Answer:
Acceleration =
centripetal acceleration
\[a=\,\frac{{{\upsilon
}^{2}}}{R}\]
But \[\upsilon =\omega R\]
\[\therefore \] \[\,a\,=\,{{\omega
}^{2}}R=\frac{4{{\pi }^{2}}R}{{{T}^{2}}}\,\]
\[=\,\frac{4\,\times
\,{{(3.14)}^{2}}\times \,6.4\,\times \,{{10}^{6}}}{{{(24\times \,3600)}^{2}}}\]
\[=3.38\times
{{10}^{2}}m{{s}^{2}}\]
Hence, \[\frac{a}{g}\,=\,\frac{3.38\,\times
\,{{10}^{-2}}}{9.8}\,=\,3.45\,\times \,{{10}^{-3}}\]
At latitude \[\theta
,\,a\,=g\,\left( 1-\frac{R{{\omega }^{2}}\,{{\cos }^{2}}\theta }{g} \right)\]
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