# Artificial Gravity and a Rotating Space Station

1. May 31, 2009

### whaaat919

lakjdlkasdlkas

Last edited: May 31, 2009
2. May 31, 2009

### HallsofIvy

Staff Emeritus

This can be modeled with the equations [/itex]x= 500 cos(\omega t)[/itex], $y= 500 sin(\omega t)$ where $\omega$ is the angular velocity. Differentiating to get the velocity, $v_x= -500\omega sin(\omega t)$, $v_y= 500\omega sin(\omega t)$. The acceleration is the derivative of that, $a_x= -500 \omega^2 cos(\omega t)$, $a_y= -500 \omega^2 sin(\omega t)$. The magnitude of the acceleration is $\sqrt{(-500 \omega^2 cos^2(\omega t))^2+ (500 \omega^2sin^2(\omega t)}$$= 500\omega^2$ and that must be equal to the gravity of earth, 9.81 m/s2. Set them equal and solve for $\omega$. To find the diameter at which to set the inner ring, go back to the original equation with that value for
$\omega$, a variable, r, in place of the 500, do the same calculations, set equal to the gravitational acceleration on mars and solve for r.

3. May 31, 2009

### nrqed

I woudl make a free body diagram of a person standing on the space station. There is only one force acting on the person. for a person to feel as if they were on Earth, that force must be equal to mg. Now use Newton's second law, so you get mg = ma.
For the acceleration, use the formula for centripetal acceleration. You will then find the speed easily