The rotation of the Earth,having radius R,abo | vedclass

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

Question

(D) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - R \omega^2 \cos^2 \lambda$.For a man to feel weightle

Vedclass · Accepted Answer

$4 \pi \sqrt{\frac{R}{g}}$

Vedclass · Answer

(D) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - R \omega^2 \cos^2 \lambda$.For a man to feel weightlessness,the effective gravity must be zero,so $g' = 0$.Substituting the given values,$0 = g - R \omega^2 \cos^2 60^{\circ}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have $0 = g - R \omega^2 (\frac{1}{2})^2$.$g = R \omega^2 (\frac{1}{4}) \Rightarrow \omega^2 = \frac{4g}{R} \Rightarrow \omega = 2 \sqrt{\frac{g}{R}}$.The time period $T$ is given by $T = \frac{2 \pi}{\omega}$.Substituting $\omega$,$T = \frac{2 \pi}{2 \sqrt{g/R}} = \pi \sqrt{\frac{R}{g}} 	imes 2 = 2 \pi \sqrt{\frac{R}{g}} 	imes 2 = 4 \pi \sqrt{\frac{R}{g}}$.

The rotation of the Earth,having radius $R$,about its axis speeds up to a value such that a man at latitude angle $60^{\circ}$ feels weightlessness. The duration of the day in such a case is:

Similar Questions

At a height of $10 \,km$ above the surface of the earth,the value of acceleration due to gravity is the same as that at a particular depth below the surface of the earth. Assuming uniform mass density for the earth,the depth is ............. $km$.

The gravitational potential at a point above the surface of the Earth is $-5.12 \times 10^7 \,J/kg$ and the acceleration due to gravity at that point is $6.4 \,m/s^2$. Assume that the mean radius of the Earth is $6400 \,km$. The height of this point above the Earth's surface is: (in $\,km$)

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.

If the earth suddenly shrinks (without changing mass) to half of its present radius,the acceleration due to gravity will be

Two planets $A$ and $B$ have densities $\varrho_1$ and $\varrho_2$ and have radii $r_1$ and $r_2$ respectively. The ratio of acceleration due to gravity on $A$ to that of $B$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module