 days ago
9
9
In an episode of the show "The Expanse", a phrase similar to "in the core where the Coriolis is really bad" is used.
– Davis Yoshida
yesterday
In an episode of the show "The Expanse", a phrase similar to "in the core where the Coriolis is really bad" is used.
– Davis Yoshida
yesterday
5
5
Wikipedia has dug out a rule of thumb (belief?) that at 2RPM or below the Coriolis force would be tolerable. 2RPM comes to about $0.2$ radians per second. Meaning that $1g$ or $10 m/s^2$ requires a station with a radius of $250$ meters.
– Jyrki Lahtonen
yesterday
Wikipedia has dug out a rule of thumb (belief?) that at 2RPM or below the Coriolis force would be tolerable. 2RPM comes to about $0.2$ radians per second. Meaning that $1g$ or $10 m/s^2$ requires a station with a radius of $250$ meters.
– Jyrki Lahtonen
yesterday
6
6
@Dithermaster "So, much like changing eyeglasses - the discomfort isn't physical" it is absolutely physical, for both cases. Changing glasses results in the lens muscles having to work in different ways, which tires them. Having uneven forces on your body is also physical.
– UKMonkey
yesterday
@Dithermaster "So, much like changing eyeglasses - the discomfort isn't physical" it is absolutely physical, for both cases. Changing glasses results in the lens muscles having to work in different ways, which tires them. Having uneven forces on your body is also physical.
– UKMonkey
yesterday
|
show 5 more comments
up vote
23
down vote
I'm speculating, but the speculation is based on actual physics :).
Your physical experience of gravity on a planet and artificial gravity at the outside of a rotating wheel might be different based on the following.
The force you feel from a planet is $G*m_{you}*M_{planet}/r^2$ (Gravitational constant times your mass times the mass of the planet, divided by the distance $r$ from you to the center of the planet, squared.
The force you feel from the rotating wheel is $m_{you}*omega^2r$ (your mass times the angular velocity (squared) times $r$, the distance from you to the center of the wheel).
So, suppose you are on a planet (which would normally have a very large value of $r$--meaning, you are a long way from its center), and you are seated, then you stand up. Your head has moved from $r$ meters to $r+1$ meters (your head is now 1 meter farther from the center of the planet). So, on earth, you've moved from about 6.4 million meters away to about 6.4 million meters...plus one! That's going to make a change in the force on your head that's probably way too small for you to notice.
On a man-made rotating wheel, you're going to have a much smaller value of $r$ (assuming the wheel is way less than the size of a planet). So $r-1$ meters (keep in mind, when you stand up inside the rotating wheel, your head is closer to the hub of the wheel, so it's a change to $r-1$ instead of $r+1$ as it would be on the planet) might be different enough from $r$ meters to be something you feel, and, if you spent a lot of time there, or were born there, or whatever, you would get used to things (like your head) being "lighter" when you stand up. If that was your "normal", then it might feel really strange to you when that didn't happen in Earth's gravity.
New contributor
1
Isn't this why such craft have to be pretty large?
– RonJohn
yesterday
3
The term of art for the effects you're talking about is tidal forces.
– Michael Seifert
yesterday
2
@RonJohn Yes, but there's an economy to consider. E.g. it would be nice if trips to space didn't require such high acceleration as in modern rockets, but it's more economical to train a few specialists to handle those accelerations than to fly rockets at lower accelerations. The same way, the rotating ships would be built as small as possible for a given tolerable level of discomfort for most of their users. Maybe at a radius of 200 meters, noöne would notice the rotation - but 200 meters is a pretty bulky ship (and it would only work on the outer edge anyway!).
– Luaan
yesterday
You should also account for "running widdershins" making your heavier. The effect seems to scale down with the square root of the radius, so it might persist longer than the linear height change impact.
– Yakk
20 hours ago
add a comment |
up vote
23
down vote
I'm speculating, but the speculation is based on actual physics :).
Your physical experience of gravity on a planet and artificial gravity at the outside of a rotating wheel might be different based on the following.
The force you feel from a planet is $G*m_{you}*M_{planet}/r^2$ (Gravitational constant times your mass times the mass of the planet, divided by the distance $r$ from you to the center of the planet, squared.
The force you feel from the rotating wheel is $m_{you}*omega^2r$ (your mass times the angular velocity (squared) times $r$, the distance from you to the center of the wheel).
So, suppose you are on a planet (which would normally have a very large value of $r$--meaning, you are a long way from its center), and you are seated, then you stand up. Your head has moved from $r$ meters to $r+1$ meters (your head is now 1 meter farther from the center of the planet). So, on earth, you've moved from about 6.4 million meters away to about 6.4 million meters...plus one! That's going to make a change in the force on your head that's probably way too small for you to notice.
