In order to make the effective acceleration d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$.At the equator,the latitud

Vedclass · Accepted Answer

$\frac{1}{800}\,rad/s$

Vedclass · Answer

(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$.At the equator,the latitude $\lambda = 0^\circ$,so $\cos \lambda = 1$.Thus,the effective acceleration becomes $g' = g - \omega^2 R$.For the effective acceleration to be zero $(g' = 0)$,we have $0 = g - \omega^2 R$.This implies $\omega^2 = \frac{g}{R}$.Given $g = 10\,m/s^2$ and $R = 6400\,km = 6.4 	imes 10^6\,m$.$\omega = \sqrt{\frac{10}{6.4 	imes 10^6}} = \sqrt{\frac{1}{6.4 	imes 10^5}} = \sqrt{\frac{1}{64 	imes 10^4}} = \frac{1}{8 	imes 10^2} = \frac{1}{800}\,rad/s$.

In order to make the effective acceleration due to gravity equal to zero at the equator,the angular velocity of rotation of the earth about its axis should be $(g = 10\,m/s^2$ and radius of earth is $6400\,km)$.

Similar Questions

How can we conclude that the Earth is not a perfect sphere based on the acceleration due to gravity?

What is the height from the surface of earth,where acceleration due to gravity will be $\frac{1}{4}$ of that of the earth (in $km$)? $(R_E = 6400 \ km)$

If a planet has a mass and radius both half that of the Earth,the acceleration due to gravity at its surface would be ......... $m/s^2$ ($g$ on Earth $= 9.8\, m/s^2$)

Weight of a body decreases by $1.5 \%$ when it is raised to a height $h$ above the surface of the Earth. When the same body is taken to the same depth $h$ in a mine,its weight will show ........

The difference in the acceleration due to gravity at the pole and equator is ( $g=$ acceleration due to gravity,$R=$ radius of earth,$\theta=$ latitude,$\omega=$ angular velocity,$\cos 0^{\circ}=1, \cos 90^{\circ}=0$ ).

Vedclass Test Series

Exam Paper Generator

Online Exam Module