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

The correct option is A $1.25\times {10}^{-3}$ rad/s

We know,

$g^{\prime }=g(1-\frac{R{\omega }^2}{g}co{s}^2\lambda )$

At the equator,

$\lambda =0$ and $cos\lambda =1$

Let $g$ be the acceleration due to gravity when the earth were supposed to be at rest. In presence of earth's rotation on acceleration due to gravity at the equator is given by,

$g^{\prime }=g-R_e{\omega }^2$

where $\omega$ is the angular velocity of the earth. For $g^{,}=0$, the above relation becomes,

$\omega =\sqrt{\frac{g}{R_e}}=\sqrt{\frac{10m{s}^{-2}}{(6400\times {10}^3)m}}=1.25\times {10}^{-3}rad/s$