The acceleration due to gravity on the earth' | vedclass

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

Question

(D) The effective acceleration due to gravity at the equator $(g_e)$ is given by $g_e = g - R\omega^2$,where $g$ is the acceleration due to gravity at

Vedclass · Accepted Answer

$\frac{R^{2} \omega^{2}}{2g}$

Vedclass · Answer

(D) The effective acceleration due to gravity at the equator $(g_e)$ is given by $g_e = g - R\omega^2$,where $g$ is the acceleration due to gravity at the poles.The acceleration due to gravity at a height $h$ above the poles $(g_h)$ is given by $g_h = g(1 - \frac{2h}{R}) = g - \frac{2gh}{R}$.Given that the weights are the same at both locations,the effective acceleration due to gravity must be equal: $g_e = g_h$.Substituting the expressions: $g - R\omega^2 = g - \frac{2gh}{R}$.Simplifying the equation: $R\omega^2 = \frac{2gh}{R}$.Solving for $h$: $h = \frac{R^2\omega^2}{2g}$.

Similar Questions

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 $\omega$)

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

At which location,the equator or the poles of the Earth,is the value of acceleration due to gravity $g$ higher? Why?

If the radius of the earth were to shrink by $1\%$ with its mass remaining the same,the acceleration due to gravity on the earth's surface would

An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size,can he hope to detect gravity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module