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One of the possible ways to simulate gravity in outer space is to have a rotating spaceship, so that the centrifugal force experienced provides a gravity-like force.

My question is: shouldn't this only work when our feet are touching the floor of the spaceship? Only in that case the floor is providing a contact force to balance the centrifugal force.

If we jumped, there is no gravity within the spacecraft so what is it that would make us come back down?

Also: imagine we had a shower: what would make the water fall down?

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If we jumped we would travel in a straight line, but the station floor would orbit (travel in a circle) and thus catch up with us, appearing as if we were pulled down towards the floor. –  ja72 May 12 at 13:19
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Ender's Game. B-) –  jnm2 May 12 at 17:51
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On the Earth you can recreate weightlessness, but then you usually run into the Earth again and regain your weight when you hit the Earth. –  Walter May 12 at 23:12
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How does "Gravitron" or "The Rotor" work at an amusement park? –  Octopus May 12 at 23:55
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Can I blame the absence of "Mary Go Rounds" for peoples lack of intuition when it comes to rotating frames? Since they have disappeared from playgrounds the past 25 years kids grown up with very little experience in rotating dynamics. –  ja72 May 13 at 18:47

5 Answers 5

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If you jumped "straight up", you would still have a horizontal component of velocity (relative to a nonrotating frame), so you would still end up coming "back down".

Likewise, the shower water is moving horizontally in a nonrotating frame, which makes it collide with the floor eventually (since the floor is curving upwards in the nonrotating frame). But to a person on the ship, it looks as if the water was moving downwards, rather than the floor (and you) moving upwards.

More dangerous would be if you were to try to run in the opposite direction of the rotation; if you ran fast enough, you would eventually find that you had become weightless. This would also mean that your feet would no longer be touching the ground, the world would be spinning underneath you, and you'd have no way of getting back down again.

Fortunately, since the air is also moving due to the rotation, the "wind" would eventually "slow you down" (technically it would actually speed you up) and you would eventually regain "gravity" and fall to the ground.

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Interesting comment on running in the wrong direction. For a better workout, run the "right" way. But I don't quite follow your comment on "horizontal component of acceleration". Can you comment further? –  garyp May 12 at 14:20
    
I agree. I don't think you can avoid the "real" centripetal force no matter which way you run. If you go the wrong way, your own tangential velocity will run you into the floor/wall. –  Carl Witthoft May 12 at 14:50
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@garyp: Oops, I meant to say horizontal component of velocity (relative to a nonrotating frame), rather than horizontal component of acceleration, corrected it. –  DumpsterDoofus May 12 at 14:59
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@CarlWitthoft: The centripetal acceleration depends on your body's angular velocity $\dot{\theta}$ as you rotate along with the spaceship. But if you run fast enough in the opposite direction as the spaceship rotates, you can make your angular velocity zero, and so your centripetal force vanishes. In essence, you're stationary (relative to a nonrotating frame) and the spaceship is rotating underneath your feet. –  DumpsterDoofus May 12 at 15:05
    
OK, I see what you meant. –  Carl Witthoft May 12 at 16:51

Tsiolkovsky figured it all out more than 100 years ago :)

Tsiolkovsky describes a cylindrical space ship: 100 meters long, 4 meters in diameter, rotating end-over-end about its "central transverse diameter", with an endpoint velocity between 1 and 10 meters per second, producing an angular velocity between (approximately) 0.2 and 2.0 rotations per minute, and a gravity level between 0.002 and 0.2 g. These numbers were chosen to illustrate a concept, and should not be taken too seriously. Nevertheless, they show that Tsiolkovsky understood the problems associated with high angular velocities, and the practicality of artificial gravity levels of less than one full g.

When you stand on the ground of the station, you are moving at the speed of the endpoint velocity, let's say 10 m/s. When you jump at the speed of 1 m/s, your velocity becomes $\sqrt{1+100}$, but its direction is slightly inwards now. So, you're going to hit the floor again, but slightly off the point from where you jumped. So, you really are not pulled by the ground, but rather get hit by the ground.

UPDATE: here's the code and a plot to demonstrate what happens when a man jumps vertically at speed of 1 m/s, inside the spaceship of 100 m radius and endpoint velocity 10 m/s. The result is that a man will land 13 cm off the point he was initially in about 2 secs. enter image description here

R=100;

ax0=0
ay0=-R;

vy=1;
vx=10;


phi=vx/(R) % angular speed
theta = (pi-2*( pi/2-atan(vy/vx)) ) % angle at hitting floor

bxf = R*sin(theta);
tau = bxf / vx % time to hit floor

axf = tau*phi; % angle the weel turns
d=(theta-axf)*R %distance from hit

t=(0:0.01:tau);
bx=ax0+vx*t;
by=ay0+vy*t;
plot(bx,by,'r')

hold on

ax=R*sin(phi*t);
ay=-R*cos(phi*t);
plot(ax,ay)


tau2 = theta/phi 

t=(tau:0.01:2*tau2);
ax=R*sin(phi*t);
ay=-R*cos(phi*t);
plot(ax,ay,'-.c')
xlabel 'x'
ylabel 'y'
title 'R=100;v_x=10,v_y=1'
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A man jumps. A man will land. Valar morghulis. –  Michael Borgwardt May 13 at 14:16

Don't forget that not only floor is spinning, but also the rest of the ship (including water in pipes and you).

