The speed of earth 's rotation about its axis is $ω$ .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 ( $g = 10 m s^{- 2}$ , $R = 6400$ km)

1

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8.5

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17

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34

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Solution

The correct option is C $17$
Earth rotates about its own axis in $24$ hours. $T = 24 \times 60 \times 60$ s. So angular speed of the earth about its own axis $= \frac{2 π}{T} = \frac{2 π}{24 \times 60 \times 60} m / s$
At equator $g^{1} = g - R^{2} g^{1} = g - R_{e} w^{12}$
or $O = g - R_{e} w^{12} w^{1} = \sqrt{\frac{g}{R_{e}}} = \frac{2 π}{84.6 \times 60}$
$∴$ $\frac{w^{1}}{w} = \frac{24 \times 60 \times 60}{84.6 \times 60} \approx 17$
$w^{1} = 17 w$

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The speed of earth 's rotation about its axis is ω.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 (g=10ms−2, R=6400 km)

The correct option is C 17Earth rotates about its own axis in 24 hours. T=24×60×60 s. So angular speed of the earth about its own axis =2πT=2π24×60×60m/sAt equator g1=g−R2g1=g−Rew12or O=g−Rew12w1=√gRe=2π84.6×60∴ w1w=24×60×6084.6×60≈17w1=17w

The speed of earth 's rotation about its axis is $ω$ .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 ( $g = 10 m s^{- 2}$ , $R = 6400$ km)