On a man-made rotating wheel, you're going to have a much smaller value of $r$ (assuming the wheel is way less than the size of a planet). So $r-1$ meters (keep in mind, when you stand up inside the rotating wheel, your head is closer to the hub of the wheel, so it's a change to $r-1$ instead of $r+1$ as it would be on the planet) might be different enough from $r$ meters to be something you feel, and, if you spent a lot of time there, or were born there, or whatever, you would get used to things (like your head) being "lighter" when you stand up. If that was your "normal", then it might feel really strange to you when that didn't happen in Earth's gravity.
New contributor
1
Isn't this why such craft have to be pretty large?
– RonJohn
yesterday
3
The term of art for the effects you're talking about is tidal forces.
– Michael Seifert
yesterday
2
@RonJohn Yes, but there's an economy to consider. E.g. it would be nice if trips to space didn't require such high acceleration as in modern rockets, but it's more economical to train a few specialists to handle those accelerations than to fly rockets at lower accelerations. The same way, the rotating ships would be built as small as possible for a given tolerable level of discomfort for most of their users. Maybe at a radius of 200 meters, noöne would notice the rotation - but 200 meters is a pretty bulky ship (and it would only work on the outer edge anyway!).
– Luaan
yesterday
You should also account for "running widdershins" making your heavier. The effect seems to scale down with the square root of the radius, so it might persist longer than the linear height change impact.
– Yakk
20 hours ago
add a comment |
up vote
23
down vote
up vote
23
down vote
I'm speculating, but the speculation is based on actual physics :).
Your physical experience of gravity on a planet and artificial gravity at the outside of a rotating wheel might be different based on the following.
The force you feel from a planet is $G*m_{you}*M_{planet}/r^2$ (Gravitational constant times your mass times the mass of the planet, divided by the distance $r$ from you to the center of the planet, squared.
The force you feel from the rotating wheel is $m_{you}*omega^2r$ (your mass times the angular velocity (squared) times $r$, the distance from you to the center of the wheel).
So, suppose you are on a planet (which would normally have a very large value of $r$--meaning, you are a long way from its center), and you are seated, then you stand up. Your head has moved from $r$ meters to $r+1$ meters (your head is now 1 meter farther from the center of the planet). So, on earth, you've moved from about 6.4 million meters away to about 6.4 million meters...plus one! That's going to make a change in the force on your head that's probably way too small for you to notice.
On a man-made rotating wheel, you're going to have a much smaller value of $r$ (assuming the wheel is way less than the size of a planet). So $r-1$ meters (keep in mind, when you stand up inside the rotating wheel, your head is closer to the hub of the wheel, so it's a change to $r-1$ instead of $r+1$ as it would be on the planet) might be different enough from $r$ meters to be something you feel, and, if you spent a lot of time there, or were born there, or whatever, you would get used to things (like your head) being "lighter" when you stand up. If that was your "normal", then it might feel really strange to you when that didn't happen in Earth's gravity.
New contributor
I'm speculating, but the speculation is based on actual physics :).
Your physical experience of gravity on a planet and artificial gravity at the outside of a rotating wheel might be different based on the following.
The force you feel from a planet is $G*m_{you}*M_{planet}/r^2$ (Gravitational constant times your mass times the mass of the planet, divided by the distance $r$ from you to the center of the planet, squared.
The force you feel from the rotating wheel is $m_{you}*omega^2r$ (your mass times the angular velocity (squared) times $r$, the distance from you to the center of the wheel).
So, suppose you are on a planet (which would normally have a very large value of $r$--meaning, you are a long way from its center), and you are seated, then you stand up. Your head has moved from $r$ meters to $r+1$ meters (your head is now 1 meter farther from the center of the planet). So, on earth, you've moved from about 6.4 million meters away to about 6.4 million meters...plus one! That's going to make a change in the force on your head that's probably way too small for you to notice.
On a man-made rotating wheel, you're going to have a much smaller value of $r$ (assuming the wheel is way less than the size of a planet). So $r-1$ meters (keep in mind, when you stand up inside the rotating wheel, your head is closer to the hub of the wheel, so it's a change to $r-1$ instead of $r+1$ as it would be on the planet) might be different enough from $r$ meters to be something you feel, and, if you spent a lot of time there, or were born there, or whatever, you would get used to things (like your head) being "lighter" when you stand up. If that was your "normal", then it might feel really strange to you when that didn't happen in Earth's gravity.
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New contributor
answered yesterday
msouth
33114
33114
New contributor
New contributor
1
Isn't this why such craft have to be pretty large?
– RonJohn
yesterday
3
The term of art for the effects you're talking about is tidal forces.
