Question

A cylindrical space colony 8 km in diameter and 30 km long has been proposed as living quarters for future space explorers.

Such a habitat would have cities, land and lakes on the inside surface and air and clouds in the center. All this would be held in place by the rotation of the cylinder about the long axis.

How fast would such a cylinder have to rotate to produce a 1-g gravitational field at the walls of the cylinder?

Answer 1

`\omega=0.05rad/s`

Step-by-step explanation:

The gravitational field (fictitious acceleration) produced would be equal to the centripetal acceleration needed to move as a circle (acceleration of the system). The equation for centripetal acceleration is
`a_(cp)=r\omega^2`, which means that the angular velocity of the cylindrical space colony, which has a radius r=4km and needs to have a gravitational field of 1-g (
`9.8m/s^2`) needs to be:

`\omega=\sqrt{(a_(cp))/(r)}=\sqrt{(9.8m/s^2)/(4000m)}=0.05rad/s`