Question

If the earth were a perfect sphere of radius $6.4\times {10}^6\text{m}$ rotating about its axis with the period of one day $(8.64\times {10}^4\text{sec})$, what is the difference in acceleration due to gravity on poles and the equator?

• A. $338\times {10}^{-6}{\text{m/s}}^2$
• B. $338\times {10}^{-4}{\text{m/s}}^2$
• C. $338\times {10}^{-2}{\text{m/s}}^2$
• D. None of these

Answer 1

The correct option is B $338\times {10}^{-4}{\text{m/s}}^2$


Let suppose $g_p$ and $g_e$ are the acceleration due to gravity on poles and equator of earth.

Radius of earth, $R=6.4\times {10}^6\text{m}$

Time period of rotation, $T=8.64\times {10}^4\text{s}$

Variation of $g$ with latitude due to rotational motion of earth is given by

$g^{\prime }=(g-{\omega }^{2}R{\cos }^2\varphi )$

where, $\varphi$ is latitude.

At pole, $\varphi ={90}^{\circ }$

$\therefore g_p=g$

And at the equator, $\varphi =0^{\circ }$

$\therefore g_e=g-{\omega }^{2}R$

So, difference in acceleration will be

$g_p-g_e=g-(g-{\omega }^{2}R)={\omega }^{2}R$

$\Rightarrow g_p-g_e=(\frac{2\pi }{T}{)}^{2}R={\left(\frac{2\pi }{8.64\times {10}^4}\right)}^2\times 6.4\times {10}^6$

$\Rightarrow g_p-g_e=3.38\times {10}^{-2}=338\times {10}^{-4}{\text{m/s}}^2$

Hence, option $\left(b\right)$ is correct.