– Michael Seifert
yesterday
2
@RonJohn Yes, but there's an economy to consider. E.g. it would be nice if trips to space didn't require such high acceleration as in modern rockets, but it's more economical to train a few specialists to handle those accelerations than to fly rockets at lower accelerations. The same way, the rotating ships would be built as small as possible for a given tolerable level of discomfort for most of their users. Maybe at a radius of 200 meters, noöne would notice the rotation - but 200 meters is a pretty bulky ship (and it would only work on the outer edge anyway!).
– Luaan
yesterday
You should also account for "running widdershins" making your heavier. The effect seems to scale down with the square root of the radius, so it might persist longer than the linear height change impact.
– Yakk
20 hours ago
add a comment |
1
Isn't this why such craft have to be pretty large?
– RonJohn
yesterday
3
The term of art for the effects you're talking about is tidal forces.
– Michael Seifert
yesterday
2
@RonJohn Yes, but there's an economy to consider. E.g. it would be nice if trips to space didn't require such high acceleration as in modern rockets, but it's more economical to train a few specialists to handle those accelerations than to fly rockets at lower accelerations. The same way, the rotating ships would be built as small as possible for a given tolerable level of discomfort for most of their users. Maybe at a radius of 200 meters, noöne would notice the rotation - but 200 meters is a pretty bulky ship (and it would only work on the outer edge anyway!).
– Luaan
yesterday
You should also account for "running widdershins" making your heavier. The effect seems to scale down with the square root of the radius, so it might persist longer than the linear height change impact.
– Yakk
20 hours ago
1
1
Isn't this why such craft have to be pretty large?
– RonJohn
yesterday
Isn't this why such craft have to be pretty large?
– RonJohn
yesterday
3
3
The term of art for the effects you're talking about is tidal forces.
– Michael Seifert
yesterday
The term of art for the effects you're talking about is tidal forces.
– Michael Seifert
yesterday
2
2
@RonJohn Yes, but there's an economy to consider. E.g. it would be nice if trips to space didn't require such high acceleration as in modern rockets, but it's more economical to train a few specialists to handle those accelerations than to fly rockets at lower accelerations. The same way, the rotating ships would be built as small as possible for a given tolerable level of discomfort for most of their users. Maybe at a radius of 200 meters, noöne would notice the rotation - but 200 meters is a pretty bulky ship (and it would only work on the outer edge anyway!).
– Luaan
yesterday
@RonJohn Yes, but there's an economy to consider. E.g. it would be nice if trips to space didn't require such high acceleration as in modern rockets, but it's more economical to train a few specialists to handle those accelerations than to fly rockets at lower accelerations. The same way, the rotating ships would be built as small as possible for a given tolerable level of discomfort for most of their users. Maybe at a radius of 200 meters, noöne would notice the rotation - but 200 meters is a pretty bulky ship (and it would only work on the outer edge anyway!).
– Luaan
yesterday
You should also account for "running widdershins" making your heavier. The effect seems to scale down with the square root of the radius, so it might persist longer than the linear height change impact.
– Yakk
20 hours ago
You should also account for "running widdershins" making your heavier. The effect seems to scale down with the square root of the radius, so it might persist longer than the linear height change impact.
– Yakk
20 hours ago
add a comment |
up vote
7
down vote
For a non-technical answer, remember when you were a kid on the playground? (Yes, I know I'm making what's perhaps a parochial assumption.) If you sat on the merry-go-round (this: https://en.wikipedia.org/wiki/Roundabout_(play) ) and got the other kids to push it around really fast, you could feel the "gravity" pulling you outwards. But because you were also going around in a tight circle, the fluid in your ears sloshed around, and so you got dizzy.
Now scale this up to a moderately-sized space station. You might still have some effect on the ears from rotation (how much depends on the size), but because you've been there a long time, your body has adapted to this as being normal. When you shift to "real" gravity, the rotation effect goes away, but to your body this is now NOT normal.
(Whether this would actually happen I can't say: AFAIK no one has tried it, but it's certainly plausible enough for SF :-))
The distance scale could be such that the rotation rate is very small, say once per 24 hours. Ear-related effects would then be too small to matter.
– Andrew Steane
yesterday
2
@AndrewSteane It depends on two things: 1) how big your habitat is and 2) how fast its spinning. The smaller it is, the faster it has to spin in order to generate 1 G of gravity on the outer surface as well as causing a steeper gradient (i.e. if your habitat is 12 feet in diameter, then your head experiences 0 G and your feet 1 G; an extreme scenario).
– Draco18s
yesterday
7
@AndrewSteane one rotation per 24 hours would require a radius of ~2 million km for 1G
– trapper
yesterday
add a comment |
up vote
7
down vote
For a non-technical answer, remember when you were a kid on the playground? (Yes, I know I'm making what's perhaps a parochial assumption.) If you sat on the merry-go-round (this: https://en.wikipedia.org/wiki/Roundabout_(play) ) and got the other kids to push it around really fast, you could feel the "gravity" pulling you outwards. But because you were also going around in a tight circle, the fluid in your ears sloshed around, and so you got dizzy.
