Acceleration Due To Gravity And Its Variation With Height, Depth And Latitude Angle (Board Level)

∴The earth is rotating about its axis from west to east. So, the earth is a non-inertial frame of reference. Every body on its surface experiences a centrifugal force mRw² cosa. Where a is latitude of the place.

The net force on a particle on the surface of the earth is the resultant of these two forces which can be calculated by using law of parallelogram.

F = $\sqrt{{{{m}^{2}}{{g}^{2}}+{{m}^{2}}{{\omega }^{4}}{{R}^{2}}{{{\cos }}^{2}}\alpha \,+\,2\,(mg)\,(m{{\omega }^{2}}R\cos \alpha )\,[\cos \,(180-\alpha )]}}$ (the angle between the two forces is 180 -a as clear from the diagram)

This net force produces the weight of the body hence F=mg’ ( where g’ is the net gravity at a latitude a

Therefore,    mg’ = $\sqrt{{{{m}^{2}}{{g}^{2}}+{{m}^{2}}{{\omega }^{4}}{{R}^{2}}{{{\cos }}^{2}}\alpha \,+\,2\,(mg)\,(m{{\omega }^{2}}R\cos \alpha )\,[\cos \,(180-\alpha )]}}$

taking m² as common from RHS and cancelling m from both sides we get  $g’=\sqrt{{{{g}^{2}}+{{\omega }^{4}}{{R}^{2}}{{{\cos }}^{2}}\alpha \,+\,2\,g\,{{\omega }^{2}}R\cos \alpha )\,[\cos \,(180-\alpha )]}}$

hence $g’=\sqrt{{{{g}^{2}}+{{\omega }^{4}}{{R}^{2}}{{{\cos }}^{2}}\alpha \,-\,2\,g\,{{\omega }^{2}}R{{{\cos }}^{2}}\alpha )]}}$

taking g² as common on RHS

$g’=\sqrt{{{{g}^{2}}\left[ {1+\frac{{{{\omega }^{4}}{{R}^{2}}{{{\cos }}^{2}}\alpha }}{{{{g}^{2}}}}\,-\frac{{2\,g\,{{\omega }^{2}}R{{{\cos }}^{2}}\alpha }}{{{{g}^{2}}}}\,} \right]}}$ $\Rightarrow g’=g\sqrt{{\left[ {1+\frac{{{{\omega }^{4}}{{R}^{2}}{{{\cos }}^{2}}\alpha }}{{{{g}^{2}}}}\,-\frac{{2\,{{\omega }^{2}}R{{{\cos }}^{2}}\alpha }}{g}\,} \right]}}$ 

here w is the angular velocity of rotation of earth about its own axis which can be given as

$\displaystyle \omega =\frac{{2\pi }}{T}=\frac{{2\pi }}{{24hrs}}=\frac{{2\pi }}{{24\times 60\times 60\sec }}=0.0000727rad/\sec$. hence it Is a very small number therefore w⁴ becomes negligible so the whole term $\displaystyle \frac{{{{\omega }^{4}}{{R}^{2}}{{{\cos }}^{2}}\alpha }}{{{{g}^{2}}}}$ from the above equation of g’ can be neglected hence we get   $g’=g{{\left[ {1-\frac{{2\,{{\omega }^{2}}R{{{\cos }}^{2}}\alpha }}{g}\,} \right]}^{{1/2}}}$

again here   $\displaystyle \frac{{{{\omega }^{2}}R}}{{{{g}^{{}}}}}$is a small term so we can use binomial approximation and can write the above equation of g’ as follows

$g’=g\left[ {1-\frac{1}{2}\times \frac{{2\,{{\omega }^{2}}R{{{\cos }}^{2}}\alpha }}{g}\,} \right]$

$\Rightarrow g’=g\left[ {1-\frac{{\,{{\omega }^{2}}R{{{\cos }}^{2}}\alpha }}{g}\,} \right]$

hence finally we get   $g’=g-\,{{\omega }^{2}}R{{\cos }^{2}}\alpha \,$                                         

 Important points to remember

(i)   at the pole, a = 90°cos a = minimum = 0 hence g’ is maximum and is equal to g

(ii)  at equator a = 0°, so g’ is minimum and is equal to g-w²R

(iii) due to rotation of earth the gravity on the surface is decreased except at poles.