Question

The imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero is equal to

• A. $1.25\times 10^{-3}$ rad/s
• B. $2.50\times 10^{-3}$ rad/s
• C. $3.75\times 10^{-3}$ rad/s
• D. $5.0\times 10^{-3}$ rad/s

Answer 1

Answer: (B)

Acceleration due to gravity $g=g-{\omega }^{2}R{\cos }^2\lambda$
$0=g-{\omega }^{2}R{\cos }^{2}60^{\circ }0=g-\frac{{\omega }^{2}R}{4}$
$\omega =2\sqrt{\frac{g}{R}}=2\sqrt{\frac{10}{6400\times 100}}$
Angular velocity $\omega =\frac{1}{400}$
$=2.5\times 10^{-3}rad/s$ (Take $g=10m/s^2$
for the acceleration due to gravity if the earth were at rest and radius of earth
equal to $6400km$.)