91影视

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

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

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

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
$a_c=\frac{v^2}{r}$ 鈥�(颈)

Where, v is the velocity and r is the radius

The linear velocity for a rotation object is given by
$v=蝇r$ 鈥�(颈颈)

Where ,$蝇$ is angular velocity

From equation (i) and equation (ii)

$$\begin{gathered} a_c=\frac{{\left(蝇r\right)}^2}{r} \\ ={蝇}^{2}r \end{gathered}$$ 鈥�(颈颈颈)

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

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

Substitute all the values

role="math" localid="1663742402891" $$\begin{gathered} 蝇=\sqrt{\frac{2脳9.80m/s^2}{800m}}\frac{rad}{s}脳\frac{1rev}{2\mathrm{蟺谤补诲}}脳\frac{60s}{1min} \\ =1.5rad/min \end{gathered}$$

So the revolutions per minute that are needed for the 鈥渁rtificial gravity鈥� is 1.5 rad/min

Using the same approach as used in part (a)

From equation (iii)

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

Substitute all the values

$$\begin{gathered} 蝇=\sqrt{\frac{2脳3.7m/s2}{800m}}\frac{rad}{s}脳\frac{1rev}{2\mathrm{蟺谤补诲}}脳\frac{60s}{1min} \\ =0.92rad/min \end{gathered}$$

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