Calculating Artificial Gravity for a Rotating Space Station

[Paul Wilson]

Hi

As I'm sure we're all aware, there are many planned missions to Mars.

Given its distance (some 3 years away given today's tech) by the time any humans make it there, their muscles would be completely useless on account of wastage as a result of no gravity.

Therefore, artificial gravity is proposed in the form of a rotating spaceship.

Though that got me thinking; would it actually work?

I think it wouldn't.

Let's say a ship is moving at a constant along a vector (x axis) at 5000 m/s relative to the launch pad. Everyone in that ship, relative to the ship, is moving at 0 m/s. Then someone jumps along the same vector at 1 m/s. That person is then moving at 5001 m/s relative to the launch pad.

The Captain decides to begin revolving the ship at 10rpm along the x axis.

During that acceleration period from revolving at 0rpm to 10rpm, everyone is 'stuck to the floor' as the ship has greater velocity than they do.

And this is where my thinking comes into play.

The ship has reached its maximum revs/minute along the x-axis, at 10rpm, its velocity still at 5000 m/s.

Surely anyone 'stuck to the floor' is now ALSO moving 10 m/s relative to the ship and thus if they were to jump at 1/ms would then be moving at 1 m/s relative to the ship thus cancel out the 'artificial-gravity'

Or in other words, artificial gravity is impossible unless the ship is under constant acceleration which, as we know, is impossible given the effects of mass at great speeds.

Thoughts?

 

[ash64449]

Hai Paul Wilson,
I don't know much about artificial gravity but according to equivalence principle, acceleration is indistinguishable from gravitational field. So if the rocket were to accelerate at a constant rate of 10m/s^2,then the persons in the rocket will feel just like gravity on surface of the earth. So do you think " by the time any humans make it there(mars), their muscles would be completely useless"?

 

[ash64449]

Or in other words, artificial gravity is impossible unless the ship is under constant acceleration which, as we know, is impossible given the effects of mass at great speeds.

Yes,keeping the acceleration constant is impossible for now because no fuel has been invented that can help the rocket reach higher velocity that is required to reach the mars.

 

[Paul Wilson]

Indeed if a constant acceleration of 10m/s^2 were POSSIBLE then gravity would be felt. But we all know that that is not possible over such a great distance. The energy does not exist to constantly accelerate at that rate, therefore the ship must stop accelerating at some point in order to A) conserve energy or B) it has completely run out of it.

The theory I saw proposed revolving at a constant rate to create it. Which I derived to be nonsense. That same theory I described.

EDIT: Ash got there microseconds before me!

 

[ash64449]

Hi

As I'm sure we're all aware, there are many planned missions to Mars.

Given its distance (some 3 years away given today's tech) by the time any humans make it there, their muscles would be completely useless on account of wastage as a result of no gravity.

Therefore, artificial gravity is proposed in the form of a rotating spaceship.

Though that got me thinking; would it actually work?

I think it wouldn't.

Let's say a ship is moving at a constant along a vector (x axis) at 5000 m/s relative to the launch pad. Everyone in that ship, relative to the ship, is moving at 0 m/s. Then someone jumps along the same vector at 1 m/s. That person is then moving at 5001 m/s relative to the launch pad.

The Captain decides to begin revolving the ship at 10rpm along the x axis.

During that acceleration period from revolving at 0rpm to 10rpm, everyone is 'stuck to the floor' as the ship has greater velocity than they do.

And this is where my thinking comes into play.

The ship has reached its maximum revs/minute along the x-axis, at 10rpm, its velocity still at 5000 m/s.

Surely anyone 'stuck to the floor' is now ALSO moving 10 m/s relative to the ship and thus if they were to jump at 1/ms would then be moving at 1 m/s relative to the ship thus cancel out the 'artificial-gravity'

How that cancel out artificial gravity? Acceleration is identifiable in every reference frame.

