Question 287: Gravitation

Question

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

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
\mathrm{g}^{\prime} & =\mathrm{g}-\mathrm{R} \omega^2 \cos ^2 	heta \
	herefore \quad \mathrm{~g}^{\prime} & =\mathrm{g}-\mathrm{R} \omega^2\quad\ldots .(\because 	heta=0)
\end{aligned}\
&	ext { Given } \mathrm{g}^{\prime}=\frac{3 \mathrm{~g}}{5}\
&	herefore \quad \omega^2=\frac{2 g}{5 R} 	ext { or } \omega=\sqrt{\frac{2 g}{5 R}}
\end{aligned}$$

Answer

$\sqrt{\frac{2 g}{5 R}}$

Answer

$\sqrt{\frac{3 \mathrm{~g}}{5 \mathrm{R}}}$

Answer

$\sqrt{\frac{5 R}{3 g}}$

Answer

$\sqrt{\frac{3}{5}} \mathrm{gR}$

Gravitation

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

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