# JEE Main & Advanced Physics Gravitation Variation in g Due to Rotation of Earth

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 :

• The latitude at a point on the surface of the earth is defined as the angle, which the line joining that point to the centre of earth makes with equatorial plane. It is denoted by $\lambda$.
• For the poles $\lambda ={{90}^{o}}$ and for equator$\lambda ={{0}^{o}}$

(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 :

• This time is about $\left[ \text{As}\,\,\,T=\frac{2\pi }{\omega } \right]$ times the present time period of earth. Therefore if earth starts rotating 17 times faster then all objects on equator will become weightless.
• If earth stops rotation about its own axis then at the equator the value of $g$ increases by ${{\omega }^{2}}R$ and consequently the weight of body lying there increases by $m{{\omega }^{2}}R$.
• After considering the effect of rotation and elliptical shape of the earth, acceleration due to gravity at the poles and equator are related as

${{g}_{p}}={{g}_{e}}+0.034+0.018m/{{s}^{2}}$    $\therefore \ {{g}_{p}}={{g}_{e}}+0.052m/{{s}^{2}}$

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