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Physics Question on Gravitation

The angular speed of earth, so that the object on equator may appear weight less, is (g=10) m/s2 (g=10)\,m/s^{2} , radius of earth 6400 km6400\,km

A

1.25×10−3rad/s 1.25\times 10^{-3}rad/s

B

1.56×10−3rad/sec⁡ 1.56\times 10^{-3}rad/ \sec

C

1.25×10−1rad/s 1.25\times 10^{-1}rad/s

D

1.56 rad/sec⁡1.56\,rad/ \sec

Answer

1.25×10−3rad/s 1.25\times 10^{-3}rad/s

Explanation

Solution

Using the relation W=m(g−Rω2)W =m\left(g-R \omega^{2}\right) 0=m(g−Rω2)0 =m\left(g-R \omega^{2}\right) So, ω=gR=106.4×106\omega=\sqrt{\frac{g}{R}}=\sqrt{\frac{10}{6.4 \times 10^{6}}} =1800=\frac{1}{800} or ω=1.25×10−3rad/sec⁡\omega=1.25 \times 10^{-3} rad / \sec