 

[Paul Wilson]

How that cancel out artificial gravity? Acceleration is identifiable in every reference frame.

How? Explain

 

[ash64449]

EDIT: Ash got there microseconds before me!

HAHA! yeah!

 

[ash64449]

How? Explain

What should i explain? acceleration is identifiable in every reference frame?

 

[Paul Wilson]

What should i explain? acceleration is identifiable in every reference frame?

If you are moving at a constant 10m/s in a spaceship and jump at 1m/s you are then moving at 1m/s relative to the spaceship and are therefore no longer under acceleration.

The theory I saw did not propose constant acceleration, merely a constant velocity.

 

[ash64449]

If you are asking how acceleration is identifiable in every reference frame,let me give you an example.

Suppose an object is moving at 10 meters per second relative to me and 20 meters per second relative to you. IF object were to accelerate and increase its speed to 20 meters per second relative to me,then it would also accelerate and reach at 30 meters per second relative to you. So i know that object accelerated at 10 meters per second per second and you know that object accelerated at 10 meters per second per second. So in any reference frame,object will accelerate at 10 meters per second per second,SO we can say that acceleration is identifiable in every reference frame.

And if rocket were to accelerate,person inside would feel same as gravity. say in Earth that person jumped at 1m/s. wouldn't the person fall to the ground afterwards? Does the jumping cancel out Earth's gravity? NO.right? since acceleration is identifiable and acceleration is indistinguishable from gravitational field,same would be the case if rocket were to accelerate.

 

[ash64449]

If you are moving at a constant 10m/s in a spaceship and jump at 1m/s you are then moving at 1m/s relative to the spaceship and are therefore no longer under acceleration.

The theory I saw did not propose constant acceleration, merely a constant velocity.

I agree with you on that.that person is moving 1m/s relative to spaceship.

Sorry,i don't know much about the gravity that is created when an object rotates and i do know that Einstein's general theory of relativity does say that a rotating body can itself create a feeling of gravity. I think other posters would give answer to your post.

 

[Nugatory]

Or in other words, artificial gravity is impossible unless the ship is under constant acceleration

Rotation IS constant acceleration, which is why it does produce a satisfactory artificial gravity.

Google around for the carnival rides called "Rotor" and "Graviton", or try this article on wikipedia: http://en.wikipedia.org/wiki/Rotor_(ride). These are devices that use rotation to apply constant acceleration to people

You may object that these devices are spinning but not moving... but they are moving, at many miles a second, because they're attached to the Earth which is orbiting the sun, which is rotating around the Milky Way galaxy, which is itself moving through space.

 

[Paul Wilson]

If you are asking how acceleration is identifiable in every reference frame,let me give you an example.

Suppose an object is moving at 10 meters per second relative to me and 20 meters per second relative to you. IF object were to accelerate and increase its speed to 20 meters per second relative to me,then it would also accelerate and reach at 30 meters per second relative to you. So i know that object accelerated at 10 meters per second per second and you know that object accelerated at 10 meters per second per second. So in any reference frame,object will accelerate at 10 meters per second per second,SO we can say that acceleration is identifiable in every reference frame.

And if rocket were to accelerate,person inside would feel same as gravity. say in Earth that person jumped at 1m/s. wouldn't the person fall to the ground afterwards? Does the jumping cancel out Earth's gravity? NO.right? since acceleration is identifiable and acceleration is indistinguishable from gravitational field,same would be the case if rocket were to accelerate.

Indeed, I agree, to quote your last sentence, if rocket were to accelerate but I reiterate myself AGAIN, the theory I saw did not propose constant acceleration, it theorized a constant velocity. It was not under constant acceleration but instead a constant velocity.

Therefore, there would be no artificial gravity to those on board.

Please re-read; I made it quite clear the ship was not under acceleration relative to the launchpad once it had finished its revolution sequence but was instead moving at a constant rate. Thus there would be no apparent effects of gravity to anyone onboard.