Now scale this up to a moderately-sized space station. You might still have some effect on the ears from rotation (how much depends on the size), but because you've been there a long time, your body has adapted to this as being normal. When you shift to "real" gravity, the rotation effect goes away, but to your body this is now NOT normal.
(Whether this would actually happen I can't say: AFAIK no one has tried it, but it's certainly plausible enough for SF :-))
The distance scale could be such that the rotation rate is very small, say once per 24 hours. Ear-related effects would then be too small to matter.
– Andrew Steane
yesterday
2
@AndrewSteane It depends on two things: 1) how big your habitat is and 2) how fast its spinning. The smaller it is, the faster it has to spin in order to generate 1 G of gravity on the outer surface as well as causing a steeper gradient (i.e. if your habitat is 12 feet in diameter, then your head experiences 0 G and your feet 1 G; an extreme scenario).
– Draco18s
yesterday
7
@AndrewSteane one rotation per 24 hours would require a radius of ~2 million km for 1G
– trapper
yesterday
add a comment |
up vote
7
down vote
up vote
7
down vote
For a non-technical answer, remember when you were a kid on the playground? (Yes, I know I'm making what's perhaps a parochial assumption.) If you sat on the merry-go-round (this: https://en.wikipedia.org/wiki/Roundabout_(play) ) and got the other kids to push it around really fast, you could feel the "gravity" pulling you outwards. But because you were also going around in a tight circle, the fluid in your ears sloshed around, and so you got dizzy.
Now scale this up to a moderately-sized space station. You might still have some effect on the ears from rotation (how much depends on the size), but because you've been there a long time, your body has adapted to this as being normal. When you shift to "real" gravity, the rotation effect goes away, but to your body this is now NOT normal.
(Whether this would actually happen I can't say: AFAIK no one has tried it, but it's certainly plausible enough for SF :-))
For a non-technical answer, remember when you were a kid on the playground? (Yes, I know I'm making what's perhaps a parochial assumption.) If you sat on the merry-go-round (this: https://en.wikipedia.org/wiki/Roundabout_(play) ) and got the other kids to push it around really fast, you could feel the "gravity" pulling you outwards. But because you were also going around in a tight circle, the fluid in your ears sloshed around, and so you got dizzy.
Now scale this up to a moderately-sized space station. You might still have some effect on the ears from rotation (how much depends on the size), but because you've been there a long time, your body has adapted to this as being normal. When you shift to "real" gravity, the rotation effect goes away, but to your body this is now NOT normal.
(Whether this would actually happen I can't say: AFAIK no one has tried it, but it's certainly plausible enough for SF :-))
answered yesterday
jamesqf
33524
33524
The distance scale could be such that the rotation rate is very small, say once per 24 hours. Ear-related effects would then be too small to matter.
– Andrew Steane
yesterday
2
@AndrewSteane It depends on two things: 1) how big your habitat is and 2) how fast its spinning. The smaller it is, the faster it has to spin in order to generate 1 G of gravity on the outer surface as well as causing a steeper gradient (i.e. if your habitat is 12 feet in diameter, then your head experiences 0 G and your feet 1 G; an extreme scenario).
– Draco18s
yesterday
7
@AndrewSteane one rotation per 24 hours would require a radius of ~2 million km for 1G
– trapper
yesterday
add a comment |
The distance scale could be such that the rotation rate is very small, say once per 24 hours. Ear-related effects would then be too small to matter.
– Andrew Steane
yesterday
2
@AndrewSteane It depends on two things: 1) how big your habitat is and 2) how fast its spinning. The smaller it is, the faster it has to spin in order to generate 1 G of gravity on the outer surface as well as causing a steeper gradient (i.e. if your habitat is 12 feet in diameter, then your head experiences 0 G and your feet 1 G; an extreme scenario).
– Draco18s
yesterday
7
@AndrewSteane one rotation per 24 hours would require a radius of ~2 million km for 1G
– trapper
yesterday
The distance scale could be such that the rotation rate is very small, say once per 24 hours. Ear-related effects would then be too small to matter.
– Andrew Steane
yesterday
The distance scale could be such that the rotation rate is very small, say once per 24 hours. Ear-related effects would then be too small to matter.
– Andrew Steane
yesterday
2
2
@AndrewSteane It depends on two things: 1) how big your habitat is and 2) how fast its spinning. The smaller it is, the faster it has to spin in order to generate 1 G of gravity on the outer surface as well as causing a steeper gradient (i.e. if your habitat is 12 feet in diameter, then your head experiences 0 G and your feet 1 G; an extreme scenario).
