Gravitation Ques 11
- The numerical value of the angular velocity of rotation of the earth should be $\mathrm{rad} / \mathrm{s}$ in order to make the effective acceleration due to gravity at equator equal to zero.
$(1984,2 \mathrm{M})$
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Answer:
Correct Answer: 11.$1.24 \times 10^{-3} $
Solution:
Formula:
Effective Gravitational Acceleration
- $g^{\prime}=g-R \omega^{2} \cos ^{2} \varphi$
At equator
$\varphi=0$
$\therefore \quad g^{\prime}=g-R \omega^{2}$
$\Rightarrow \quad 0=g-R \omega^{2}$
$$ \therefore \quad \omega=\sqrt{\frac{g}{R}}=\sqrt{\frac{9.8}{6400 \times 10^{3}}} $$
$$ =1.24 \times 10^{-3} \mathrm{rad} / \mathrm{s} $$
BETA
