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Today, 01:14

The uniform circular motion of a space station can be used to create artificial gravity. This can be done by adjusting the angular speed of the space station, so the centripetal acceleration at an astronaut's feet equals g, the magnitude of the acceleration due to the earth's gravity. If such an adjustment is made, what will be true about the acceleration at the astronaut's head due to the artificial gravity?

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  1. Today, 02:56
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    The acceleration at the astronaut's head decreases.

    Explanation:

    Since the centripetal acceleration equals acceleration due to gravity,

    a = g = GM/R². since a changes infinitesimally from his foot to his head, we differentiate a with respect to r to get da/dr = - 2GM/R³.

    So, da, the change in acceleration = - 2GMdR/R³ = - 2gdR/R = - 2 * 9.8/6.4 * 10⁶ m = - 3.0625 * 10⁻⁶dR m/s².

    Since dR = height of astronaut = 1.80 m, da = - 3.0625 * 10⁻⁶ * 1.8 = - 5.5125 * 10⁻⁶ m/s².

    So the acceleration at the astronaut's head is g + da = 9.8 - 0.0000055125 = 9.7999944875 m/s².

    So the acceleration at the astronaut's head decreases.
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