If the earth stops rotating on its own axis,t | vedclass

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

Question

(C) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$,where $\omega$ is the angu

Vedclass · Accepted Answer

Poles

Vedclass · Answer

(C) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$,where $\omega$ is the angular velocity of the earth's rotation and $R$ is the radius of the earth.At the poles,the latitude $\lambda = 90^\circ$. Substituting this into the formula,we get $g' = g - \omega^2 R \cos^2(90^\circ) = g - 0 = g$.Since the value of $g'$ at the poles is independent of the angular velocity $\omega$,the weight of a body $(w = mg')$ will not change if the earth stops rotating on its own axis.Therefore,there will be no variation in the weight of our bodies at the poles.

If the earth stops rotating on its own axis,there will be no variation in the weight of our bodies at:

Similar Questions

$A$ planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing $W$ on earth will weigh on that planet:

At what depth does the acceleration due to gravity become $\frac{g}{4}$? ($R$ = radius of the Earth)

Write the equation of acceleration due to gravity at a height $h$ from the surface of the Earth,where $h << R_e$.

Assuming the Earth to be a sphere of uniform density, the acceleration due to gravity:

The linear speed of a particle at the equator of the earth due to its spin motion is $V$. The linear speed of the particle at latitude $30^{\circ}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module