The angular speed of rotation of the Earth ab | vedclass

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

Question

(B) The effective acceleration due to gravity at the equator $(g_E)$ is given by $g_E = g_P - \omega^2 R$,where $g_P$ is the acceleration due to gravi

Vedclass · Accepted Answer

$8.75 	imes 10^{-4} \ rad \ s^{-1}$

Vedclass · Answer

(B) The effective acceleration due to gravity at the equator $(g_E)$ is given by $g_E = g_P - \omega^2 R$,where $g_P$ is the acceleration due to gravity at the poles,$\omega$ is the angular speed of the Earth,and $R$ is the radius of the Earth.Given that the weight at the equator is half the weight at the poles,we have $g_E = \frac{g_P}{2}$.Substituting this into the equation: $\frac{g_P}{2} = g_P - \omega^2 R$.Rearranging for $\omega$: $\omega^2 R = g_P - \frac{g_P}{2} = \frac{g_P}{2}$.Thus,$\omega = \sqrt{\frac{g_P}{2R}}$.Using $g_P = 9.8 \ m/s^2$ and $R = 6400 \ km = 6.4 	imes 10^6 \ m$:$\omega = \sqrt{\frac{9.8}{2 	imes 6.4 	imes 10^6}} = \sqrt{\frac{9.8}{12.8 	imes 10^6}} = \sqrt{0.7656 	imes 10^{-6}} \approx 8.75 	imes 10^{-4} \ rad/s$.

The angular speed of rotation of the Earth about its axis,at which the weight of a man standing on the equator becomes half of his weight at the poles,is given by:

Similar Questions

The angular speed of the Earth,such that an object on the equator may appear weightless,is ($g = 10\,m/s^2$,radius of the Earth $R = 6400\,km$).

The depth at which acceleration due to gravity becomes $\frac{g}{2n}$ is ($R=$ radius of earth,$g=$ acceleration due to gravity on earth's surface,$n$ is an integer).

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

What is the acceleration due to gravity at a distance of $2R$ from the surface of the Earth,where $R$ is the radius of the Earth?

Find the angular speed of rotation of the Earth,so that the apparent $g$ at the equator becomes $(1/6)$th of its original value. $(R = 6.4 \times 10^6 \ m)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module