A body weights 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?
[Use $g = {{GM} \over {{R^2}}}$ = 9.8 ms$-$2 and radius of earth, R = 6400 km.]
Solution
Given, weight of body at North pole,<br/><br/>w<sub>p</sub> = mg = 49 N<br/><br/>Radius of Earth, R = 6400 km<br/><br/>Let weight of body at equator be w<sub>e</sub>.<br/><br/>At equator, g<sub>e</sub> = g $-$ R$\omega$<sup>2</sup><br/><br/>$\therefore$ w<sub>e</sub> = mg<sub>e</sub> = m(g $-$ R$\omega$<sup>2</sup>)<br/><br/>Since, w<sub>p</sub> > w<sub>e</sub> $\Rightarrow$ w<sub>e</sub> < 49 N<br/><br/>Hence, above condition is satisfied by only option (b).
About this question
Subject: Physics · Chapter: Gravitation · Topic: Kepler's Laws
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