– Draco18s
yesterday
@AndrewSteane It depends on two things: 1) how big your habitat is and 2) how fast its spinning. The smaller it is, the faster it has to spin in order to generate 1 G of gravity on the outer surface as well as causing a steeper gradient (i.e. if your habitat is 12 feet in diameter, then your head experiences 0 G and your feet 1 G; an extreme scenario).
– Draco18s
yesterday
7
7
@AndrewSteane one rotation per 24 hours would require a radius of ~2 million km for 1G
– trapper
yesterday
@AndrewSteane one rotation per 24 hours would require a radius of ~2 million km for 1G
– trapper
yesterday
add a comment |
up vote
4
down vote
You would be unlikely to notice any difference unless the spacecraft is fairly small.
For example with 50m radius there is only a 2% difference between 50m and 49m. The station in this case would be spinning at 4.25 rpm to generate 1G.
New contributor
2
2% per metre is quite a lot. A 2m tall, 80kg person upon standing up, would be thrown forwards with a 3kg force and vertical as sensed by their inner ear would vary by up to 18 degrees as you did so, depending upon which way you were facing relative to the direction of travel. That should be enough to stumble or fall if you expected it to go one way and it went the other.
– Robert Frost
12 hours ago
I can't even imagine what kind of math lead you to those conclusions.
– trapper
11 hours ago
1
Sine X approximates X for small X so simply multiply mass by percentage difference for an instant approximation. Simples.
– Robert Frost
11 hours ago
1
You can’t just multiply numbers randomly though. 3kg is not a ‘force’, and your centre of mass while standing is at hip level, not 2m off the ground.
– trapper
11 hours ago
1
You're obviously right re kg not being a force but if rotation is generating 1g at the circumference then mass at the circumference is isometric with weight on earth, so I was talking in terms of the weight of 3kg on Earth.
– Robert Frost
8 hours ago
add a comment |
up vote
4
down vote
You would be unlikely to notice any difference unless the spacecraft is fairly small.
For example with 50m radius there is only a 2% difference between 50m and 49m. The station in this case would be spinning at 4.25 rpm to generate 1G.
New contributor
2
2% per metre is quite a lot. A 2m tall, 80kg person upon standing up, would be thrown forwards with a 3kg force and vertical as sensed by their inner ear would vary by up to 18 degrees as you did so, depending upon which way you were facing relative to the direction of travel. That should be enough to stumble or fall if you expected it to go one way and it went the other.
– Robert Frost
12 hours ago
I can't even imagine what kind of math lead you to those conclusions.
– trapper
11 hours ago
1
Sine X approximates X for small X so simply multiply mass by percentage difference for an instant approximation. Simples.
– Robert Frost
11 hours ago
1
You can’t just multiply numbers randomly though. 3kg is not a ‘force’, and your centre of mass while standing is at hip level, not 2m off the ground.
– trapper
11 hours ago
1
You're obviously right re kg not being a force but if rotation is generating 1g at the circumference then mass at the circumference is isometric with weight on earth, so I was talking in terms of the weight of 3kg on Earth.
– Robert Frost
8 hours ago
add a comment |
up vote
4
down vote
up vote
4
down vote
You would be unlikely to notice any difference unless the spacecraft is fairly small.
For example with 50m radius there is only a 2% difference between 50m and 49m. The station in this case would be spinning at 4.25 rpm to generate 1G.
New contributor
You would be unlikely to notice any difference unless the spacecraft is fairly small.
For example with 50m radius there is only a 2% difference between 50m and 49m. The station in this case would be spinning at 4.25 rpm to generate 1G.
New contributor
New contributor
answered yesterday
trapper
1411
1411
New contributor
New contributor
2
2% per metre is quite a lot. A 2m tall, 80kg person upon standing up, would be thrown forwards with a 3kg force and vertical as sensed by their inner ear would vary by up to 18 degrees as you did so, depending upon which way you were facing relative to the direction of travel. That should be enough to stumble or fall if you expected it to go one way and it went the other.
– Robert Frost
12 hours ago
I can't even imagine what kind of math lead you to those conclusions.
– trapper
11 hours ago
1
Sine X approximates X for small X so simply multiply mass by percentage difference for an instant approximation. Simples.
– Robert Frost
11 hours ago
1
You can’t just multiply numbers randomly though. 3kg is not a ‘force’, and your centre of mass while standing is at hip level, not 2m off the ground.
– trapper
11 hours ago
1
You're obviously right re kg not being a force but if rotation is generating 1g at the circumference then mass at the circumference is isometric with weight on earth, so I was talking in terms of the weight of 3kg on Earth.
