# Jumping in artificial gravity

1. Dec 21, 2009

### Null-Set

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 $$v_{t}$$. The jumper then jumps and leaves the ground with radial velocity $$v_{r}$$, 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 $$\sqrt{v^{2}_{t} + v^{2}_{r}}$$ and the starting point is moving along a curved path at a speed $$|v_{t}|$$. 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?

2. Dec 21, 2009

### willem2

You're completely right. Another way to see it is conservation of angular momentum/ $r v_t$ is constant for you and for the cylinder. Since your r is smaller while you jump, $v_t$ must be larger, and your angular speed, wich is $v_t/r$ 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. Dec 23, 2009

### Cleonis

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

Last edited by a moderator: May 4, 2017