Question

the empire's space station is a long way from any star. it is circular and has a radius of 5.10 km. the angular velocity that is needed to give the station an artificial gravity of 9.80 m/s2 at its circumference is

Answer 1

Answer:

Approximately $0.0438$ radians per second.

Explanation:

In standard units, the radius of this space station will be $r = 5.10\; {\rm km} = 5.10 \times 10^{3}\; {\rm m}$.

Objects at the circumference of this space station will be in a circular motion. Consider an object of mass $m$. As a result of this centripetal motion, the net force on this object will be:

$F_{\text{net}} = m\, \omega^{2}\, r$,

Where:

• $r = 5.10 \times 10^{3}\; {\rm m}$ is the radius of the circular path (same as the radius of the space station), and
• $\omega$ is the angular velocity of the object (same as that of the space station).

If an object of mass $m$ is in a free fall in a gravitational field of magnitude $g = 9.80\; {\rm m\cdot s^{-2}}$, the net force on that object will be equal to $m\, g$, same as the weight of the object.

To emulate the effect of gravity, for objects that are not attached to the walls of the space station, the net force on an object $m$ should also be equal to $m\, g$. In other words:

$F_{\text{net}} = m\, g$.

Equate the two expressions for $F_{\text{net}}$ and solve for angular velocity $\omega$:

$m\, g = F_{\text{net}} = m\, \omega^{2}\, r$.

$\begin{aligned}\omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.80\; {\rm m\cdot s^{-2}}}{5.10 \times 10^{3}\; {\rm m}}} \\ &\approx 0.0438\; {\rm s^{-1}}\end{aligned}$.

(Unit: radians per second.)