Question

A circular space station of radius 100 m rotates twice each minute.

Answer 1

The space station completes 2 revolutions each minute, so that it traverses a distance of 2π (100 m) = 200π m each minute, giving it a linear/tangential speed of

v = (200π m) / (60 s) ≈ 10.472 m/s

(a) The astronaut would experience an acceleration of

a = v ² / (100 m) ≈ 1.09662 m/s² ≈ 0.1119g ≈ 0.11g

(b) Now you want to find the period T such that a = g. This would mean the astronaut has a tangential speed of

v = (200π m) / T

so that her centripetal/radial acceleration would match g :

a = g = ((200π m) / T )² / (100 m)

Solve for T :

(100 m) g = (400π ² m²) / T ²

T ² = (400π ² m²) / ((100 m) g) = (4π ² m)/g

T = √((4π ² m) / (9.8 m/s²)) ≈ 2π √(0.102 s²) ≈ 2.007 s ≈ 2.0 s