– Robert Frost
8 hours ago
add a comment |
2
2% per metre is quite a lot. A 2m tall, 80kg person upon standing up, would be thrown forwards with a 3kg force and vertical as sensed by their inner ear would vary by up to 18 degrees as you did so, depending upon which way you were facing relative to the direction of travel. That should be enough to stumble or fall if you expected it to go one way and it went the other.
– Robert Frost
12 hours ago
I can't even imagine what kind of math lead you to those conclusions.
– trapper
11 hours ago
1
Sine X approximates X for small X so simply multiply mass by percentage difference for an instant approximation. Simples.
– Robert Frost
11 hours ago
1
You can’t just multiply numbers randomly though. 3kg is not a ‘force’, and your centre of mass while standing is at hip level, not 2m off the ground.
– trapper
11 hours ago
1
You're obviously right re kg not being a force but if rotation is generating 1g at the circumference then mass at the circumference is isometric with weight on earth, so I was talking in terms of the weight of 3kg on Earth.
– Robert Frost
8 hours ago
2
2
2% per metre is quite a lot. A 2m tall, 80kg person upon standing up, would be thrown forwards with a 3kg force and vertical as sensed by their inner ear would vary by up to 18 degrees as you did so, depending upon which way you were facing relative to the direction of travel. That should be enough to stumble or fall if you expected it to go one way and it went the other.
– Robert Frost
12 hours ago
2% per metre is quite a lot. A 2m tall, 80kg person upon standing up, would be thrown forwards with a 3kg force and vertical as sensed by their inner ear would vary by up to 18 degrees as you did so, depending upon which way you were facing relative to the direction of travel. That should be enough to stumble or fall if you expected it to go one way and it went the other.
– Robert Frost
12 hours ago
I can't even imagine what kind of math lead you to those conclusions.
– trapper
11 hours ago
I can't even imagine what kind of math lead you to those conclusions.
– trapper
11 hours ago
1
1
Sine X approximates X for small X so simply multiply mass by percentage difference for an instant approximation. Simples.
– Robert Frost
11 hours ago
Sine X approximates X for small X so simply multiply mass by percentage difference for an instant approximation. Simples.
– Robert Frost
11 hours ago
1
1
You can’t just multiply numbers randomly though. 3kg is not a ‘force’, and your centre of mass while standing is at hip level, not 2m off the ground.
– trapper
11 hours ago
You can’t just multiply numbers randomly though. 3kg is not a ‘force’, and your centre of mass while standing is at hip level, not 2m off the ground.
– trapper
11 hours ago
1
1
You're obviously right re kg not being a force but if rotation is generating 1g at the circumference then mass at the circumference is isometric with weight on earth, so I was talking in terms of the weight of 3kg on Earth.
– Robert Frost
8 hours ago
You're obviously right re kg not being a force but if rotation is generating 1g at the circumference then mass at the circumference is isometric with weight on earth, so I was talking in terms of the weight of 3kg on Earth.
– Robert Frost
8 hours ago
add a comment |
up vote
1
down vote
Experiencing rotational forces and fixed direction gravity at the same time would be weird.
A person under the influence of gravity experiences a constant acceleration. A person in a rotating reference frame experiences a constant magnitude acceleration, but the direction is changing constantly.
This means that if you are experiencing both at once, and the axis of rotation is not parallel to the direction of gravity, the total acceleration that you feel will be constantly fluctuating. It's more or less equivalent to the fact that if you swing a bucket on a rope in a vertical circle, the tension in the rope is higher when the bucket is near the ground than when it is at the top of the swing.
Depending on how fast the rotation of your station is, this could make the transition period feel like a rollercoaster.
Of course, the logical way to transition reference frames would be to leave one, enter zero-g, then enter the second. That would avoid the roller coaster effect. But if they skipped that process then I could easily see people emptying their stomachs during the process.
New contributor
1
Sorry, but this is incorrect. Imagine swinging a bucket on a rope in a horizontal circle.
– Beta
yesterday
@Beta, well, it depends on which way the station is rotating. You could organize the transition in a logical, non-rollercoaster manner. But you don't have to.
– Arcanist Lupus
yesterday
2
I hope you aren't referring to orientation relative to a planet -- the only linear acceleration on the station would be due to its translational rocket engine.
– amI
yesterday
The tension on a rope on a bucket increases and decreases because you are standing on a planet experiencing its gravitational field. That does not apply in this situation.
– msouth
3 hours ago
add a comment |
up vote
1
down vote
Experiencing rotational forces and fixed direction gravity at the same time would be weird.
A person under the influence of gravity experiences a constant acceleration. A person in a rotating reference frame experiences a constant magnitude acceleration, but the direction is changing constantly.
