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If the angular speed of the Earth is 7.2...

If the angular speed of the Earth is `7.26xx10^(-5)` rad/s and radius of the Earth is 6,400 km, calculate the change in weight of 1 kg of mass taken from equator to pole.

Text Solution

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Given, angular speed of Earth `(omega)=7.26xx10^(-5)` rad/s
radius of earth (R) `=6400km=6.4xx10^(6)m`
mass of body (m)=1 kg
At poles, `phi=90^(@)`
`:.cos90^(@)=0`
Now,
Change in weight `(DeltaW)=mDeltag`
`=m(Romega^(2))`
`=1xx6.4xx10^(6)xx(7.26xx10^(-5))^(2)`
`=3.373xx10^(-2)N
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Knowledge Check

  • The mass of the Earth is 6 xx 10^(24) kg and its radius is 6400 km. Find the acceleration due to gravity on the surface of the Earth.

    A
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  • Calculate angular velocity of the earth so that acceleration due to gravity at 60^(@) latitude becomes zero (radius of the earth = 6400 km, gravitational acceleration at poles = 10 m//s^(2) , cos60^(@) = 0.5 )

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