The speed of the earth's rotation about its axis is $ \omega $. Its speed is increased to $ x $ times to make the effective acceleration due to gravity equal to zero at the equator, then $ x $ is around $ \left(g=10 \mathrm{~m} \mathrm{~s}^{-2}, R=6400 \mathrm{~km}\right) $

Question

The speed of the earth's rotation about its axis is $ \omega $. Its speed is increased to $ x $ times to make the effective acceleration due to gravity equal to zero at the equator, then $ x $ is around $ \left(g=10 \mathrm{~m} \mathrm{~s}^{-2}, R=6400 \mathrm{~km}\right) $, $ 1 $, $ 8.5 $, $ 17 $, $ 34 $

MARKS App · Accepted Answer

At the equator, g ′ = g - R ω 2 When angular velocity be ω ′ = x ω then, 0 = g - R ω ′ 2 or ω = g R present angular velocity of the earth ω 0 = 2 π 24 × 60 × 60 ω ω 0 = 17

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