Question 333: Gravitation

Question

The radius of earth is $$6400 \mathrm{~km}$$ and acceleration due to gravity $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$. For the weight of body of mass $$5 \mathrm{~kg}$$ to be zero on equator, rotational velocity of the earth must be (in $$\mathrm{rad} / \mathrm{s}$$ )

Accepted Answer

At equator, for the weight to be zero, the gravitational force must be equal to centrifugal force.
$$\begin{aligned}
& \mathrm{mR} \omega^2=\mathrm{mg} \
& \omega^2=\frac{\mathrm{g}}{\mathrm{R}} \
& \omega=\sqrt{\frac{\mathrm{g}}{\mathrm{R}}} \
& \omega=\sqrt{\frac{10}{6.4 	imes 10^6}} \
& \omega=\frac{1}{800} \frac{\mathrm{rad}}{\mathrm{s}} .
\end{aligned}$$

Answer

$$\frac{1}{80}$$

Answer

$$\frac{1}{400}$$

Answer

$$\frac{1}{800}$$

Answer

$$\frac{1}{1600}$$

Gravitation

The radius of earth is $6400 \mathrm{~km} and acceleration due to gravity \mathrm{g}=10 \mathrm{~ms}^{-2}. For the weight of body of mass 5 \mathrm{~kg} to be zero on equator, rotational velocity of the earth must be (in \mathrm{rad} / \mathrm{s}$ )

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