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Question: The angular speed of earth, so that the object on equator may appear weightless, is <img src="https:...

The angular speed of earth, so that the object on equator may appear weightless, is , radius of earth 6400 km)

A

1.25×10−3rad/sec1.25 \times 10 ^ { - 3 } \mathrm { rad } / \mathrm { sec }

B

1.56×10−3rad/sec1.56 \times 10 ^ { - 3 } \mathrm { rad } / \mathrm { sec }

C

1.25×10−1rad/sec1.25 \times 10 ^ { - 1 } \mathrm { rad } / \mathrm { sec }

D

1.56 rad/sec

Answer

1.25×10−3rad/sec1.25 \times 10 ^ { - 3 } \mathrm { rad } / \mathrm { sec }

Explanation

Solution

For condition of weightlessness of equator

ω=gR=1800=1.25×10−3rads\omega = \sqrt { \frac { g } { R } } = \frac { 1 } { 800 } = 1.25 \times 10 ^ { - 3 } \frac { \mathrm { rad } } { \mathrm { s } }