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Calculate angular velocity of the earth ...

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`)

A

`7.8 xx 10^(-2)` rad/s

B

`0.5 xx 10^(-3)` rad/s

C

`1 xx 10^(-3)` rad/s

D

`2 . 5 xx 10^(-3)` rad/s

Text Solution

Verified by Experts

The correct Answer is:
D
New acceleration due to gravity g' is given by
`g' g - R_(e) omega^(2) cos^(2) lambda `
`0 = 10 - 6.4 xx 10^(6) omega^(2) cos^(2) 60^(@)`
`implies omega^(2) = (10)/(6.4 xx 10^(6) [0.5]^(2)) = 2.5 xx 10^(-3) rad//s`
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Knowledge Check

  • The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on 60^@ latitude becomes zero is

    A
    `2.5xx10^(-3) rads^(-1)`
    B
    `1.5xx10^(-3) rad s^(-1)`
    C
    `4.5xx10^(-3) rad s^(-1)`
    D
    `0.5xx10^(-3) rad s^(-1)`

  • Altitude at which acceleration due to gravity decreases by 0.1% approximately : (Radius of earth = 6400 km)

    A
    3 . 2 km
    B
    6 . 4 km
    C
    2 . 4 km
    D
    1 . 6 km

  • What should be the angular velocity of earth, if the apparent value of acceleration due to gravity at earth's surface on equatorial plane is zero ? Radius of earth is 6400 km and g at earth's surface is 10m//s^(2) .

    A
    125 rad/s
    B
    1.25 rad/s
    C
    `1.25xx10^(-3)"rad/s"`
    D
    zero

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    At what height from the surface of earth, the acceleration due to gravity decreases by 2% ? [Radius of earth = 6,400 km.]

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    Find the magnitude of the acceleration due to gravity at the surface of the earth. ( M = 6 xx 10^(24) Kg, R = 6400 Km)

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