The difference in the acceleration due to gra | 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 a latitude $	heta$ due to the rotation of the Earth is given by $g^{\prime} = g - R \omega^2 \cos^2

Vedclass · Accepted Answer

$R \omega^2$

Vedclass · Answer

(B) The effective acceleration due to gravity at a latitude $	heta$ due to the rotation of the Earth is given by $g^{\prime} = g - R \omega^2 \cos^2 	heta$.At the equator,the latitude $	heta = 0^{\circ}$,so $\cos 0^{\circ} = 1$.Thus,the acceleration due to gravity at the equator is $g_e = g - R \omega^2$.At the poles,the latitude $	heta = 90^{\circ}$,so $\cos 90^{\circ} = 0$.Thus,the acceleration due to gravity at the poles is $g_p = g$.The difference in the acceleration due to gravity between the pole and the equator is $g_p - g_e = g - (g - R \omega^2) = R \omega^2$.

The difference in the acceleration due to gravity at the pole and equator is ( $g=$ acceleration due to gravity,$R=$ radius of earth,$\theta=$ latitude,$\omega=$ angular velocity,$\cos 0^{\circ}=1, \cos 90^{\circ}=0$ ).

Similar Questions

What should be the angular speed of the Earth so that a body lying on the equator may appear weightless? $(g = 10\,m/s^2, R = 6400\,km)$

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

The weight of a body on the surface of the earth is $63 \ N$. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth? (in $N$)

$Assertion$ : In a free fall,the weight of a body becomes effectively zero.
$Reason$ : The acceleration due to gravity acting on a body in free fall is zero.

An astronaut on a strange planet finds that the acceleration due to gravity is twice that on the surface of the Earth. Which of the following could explain this?

Vedclass Test Series

Exam Paper Generator

Online Exam Module