This means that if you are experiencing both at once, and the axis of rotation is not parallel to the direction of gravity, the total acceleration that you feel will be constantly fluctuating. It's more or less equivalent to the fact that if you swing a bucket on a rope in a vertical circle, the tension in the rope is higher when the bucket is near the ground than when it is at the top of the swing.
Depending on how fast the rotation of your station is, this could make the transition period feel like a rollercoaster.
Of course, the logical way to transition reference frames would be to leave one, enter zero-g, then enter the second. That would avoid the roller coaster effect. But if they skipped that process then I could easily see people emptying their stomachs during the process.
New contributor
1
Sorry, but this is incorrect. Imagine swinging a bucket on a rope in a horizontal circle.
– Beta
yesterday
@Beta, well, it depends on which way the station is rotating. You could organize the transition in a logical, non-rollercoaster manner. But you don't have to.
– Arcanist Lupus
yesterday
2
I hope you aren't referring to orientation relative to a planet -- the only linear acceleration on the station would be due to its translational rocket engine.
– amI
yesterday
The tension on a rope on a bucket increases and decreases because you are standing on a planet experiencing its gravitational field. That does not apply in this situation.
– msouth
3 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
Experiencing rotational forces and fixed direction gravity at the same time would be weird.
A person under the influence of gravity experiences a constant acceleration. A person in a rotating reference frame experiences a constant magnitude acceleration, but the direction is changing constantly.
This means that if you are experiencing both at once, and the axis of rotation is not parallel to the direction of gravity, the total acceleration that you feel will be constantly fluctuating. It's more or less equivalent to the fact that if you swing a bucket on a rope in a vertical circle, the tension in the rope is higher when the bucket is near the ground than when it is at the top of the swing.
Depending on how fast the rotation of your station is, this could make the transition period feel like a rollercoaster.
Of course, the logical way to transition reference frames would be to leave one, enter zero-g, then enter the second. That would avoid the roller coaster effect. But if they skipped that process then I could easily see people emptying their stomachs during the process.
New contributor
Experiencing rotational forces and fixed direction gravity at the same time would be weird.
A person under the influence of gravity experiences a constant acceleration. A person in a rotating reference frame experiences a constant magnitude acceleration, but the direction is changing constantly.
This means that if you are experiencing both at once, and the axis of rotation is not parallel to the direction of gravity, the total acceleration that you feel will be constantly fluctuating. It's more or less equivalent to the fact that if you swing a bucket on a rope in a vertical circle, the tension in the rope is higher when the bucket is near the ground than when it is at the top of the swing.
Depending on how fast the rotation of your station is, this could make the transition period feel like a rollercoaster.
Of course, the logical way to transition reference frames would be to leave one, enter zero-g, then enter the second. That would avoid the roller coaster effect. But if they skipped that process then I could easily see people emptying their stomachs during the process.
New contributor
edited yesterday
New contributor
answered yesterday
Arcanist Lupus
1113
1113
New contributor
New contributor
1
Sorry, but this is incorrect. Imagine swinging a bucket on a rope in a horizontal circle.
– Beta
yesterday
@Beta, well, it depends on which way the station is rotating. You could organize the transition in a logical, non-rollercoaster manner. But you don't have to.
– Arcanist Lupus
yesterday
2
I hope you aren't referring to orientation relative to a planet -- the only linear acceleration on the station would be due to its translational rocket engine.
– amI
yesterday
The tension on a rope on a bucket increases and decreases because you are standing on a planet experiencing its gravitational field. That does not apply in this situation.
– msouth
3 hours ago
add a comment |
1
Sorry, but this is incorrect. Imagine swinging a bucket on a rope in a horizontal circle.
– Beta
yesterday
@Beta, well, it depends on which way the station is rotating. You could organize the transition in a logical, non-rollercoaster manner. But you don't have to.
– Arcanist Lupus
yesterday
2
I hope you aren't referring to orientation relative to a planet -- the only linear acceleration on the station would be due to its translational rocket engine.
– amI
yesterday
The tension on a rope on a bucket increases and decreases because you are standing on a planet experiencing its gravitational field. That does not apply in this situation.
– msouth
3 hours ago
1
1
Sorry, but this is incorrect. Imagine swinging a bucket on a rope in a horizontal circle.
– Beta
yesterday
Sorry, but this is incorrect. Imagine swinging a bucket on a rope in a horizontal circle.
– Beta
yesterday
@Beta, well, it depends on which way the station is rotating. You could organize the transition in a logical, non-rollercoaster manner. But you don't have to.
– Arcanist Lupus
yesterday
@Beta, well, it depends on which way the station is rotating. You could organize the transition in a logical, non-rollercoaster manner. But you don't have to.
– Arcanist Lupus
yesterday
2
2
I hope you aren't referring to orientation relative to a planet -- the only linear acceleration on the station would be due to its translational rocket engine.
