Calculate the difference in the value of g at | vedclass

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

Question

(A) The acceleration due to gravity at the poles $(g_{p})$ is given by $g_{p} = G M / R^{2}$.At the equator,the effective acceleration due to gravity

Vedclass · Accepted Answer

$R_{e} \omega^{2}$

Vedclass · Answer

(A) The acceleration due to gravity at the poles $(g_{p})$ is given by $g_{p} = G M / R^{2}$.At the equator,the effective acceleration due to gravity $(g_{e})$ is given by $g_{e} = g - R_{e} \omega^{2}$,where $R_{e}$ is the radius of the Earth and $\omega$ is the angular velocity of the Earth's rotation.The difference in the value of $g$ between the poles and the equator is $\Delta g = g_{p} - g_{e}$.Substituting the expressions: $\Delta g = g - (g - R_{e} \omega^{2})$.Therefore,$\Delta g = R_{e} \omega^{2}$.

Calculate the difference in the value of $g$ at the equator and at the poles due to the rotation of the Earth.

Similar Questions

$A$ simple pendulum is oscillating with frequency $F$ on the surface of the earth. It is taken to a depth $R/3$ below the surface of the earth ($R =$ radius of earth). The frequency of oscillation at depth $R/3$ is

Derive the equation for the variation of $g$ due to height from the surface of the Earth.

Two spherical planets $A$ and $B$ have the same mass,but their densities are in a ratio $8:1$. For these planets,the ratio of acceleration due to gravity at the surface of $A$ to its value at the surface of $B$ is:

The acceleration due to gravity on a planet is $1.96 \, m/s^2$. If it is safe to jump from a height of $3 \, m$ on the Earth,the corresponding height on the planet will be ........ $m$.

If the radius of earth shrinks by $2 \%$ while its mass remains the same,the acceleration due to gravity on the earth's surface will approximately:

Vedclass Test Series

Exam Paper Generator

Online Exam Module