Question

How many time faster than the present speed would the earth have to rotate about its axis so that the apparent weight of the bodies on the equator becomes zero? (The radius of the earth $R=6.4\times 10^{6} \, m$ )

• A. $5times$
• B. $9times$
• C. $13times$
• D. $17times$

Answer 1

Answer: (D)

Let $\omega _{0}$ be the new angular velocity.
At the equator $g_{E}=g-R\omega ^{2}$
The weight $mg_{E}$ should be zero. $g_{E}=0$
$\therefore R\omega _{0}^{2}=g$ or $\omega _{0}=\sqrt{\frac{g}{R}}$
$\therefore \omega _{0}=\sqrt{\frac{9.8}{6.4 \times 1 0^{6}}}=1.24\times 10^{- 3}\frac{r a d}{s}$
The present angular velocity of the earth is $\omega =\frac{2 \pi }{T}=\frac{2 \times 3.14}{24 \times 60 \times 60}$
$\therefore $ $\omega =\frac{3.14 \times 1 0^{- 2}}{12 \times 36}=7.27\times 10^{- 5}$
$\therefore \frac{\omega _{0}}{\omega }=\frac{1.24 \times 1 0^{- 3}}{7.27 \times 1 0^{- 5}}=17$
$\therefore \omega _{0}=17\omega $
It should rotate faster by 17 times the present value.