Physics - Gravitation Question with Solution | TestHub

True weight at equator, $W=mg$ Observed weight at equator, $${\mathrm{W}}^{\prime }={\mathrm{mg}}^{\prime }=\frac{3}{5}\mathrm{mg}$$ At equator, latitude $\lambda =0$ Using the formula, $$\begin{array}{l}{\mathrm{mg}}^{\prime }=\mathrm{mg}-\mathrm{mR}{\omega }^2{\cos }^2\lambda \\ \frac{3}{5}\mathrm{mg}=\mathrm{mg}-\mathrm{mR}{\omega }^2{\cos }^{2}0=\mathrm{mg}-\mathrm{mR}{\omega }^2\\ \Rightarrow \mathrm{mR}{\omega }^2=\mathrm{mg}-\frac{3}{5}\mathrm{mg}=\frac{2}{5}\mathrm{mg}\\ \therefore \omega ={\left(\frac{2}{5}\frac{\mathrm{g}}{\mathrm{R}}\right)}^{1/2}\\ ={\left(\frac{2\times 9.8}{5\times 6.4\times 10^6}\right)}^{1/2}=7.8\times 10^{-4}\mathrm{rad}/\mathrm{s}\end{array}$$