Question

The angular velocity of rotation of the earth in order to make the effective acceleration due to gravity equal to zero at equator should be $g=10m/s^2,R=6400km$.

• A. $2.15\times {10}^3$ rad/sec
• B. $2.25\times {10}^{-3}$ rad/sec
• C. $1.25\times {10}^3$ rad/sec
• D. $1.25\times {10}^{-3}$ rad/sec

Answer 1

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

$g^{\prime }=g-{\omega }^{2}Rco{s}^2\theta$
At equator $g^{\prime }=0$
$cos\theta =0$
$g={\omega }^{2}R$
${\omega }^2=\frac{g}{R}$
$\omega =\sqrt{\frac{g}{R}}=\sqrt{\frac{10}{6400\times 1000}}$
$=\sqrt{\frac{1}{640000}}=\frac{1}{800}=1.25\times {10}^{-3}rad/sec.$