 

[PeterDonis]

Or in other words, artificial gravity is impossible unless the ship is under constant acceleration

This is not correct. Spinning the ship can indeed create an effect that is similar to the effect gravity creates when you stand on the surface of the Earth.

Suppose the ship is moving at a constant velocity in the x direction (so that there is no linear acceleration present). Suppose the ship is disc-shaped, and is oriented perpendicular to its linear motion (i.e., the outer rim of the ship is a circle in the y-z plane). And suppose the ship is spinning about the x-axis (meaning the x-axis goes through the center of the disc and is perpendicular to it), so it is rotating in the y-z plane.

Then a person inside the ship can stand on the inner surface of its outer rim just as they could on the surface of the Earth, and feel the same weight there, if the ship's diameter and rotation rate are set appropriately. This is because, relative to someone sitting at the center of the disc (i.e., on the x-axis), a person standing on the outer rim will be accelerating inward, towards the center, because the outer rim of the ship is pushing on them (just as the surface of the Earth pushes up on you when you're standing on it). Note that a person at the x-axis will *not* feel weight; they will be in free fall.

 

[Paul Wilson]

Rotation IS constant acceleration, which is why it does produce a satisfactory artificial gravity.

Google around for the carnival rides called "Rotor" and "Graviton", or try this article on wikipedia: http://en.wikipedia.org/wiki/Rotor_(ride). These are devices that use rotation to apply constant acceleration to people

You may object that these devices are spinning but not moving... but they are moving, at many miles a second, because they're attached to the Earth which is orbiting the sun, which is rotating around the Milky Way galaxy, which is itself moving through space.

Indeed. I didn't take into account the effects of relativity from the Sun, Solar System, etc.

But do those carnival rides not rely upon the gravity of earth? Which of course, once we leave the Earth's gravitational influence, the Sun's, assuming we are in orbit around it, becomes relatively little if in the same carnival ride...?

PS, I'm off to bed now. No replies until tomorrow from me.

 

[Nugatory]

But do those carnival rides not rely upon the gravity of earth?

Only to the extent that if gravity didn't hold them on the surface of the Earth they'd float off into space and then the operators wouldn't be able to make money by selling rides to people on the surface of the earth. The force that pins you to the walls of a Rotor is completely independent of the Earth's gravity; it just depends on the diameter and rotational speed of the machine.

 

[Wes Tausend]

Paul,

Suppose that there was a merry-go-round (round-about) attached in this spaceship. On earth, passengers could ride around on it and experience an addition to the real gravity they were feeling as the ship was parked. They would literally seem to weigh more than their real weight that same morning standing "still" upon a bathroom scale.

If you have not experienced an aggressive merry-go-round ride as a child (example, someone did not push it so fast you nearly slung off) then you are missing an important part of lifes experiences, one that would namely let you see how rotational centrifugal force emulates that of gravity. The artificial gravity may be stronger or weaker than real gravity (depending on rpms), but there will be the essence just the same.

On this spaceship merry-go-round, a ride may be taken in outer space when the passenger is normally totally weightless. But when he is spun around, he will be tend to be thrown to the outside, and suddenly the sensation will be familiar; that of gravity. If the merry-go-round is spun at exactly the right rpm, he will seem to weigh exactly what he weighed on earth, only outward rather than downward. But why use a small merry-go-round when a large portion of the ship may be spun. Then everyone can enjoy the artificial gravitational benefits at once.

The spin is everything in space; the steady forward motion towards Mars makes no difference at all, just like when the ship was steadily parked before take-off. Granted, any ship acceleration will briefly interfere, but in most cases it will simply add to the total artificial gravity. This no different than when we sit in a car and it accelerates. We weigh both the same as real gravity, plus weigh a little more, because of the additional acceleration.

