R is the radius of the Earth,omega is its ang | vedclass

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

Question

(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g_p - R\omega^2 \cos^2 \lambda$.Given that t

Vedclass · Accepted Answer

$g_p - \frac{1}{4}R\omega^2$

Vedclass · Answer

(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g_p - R\omega^2 \cos^2 \lambda$.Given that the latitude $\lambda = 60^\circ$,we substitute this value into the equation.Since $\cos 60^\circ = \frac{1}{2}$,we have $\cos^2 60^\circ = (\frac{1}{2})^2 = \frac{1}{4}$.Substituting this into the formula: $g' = g_p - R\omega^2 (\frac{1}{4})$.Therefore,the effective value of $g$ is $g_p - \frac{1}{4}R\omega^2$.

$R$ is the radius of the Earth,$\omega$ is its angular velocity,and $g_p$ is the value of acceleration due to gravity at the poles. The effective value of $g$ at latitude $\lambda = 60^\circ$ will be equal to:

Similar Questions

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

Considering Earth to be a sphere of radius $R$ having uniform density $\rho$,the value of acceleration due to gravity $g$ in terms of $R$,$\rho$,and $G$ is:

$A$ body has weight $W$ newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be:

The weight of a body of mass $m$ decreases by $1\%$ when it is raised to a height $h$ above the Earth's surface. If the body is taken to a depth $h$ in a mine,the change in its weight is:

The acceleration due to gravity is $g$ at a point distant $r$ from the centre of the Earth of radius $R$. If $r < R$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module