 # One problem for humans living in outer space is that they are apparent kramtus51 2021-12-19 Answered
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 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be $9.80\frac{m}{{s}^{2}}$? (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 $\left(3.70\frac{m}{{s}^{2}}\right)$. How many revolutions per minute are needed in this case?
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Step 1
Given:
${g}_{E}=9.80\frac{m}{{s}^{2}}$
${g}_{M}=3.70\frac{m}{{s}^{2}}$
Step 2
We know that the centripetal acceleration is given by
${a}_{r}=\frac{{v}^{2}}{R}$
and we also know that the linear velocity for a rotation object is given by
$v=\omega R$
Hence, ${a}_{r}=\frac{{\left(\omega R\right)}^{2}}{R}=\frac{{\omega }^{2}{R}^{2}}{R}$
${a}_{r}={\omega }^{2}R\left(1\right)$
And this is a general formula for both cases in this problem.
Step 3
a) We need to make the centripetal acceleration equals the free-fall acceleration of Earth, so, from (1);
${g}_{E}={\omega }^{2}R$
solving for $\omega$;
${\omega }^{2}=\frac{{g}_{E}}{R}$
$\omega =\sqrt{\frac{{g}_{E}}{R}}$
Plug the given;
$\omega =\sqrt{\frac{9.80}{400}}\frac{rad}{s}\left(2\right)$
We need to find it in rev/min, so we will use some converting factors.
Noting that $1.0rev=2\pi rad$, hence
$\frac{1.0rev}{2\pi rad}=1.0\left(A\right)$
and $60s=1.0min$, hence
$\frac{60s}{1.0min}=1.0\left(B\right)$
Step 4
Multiply (2) by the two converting factors (A) and (B).
$\omega =\sqrt{\frac{9.80}{400}}\frac{rad}{s}×\frac{1.0rev}{2\pi rad}×\frac{60s}{1.0min}$
rad ans s are cancelled.
$\omega =1.49re\frac{v}{min}$
Step 5
b)Using the same approach above but changing the free-fall acceleration of Earth ${g}_{E}$ to the free-fall acceleration of Mars ${g}_{M}$.
Hence, from (1) ${g}_{M}={\omega }^{2}R$
solving for $\omega$;
${\omega }^{2}=\frac{{g}_{M}}{R}$
$\omega =\sqrt{\frac{{g}_{M}}{R}}$
Plug the given;
$\omega =\sqrt{\frac{3.70}{400}}\frac{rad}{s}\left(3\right)$
Multiply (3) by the two converting factors (A) and (B).

###### Not exactly what you’re looking for? porschomcl

A) The circumference can be used to obtain the radius of the space station

The acceleration a for a rotational motion in terms of angular velocity $\omega$
$a=g\left(\text{in this case}\right)=\frac{{v}^{2}}{r}=r{\omega }^{2}$
$\omega =\sqrt{\frac{g}{r}}=0.1807\frac{radians}{\mathrm{sec}}$
The number of revolution required is the frequency of $2\pi$ rotation in a minute
$N=\omega \cdot 60\frac{\mathrm{sec}}{2\pi }=0.1807\cdot \frac{60}{6.28}=1.726re\frac{v}{min}$
B) We adopt the same procedure as we did for Part A except that the acceleration in now ${a}_{mars}=3.7\frac{m}{{\mathrm{sec}}^{2}}$.
$\omega =\sqrt{\frac{g}{r}}=0.111\frac{radians}{\mathrm{sec}}$
$N=\omega \cdot \frac{60\mathrm{sec}}{2\pi }=0.1807\cdot \frac{60}{6.28}=1.06re\frac{v}{min}$

###### Not exactly what you’re looking for? nick1337

Explanation:
In space station artificial gravity is created by rotational motion of space station
Centifugal  acceleration creates artificial gravity
${\omega }^{2}r=g$
${\omega }^{2}×400=9.8$
${\omega }^{2}=9.8/400$
NSK
$\omega =.1565rad/s$
no of turns per min $\frac{\omega }{2\pi }$ per sec $=1.56$ rev per min.
For Mars
Same theory will apply
${\omega }^{2}r=g$
${\omega }^{2}×400=3.7$
${\omega }^{2}=\frac{3.7}{400}$
$\omega =.096rad/s$
$n=\frac{\omega }{2\pi }$
$=\frac{.096}{2\pi }$
$=.01528$ per sec
per min
$=.9168$ per min.