Chapter 5, Problem 13P

To determine

The rotation speed (revolutions per day) if an effect nearly equal to gravity at the surface of the Earth (say 0.90g) is to be felt.

Answer to Problem 13P

Solution:

The space station of a circular tube that will rotate about its center to produce acceleration nearly equal to gravity at the surface of the Earth $\left(\text{say}\text{0}\text{.90g}\right)=1742$ revolution per day.

Explanation of Solution

Given:

The diameter of the space station $\text{=1}\text{.1}\text{km}$

Centripetal acceleration $\text{=}\text{0}\text{.90}\text{g}$

Formula Used:

Centripetal acceleration $={\omega }^{2}r$

  $\omega$ : angular speed

r : radius of space station.

Calculation:

  $\begin{array}{l}\text{radius }=\frac{\text{diameter}}{\text{2}}\\ =\frac{1.1\times 1000\text{m}}{2}\\ =550\text{m}\end{array}$

Centripetal acceleration $={\omega }^{2}r$

  $\begin{array}{l}0.90g={\omega }^{2}r\\ \omega =\sqrt{\frac{0.90\times 9.8}{550}}\text{ rad/s}\\ =\sqrt{\frac{0.90\times 9.8}{550}}\times \frac{1}{2\pi }\times \left(24\times 60\times 60\right)\\ =1742\text{ rpd}\end{array}$

Conclusion:

Space station must be rotated about its center to produce virtual gravity so that astronaut does not feel weightlessness.