Question

The space station Babylon 5 has a diameter of 1 miles, and rotated around its long axis to produce artificial gravity for its inhabitants. It completes 1 revolutions/min. Calculate the effective gravitational strength at the outer rim of the station, in units of g. (use g = 9.81 m/s^2)

Answer 1

This question involves the concept of artificial gravity.

The effective gravitational strength at outer rims is "0.9 g".

Artificial Gravity

The effective gravitational strength at the outer rims can be given by the following formula:

$v=\frac{1}{2\pi}\sqrt{\frac{a_c}{r}}\\\\a_c=4\pi^2v^2r$

where,

• $a_c$ = effective gravitational strength = ?
• v = rotational speed = 1 rev/min = 0.0167 rev/s
• r = radius = diameter/2 = 1 mile/2 = 0.5 miles = 804.672 m

Therefore,

$a_c = 4\pi^2(0.0167\ rev/s)^2(804.672\ m)\\a_c =8.82\ m/s^2\\\\In\ terms\ of\ g:\\\\a_c=(8.82\ m/s^2)(\frac{g}{9.81\ m/s^2})\\\\a_c = 0.9g$

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