Variation in g Due to Rotation of Earth
Category : JEE Main & Advanced
As the earth rotates, a body placed on its surface moves along the circular path and hence experiences centrifugal force, due to it, the apparent weight of the body decreases.
Since the magnitude of centrifugal force varies with the latitude of the place, therefore the apparent weight of the body varies with latitude due to variation in the magnitude of centrifugal force on the body.
If the body of mass m lying at point P, whose latitude is \[\lambda ,\] then due to rotation of earth its apparent weight can be given by \[\frac{g\,{{R}^{2}}{{T}^{2}}}{4{{\pi }^{2}}}={{\left( R+h \right)}^{3}}\]
or \[m{g}'=\sqrt{{{(mg)}^{2}}+{{({{F}_{c}})}^{2}}+2mg\ {{F}_{c}}\ \cos (180-\lambda )}\]
\[\Rightarrow \] \[m{g}'=\sqrt{{{(mg)}^{2}}+{{(m{{\omega }^{2}}R\cos \lambda )}^{2}}+2mg\ m{{\omega }^{2}}R\cos \lambda \ (-\cos \lambda )}\]
[As \[{{F}_{c}}=m{{\omega }^{2}}r=m{{\omega }^{2}}R\,\cos \lambda \]]
By solving we get \[{g}'=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda \]
Note :
(i) Substituting \[\lambda ={{90}^{o}}\] in the above expression we get \[=2\pi \,\,\sqrt{\frac{{{(R+h)}^{3}}}{g{{R}^{2}}}}\]
\[\therefore \] \[{{g}_{pole}}=g\] ...(i)
i.e., there is no effect of rotational motion of the earth on the value of \[g\] at the poles.
(ii) Substituting \[\lambda ={{0}^{o}}\] in the above expression we get \[T=2\pi \,\sqrt{\frac{{{r}^{3}}}{GM}}\]
\[\therefore \] \[{{g}_{equator}}=g-{{\omega }^{2}}R\] ...(ii)
i.e., the effect of rotation of earth on the value of \[g\] at the equator is maximum.
From equation (i) and (ii)
\[{{g}_{pole}}-{{g}_{equator}}=R{{\omega }^{2}}=0.034m/{{s}^{2}}\]
(iii) When a body of mass \[m\] is moved from the equator to the poles, its weight increases by an amount
\[m({{g}_{p}}-{{g}_{e}})=m{{\omega }^{2}}R\]
(iv) Weightlessness due to rotation of earth : As we know that apparent weight of the body decreases due to rotation of earth. If \[\omega \] is the angular velocity of rotation of earth for which a body at the equator will become weightless
\[{g}'=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda \]
\[\Rightarrow \] \[0=g-{{\omega }^{2}}R{{\cos }^{2}}{{0}^{o}}\] [As \[\lambda ={{0}^{o}}\] for equator]
\[\Rightarrow \] \[g-{{\omega }^{2}}R=0\]
\[\therefore \] \[T=\,\infty \]
or time period of rotation of earth \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{R}{g}}\]
Substituting the value of \[R=6400\times {{10}^{3}}m\] and \[{{\omega }_{S}}={{\omega }_{E}}\] we get
\[\omega =\frac{1}{800}=1.25\times {{10}^{-3}}\frac{rad}{sec}\] and \[T=5026.5\ sec=1.40\ hr.\]
Note :
\[{{g}_{p}}={{g}_{e}}+0.034+0.018m/{{s}^{2}}\] \[\therefore \ {{g}_{p}}={{g}_{e}}+0.052m/{{s}^{2}}\]
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