A circular SPACE HOTEL in orbit around the earth has a diameter of 355 m. In ord

Question

A circular SPACE HOTEL in orbit around the earth has a diameter of 355 m. In order to produce fake gravity along the outer rim, it is desired to rotate it at a speed that will produce a centripetal acceleration of 9.81 m/s^2. Note the geometry: people walking on the inside of the outer rim will weigh the same as if they were on earth. A GOOD DRAWING WOULD HELP.

(a) Find the tangential speed of a point on the rim when the station is producing the required centripetal acceleration.

(b) Find the station's angular velocity under those conditions, in radians per second.

(c) If you are below deck at a point 77 m from the outer rim, how much gravity will you experience?

(d) If you start moving from the rim toward the central hub of the space station, what will it feel like? How will perception change as you move?

Explanation / Answer

radius = d / 2

= 355 / 2 = 177.5 m

a = v2/ R

v = sqrt(a * R) = sqrt(9.81 * 177.5)

= 41.73 m/s

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w = v/R = 41.73 / 177.5

= 0.235 rad/s

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c) a = v2/ r

= 41.732/ (177.5 - 77)

= 17.32 m/s2

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d) the person feels that his weight is increasing

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