– amI
yesterday
I hope you aren't referring to orientation relative to a planet -- the only linear acceleration on the station would be due to its translational rocket engine.
– amI
yesterday
The tension on a rope on a bucket increases and decreases because you are standing on a planet experiencing its gravitational field. That does not apply in this situation.
– msouth
3 hours ago
The tension on a rope on a bucket increases and decreases because you are standing on a planet experiencing its gravitational field. That does not apply in this situation.
– msouth
3 hours ago
add a comment |
up vote
0
down vote
In my opinion there is mechanical difference in which the rotation affects you in those two cases (you rotate on planet while not on poles). On planets surface the mass pulls you inward and the planetary rotation lessens the force applied to you. On the station the rotation works the other way, basically creating gravity from nothing.
Have a nice day.
New contributor
add a comment |
up vote
0
down vote
In my opinion there is mechanical difference in which the rotation affects you in those two cases (you rotate on planet while not on poles). On planets surface the mass pulls you inward and the planetary rotation lessens the force applied to you. On the station the rotation works the other way, basically creating gravity from nothing.
Have a nice day.
New contributor
add a comment |
up vote
0
down vote
up vote
0
down vote
In my opinion there is mechanical difference in which the rotation affects you in those two cases (you rotate on planet while not on poles). On planets surface the mass pulls you inward and the planetary rotation lessens the force applied to you. On the station the rotation works the other way, basically creating gravity from nothing.
Have a nice day.
New contributor
In my opinion there is mechanical difference in which the rotation affects you in those two cases (you rotate on planet while not on poles). On planets surface the mass pulls you inward and the planetary rotation lessens the force applied to you. On the station the rotation works the other way, basically creating gravity from nothing.
Have a nice day.
New contributor
New contributor
answered 20 hours ago
Martin Hasa
1
1
New contributor
New contributor
add a comment |
add a comment |
up vote
0
down vote
Please have a look at "Why don't spaceships have artificial gravity" by SciShow Space on YouTube. It explains the topic way better than I could ever do.
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1
Hi and welcome to Physics Stack Exchange. However, please note that 'link only' answers such as this are generally frowned upon. A good answer should be able to stand 'on its own two feet'. This would probably be more appropriate for a comment.
– Time4Tea
1 hour ago
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
– Time4Tea
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Please have a look at "Why don't spaceships have artificial gravity" by SciShow Space on YouTube. It explains the topic way better than I could ever do.
New contributor
1
Hi and welcome to Physics Stack Exchange. However, please note that 'link only' answers such as this are generally frowned upon. A good answer should be able to stand 'on its own two feet'. This would probably be more appropriate for a comment.
– Time4Tea
1 hour ago
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
– Time4Tea
1 hour ago
add a comment |
up vote
0
down vote
up vote
0
down vote
Please have a look at "Why don't spaceships have artificial gravity" by SciShow Space on YouTube. It explains the topic way better than I could ever do.
New contributor
Please have a look at "Why don't spaceships have artificial gravity" by SciShow Space on YouTube. It explains the topic way better than I could ever do.
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New contributor
answered 3 hours ago
elPolloLoco
101
101
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New contributor
1
Hi and welcome to Physics Stack Exchange. However, please note that 'link only' answers such as this are generally frowned upon. A good answer should be able to stand 'on its own two feet'. This would probably be more appropriate for a comment.
– Time4Tea
1 hour ago
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
– Time4Tea
1 hour ago
add a comment |
1
Hi and welcome to Physics Stack Exchange. However, please note that 'link only' answers such as this are generally frowned upon. A good answer should be able to stand 'on its own two feet'. This would probably be more appropriate for a comment.
– Time4Tea
1 hour ago
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
– Time4Tea
1 hour ago
1
1
Hi and welcome to Physics Stack Exchange. However, please note that 'link only' answers such as this are generally frowned upon. A good answer should be able to stand 'on its own two feet'. This would probably be more appropriate for a comment.
– Time4Tea
1 hour ago
Hi and welcome to Physics Stack Exchange. However, please note that 'link only' answers such as this are generally frowned upon. A good answer should be able to stand 'on its own two feet'. This would probably be more appropriate for a comment.
– Time4Tea
1 hour ago
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
– Time4Tea
1 hour ago
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
– Time4Tea
1 hour ago
add a comment |
Mark Cassidy is a new contributor. Be nice, and check out our Code of Conduct.
Mark Cassidy is a new contributor. Be nice, and check out our Code of Conduct.
Mark Cassidy is a new contributor. Be nice, and check out our Code of Conduct.
Mark Cassidy is a new contributor. Be nice, and check out our Code of Conduct.
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I deleted some off-topic comments.
– David Z♦
14 hours ago