Were you aware that you seemed to weigh more in an accelerating car? It's true. You could put a swing in the back of a pick-up and stand on a scale placed on the seating board while parked. You will weigh the same as in the house. But when the pick-up takes off, the swing will swing back, and you will measure an additional weight during a period of acceleration (or braking, deceleration!). The same thing would occur if you swung around in a circle from a maypole and weighed yourself with a scale because it is just like a merry-go-round. I hope you can see this now. Good luck in solving this for yourself.

Wes
...

 

[ash64449]

Indeed, I agree, to quote your last sentence, if rocket were to accelerate but I reiterate myself AGAIN, the theory I saw did not propose constant acceleration, it theorized a constant velocity. It was not under constant acceleration but instead a constant velocity.

Therefore, there would be no artificial gravity to those on board.

Please re-read; I made it quite clear the ship was not under acceleration relative to the launchpad once it had finished its revolution sequence but was instead moving at a constant rate. Thus there would be no apparent effects of gravity to anyone onboard.

But rotation is itself constant acceleration. it is indistinguishable from gravity.

 

[ash64449]

Rotation IS constant acceleration, which is why it does produce a satisfactory artificial gravity.

Google around for the carnival rides called "Rotor" and "Graviton", or try this article on wikipedia: http://en.wikipedia.org/wiki/Rotor_(ride). These are devices that use rotation to apply constant acceleration to people

You may object that these devices are spinning but not moving... but they are moving, at many miles a second, because they're attached to the Earth which is orbiting the sun, which is rotating around the Milky Way galaxy, which is itself moving through space.

This is a satisfactory answer. But i have a doubt.

if we are present in a body(for example a big disk like shaped one) and if it were to rotate,won't the person in that body feel a force pushing outwards(Centrifugal force)?

isn't that effect opposite of gravity? Isn't the same case as that of introduced by poster?

 

[ash64449]

This is because, relative to someone sitting at the center of the disc (i.e., on the x-axis), a person standing on the outer rim will be accelerating inward, towards the center, because the outer rim of the ship is pushing on them

i couldn't understand this point.if the outer rim pushed the person outward how can be accelerating inwards?( isn't Centrifugal force taking here when a body rotates?)

 

[Nugatory]

This is a satisfactory answer. But i have a doubt.

if we are present in a body(for example a big disk like shaped one) and if it were to rotate,won't the person in that body feel a force pushing outwards(Centrifugal force)?

isn't that effect opposite of gravity? Isn't the same case as that of introduced by poster?

Inside a rotating space station, a dropped object will fall towards the outside instead of towards the center. If you are inside, standing with your feet on the skin of the station and your head pointing inwards towards the center, you will feel as if you are standing on the surface of a planet or the floor of Einstein's accelerating elevator.

 

[ash64449]

Inside a rotating space station, a dropped object will fall towards the outside instead of towards the center. If you are inside, standing with your feet on the skin of the station and your head pointing inwards towards the center, you will feel as if you are standing on the surface of a planet or the floor of Einstein's accelerating elevator.

yes.nugatory.. I agree.. But i couldn't grasp the second point.. Nor can i understand it.. Can you tell me an easy familiar example and some other application of physics to explain me that fact?

 

[Bandersnatch]

i couldn't understand this point.if the outer rim pushed the person outward how can be accelerating inwards?( isn't Centrifugal force taking here when a body rotates?)

The outer rim does not push outwards, but inwards. The push provides the centripetal force needed to keep the person standing there in a circular trajectory around the centre of rotation.

The centrifugal force only appears if you choose your reference frame to rotate along with the rim(it's a fictitious force; it disappears in inertial reference frames). I.e., now you're the person standing on the rim. From this perspective the whole shebang is not rotating, but you do feel a force pushing on your feet(upwards). You know that you're not moving due to this push, so you reckon there must be another force pushing you downwards and cancelling the previous one - you call it the centrifugal force.

Switch back to the non-rotating frame(e.g., outside the ship) and now the centrifugal force dissapears, and the only force acting on the person standing on the rim is the rim pushing on their feet. From this point of view you can clearly see that this force does actually cause movement, as the tangential velocity around the centre of rotation is constantly changing direction.

