Question

If the earth stops rotating.then the value of acceleration due to gravity at a latitude ${45}^o$ will increases by.

Answer 1

REF.Image

We know, the acceleration due to gravity due to rotation of earth

$g^{\prime }=g-{\omega }^{2}Rco{s}^2\varphi$

According to the question

$\varphi ={45}^{\circ }$

So, acceleration due to gravity at ${45}^{\circ }$ latitude

$g^{\prime }=g-{\omega }^{2}Rco{s}^2{45}^{\circ }$

$g^{\prime }=g-\frac{{\omega }^{2}R}{2}$

When earth steps rotating i.e., $\omega =0$ the acceleration due to gravity at ${45}^{\circ }$ latitude will increase by $\frac{{\omega }^{2}R}{2}$ $[\omega =7.292\times {10}^{-5}rad/secR=6317km]$

$\Rightarrow \frac{{\omega }^{2}R}{2}=\frac{(7.292\times {10}^{-5}{)}^2\times 6371\times {10}^3}{2}=0.016938m/s^2$