Q53E - University Physics with Modern Physics

Question

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is , how many revolutions per minute are needed for the “artificial gravity” acceleration to be $\mathbf{9}\mathbf{.}\mathbf{80}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$ ? (b) If the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravity on the Martian surface (${\mathbf{\text{3.70 m/s}}}^{\mathbf{2}}$). How many revolutions per minute are needed in this case?

Step-by-Step Solution

Answer

(a) $1.5 rad/min$

(b) 0.92rad/min

Step 1: Given data

Diameter of the space station is $d=800m$

Acceleration due gravity on earth  $g_E=9.80 m/s^2$

Acceleration due gravity on moon $g_M=3.70m/s^2$

Step 2: Concept of centripetal acceleration

Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's centre.

Centripetal acceleration is given by
$a_c=\frac{v^2}{r}$                                                                                                                       …(i)

Where,  v is the velocity and r is the radius

Step 3: (a) Determining the revolutions per minute that are needed for the “artificial gravity”

The linear velocity for a rotation object is given by
 $v=\omega r$                                                                                                                         …(ii)

Where , $\omega$ is angular velocity

From equation (i) and equation (ii)

   $$\begin{gathered} a_c=\frac{{\left(\omega r\right)}^2}{r} \\ ={\omega }^{2}r \end{gathered}$$                                                                                                                …(iii)

So from equation (iii) equal the centripetal acceleration and the free-fall acceleration of earth

$$\begin{gathered} g_E={\omega }^{2}r \\ ={\omega }^2\times \left(\frac{d}{2}\right) \\ \omega =\sqrt{\frac{2g_E}{d}} \end{gathered}$$

Substitute all the values

  $$\begin{gathered} \omega =\sqrt{\frac{2\times 9.80 m/s^2}{800m}}\frac{rad}{s}\times \frac{1rev}{2\mathrm{\pi rad}}\times \frac{60s}{1min} \\ =1.5 rad/min \end{gathered}$$

So the revolutions per minute that are needed for the “artificial gravity” is 1.5 rad/min

Step 4: (b) Determining the revolutions per minute that are needed for gravity on the Martian surface ( 3 . 70 m / s 2 )

Using the same approach as used in part (a)

From equation (iii)

$$\begin{gathered} g_M={\omega }^{2}r \\ ={\omega }^2\times \left(\frac{d}{2}\right) \\ \omega =\sqrt{\frac{2g_M}{d}} \end{gathered}$$

Substitute all the values

$$\begin{gathered} \omega =\sqrt{\frac{2\times 3.7 m/s2}{800m}}\frac{rad}{s}\times \frac{1rev}{2\mathrm{\pi rad}}\times \frac{60s}{1min} \\ =0.92rad/min \end{gathered}$$

So the revolutions per minute that are needed for gravity on the Martian surface ($3.70m/s^2$) is 0.92rad/min