 

[Dale]

This is a satisfactory answer. But i have a doubt.

if we are present in a body(for example a big disk like shaped one) and if it were to rotate,won't the person in that body feel a force pushing outwards(Centrifugal force)?

isn't that effect opposite of gravity? Isn't the same case as that of introduced by poster?

i couldn't understand this point.if the outer rim pushed the person outward how can be accelerating inwards?( isn't Centrifugal force taking here when a body rotates?)

They were describing the same system from two different reference frames.

In the ROTATING frame: the spacecraft is stationary and there exists a fictitious force pointing outwards. This force is exactly canceled out by the inwards normal force acting on the astronaut, so he is at rest also. The force is not canceled out on the ball that the astronaut dropped so it accelerates towards the floor.

In the INERTIAL frame: the spacecraft is rotating and there are no fictitious forces pointing outwards. The inwards normal force acting on the astronaut is not canceled, so the astronaut is rotating along with the spacecraft also. There are no forces acting on the ball so it travels in a straight line at a constant velocity and the floor accelerates towards it.

 

[ash64449]

They were describing the same system from two different reference frames.

In the ROTATING frame: the spacecraft is stationary and there exists a fictitious force pointing outwards. This force is exactly canceled out by the inwards normal force acting on the astronaut, so he is at rest also. The force is not canceled out on the ball that the astronaut dropped so it accelerates towards the floor.

In the INERTIAL frame: the spacecraft is rotating and there are no fictitious forces pointing outwards. The inwards normal force acting on the astronaut is not canceled, so the astronaut is rotating along with the spacecraft also. There are no forces acting on the ball so it travels in a straight line at a constant velocity and the floor accelerates towards it.

i cannot visualise the following fact. How floor accelerate towards the ball?
If i have a ball,i and the ball are rotating with the spaceship.if i release the ball,how floor accelerate towards ball if the floor is rotating?

 

[Dale]

i cannot visualise the following fact. How floor accelerate towards the ball?
If i have a ball,i and the ball are rotating with the spaceship.if i release the ball,how floor accelerate towards ball if the floor is rotating?

In the INERTIAL frame: at the moment of release the ball no longer has any forces acting on it so it continues in a straight line at a constant velocity. That initial velocity is tangent to the circle. Initially, the floor and the ball are both traveling in the same direction (tangent) but the floor is undergoing uniform circular acceleration, so its path curves inward and collides with the ball.

 

[ash64449]

In the INERTIAL frame: at the moment of release the ball no longer has any forces acting on it so it continues in a straight line at a constant velocity. That initial velocity is tangent to the circle. Initially, the floor and the ball are both traveling in the same direction (tangent) but the floor is undergoing uniform circular acceleration, so its path curves inward and collides with the ball.

i think i have a misunderstanding about centripetal force and centrifugal force. I didn't think them in terms of upwards and downwards instead i thought them as pushing inwards and pushing outwards. So only thinking them in terms of pushing upwards and pushing downwards can help explain this? Correct?

 

[DrStupid]

But rotation is itself constant acceleration. it is indistinguishable from gravity.

It is distinguishable by the Coriolis force. Apart from negligible relativistic effects gravity is independent from velocity.

 

[Nugatory]

i think i have a misunderstanding about centripetal force and centrifugal force. I didn't think them in terms of upwards and downwards instead i thought them as pushing inwards and pushing outwards. So only thinking them in terms of pushing upwards and pushing downwards can help explain this? Correct?

"Down" is defined as the direction in which a dropped object will fall. So if you're standing on the surface of the earth, "down" points towards the center of the earth, and of you drop something falls towards the surface.

If you're standing on the inside skin of a rotating space station, "down" points outward from the center because that's the direction a dropped object will fall.

 

[ash64449]

"Down" is defined as the direction in which a dropped object will fall. So if you're standing on the surface of the earth, "down" points towards the center of the earth, and of you drop something falls towards the surface.

