Question

If earth were to rotate on its own axis such that the weight of a person at the equator becomes half the weight at the poles, then its time period of rotation is
(g = acceleration due to gravity near the poles and R is the radius of earth) (Ignore equatorial bulge) [EAMET 2013]

Difficult

Solution

Here, $g_{0} = g^{'} - R ω^{2}$
Given, $g_{0} = \frac{g}{2} and g^{'} = g$
So, $\frac{g}{2} = g - R ω^{2} \Rightarrow ω = \sqrt{\frac{g}{2 R}}$
We have, $T = 2 π (\frac{r}{v}) \Rightarrow T = \frac{2 π}{ω} (∵ ω = \frac{v}{r})$
Hence, $T = 2 π \sqrt{\frac{2 R}{g}}$

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