What should be the period of rotation of earth so as to make any object on the equator weigh half of its present value?
Answer & explanation
Correct answer: option 1
$g_e=g_0-R \omega^2 \Rightarrow g_0 / 2=g_0-R \omega^2$
$\Rightarrow \omega^2 \mathrm{R}=\frac{\mathrm{g}_0}{2}$
$\Rightarrow \omega=\sqrt{\frac{g_0}{2 R}}$
$\Rightarrow \mathrm{T}=2 \pi \sqrt{\frac{2 \mathrm{R}}{\mathrm{g}_0}}$, putting $\mathrm{R}=6.4 \times 10^6 \mathrm{~m}$ and $\mathrm{g}_0=9.8 \mathrm{~m} / \mathrm{sec}^2$, we obtain,
$\mathrm{T}=1.99 \mathrm{~hrs} \approx 2 \mathrm{~hrs}$