If you're standing on the inside skin of a rotating space station, "down" points outward from the center because that's the direction a dropped object will fall.

oh.. This does helps to clear my doubt! thank you! You are a great adviser too!

 

[Dale]

i think i have a misunderstanding about centripetal force and centrifugal force. I didn't think them in terms of upwards and downwards instead i thought them as pushing inwards and pushing outwards. So only thinking them in terms of pushing upwards and pushing downwards can help explain this? Correct?

The term "down" means "in the direction of the fictitious force". Since the fictitious force is "outwards" that means that "outwards" IS "down".

Edit: I see Nugatory answered essentially the same.

 

[ash64449]

The term "down" means "in the direction of the fictitious force". Since the fictitious force is "outwards" that means that "outwards" IS "down".

Edit: I see Nugatory answered essentially the same.

yes.. i was waiting for this! thank you DaleSpam for keeping patience.. i thought you would go out of patience by my questions...

 

[ash64449]

The term "down" means "in the direction of the fictitious force". Since the fictitious force is "outwards" that means that "outwards" IS "down".

Edit: I see Nugatory answered essentially the same.

Actually,i also figured it out when i was offline and went to say it to you.But i saw nugatory say the same!

 

[BugsBunny]

Ok Sorry but this all sounds totally looney-toons:

First, even if you could make a large enough space object where someone could stand up straight while it was turning underneath them, (anyone who ever spun a bicycle wheel by it's axle knows how hard it is to keep the wheel spinning in one plane.) Should there be a shift in MASS ie people moving around, especially with no friction this object would certainly become unstable in short order and simply disintegrate under all the extreme forces

Secondly, a merry-go-round on Earth already has gravity keeping it's occupants seated ... in space, objects while possibly moving in the same direction are not going to magically interact as if there were gravity ... the expectation would be in the case of a hollow circular tube that if you started the tub turning, any objects not tied down would not move with respect to the tube ... therefore any objects and or people would not be able to experience any of the artificial forces unless they were attached ... some might suggest magnetic boots could be used to achieve this

Third, even if you could build a ginormous spinning object big enough where the movement of people would be insignificant to its balance, there would be other issues of motion that a person would need to overcome ... moving in the direction of spin would probably be fine but any motion against the direction of spin would most likely be very disorienting ... being on a ship in heavy seas comes to mind

 

[Nugatory]

(anyone who ever spun a bicycle wheel by it's axle knows how hard it is to keep the wheel spinning in one plane.) Should there be a shift in MASS ie people moving around, especially with no friction this object would certainly become unstable in short order and simply disintegrate under all the extreme forces

What keeps the Earth spinning stably (stably enough) on its axis then?

The difficulty with holding a spinning bicycle wheel in one plane comes from precessional effects that are caused by the Earth's gravity acting on the wheel. In free fall, the spinning wheel would tend to remain in one plane and resist any effort to move it out of that plane.

ts while possibly moving in the same direction are not going to magically interact as if there were gravity ... the expectation would be in the case of a hollow circular tube that if you started the tub turning, any objects not tied down would not move with respect to the tube ... therefore any objects and or people would not be able to experience any of the artificial forces unless they were attached ... some might suggest magnetic boots could be used to achieve this

Any object that has any radial velocity at all will eventually move into contact with the wall. When it does, friction between object and the rotating wall will accelerate the object tangentially until it has the same velocity as the wall and the centripetal acceleration will hold the object there in the same way that gravity holds an object on the surface of the earth.

[quote[moving in the direction of spin would probably be fine but any motion against the direction of spin would most likely be very disorienting [/QUOTE]

Why? In fact, what experiment might you use to discover if you were moving with the spin instead of against it? Before you answer this question, you might want to consider whether you feel more disorientation moving to the west against the Earth's rotation, than you do when moving to the east with the Earth's rotation.