Question

The place(s) where the value of 'g' is unaffected by the increase (or) decrease in the speed of rotation of the earth about its own axis is :

• A. The north pole
  
• B. The south pole
  
• C. The equator
  
• D. None of the above
  

Answer 1

Answer: (A), (B)

The correct options are
A The north pole
B The south pole

From parallelogram law of vector addition

${PB}^2={PO}^2+{PA}^2+2POPA\cos ({180}^{\circ }-\varphi )$

$\Rightarrow {\left({mg}^1\right)}^2={\left(mg\right)}^2+{\left({mw}^{2}R\cos \varphi \right)}^2+2mg{mw}^{2}R\cos \varphi (-\cos \varphi )$

$\Rightarrow m^2{g}^{1^2}=m^2{g}^2+m^2{w}^4{R}^2{\cos }^2\varphi -2m^2{gw}^{2}R{\cos }^2\varphi$

$g^{1^2}=g^2(1+\frac{w^4{R}^2{\cos }^2\varphi }{g^2}-\frac{{2w}^{2}R{\cos }^2\varphi }{g})$

$w$ is very small

so $w^2$ can be neglected and the term

containing $w^4$ can be neglected

$\Rightarrow g^{1^2}=g^2(1-\frac{{2w}^{2}R{\cos }^2\varphi }{g})$

or, $g^1=g{(1-\frac{{2w}^{2}R{\cos }^2\varphi }{g})}^{\frac{1}{2}}$

Applying binomial expansion and neglecting higher powers we get,

$g^1=g(1-\frac{1}{2}\times \frac{{2w}^{2}R{\cos }^2\varphi }{g})$

$\Rightarrow g^1=g(1-\frac{w^{2}R{\cos }^2\varphi }{g})$

$\Rightarrow g^1=g-w^{2}R{\cos }^2\varphi$

At poles, $\varphi ={90}^0$

$\therefore g^1=g$ which is independent of $w$

Thus the places where the value of ${}^{\prime }{g}^{\prime }$ is unaffected by the increase/decrease in the speed of rotation of the earth about its own axis are the north and south poles.