If the earth rotates faster than its present | vedclass

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

Question

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

Vedclass · Accepted Answer

Decrease at the equator but remain unchanged at the poles

Vedclass · Answer

(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g - \omega^2 R \cos^2 \lambda$,where $\omega$ is the angular velocity of the Earth and $R$ is the radius of the Earth.At the equator,the latitude $\lambda = 0^o$,so $\cos 0^o = 1$. Thus,$g' = g - \omega^2 R$. If the Earth rotates faster,$\omega$ increases,which leads to a decrease in the effective gravity $g'$,and consequently,the weight of an object decreases at the equator.At the poles,the latitude $\lambda = 90^o$,so $\cos 90^o = 0$. Thus,$g' = g$. Since the term involving $\omega$ becomes zero,the weight of an object remains unchanged at the poles regardless of the speed of rotation.

If the earth rotates faster than its present speed,the weight of an object will

Similar Questions

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{2}$ below the surface of the Earth,where $R$ is the radius of the Earth,what is the new frequency of oscillations at this depth?

Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.
Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.

By approximately how many meters is the radius of the Earth at the equator greater than the radius of the Earth at the poles (in $,000 \ m$)?

If the density of a small planet is the same as that of earth, while the radius of the planet is $0.2$ times that of the earth, the gravitational acceleration on the surface of that planet is (in $\,g$)

Two planets of radii in the ratio $2 : 3$ are made from the material of density in the ratio $3 : 2$. Then the ratio of acceleration due to gravity $g_1/g_2$ at the surface of the two planets will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module