The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh 6 1 th as much as at present is ( g = gravitational acceleration, R = equatorial radius

Question

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh 6 1 ​ th as much as at present is ( g = gravitational acceleration, R = equatorial radius of the earth)

TestHub · Accepted Answer

The apparent weight of a person at the equator is given by:

$W^{'} = m (g - R ω^{2})$

Given that the new weight is $\frac{1}{6}$ of the present weight. Assuming the present weight is approximately $m g$ (since the current centrifugal acceleration is negligible):

$W^{'} = \frac{1}{6} m g$

Equating the two expressions for apparent weight:

$m (g - R ω^{2}) = \frac{1}{6} m g$

$g - R ω^{2} = \frac{1}{6} g$

$R ω^{2} = g - \frac{1}{6} g$

$R ω^{2} = \frac{5}{6} g$

$ω^{2} = \frac{5}{6} \frac{g}{R}$

$ω = \frac{5}{6} \frac{g}{R}$

Answer: $\frac{5}{6} \frac{g}{R}$

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