This causes that the water is pulled "to the ground". Since the acceleration is given by $a=v^2/r$ and $v=\omega*r$ you got $a=(\omega*r)^2/r$. That gives you $a=\omega^2*r$.

This means that anything spinning around the axis of the ship is accelerated outwards proportionally to the distance from the axis.

So if the shower is on the ceiling, water will accelerate downwards, but with increasing acceleration.

And remember, when you jump, you still have the velocity given you by the previous contact with floor, so after jump, you don't go just "up" but also "forwards" that means you're still spinning around the axis.

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Einstein said it: Acceleration is gravity and gravity is acceleration. When in contact with the floor you are accelerating (rotating) with respect to an inertial frame of reference and one can interpret that as a gravitational field. That's why you feel the force of your mass at your feet. You can juggle balls, do an experiment or make a measurement, everything just like in gravity.

When you jump you are free falling that's why you feel weightlessness. Your velocity is constant with respect to a non rotating frame. But your spaceship is still accelerating so it catches up with you. That is unless the force you apply when you jump is carefully chosen so that you cancel the rotation speed of the ship at the jump point. Then w.r.to the non rotating frame you are moving with constant velocity in a straight line towards the center of rotation. Then you will effortlessly reach it soon enough and the whole spaceship will be rotating around you!

Similarly with water exiting a rotating pipe: It's in free fall but you are accelerating into the water. Of course it will come down at an angle that depends on the speed of rotation but if you position yourself accordingly you will get a full shower!

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So you aren't really simulating gravity, you are making it then? –  PyRulez May 13 at 21:52
    
Gravity is a force which causes acceleration. Any force which causes acceleration will "feel" like gravity. However, gravity is usually defined as a force of attraction between two masses. The effect is the same, but the cause is different. –  daviewales May 14 at 4:53

Let's say you're in a large sphere or cylinder that's rotating. You're floating at the center, and there's no atmosphere to drag you in any direction. Assume that the structure is light enough that gravitational pull due to mass is negligible (and would cancel out if symmetrical). I would assume that you're not going to be moved in any particular direction. Now, with a short burst of thruster, you slowly start moving towards the outer wall. Do you experience any additional acceleration due to the "artificial gravity" before you make contact with the outer wall? I would think not. Do you suddenly accelerate sideways at contact?

If I spin a bucket of water, the water is moving instantaneously tangential to me (and would fly off in a more or less straight line if I let go). I am applying centripetal force to the bucket and water, to accelerate it towards me, and the reactive centrifugal force of the water keeps it in the bucket. So, once I am in contact with the spinning outer wall, I would be constantly accelerated towards the center, and my equal and opposite reaction would feel like I'm being pressed against the outer wall/floor ("gravity").

So if I jump up, what happens? What is accelerating me outwards back towards the outer wall ("floor")? I will of course have some sideways (tangential) velocity from the spinning wall, which will make me drift in the direction of the wall's spin (Coriolis Effect) and eventually I should hit the floor again, but not in the same place I jumped from. That is, "down" is not well defined.

This would also apply to water released from a showerhead towards "down" (the outer wall). If a blob of water seeped out of the showerhead with no velocity in any particular direction, what (if anything) would make it move "down"? In a microgravity environment such as the ISS, they use fans to ensure movement. If it initially had some velocity "down", I would expect it to not accelerate downwards because there is nothing acting on it once it leaves the showerhead. It would be moving "too slowly" and curve off to the side of the shower stall due to Coriolis Effect, but other than that?

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You forget that the water droplet will travel at a tangent to the rotating spaceship. In order for a water droplet to have 0 velocity, you would have to throw it in the opposite direction of the ship's spin. Then it would have no "downwards" acceleration, but it would quickly collide with doors / walls, and then it would "learn" where down was again. –  daviewales May 14 at 5:02
    
So we have a water droplet that dribbled out of a leaky showerhead and has detached due to evaporation or something. It was rotating with the ship while attached to the showerhead. Neglecting effects of atmosphere sweeping it along or otherwise modifying its trajectory, where is it going to go? Will it keep on a straight line and collide with the wall of the shower stall, once it's detached and the showerhead can no longer drag it along (apply a force)? I'm trying to understand the concepts here one small step at a time. Maybe a grain of dust in a vacuum would be a better starting point. –  Phil Perry May 14 at 13:57
    
Objects will always travel in straight lines, unless acted on by an external force. (Newton's 1st law) This means that things will only travel in a circle if they are forced to do so. In our shower, the force of surface tension causes the water drop to stay on the shower head, and the force of the shower head causes the drop to travel in circular motion. If the water drop grows too large, the force of surface tension is not strong enough to hold the water droplet, so it is released to travel in a straight line. This straight line intersects with the wall of the ship. –  daviewales May 14 at 14:55
    
As the water droplet was launched from a small circle closer to the centre of the ship, as it travels in a straight line, it moves into larger and larger circles until it reaches the outside wall of the ship. All the circles have the same angular velocity, but the outer circles have greater tangential velocity, because they have to move further in the same amount of time, because they are larger. Once the drop leaves the shower, it will not accelerate when compared with the stationary frame. However, in relation to the rotating frame, it will appear to accelerate towards the outside wall. –  daviewales May 14 at 15:09
    
In addition to travelling "down", it will also travel slightly "sideways", because its initial tangential velocity is less than the the tangential velocity at the edge of the ship. –  daviewales May 14 at 15:16

protected by Qmechanic May 13 at 18:34

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