If the earth stops rotating,the value of g at | vedclass

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

Question

(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g - \omega^2 R \cos^2 \lambda$. At the equat

Vedclass · Accepted Answer

Increase

Vedclass · Answer

(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g - \omega^2 R \cos^2 \lambda$. At the equator,the latitude $\lambda = 0^\circ$,so $\cos 0^\circ = 1$. Thus,the effective gravity at the equator is $g_{eq} = g - \omega^2 R$. If the earth stops rotating,the angular velocity $\omega$ becomes $0$. Consequently,the term $\omega^2 R$ becomes $0$,and the value of $g$ at the equator becomes $g_{eq} = g$. Since $g$ is greater than $(g - \omega^2 R)$,the value of $g$ at the equator will increase.

If the earth stops rotating,the value of $g$ at the equator will

Similar Questions

Assuming Earth to be a sphere of uniform density,what is the value of gravitational acceleration in a mine $100 \, km$ below the Earth's surface? (Given $R = 6400 \, km$,$g = 9.8 \, m/s^2$)

At what height from the surface of the earth will the value of acceleration due to gravity fall to half of its value on the surface of the earth?

The value of acceleration due to gravity at a depth of $ 1600 \,km $ is equal to: (Radius of Earth $ = 6400 \,km $) (in $\,ms^{-2}$)

The acceleration due to gravity at a height of $6400 \,km$ from the surface of the earth is $2.5 \,ms^{-2}$. The acceleration due to gravity at a height of $12800 \,km$ from the surface of the earth is (Radius of the earth $= 6400 \,km$) (in $\,ms^{-2}$)

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 $= 9.8\, m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module