The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh 5 3 th as much as at present weight is ( g = gravitational acceleration, R = equatorial

Question

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh 5 3 ​ th as much as at present weight 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 $W^{'} = m (g - R ω^{2})$ , where $ω$ is the angular speed of the earth.

The present weight of the person is approximately $W = m g$ (since the present angular speed of the earth is very small, the term $R ω^{2}$ is negligible).

Given that the new weight is $\frac{3}{5}$ of the present weight:

$m (g - R ω^{2}) = \frac{3}{5} m g$

$g - R ω^{2} = \frac{3}{5} g$

$R ω^{2} = g - \frac{3}{5} g$

$R ω^{2} = \frac{2 g}{5}$

$ω^{2} = \frac{2 g}{5 R}$

$ω = \frac{2 g}{5 R}$

Answer: $\frac{2 g}{5 R}$

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