Question

What will be angular velocity of earth for which a particle at equator just flies off

• A. $17$ times of present value
• B. $18$ times of present value
• C. $19$ times of present value
• D. $20$ times of present value

Answer 1

The correct option is A $17$ times of present value

$g_{\lambda }=g-R{w}^{2}co{s}^2\lambda$ where $\lambda$ is the latitude.

Now at equator, $\lambda =0^o⟹g_e=g-R{w}^2$

For a particle at equator to fly off, $g_e$ must be zero i-e, $g=R{w}^2$

As $g=9.8m/s^{2}R=6400km$

$9.8=6400\times 1000w^2⟹w=1.237\times {10}^{-3}rad/s$

As at present $T=24h$ so, $w_o=\frac{2\pi }{T}=\frac{2\pi }{24\times 3600}=7.268\times {10}^{-5}rad/s$

Thus $\frac{w}{w_o}=17$