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Question: The speed of Earth’s rotation about its axis is \(\varpi \) . Its speed is increased to \(x\) times ...

The speed of Earth’s rotation about its axis is ϖ\varpi . Its speed is increased to xx times to make the effective acceleration due to gravity equal to zero at the equator, then xx is around (g=10ms2,R=6400kmg=10m{{s}^{-2}},R=6400km)
A. 11
B. 8.58.5
C. 1717
D. 3434

Explanation

Solution

Here we will clear our concept related to variation of gravitational force in difference in latitude. We will know which force will be acting on Earth. Earth is under the influence of centripetal and centrifugal force as it rotates on its own axis.

Complete answer:
The centripetal and centrifugal force come in picture when an object is in circular motion and so is our Earth. Let us consider a body of mass mm at the latitude, which makes an angle ϕ\phi at the center of the Earth. Now, as Earth rotates on its axis, then mass m will also rotate and form a circle about the axis of rotation.

And, as the body rotates in a circular motion then, there is a centrifugal force which acts outside away from the center of the circle. Hence the effective gravity will be g=gϖ2Rcos2ϕg'=g-{{\varpi }^{2}}R{{\cos }^{2}}\phi . When the body is at equator, cosϕ=cos0=1\cos \phi =\cos 0=1 thus the net acceleration will be g=gϖ2Rg'=g-{{\varpi }^{2}}R

We know that Earth takes 24 hours to rotate about its own axis. So,
T=24×60×60sT=24\times 60\times 60\,s
Angular speed of Earth about its own axis will be:
ϖ=2πT=2π24×60×60rads1\varpi =\dfrac{2\pi }{T}=\dfrac{2\pi }{24\times 60\times 60}rad{{s}^{-1}}
At equator, g=gRe2g=gReϖ2g'=g-{{R}_{e}}^{2}g'=g-{{R}_{e}}\varpi {{'}^{2}} or 0=gReϖ20=g-{{R}_{e}}\varpi {{'}^{2}}
So, ϖ=gRe=2π84.6×60\varpi '=\sqrt{\dfrac{g}{{{R}_{e}}}}=\dfrac{2\pi }{84.6\times 60}
When we take their ratio we get
ϖϖ=24×60×6084.6×6017\dfrac{\varpi '}{\varpi }=\dfrac{24\times 60\times 60}{84.6\times 60}\approx 17
ϖ=17ϖ\therefore \varpi '=17\varpi

Hence, the correct answer is option C.

Note: One must know that when the body is placed at pole the angle between center of Earth and the body will be ϕ=90\phi =90{}^\circ cosϕ=cos90=0\Rightarrow \cos \phi =\cos 90=0, therefore the effective gravity will be g=gg'=g. Hence, we have the maximum gravitational acceleration at the poles and minimum at the equator.