
The speed of Earth’s rotation about its axis is $\varpi $ . Its speed is increased to $x$ times to make the effective acceleration due to gravity equal to zero at the equator, then $x$ is around ($g=10m{{s}^{-2}},R=6400km$)
A. $1$
B. $8.5$
C. $17$
D. $34$
Answer
419.4k+ views
Hint: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 $m$ at the latitude, which makes an angle $\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'=g-{{\varpi }^{2}}R{{\cos }^{2}}\phi $. When the body is at equator, $\cos \phi =\cos 0=1$ thus the net acceleration will be $g'=g-{{\varpi }^{2}}R$
We know that Earth takes 24 hours to rotate about its own axis. So,
$T=24\times 60\times 60\,s$
Angular speed of Earth about its own axis will be:
$\varpi =\dfrac{2\pi }{T}=\dfrac{2\pi }{24\times 60\times 60}rad{{s}^{-1}}$
At equator, $g'=g-{{R}_{e}}^{2}g'=g-{{R}_{e}}\varpi {{'}^{2}}$ or $0=g-{{R}_{e}}\varpi {{'}^{2}}$
So, $\varpi '=\sqrt{\dfrac{g}{{{R}_{e}}}}=\dfrac{2\pi }{84.6\times 60}$
When we take their ratio we get
$\dfrac{\varpi '}{\varpi }=\dfrac{24\times 60\times 60}{84.6\times 60}\approx 17$
$\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 $\phi =90{}^\circ $$\Rightarrow \cos \phi =\cos 90=0$, therefore the effective gravity will be $g'=g$. Hence, we have the maximum gravitational acceleration at the poles and minimum at the equator.
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 $m$ at the latitude, which makes an angle $\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'=g-{{\varpi }^{2}}R{{\cos }^{2}}\phi $. When the body is at equator, $\cos \phi =\cos 0=1$ thus the net acceleration will be $g'=g-{{\varpi }^{2}}R$

We know that Earth takes 24 hours to rotate about its own axis. So,
$T=24\times 60\times 60\,s$
Angular speed of Earth about its own axis will be:
$\varpi =\dfrac{2\pi }{T}=\dfrac{2\pi }{24\times 60\times 60}rad{{s}^{-1}}$
At equator, $g'=g-{{R}_{e}}^{2}g'=g-{{R}_{e}}\varpi {{'}^{2}}$ or $0=g-{{R}_{e}}\varpi {{'}^{2}}$
So, $\varpi '=\sqrt{\dfrac{g}{{{R}_{e}}}}=\dfrac{2\pi }{84.6\times 60}$
When we take their ratio we get
$\dfrac{\varpi '}{\varpi }=\dfrac{24\times 60\times 60}{84.6\times 60}\approx 17$
$\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 $\phi =90{}^\circ $$\Rightarrow \cos \phi =\cos 90=0$, therefore the effective gravity will be $g'=g$. Hence, we have the maximum gravitational acceleration at the poles and minimum at the equator.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Why was the Vernacular Press Act passed by British class 11 social science CBSE

Arrange Water ethanol and phenol in increasing order class 11 chemistry CBSE

Name the nuclear plant located in Uttar Pradesh class 11 social science CBSE

What steps did the French revolutionaries take to create class 11 social science CBSE

How did silk routes link the world Explain with three class 11 social science CBSE

What are the various challenges faced by political class 11 social science CBSE
