Question

How acceleration due to gravity varies due to rotation of the earth?

Answer 1

Consider the earth to be a spherical ball of mass ‘M’ and radius ‘R’. An object of mass ‘m’ is at point P at latitude $\phi$. When the earth is not rotating the weight of the object is mg. But the earth is rotating with angular velocity $\omega$. So, the object is moving in a circular path of radius ‘r’ as shown in the figure. The object experiences centrifugal force,
$f_f=m{\omega }^{2}r=m{\omega }^{2}Rcos\phi$

The object is being acted by two forces $mg$ and $f_f$. The resultant of these two forces gives the apparent weight of the object $m{g}^{\prime }$. The resultant of the two forces is given by the diagonal of the parallelogram OPAB.
From parallelogram law of vector addition:
$P{B}^2=P{O}^2+P{A}^2+2PO\times PAcos(180-\phi )(m{g}^{\prime }{)}^2=(mg{)}^2+(m{\omega }^{2}Rcos\phi {)}^2+2mgm{\omega }^{2}Rcos\phi (-cos\phi )m^2{g}^{\prime 2}=m^2{g}^2+m^2{\omega }^4{R}^{2}co{s}^2\phi -2m^{2}g{\omega }^{2}Rco{s}^2\phi g^{\prime 2}=g^2(1+\frac{{\omega }^4{R}^{2}co{s}^2\phi }{g^2}-\frac{2{\omega }^{2}Rco{s}^2\phi }{g})$
The value of $\omega$ is in the order of ${10}^{-4}$. So, the term containing ${\omega }^{-4}$ is much smaller than 1 and ${\omega }^2$ and can be neglected.
$g^{\prime 2}=g^2(1-\frac{2{\omega }^{2}Rco{s}^2\phi }{g})g^{\prime }=g(1-\frac{2{\omega }^{2}Rco{s}^2\phi }{g}{)}^{\frac{1}{2}}$
Applying binomial expansion and neglecting higher powers we get,
$g^{\prime }=g(1-\frac{1}{2}\times \frac{2{\omega }^{2}Rco{s}^2\phi }{g})=g(1-\frac{{\omega }^{2}Rco{s}^2\phi }{g})\therefore g^{\prime }=g-{\omega }^{2}Rco{s}^2\phi$
This gives the acceleration due to gravity due to rotaion of earth.