Question

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 9.80m/s2? (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 (3.70m/s2). How many revolutions per minute are needed in this case?

Answer 1

Answer:

Explanation:

In space station artificial gravity is created by rotational motion of space station

Centifugal  acceleration creates artificial gravity

ω²r = g

ω² x 400 = 9.8

ω² = 9.8 /400

ω = .1565 rad /s

no of turns per min ω / 2π per sec   = 1.56 rev per min .

For Mars

Same theory will apply

ω²r = g

ω² x 400 = 3.7

ω² = 3.7 /400

ω = . 096  rad /s

n = ω / 2π

= .096 /  2π

=.01528 per sec

per min

= .9168  per min .