Jumping in artificial gravity

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Main Question or Discussion Point

Gravity in space is best simulated by rotation. If I were inside a rotating cylinder facing the direction of rotation (i.e. a window at my feet would show objects coming into view from the top of the window) and jumped straight into the air, would I land ahead of where I started or behind where I started?

I read that you should land behind where you started, which does not make sense to me. I think that you do drift as you jump due to Coriolis acceleration, but I never learned about that so I tried to work it all out in the inertial frame. The jumper and the space station start with a tangential velocity [tex]v_{t}[/tex]. The jumper then jumps and leaves the ground with radial velocity [tex]v_{r}[/tex], from his perspective. Now that he is in free fall he will travel in a straight path until he intersects the edge of the cylinder again. Meanwhile, the cylinder rotates beneath him. The jumper is following a straight path at a speed [tex]\sqrt{v^{2}_{t} + v^{2}_{r}}[/tex] and the starting point is moving along a curved path at a speed [tex]|v_{t}|[/tex]. The jumper will intersect the cylinder again before the starting point rotates to that point, because the jumper is moving on a shorter path at a greater speed. Thus, the jumper lands ahead of where he started after jumping straight up.

Did I make any mistakes there?
 

Answers and Replies

  • #2
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You're completely right. Another way to see it is conservation of angular momentum/ [itex] r v_t [/itex] is constant for you and for the cylinder. Since your r is smaller while you jump, [itex] v_t [/itex] must be larger, and your angular speed, wich is [itex] v_t/r [/itex] is also larger than that of the cylinder.

If you jump on the rotating earth, you will end up a little bit behind, because your r will be larger while you jump.
 
  • #3
Cleonis
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Gravity in space is best simulated by rotation. If I were inside a rotating cylinder facing the direction of rotation (i.e. a window at my feet would show objects coming into view from the top of the window) and jumped straight into the air, would I land ahead of where I started or behind where I started?

I read that you should land behind where you started, which does not make sense to me. I think that you do drift as you jump due to Coriolis acceleration, but I never learned about that so I tried to work it all out in the inertial frame. The jumper and the space station start with a tangential velocity [tex]v_{t}[/tex]. The jumper then jumps and leaves the ground with radial velocity [tex]v_{r}[/tex], from his perspective. Now that he is in free fall he will travel in a straight path until he intersects the edge of the cylinder again. Meanwhile, the cylinder rotates beneath him. The jumper is following a straight path at a speed [tex]\sqrt{v^{2}_{t} + v^{2}_{r}}[/tex] and the starting point is moving along a curved path at a speed [tex]|v_{t}|[/tex]. The jumper will intersect the cylinder again before the starting point rotates to that point, because the jumper is moving on a shorter path at a greater speed. Thus, the jumper lands ahead of where he started after jumping straight up.
You can verify your reasoning visually with the following Java applet that is on my website:
http://www.cleonis.nl/physics/ejs/spacestation_vertical_throw_simulation.php" [Broken]
Which has exactly the setup you are describing here
The applet allows you to vary several settings, so you can explore a range of cases.

Your reasoning is correct: if the jumper jumps up perpendicular to the local cilinder wall, then his velocity relative to the inertial frame is the vector sum of the tangential velocity of co-rotating with the cilinder, and the radial velocity of the jump. Hence with such a jump you will allways land ahead of the spot where you jumped from.

Cleonis
http://www.cleonis.nl
 
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