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

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(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

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Poles

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(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.

Column-$I$	Column-$II$
$(1)$ Maximum value of acceleration due to gravity $g$	$(a)$ At the center of the Earth
$(2)$ Minimum value of acceleration due to gravity $g$	$(b)$ At the poles
$(3)$ Zero value of acceleration due to gravity $g$	$(c)$ At the equator

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

Similar Questions

Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.
Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.

The angular velocity of the Earth at present is $\omega$. With what angular velocity should it rotate so that the weight of a body at the equator appears to be zero (in $\omega$)? (Express the answer as a multiple of $\omega$)

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth

$(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles

$(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is $(R = \text{radius of the earth})$

How does the acceleration due to gravity $(g)$ at a location on Earth change with latitude?

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