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  • What should be the period of rotation of earth so as to make any object on the equator weigh half of its present value?Option: 1 <img alt="\mathrm{2\

What should be the period of rotation of earth so as to make any object on the equator weigh half of its present value?

Option: 1

\mathrm{2\: hrs}


Option: 2

\mathrm{24\: hrs}


Option: 3

\mathrm{8\: hrs}


Option: 4

\mathrm{12\: hrs}


Answers (1)

best_answer

\mathrm{g_e=g_0-R \omega^2 \Rightarrow g_0 / 2=g_0-R \omega^2}

\mathrm{\Rightarrow \quad \omega^2 \mathrm{R}=\frac{\mathrm{g}_0}{2} }

\mathrm{ \Rightarrow \quad \omega=\sqrt{\frac{g_0}{2 \mathrm{R}}} }

\mathrm{ \Rightarrow \quad \mathrm{T}=2 \pi \sqrt{\frac{2 \mathrm{R}}{\mathrm{g}_0}} \text {, putting } \mathrm{R}=6.4 \times 10^6 \mathrm{~m} \text { and } \mathrm{g}_0=9.8 \mathrm{~m} / \mathrm{sec}^2 \text {, we obtain, } }

\mathrm{ \mathrm{T}=1.99 \mathrm{hrs} \approx 2 \mathrm{hrs} }

Hence option 1 is correct.





 

Posted by

HARSH KANKARIA

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